maths basics 2010

7/30/2019 Maths Basics 2010

1/202

Mathematical Literacy Notes

Margaret Lubczonok and Sizwe Mabizela

Department of Mathematics (Pure & Applied)Rhodes University


2/202

Contents

1 Mathematics in everyday life 1

1.1 The metric system (SI) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Ratio, Proportion, and Percentage . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Real Numbers 12

2.1 Classification of real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2 Graphical representation of real numbers . . . . . . . . . . . . . . . . . . . . 15

2.3 Operations on real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Algebraic Expressions 223.1 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.2 Algebraic Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.2.1 Simplifying algebraic fractions by cancelling out common factors . . . . 27

3.2.2 Multiplication and Division of algebraic fractions . . . . . . . . . . . . 29

3.2.3 Addition and Subtraction of algebraic fractions . . . . . . . . . . . . . 32

3.3 Partial Fraction Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Linear and Quadratic Equations 43

4.1 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4.1.1 Simultaneous linear equations . . . . . . . . . . . . . . . . . . . . . . 48

4.1.2 Applications: Problems leading to systems of linear equations . . . . . 53

4.2 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.2.1 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . 55

i


3/202

4.3 Equations leading to Linear or Quadratic Equations : Rational and IrrationalEquations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.3.1 Irrational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

4.4 Applications: Problems leading to linear/quadratic equations . . . . . . . . . . 74

5 Linear and Quadratic Functions 79

5.1 Concept of Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.2 Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

5.2.1 Increasing and Decreasing Function . . . . . . . . . . . . . . . . . . . 84

5.2.2 Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.2.3 Zeroes of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 865.3 Linear Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

5.3.1 Sketching a graph of a linear function . . . . . . . . . . . . . . . . . . 89

5.4 Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.5 Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

6 Polynomials 113

6.1 Addition and Subtraction of Polynomials . . . . . . . . . . . . . . . . . . . . 114

6.2 Multiplication of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 115

6.3 Polynomial Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.4 The Remainder Theorem and Factor Theorem . . . . . . . . . . . . . . . . . 120

6.5 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

7 The Absolute Value (Modulus) Function 128

7.1 Absolute-value Equations and Inequalities . . . . . . . . . . . . . . . . . . . . 129

7.2 The functions y = a|x p| + q . . . . . . . . . . . . . . . . . . . . . . . . . 1337.3 Algebraic and graphical solution of equations and inequalities . . . . . . . . . 137

7.4 Geometric representation of Absolute Value . . . . . . . . . . . . . . . . . . 139

7.5 Miscellaneous Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

8 Trigonometry 146

ii


4/202

8.1 The radian measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

8.2 Trigonometric ratios in a right-angled triangle . . . . . . . . . . . . . . . . . 148

8.3 Trigonometric function of any angle . . . . . . . . . . . . . . . . . . . . . . 153

8.4 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

8.5 Periodic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

8.6 Parity relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

8.7 Reduction Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

8.8 More Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . 163

8.9 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . 166

8.9.1 Some important transformations which are applied to the trigonometricfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

8.10 Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

8.10.1 Solving Equations: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

9 Exponential and Logarithmic Functions 179

9.1 Exponential Laws and Definitions . . . . . . . . . . . . . . . . . . . . . . . . 179

9.1.1 Exponential Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1799.1.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

9.1.3 Surds Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

9.2 Operations on Exponents and Surds . . . . . . . . . . . . . . . . . . . . . . . 180

9.3 Exponential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

9.4 Equations with Rational Exponents . . . . . . . . . . . . . . . . . . . . . . . 182

9.5 Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

9.6 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

9.7 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

9.8 Logarithms (Definitions) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

9.9 Logarithmic Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

9.10 Change of base . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

9.11 Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

iii


5/202

9.12 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

9.12.1 Definition of Logarithmic Function . . . . . . . . . . . . . . . . . . . 196

iv


6/202

Chapter 1

Mathematics in everyday life

1.1 The metric system (SI)

Designed during the French Revolution of the 1790s, the metric system brought order out of

the conflicting and confusing traditional systems of weights and measures then being used in

Europe. Prior to the introduction of the metric system, it was common for units of length,

land area, and weight to vary, not just from one country to another but from one region

to another within the same country. As the modern nations were gradually assembled from

smaller kingdoms and principalities, confusion simply multiplied. Merchants, scientists, and

educated people throughout Europe realized that a uniform system was needed, but it was

only in the climate of a complete political upheaval that such a radical change could actually

be considered.

The metric system replaces all the traditional units, except the units of time and of angle

measure, with units satisfying three conditions:

1. One fundamental unit is defined for each quantity. These units are now defined precisely

in the International System of Units.

2. Multiples and fractions of these fundamental units are created by adding prefixes to the

names of the defined units. These prefixes denote powers of ten, so that metric unitsare always divided into tens, hundreds, thousands, etc. The original prefixes included

milli- for 1/1 000, centi- for 1/100, deci- for 1/10, deka- for 10, hecto- for 100, and

kilo- for 1,000.

3. The fundamental units are defined rationally and are related to each other in a rational

fashion.

The metric units were defined in an elegant way unlike any traditional units of measure. The

Earth itself was selected as the measuring stick. The meter was defined to be one ten-millionth

of the distance from the Equator to the North Pole. The litre was to be the volume of one

1


7/202

cubic decimeter, and the kilogram was to be the weight of a litre of pure water. It didnt turn

out quite like this, because the scientific methods of the time were not quite up to the task of

measuring these quantities precisely, but the actual metric units come very close to the design.

The metric system was first proposed in 1791. It was adopted by the French revolutionary

assembly in 1795, and the first metric standards (a standard meter bar and kilogram bar) were

adopted in 1799. There was considerable resistence to the system at first, and its use was

not made compulsory in France until 1837. The first countries to actually require use of the

metric system were Belgium, the Netherlands, and Luxembourg, in 1820.

Around 1850 a strong movement began among scientists, engineers, and businessmen in favor

of a international system of weights and measures. The scientific and technical revolution was

well underway and a global economy was developing. The need for uniformity in measurement

was becoming obvious. Furthermore, the metric system was the only real choice available.

The only possible competitor, the British Imperial system, was so closely tied to the British

Empire it was not even acceptable to the Americans, let alone to non-English speakers.

Between 1850 and 1900 the metric system made rapid progress. It was adopted throughout

continental Europe, in Latin America, and in many countries elsewhere. It became firmly

established as a key part of the language of science.

In the 1870s the French made a crucial decision to turn control of the system over to an

international body. In 1875, most of the leading industrialized countries (including the United

States, but not Britain) signed the Treaty of the Meter. The treaty established the International

Bureau of Weights and Measures, which has presided ever since over what we now call the

International System of Units. It also provided for distribution of copies of the metric standardsthroughout the world and for continuing consultation and periodic revision and improvement

of the system through regular meetings of a General Conference of Weights and Measures.

The 22nd General Conference met in October 2003.

Since 1875 the eventual triumph of the metric system in science and international commerce

has been assured, despite continuing popular opposition in Britain and the United States. In

fact, the metric system has met popular opposition in every country at the time of its adoption.

People dont want to change their customary units, which are part of how they see and control

the world. It is naturally disturbing to do so. This opposition has been largely overcome

everywhere, except in the U.S., by economic necessity: the need to participate fully in theglobal economic system. Even in the U.S., economic needs assure the continued creeping

adoption of the system in one area and then another.

Those Americans opposing adoption of metric units often argue that the metric system is

abstract and intellectual or that its use would embroil us in calculations. This is not true. The

metric system has been the customary measurement system in France for almost two centuries,

in the rest of continental Europe for at least one century, and in the rest of the world for a

least a generation or two. Most people in the world know exactly how long a kilometer is, how

large a litre is, how much a kilogram weighs, and how warm 25C is, because they use these

2


8/202

units every day of their lives in the same way Americans use miles, gallons, and pounds.

Outside Britain and the United States there is almost no need to convert metric units into

something else. In fact, the way to avoid conversion formulas is to adopt the metric system.

As long as Britons and Americans continue to use traditional units, they will have to remember

how these units relate to the metric units.

Widespread use of the metric system has not meant the complete elimination of traditional

units, nor has it stopped the continuing creation of new units, many metric but some not, to

meet new needs.

All systems of weights and measures, metric and non-metric, are linked through a network

of international agreements supporting the International System of Units. The International

System is called the SI, using the first two initials of its French name Systeme International

dUnites. The key agreement is the Treaty of the Meter (Convention du Metre), signed in Paris

on May 20, 1875. 48 nations have now signed this treaty, including all the major industrialized

countries. The United States is a charter member of this metric club, having signed the original

document back in 1875.

The SI is maintained by a small agency in Paris, the International Bureau of Weights and

Measures (BIPM, for Bureau International des Poids et Mesures), and it is updated every

few years by an international conference, the General Conference on Weights and Measures

(CGPM, for Conference Generale des Poids et Mesures), attended by representatives of all

the industrial countries and international scientific and engineering organizations. The 22nd

CGPM met in October 2003; the next meeting will be in 2007. As BIPM states on its web site,

The SI is not static but evolves to match the worlds increasingly demanding requirementsfor measurement.

At the heart of the SI is a short list of base units defined in an absolute way without referring

to any other units. The base units are consistent with the part of the metric system called the

MKS system. In all there are seven SI base units:

the meter for distance, the kilogram for mass the second for time, the ampere for electric current, the kelvin for temperature, the mole for amount of substance, and the candela for intensity of light.

Other SI units, called SI derived units, are defined algebraically in terms of these fundamental

units. For example, the SI unit of force, the newton, is defined to be the force that accelerates

3


9/202

a mass of one kilogram at the rate of one meter per second per second. This means the newton

is equal to one kilogram meter per second squared, so the algebraic relationship is N = kgms2. Currently there are 22 SI derived units. They include:

the radian and steradian for plane and solid angles, respectively;

the newton for force and the pascal for pressure; the joule for energy and the watt for power; the degree Celsius for everyday measurement of temperature;

units for measurement of electricity: the coulomb (charge), volt (potential), farad(capacitance), ohm (resistance), and siemens (conductance);

units for measurement of magnetism: the weber (flux), tesla (flux density), and henry(inductance);

the lumen for flux of light and the lux for illuminance; the hertz for frequency of regular events and the becquerel for rates of radioactivity

and other random events;

the gray and sievert for radiation dose; and the katal, a unit of catalytic activity used in biochemistry.

The SI does not allow use of any units other than those listed above and their multiples. Inparticular, it does not allow use of any of the English traditional units (the horsepower, for

example), nor does it allow the use of any of the algebraically-derived units of the former CGS

system, such as the erg, gauss, poise, stokes, or gal. In addition, the SI does not allow use of

other traditional scientific and engineering units, such as the torr, curie, calorie, or rem.

Certain scientific fields have defined units more or less compatible with the SI, but not part

of the SI. The use of the jansky in astronomy is a good example. There is always the chance

that future meetings of the CGPM could add these units to the SI, but for the present they

are not approved.

For multiples of approved units, the SI includes a list of prefixes. This list has been extended

several times, most recently by the 19th CGPM in 1991. Prefixes now range from yotta- at

1024 (one septillion) to yocto- at 1024 (one septillionth). There seems to be some need for

another extension, but this question was not addressed at the 1999 CGPM. The SI does not

allow these prefixes to be used for binary multiples, such as the use of kilobit to mean 1024

bits instead of 1000. For binary multiples a new list of special prefixes has been established

by the International Electrotechnical Commission.

Metric Prefixes

Here are the metric prefixes, with their numerical equivalents:

4


10/202

Prefix Symbol Multiplier Numerical Exponential

yotta Y 1 000 000 000 000 000 000 000 000 1024

zetta Z 1 000 000 000 000 000 000 000 1021

exa E 1 000 000 000 000 000 000 1018

peta P 1 000 000 000 000 000 1015

tera T 1 000 000 000 000 1012

giga G 1 000 000 000 109

mega M 1 000 000 106

kilo k 1 000 103

hecto h 100 102

deca da 10 101

no prefix means: 1 100

deci d 0.1 101

centi c 0.01 102

milli m 0.001 103

micro 0.000 001 106

nano n 0.000 000 001 109

pico p 0.000 000 000 001 1012

femto f 0.000 000 000 000 001 1015

atto a 0.000 000 000 000 000 001 1018

zepto z 0.000 000 000 000 000 000 001 1021

yocto y 0.000 000 000 000 000 000 000 001 1024

5


11/202

Commonly used metric system units and symbols

Quantity measured Unit Symbol Relationship

Length, width,distance, thickness,

girth, etc.

millimetercentimetermeter

kilometer

mmcmm

km

10 mm = 1 cm100 cm = 1 m

1 km = 1000 m

Mass (weight)*

milligramgramkilogrammetric ton

mggkgt

1000 mg = 1 g1 kg = 1000 g1 t = 1000 kg

Time second s

Temperature degree Celsius C

Areasquare meterhectaresquare kilometer

mhakm2

1 ha = 10 000 m2

1km2 = 100 ha

Volume

millilitercubic centimeterlitrecubic meter

mcm3

m3

1000 m = 1 1cm3 = 1 m1000 = 1 m3

Speed, velocitymeter per second

kilometer per hour

m/s

km/h 1 km/h = 0.278 m/sDensity kilogram per cubic meter kg/m3

Force newton NPressure, stress kilopascal kPa

Powerwattkilowatt

WkW 1 kW = 1000 W

Energykilojoulemegajoulekilowatt hour

kJMJkWh

1 MJ = 1000 kJ1 kWh = 3.6 MJ

Electric current ampere A

6


12/202

1.2 Ratio, Proportion, and Percentage

Definition 1.1. (Theorem 3.2.1[2])A ratio is a comparison of two quantities.

Ratios are written with a colon between the quantities (e.g., 3 : 4) or as a quotient (e.g.,

3

4).The ratio a : b is read as a is to b.

NB: The order in which the numbers appear in a ratio is important: a : b is NOT the same

as b : a.

Examples 1.2. 1. In a certain school, the ratio of boys to girls is 3 : 2. This means that

there are three boys to every two girls in this school. Put differently, three out of every

five (= 3 + 2) pupils in this school are boys and two out of every five are girls.

2. A box contains oranges, apples and bananas in a ratio of3 : 2 : 5. This means that out

of every 10 = 3 + 2 + 5 fruit, we have 3 apples, 2 oranges and 5 bananas.

Dividing a quantity in a given ratio

Examples 1.3. 1. Divide 1360 in a ratio 2 : 3.

Soln: A ratio of2 : 3 means that every 5 units are split as 2 and 3. The first number is2

5of

1360 and the second is3

5of 1360. That is,

25 1360 = 544 and 35 1360 = 816.The number 1360 is divided into 544 and 816.

2. John, Jim and Dave have agreed to divide their bonus of R2 300 in a ratio of2 : 3 : 5,

with John getting 2 parts, Jim 3 parts, and Dave 3 parts of the bonus. How much will

each get?

Soln: A ratio of 2 : 3 : 5 means that every 10 = 2 + 3 + 5 units are split as 2, 3 and 5.

Therefore

John will get 210

2300 = 460 rand;

Jim will get3

10 2300 = 690 rand, and

Jim will get5

10 2300 = 1150 rand.

3. In a park, the ratio of ducks to geese is 12 to 8. How many of the 250 birds are ducks

and how many are geese?

7


13/202

Soln: The ratio of 12 : 8 means that, of every 12 + 8 = 20 birds, 12 are ducks and 8

are geese. That is,12

20of the birds are ducks and

8

20are geese. Therefore there are

12

20 250 = 150 ducks and 8

20 250 = 100 geese.

Definition 1.4. A proportion is an equation stating that two ratios are equivalent.

Proportion may be written with a double colon (::) or with the equality sign (=). The

proportion a : b :: c : d is read as a is to b as c is to d. The proportion a : b :: c : d is also

expressed asa

b=

c

d.

For example, 1 is to 4 as 3 is to 12, is written written as

1 : 4 :: 3 : 12 or1

4=

3

12.

In a proportion a : b :: c : d the first and last terms (a and d) are called extremes and the

second and third terms (b and c) are called the means.

In a proportion, the product of the extremes is equal to the product of the means; i.e.,a : b :: c : d if and only if ad = bc.

Examples 1.5. 1. The ratio of 5th graders to 6th graders at a school is 2 to 3. If there

are 440 5th graders, how many 6th graders are there?

2. If 3 out of every 5 people will vote in an election, how many people will vote in a

population of 40 000?

Soln: Let n be the number of people out of the population of 40 000 that will vote. Then,the given information says that

3 : 5 :: n : 40 000.

Equivalently,3

5=

n

40 000.

Therefore n =3

5 40 000 = 24 000.. That is, 24 000 people out of the population of

40 000.

Ratio and proportion are used to solve many real-life problems:

1. Maps, blueprints, and scale models- An appropriate ratio, or scale, is chosen. Large

objects or areas can then be shown in a considerably smaller size or space.

2. Medicine dosages are frequently related to the weight of the patient, so the dosage must

be increased/decreased proportionally to weight.

3. Recipes or formulas may be increased/decreased by using proportions.

8


14/202

Definition 1.6. Two quantitiesx and y are said to be

1. directly proportional if one of these is a constant multiple of the other; i.e.

y = cx, wherec is a constant. Ifx and y are directly proportional to each other, we

writex y.2. inversely or indirectly proportional if their product is a constant; i.e., xy = k,

wherek is a constant. If x and y are inversely proportional to each other, we write

x 1y

.

Definition 1.7. A percent is a ratio that compares the number to 100.

The word percent means per one hundred. We use the symbol % to denote a percentage.

For example, 15% means15

100or 0.15.

Examples 1.8. 1. Express 1740

as a percent.

Soln: We are asking the question: what number compares to 100 as 17 compares to 40? Let

n be that number. We set up the following proportion

n : 100 :: 17 : 40 equivalentlyn

100=

17

40.

It follows that n =17

40 100 = 42.5.

2. Figures from Makana Municipality show that 22 570 people visited Grahamstown during

the 2006 Arts Festival. This was an increase of 12.85% on 2005 figures. How manypeople attended the Arts Festival in 2005?

Soln: Let n be the number of people who attended the Arts Festival in 2005. From the given

information, the number of people who attended the Arts Festival in 2006 is

n + 0.1285n = 22 570, or equivalently 1.1285n = 22 570.

Therefore, solving for n, we have that

n =22 570

1.1285= 20 000.

That is, 20 000 had attended the Arts Festival in 2005.

3. A recent poll has found that 36% of those asked planned to vote for one particular

candidate for President of the SRC. If 4 350 students vote, how many students vote for

the particular candidate?

Soln: We need to find 36% of 4 350. We first convert 36% to the number 0.36. Since we

want 36% of 4 350, we multiply 0.36 and 4 350 to get

0.36 4 350 = 1 566.That is, 1 566 will vote for the particular candidate for President of the SRC.

9


15/202

Definition 1.9. A scale drawing is a drawing that has the same shape but different

size to the actual object.

The scale of the drawing is a ratio of the size of the drawing to the actual size of object. That

is,

scale = Objects model sizeObjects actual size

.

To solve a scale problem, set up a proportion

Objects model size

Objects actual size=

Scale information of the model

Scale information of the actual object

Examples 1.10. 1. A stadium is 90m long and 60m wide. If 1cm represents 10m, what

are the dimensions of the stadium if it is drawn on a sheet of paper?

Soln Let x be the length of the stadium on paper and y its width on paper. Let us establish

proportions:

x

90=

1

10or equivalently x =

1

10 90,

and so x = 9cm. That is, the length of the stadium on paper is 9cm.

y

60=

1

10or equivalently y =

1

10 60,

and so y = 6cm. That is, the width of the stadium on paper is 6cm.

2. Cartographers use scale extensively to draw maps on paper. Scales on maps relate the

distance on the map to the actual distance on the ground. A scale 1:10 000 shows that

one measurement unit on the map represents 10 000 of the same units on the ground.

For example, if units used are centimetres, then a scale of 1:10 000 means that 1cm on

the map represents 10 000 cm on the ground.

What would 7cm represent if the scale is 1:1 000?

Soln: From the given scale, 1cm represents 1 000 cm. Therefore, 7cm represents 7 000 cm(or 7km).

Exercises

1. A car travels a distance of 144 km on 12 litres of petrol. How far will it travel on 50

litres of petrol?

2. If four loaves of bread feed three people, how many loaves will you need to feed tewlve

people?

10


16/202

3. In a rectangle, the ratio of length to width is 5 : 2. The perimeter of the rectangle is 28

cm. Find the dimensions of the rectangle.

4. An amount of R1 750 is to be divided between two people in the ratio of 3 : 4. How

much does each person receive?

5. The ratio of alcohol to water in a mixture is 1 : 9. Find the percentage of alcohol in themixture.

6. A ratio of males to females in a school is 5 : 4. If there are 1 800 pupils in the school,

find the number of each sex.

7. If4 out of every 7 people are going to vote in an election, how many people will vote in

a population of 210 000?

8. A mixture of alcohol and water is in the proportion alcohol : water = 1 : 10. How much

water must be added in the mixture in order for the mixture to have 5% alcohol?

9. A sale price of an item is increased by 10% in December to a new price. In January,

this new price of the item is reduced by 10%. Is the January sale price more, less, or

the same as the November sale price? Justify your answer.

10. Is a price increase of 10% followed by a further increase of 10% equivalent to a one-off

20% price increase on the original price? Justify your answer.

11. The ratio of female to male shoppers at a department store has been found to be 9 to

5. If 5 706 female shoppers were at the store one Saturday, how many shoppers were

there in all that day?

12. On a map, 1cm represents 15km. If two cities are 7cm apart on the map, what is the

actual distance between the two cities?

13. To make green paint, one has to mix 2 parts of blue paint and 3 parts of yellow paint. If

a painter has 60 of blue paint and 100 of yellow paint, what is the maximum amount

of green paint can the painter make?

14. The results of a survey show that 23% of TV viewers in South Africa watch the sit-

com Generations. If there are approximately 7 000 000 viewers, how many people

Generations?

15. Suppose that we have 10 litres of a 20% solution and we want to add a 5% solution to

it to make a 10% solution. How much of the 5% solution do we need to add per litre

of the 20% solution?

16. Find the number of litres of5% solution that must be added to 1 litre of a 40% solution

to obtain a 30% solution.

17. Find the number of litres of water that must be added to 1 litre of a 60% solution to

obtain a 10% solution.

11


17/202

Chapter 2

Real Numbers

2.1 Classification of real numbers

We start this section by reviewing various types of numbers that make up the real number

system.

The set of natural numbers is the set {1, 2, 3, . . .}. We use the symbol N to denotethe set of natural numbers. That is,

N = {1, 2, 3, . . .}.

The set ofwhole numbers is the set {0, 1, 2, 3, . . .}. We use the symbol N0 to denotethe set of whole numbers. That is,

N0 = {0, 1, 2, 3, . . .}.

The set of integers is the set {. . . , 1, 2, 1, 0, 1, 2, 3, . . .}. We use the symbol Zto denote the set of integers numbers. That is,

Z =

{. . . ,

1,

2,

1, 0, 1, 2, 3, . . .

}.

A rational number is any number that can be written as a quotient of two integers.The set of rational numbers is denoted by Q. That is,

Q = {ab

, where a and b are integers, with b = 0}.

We can also define a rational number as a number whose decimal representation either

terminates or repeats.

An irrational number is any number that cannot be written as a quotient of two

integers. Some examples of irrational numbers are: 2, 37, , 0.101001000....

12


18/202

The decimal representation of an irrational number is nonterminating and nonrepeating.

There is no standard notation for the set of irrational numbers.

The set of real numbers is the union of rational numbers and irrational numbers.That is, R = {Rationals} {Irrationals}.

It is clear that every natural number is a whole number. This we denote by

N N0.

The set of integers includes natural numbers and their negatives. Therefore

N0 Z.

Every integer is a rational number in the sense that one can write an integer m in the form

m =

m

1 . It therefore follows thatZ Q.

We therefore have the following containments:

N N0 Z Q R.

We can represent these containments using the following diagram

&%'$

N'&$%N0

'&

$%Z

'

&

$

%Q

'

&

$

%R

Properties of Real Numbers

Let a, b, and c be real numbers.

13


19/202

Property Name of the property

(i) a + b is a real number Closure property for addition(When we add two real numbers we geta real number)

a

b is a real number Closure property for multiplication

(When we multiply two real numbers weget a real number)

(ii) a + b = b + a Commutative property for additiona b = b a Commutative property for multiplication

(iii) (a + b) + c = a + (b + c) Associative property for addition(a b) c = a (b c) Associatative property for multiplication

(iv) a(b + c) = a b + a c Distributive property(a + b)c = a c + b c Distributive property

(v) a + 0 = 0 + a = a Identity property for additiona 1 = 1 a = a Identity property for multiplication

(vi) a + (a) = (a) + a = 0 Additive Inverse propertyFor a = 0, a 1

a=

1

a a = 1 Multiplicative Inverse property

Note: The commutative and associative properties do not hold for subtraction or division.

a (b c) is not equal to (a b) ca (b c) is not equal to (a b) c

a b is not equal to b aa b is not equal to b a

Use some examples to convince yourself that commutative and associative properties may fail

for the operation of subtraction or division.

One of the main aspects of Mathematics is the ability to construct a mathematical argumentto justify (prove) a statement that one makes. A Proof is a sequence of logical statements

aimed at establishing the truth of some statement.

Let us prove that

2 is not a rational number.

Theorem 2.1. The real number

2 is irrational.

Proof. Assume that

2 is a rational number. Then

2 can be written in the forma

bwhere

a and b are integers, with b = 0. That is,

2 = ab

. ()

14


20/202

We assume that a and b have no factors in common except 1. Squarring both sides of (),

we have that

2 =a2

b2or equivalently a2 = 2b2. ()

This means that a2 is an even number. But if the square of a number is even, then the number

itself must be even. That is, a is an even number. This means that a can be written in theform a = 2k, where k is an integer. Therefore, we can write () as

a2 = 2b2 or equivalently (2k)2 = 2b2 or equivalently 4k2 = 2b2. ( )

It follows from ( ) that b2 = 2k2, which shows that b2 is an even number. This, in turn,

implies that b is an even number.

Our argument has established that both a and b are even numbers. This, of course, means

that a and b have a common factor of 2. This contradicts our assumption that a and b have

no common factors except 1. It now follows that our original assumption that 2 is rationalis false. Therefore

2 is irrational.

Note: The method of proof used in the above Theorem is called Proof by Contradiction.

2.2 Graphical representation of real numbers

Real numbers can represented as points on the number line. To every real number corresponds

one and only one point on the number line, and to every point on the real line correspondsone and only one real number.

The number line is such that all the positive and negative integers are represented by a set of

equally-spaced point on a line. All the rational numbers which are not integers also have their

appropriate places on the number line.

>

4

3

2 3

2

1 0 1

21 2 5

23

It is clear that between any two rational numbers we can insert another rational number,

namely their average. Between a and b there is the numbera + b

2. There are infintely many

rational numbers between any two rational numbers.

Example: Find a rational number between 3100

and 311000

(i.e. between 0.03 and 0.031)

Solution: One such number is

3100 + 311000 2 = 612000 = 30510000 = 0, 030515


21/202

There are still points left on the number line which represent irrational numbers.

Let us consider the irrational number

2:

1 0,the equation has two distinct and real roots. By the quadratic formula,

x =b b2 4ac

2a

=(7) (7)2 4(2)(15)

2(2)

=7 49 + 120

4

= 7 1694

= 7 134

.

Therefore

x =7 13

4=

64

=3

2or x =

7 + 13

4=

20

4= 5.

8. Solve the quadratic equation 2x2 + 3x + 1 = 0.

Soln: Here a = 2, b = 3 and c = 1. Since the discriminant

= b2

4ac = 9

8 = 1 > 0,

60


66/202

the equation has two distinct and real roots. Using the quadratic formula,

x =b b2 4ac

2a

=3

(3)2 4(2)

2(2)

=3 9 8

4

=3 1

4=

3 14

.

Therefore

x =3 1

4=

44

= 1 or x = 3 + 14

=2

4=

12

.

9. Solve the quadratic equation 2x2 + 5x 4 = 0.Soln: Here a = 2, b = 5 and c = 4. Since the discriminant

= b2 4ac = 25 + 32 = 57 > 0,the equation has two distinct and real roots. Using the quadratic formula,

x =b b2 4ac

2a

=5

(5)2 4(2)(4)

2(2)

= 5 25 + 324

=5 57

4.

Therefore

x =5 57

4or x =

5 + 574

.

10. Solve the equation 3x2 + x + 1 = 0 using the quadratic formula.

Soln: Here a = 3, b = 1, and c = 1. Since the discriminant is

= b2 4ac = 12 4(3)(1) = 1 12 = 11 < 0,the equaton has no real roots.

11. Solve the equation 3x2 + 2x + 7 = 0 using the quadratic formula.

Soln: Here a = 3, b = 2, and c = 7. Since the discriminant is

= b2 4ac = 22 4(3)(7) = 80 < 0,the equation has no real roots.

61


67/202

12. Find the values ofp for which the quadratic equation px2 + 2px 2 = 0 has

(i) distinct real roots (ii) real repeated root, (iii) non-real roots.

Soln: Here a = p, b = 2p, and c = 2. The discriminant is

= b2

4ac = (2p)2

4p(2) = 4p2

+ 8p = 4(p2

+ 2p).

(i) The equation px2 + 2px 2 = 0 has distinct real roots if and only if the discriminantis positive. That is, if and only if

> 0 4(p2 + 2p) > 0 p2 + 2p > 0 p(p + 2) > 0 p < 2 or p > 0.

In this case, the roots are given by

x = b b2

4ac2a

=2p 4(p2 + 2p)

2p

=2p 2

p2 + 2p

2p

=p

p2 + 2p

p.

Therefore

x = p p2 + 2pp

or x = p + p2 + 2pp

.

(ii) The equation px2+2px2 = 0 has a real repeated root if and only if the discriminantis zero. That is,

= 0 4(p2 + 2p) = 0 p2 + 2p = 0 p(p + 2) = 0 p = 0 or p = 2.

But p = 0. (WHY?) Therefore p = 2. In this case, the real repeated root is

x =2p

2p= 1.

(iii) The equation px2 + 2px 2 = 0 has non-real roots if the discriminant is negative.That is,

< 0

4(p2 + 2p) < 0

p2 + 2p < 0

p(p + 2) < 0

2 < p < 0.

62


68/202

Exercise 4.10. 1. Solve the following equations by Completing the Square:

(a) 4x2 7x + 1 = 0(b) 2x2 x 3 = 0(c) 10x2

x + 2 = 0

(d) 6x2 px 2p2 = 0 (Consider all cases)(e) 2x(x + e) = 3e(x + e) (Consider all cases)

2. Solve the following equations by using the Quadratic Formula:

(a) 2x2 7x 4 = 0(b) 2x2 + 8x + 3 = 0

(c) x2

6x + 2 = 0(d) 4x2 x + 2 = 0(e) 2px(x + 2q) = pqx 2 (Consider all cases.)(f) x2 + (2k 3)x + 2k + 5 = 0 (Consider all cases.)

Nature of the Roots of Quadratic Equation

Consider the quadratic equation

ax2 + bx + c = 0,

with rational coefficients. We have noted above the importance of the expression = b24acas the key to determining the nature of the roots of a given quadratic equation. is called

the discriminant of the quadratic equation ax2 + bx + c = 0.

(i) If > 0 and the square of a rational number, then the roots of the quadratic equaton

are unequal and rational (real).

(ii) If > 0 and not the square of a rational number, then the roots are unequal,

irrational (real).

(iii) If = 0, then the roots are equal, rational (double root).

(iv) If < 0, then the roots are non-real.

Examples 4.11. 1. Determine of each of the following equations and discuss the na-

ture of their roots.

(a) 6x2 7x 3 = 0.Soln: = 49 + 72 = 121 > 0.

63


69/202

Since is a perfect square and coefficients are rational, the roots of the equation

6x2 7x 3 = 0 are unequal and rational.(b) 2x2 3x 1 = 0.

Soln: = 9 + 8 = 17 > 0.

Since is not a perfect square, the roots are unequal and irrational.(c) 4x2 + 4x + 1 = 0.

Soln: = 16 16 = 0.Roots are equal and rational.

(d) 2x2 3x + 2 = 0.Soln: = 9 16 = 7 < 0, the equation has nonreal roots.

2. For the quadratic equation x2 + 2k(x + 1) = 1 2x,

(a) show that its roots are rational for all rational values of k.

(b) determine k if roots are real and equal (double root).

Soln: (a)

x2 + 2kx + 2k + 2x + 1 = 0 x2 + x(2k + 2) + 2k + 1 = 0.The discriminant is

= (2k + 2)2 4(2k + 1) = 4k2 + 8k + 4 8k 4 = 4k2 = (2k)2.

Since is a perfect square and all coefficents are rational, the roots of the quadraticequation x2 + 2k(x + 1) = 1 2x are rational.

(b) For equal roots, we need = 0. Therefore 4k2 = 0 and so k = 0. With k = 0,

the given quadratic equation becomes

x2 + 2x + 1 = 0 (x + 1)2 = 0 x = 1.

3. Find the values of the natural number k for which the roots of 2x2 6x + k = 0 arerational.

Soln: The discriminant is = 36 8k. We need to find the value(s) of k for which is aperfect square. If k = 4, then = 35 32 = 4 a perfect square.

4. Show that the roots of

k2x2 + kx + 2x2 + 2x + 1 = 0

are non-real for all real values of k.

Soln: Note first that

k2x2 + kx + 2x2 + 2x + 1 = 0

x2(k2 + 2) + x(k + 2) + 1 = 0.

64


70/202

Now, the discriminant of this quadratic equation is

= (k + 2)2 4(k2 + 2) = k2 + 4k + 4 4k2 8 = 3k2 + 4k 4.

Now, completing the square, we have

3k2 + 4k 4 = 3k2 43

k + 43 = 3k2 2 2

3 k + 2

32 2

32 + 4

3

= 3

k 23

2 4

9+

4

3

= 3

k 23

2+

8

9

.

Since the expression in the square bracket is always positive, it follows that

= 3k2 + 4k 4 = 3

k 23

2+

8

9

< 0

for all values ofk R. Therefore the roots of the equation k2x2+ kx +2x2+2x +1 = 0are non-real.

5. Determine the values of p for which 3x2 px + 3 = 0 has real roots.Soln: The equation has real roots if and only if its discriminant is nonnegative. That is,

0 p2 36 0 (p 6)(p + 6) 0

p 6 0 and p + 6 0 OR p 6 0 and p + 6 0

p 6 and p 6 OR p 6 and p 6.

Therefore p 6 or p 6.

6. If the equation x2 + 2kx + 2x = 2x 9k has equal roots, find k.

Soln: Note that

x2 + 2kx + 2x = 2x 9k x2 + x(2k + 4) + 9k = 0.

The discriminant for this equation is

= (2k + 4)2 36k = 4k2 + 16k + 16 36k = 4k2 20k + 16.

65


71/202

For the given quadratic equation to have equal roots, its discriminant must be zero.

Therefore

= 0 4k2 20k + 16 = 0 k2 5k + 4 = 0 (k 1)(k 4) = 0 k = 1 or k = 4.

If k = 1, the given quadratic equation becomes

x2 + 6x + 9 = 0 (x + 3)2 = 0 x = 3.

That is, x = 3 is a repeated root.If k = 4, the given quadratic equation becomes

x2 + 12x + 36 = 0 (x + 6)2 = 0 x = 6.

That is, x =

6 is a repeated root.

7. Show that the roots of the equation 3x2 k + 2kx 2x = 0, with k R, can never beequal.

Soln: Note that

3x2 k + 2kx 2x = 0 3x2 + x(2k 2) k = 0.The discriminant for this equation is

= (2k 2)2 + 12k = 4k2 8k + 4 + 12k = 4k2 + 4k + 4 = 4(k2 + k + 1)

= 4

x +

1

2

2+

3

4

> 0. [obtained by completing the square].

That is, the discriminant of the given equatoin is always positive. Therefore = 0 andso the given equation can never have equal roots.

Exercise 4.12. 1. Discuss the nature of the roots of the following equations (do not solve

the equations):

(a) 3x2 x + 4 = 0(b) x2 + 7x + 1 = 0

(c) 2x2 3x + 1 = 0(d) 9x2 + 6x + 1 = 0

(e) rx2 = r x ; r = 0(f) 6x2 7px 5p2 = 0 ; p N

66


72/202

2. For which value(s) of r will x2 5x = r have real roots?

3. If the roots of the equation rx2 + 2rx + 3 = 0 are real and equal, find r.

4. What is the nature of the roots of the following equations?

(a) x2

22x + 2 = 0(b)

3x2 + 4x +

2 = 0

(c)

2x2 +

8x 18 = 0

5. Ax2 + Bx + C = 0 ; B = 0; C 0; A < 0. Discuss the nature of the roots.

6. Show that the roots ofkx2 + kx k = 3x x2 are real for all real values of k.

4.3 Equations leading to Linear or Quadratic Equa-tions : Rational and Irrational Equations

In this section we solve rational/irrational equations that lead to linear or quadratic equations.

When dealing with such equations it is always important to state the appropriate restrictions

for which the expression is defined.

Examples 4.13. 1. Solve for x, ifx 3

x2 + 3x + 2 5

x2 4 =4

x + 1.

Soln: Note that

x 3x2 + 3x + 2

5x2 4 =

4x + 1

x 3(x + 1)(x + 2)

5(x 2)(x + 2) =

4x + 1

.

It is clear that x = 1, x = 2 and x = 2. Now,x 3

x2 + 3x + 2 5

x2 4 =4

x + 1 x 3

(x + 1)(x + 2) 5

(x 2)(x + 2) =4

x + 1

(x 3)(x 2)(x 2)(x + 2)(x + 1)(x + 2)

5(x + 1)(x 2)(x + 2)(x 2)(x + 2) =

4(x + 1)(x 2)(x + 2)x + 1

(x 3)(x 2) 5(x + 1) = 4(x 2)(x + 2) x2 5x + 6 5x 5 = 4x2 + 16 5x2 10x 15 = 0 5(x2 2x 3) = 0 x2 2x 3) = 0 (x + 1)(x 3) = 0 x = 1 or x = 3.

Since x = 1, we must have that x = 3.

67


73/202

2. Solve for x, if16

x2 9 +4

x2 5x + 6 +5(x + 2)

x2 + x 6 = 0.

Soln: Note that, since

16x2 9 + 4x2 5x + 6 + 5(x + 2)x2 + x 6 = 0

16(x + 3)(x + 3)

+4

(x 2)(x 3) +5(x + 2)

(x + 3)(x 2) = 0,

we must have that x = 3 x = 2.Now multiply both sides of the equations by (x 3)(x + 3)(x 2).

16

x2

9

+4

x2

5x + 6

+5(x + 2)

x2 + x

6

= 0

16(x + 3)(x + 3)

+4

(x 2)(x 3) +5(x + 2)

(x + 3)(x 2) = 0

16(x 3)(x + 3)(x 2)(x + 3)(x + 3)

+4(x 3)(x + 3)(x 2)

(x 2)(x 3) +5(x + 2)(x 3)(x + 3)(x 2)

(x + 3)(x 2) = 0

16(x 2) + 4(x + 3) + 5(x + 2)(x 3) = 0

16x

32 + 4x + 12 + 5x2

5x

30 = 0

5x2 + 15x 50 = 0 5(x2 + 3x 10) = 0

x2 + 3x 10 = 0 (x + 5)(x 2) = 0

x = 5 or x = 2.

Since, as noted above, x = 2, it follows that x = 5.

3. Solve for x, if 4x + 2

1x

= 2x 1x2 + 2x

.

Soln: Note that, since

4

x + 2 1

x=

2x 1x2 + 2x

4x + 2

1x

=2x 1

x(x + 2),

68


74/202

x = 2 and x = 0. Now,4

x + 2 1

x=

2x 1x2 + 2x

4x + 2

1x

2x 1x2 + 2x

= 0

4

x + 2

1

x

2x 1

x(x + 2)

= 0

4x (x + 2) (2x 1)x(x + 2)

= 0

4x x 2 2x + 1x(x + 2)

= 0

x 1x(x + 2)

= 0

x 1 = 0

x = 1.

Therefore x = 1 is the solution of the equation4

x + 2 1

x=

2x 1x2 + 2x

.

4. Solve for x, ifx + 6

x2 4 +1

x + 2=

2

x 2 .

Soln: Sincex + 6

x2 4 +1

x + 2=

2

x 2 x + 6

(x 2)(x + 2) +1

x + 2=

2

x 2 ,

we must have that x = 2 and x = 2.Multiply both sides by (x 2)(x + 2):

x + 6

x2 4 +1

x + 2=

2

x 2

x + 6(x 2)(x + 2) +

1

x + 2=

2

x 2

(x + 6)(x 2)(x + 2)(x

2)(x + 2)

+1(x 2)(x + 2)

x + 2=

2(x 2)(x + 2)x

2

x + 6 + x 2 = 2(x + 2) 2x + 4 = 2x + 4 0x = 0.

The last equation is true for all real numbers x. Because of our original restriction that

x = 2, it follows that the solution of the original equation consists of all real numbersexcept 2 and 2.

5. Solve for x, if3

x

1

+4x 1x + 1

=x2 + 5

(x

1)(x + 1)

5.

69


75/202

Soln: Note that x = 1 and x = 1. Multiply both sides of the equation by (x 1)(x + 1):3

x 1 +4x 1x + 1

=x2 + 5

(x 1)(x + 1) 5

3(x 1)(x + 1)

x 1+

(4x 1)(x 1)(x + 1)

x + 1

=(x2 + 5)(x 1)(x + 1)

(x 1)(x + 1) 5(x

1)(x + 1)

3(x + 1) (4x 1)(x 1) = x2 + 5 5x2 + 5

3x + 3 4x2 + 5x 1 = 4x2 + 10 8x = 8 x = 1.

But since x = 1, the given equation has no solutoin.

6. Solve for x, ifx 2ax

3a

= 3 2x2 13a2

x2

9a2

.

Soln: Noting that

x 2ax 3a = 3

2x2 13a2x2 9a2 x

x 2ax + 3a

= 3 2x2 13a2

(x 3a)(x + 3a) ,

we must have that x = 3a. Now, multiplying by (x 3a)(x + 3a):

x 2ax 3a = 3

2x2 13a2x2 9a2 x

x 2ax + 3a = 3 2x2

13a2

(x 3a)(x + 3a)

(x 2a)(x 3a)(x + 3a)x + 3a

= 3(x 3a)(x + 3a) (2x2 13a2)(x 3a)(x + 3a)

(x 3a)(x + 3a)

(x 2a)(x 3a) = 3(x2 9a2) (2x2 13a2) x2 5ax + 6a2 = 3x2 27a2 2x2 + 13a2 5ax = 20a2.

If a = 0, then x = 4a is the solution of the given equation.If a = 0, then 0x = 0. This is true of any real number x. In this case, the set of

solutions of the given equation consists of all real numbers except 3a and 3a.

4.3.1 Irrational Equations

In this subsection we focus on solving irrational equations. An irrational equation is an

equation in which the variable is inside a radical. These are equations like

px + q = r and ax + b px + q = c.

70


76/202

For simplicity, we shal confine our discussion to square roots only.

NOTE: If a2 = b2, then a = b or a = b. Most students make a mistake of saying thatif a2 = b2, then a = b. This is NOT true! That is, a2 = b2 a = b. However, if weknow that a 0 and b 0, then we can conclude that a2 = b2 = a = b. It is for thisreason that when solving an irrational equation such as

px + q = r, we must imposing the

restrictions that px + q 0 and r 0.

Method of solving irrational equations involving square roots:

1. If there is only one radical in the given equation, rewrite the equation so that the radical

is on one side of the equation and all the other terms on the other side. Then square both

sides to get rid of the radical. Solve the resulting equation. Note that the equation that

resulted from the squaring process is not equivalent to the original equation. Therefore

the solutions of the new equation (arising from the squaring process) are not necessarily

solutions of the original equations. All solutions must be checked in the original equation.

2. If there are two radicals, separate them so that one appears on one side of the equation

and the other on the other side together with the other terms. Square both sides of the

equation to get rid of the radical on the one side of the equation. The other part of

the equation may still have a radical. Proceed as in (1) above to handle the remaining

radical.

Examples 4.14. 1. Solve for x, if

5x + 6 = x.

Soln: Note that 5x + 6 0 x 65

(for the square root to be defined). Also, the

right hand side must be nonnegative as we are taking the nonnegative square root on

the left hand side. That is, x 0. From these two observations, we have that x 0.

5x + 6 = x = (5x + 6)2 = x2= 5x + 6 = x2 x2 5x 6 = 0= (x + 1(x 6)) = 0=

x =

1 or x = 6.

From the note in the beginning, we must have that x = 6 is the only solution of the

given equation.

Check: If x = 1, then LHS =

5(1) + 6 = 1 = 1 and RHS = 1. Since the left handside and the right hand side are not the same, x = 1 is not a solution of the givenequation.

If x = 6, then LHS =

5(6) + 6 =

36 = 6 and RHS = 6. Therefore x = 6 is the

only solution of the given equation.

2. Solve for x, ifx + 6 x = 4.

71


77/202

Soln: Rewrite the equation as

x + 6 = x + 4. Note that x + 6 0 x 6 (for thesquare root to be defined). Also, the right hand side must be nonnegative as we are tak-

ing the nonnegative square root on the left hand side. That is, x + 4 0 x 4.From these two observations, we have that x 4.

x + 6 x = 4 = x + 6 = x + 4

= (x + 6)2 = (x + 4)2= x + 6 = x2 + 8x + 16= x2 + 7x + 10 = 0 (x + 5)(x + 2) = 0= x = 5 and x = 2.

Since x 4, it follows that x = 2 is the only solution of the given equation.

Check: If x = 5, then LHS = 5 + 6 (5) = 1 + 5 = 6 = 4 = RHS. Since the lefthand side and the right hand side are not the same, x = 5 is not a solution of thegiven equation.

If x = 2, then LHS = 2 + 6 (2) = 4 + 2 = 4 = RHS. Therefore x = 2 isthe only solution of the given equation.

3. Solve for x, if 2

x 3 = x 3.

Soln: Note that x 3 0 (for the square root to be defined); i.e., x 3. Also, since the lefthand side is a nonnegative square root, the right hand side should also be nonnegative;

i.e. x 3.Now,

2

x 3 = x 3 = 2x 32 = (x 3)2= 4(x 3) = (x 3)2= x2 10x + 21 = 0 (x 3)(x 7) = 0= x = 3 or x = 7.

Since both these numbers are greater or equal to 3, they must both be solutions of the

given equation.

Check: If x = 3, then LHS = 2

2 3 = 0 = 3 3 = RHS. Therefore x = 3 is a solution ofthe given equation.

If x = 7, then LHS = 2

7 3 = 4 = 7 3 = RHS. Therefore x = 7 is a solution ofthe given equation.

4. Solve for x, if

6 + x 7 x = 1.

72


78/202

Soln:

6 + x 7 x = 1 6 + x = 7 x + 1

=

6 + x2

=

7 x + 12=

6 + x = 7

x + 2

7

x + 1

= 7 x = x 1= 7 x2 = (x 1)2= 7 x = x2 2x + 1= x2 x 6 = 0 (x + 2)(x 3) = 0= x = 2 or x = 3.

Check: If x = 2, then LHS = 6 2

7 (2) = 4 9 = 2 3 = 1 = 1 = RHS.Therefore x = 2 is not a solution of the given equation.If x = 3, then LHS = 6 + 3 7 3 = 9 4 = 3 2 = 1 = RHS. Thereforex = 3 is the only solution of the given equation.

5. Solve for x, if

x + 1 x + 6 = 1.

Soln:

x + 1 x + 6 = 1 x + 1 = x + 6 + 1

=

x + 12

=

x + 6 + 12

=

x + 1 = x + 6 + 2

x + 6 + 1

= x + 6 = 3.

The left hand side of the last equation is nonnegative whereas the right hand side is

negative. Therefore the equation has no solution.

Exercise 4.15. 1. Solve the following rational equations:

(a)x + 2

x + 1 3

x

2=

1

x + 1

(b) 1 +6

x + 1+

x + 2

x + 1=

x + 3

x 1 +2x 21 x2

(c)4

x 2 =3x + 6

x2 x 2 +4

x2 + 4x + 3

(d)x + 1

x2 4 +1 xx + 2

=2

5(x 2)

(e) 2x2 + 3x + 2 = 21 x2 + 1x2 + 3x + 2 + 3x2 + x 2

73


79/202

(f)x + 3

x2 4 +1

x2 + x 2 =1

2 x 7

4 x22. Solve for x, if

a

ac + bc+

a b2bx

=a + b

2bc

b

ax + bx.

3. Solve the following irrational equations:

(a)

9 + 4x = 2x 3(b)

3x 11 x = 3

(c) 3

x2 1 = 2 (Hint: a3 = b3 a = b.)(d)

x + 2 = 7 x + 9

(e)

3

2x + 3 = 2(f)

x + 3 x 2 = 7 x

(g)

2 x + x 2 = 1(h) 2x +

8x 3 = 0

(i)

2x + 9 +

3x + 16 = 17

(j)1x

2x + 27

= 0.

4.4 Applications: Problems leading to linear/quadraticequations

In this section we look at some examples of word problems that lead to linear/quadratic

equations.

Examples 4.16. 1. Find two integers whose sum is 4 and whose product is 45.

Soln: Let x and y be the two numbers. Then

x + y = 4 (1.1)

and

xy = 45 (1.2)From (1.1) we have that y = 4 x. Substituting this in (1.2), we have thatx(4x) = 45 4xx2 = 45 x24x45 = 0 (x9)(x+5) = 0.Therefore x =

5 or x = 9. If x = 5, then y = 4

(

5) = 9 and, if x = 9, then

y = 4 9 = 5. the two numbers are 5 and 9.

74


80/202

2. The sum of a number and its reciprocal is

20. Find the number. (Leave you answer

in surd form).

Soln: Let the required number be x . Then x +1

x=

20.

Now,

x +1

x=

20 x

2 + 1

x=

20

x2 + 1 =

20 x

x2

20 x + 1 = 0

x =2

5

202 4

2

x = 2

5 162

x =2

5

4

2 .

That is, x =2

5 42

=

5 2 or x = 2

5 + 4

2=

5 + 2.

3. The printed area of a rectangular page is to be 60% of the total area. If the dimensions

of the page are 50cm by 40cm, find the margin, of constant width, to leave around the

printed area.

Soln: Let x be the width of the margin.

40cm

50cm

x

x

xx' E ' E

Tc

Tc

PrintedAreaPrintedAreaPrintedArea

The printed area is

(40 2x)(50 2x) = 60% of2000 (40 2x)(50 2x) = 1200 2000 180x + 4x2 = 1200 x2 45x + 200 = 0 (x 40)(x 5) = 0.

Therefore x = 40 or x = 5. But 40 2x > 0 x < 20 and so 0 < x < 20. Thereforex = 5cm is the only solution.

75


81/202

4. A man paid R40 for paint at x rand per litre. Two weeks later the price of paint was

increased by R1 per litre and he left the shop with 2 litres less paint than the previous

time having paid R40. Determine x.

Soln: The number of litres of paint that the man bought with his R40 is40

x. Two weeks later,

the number of litres of paint that the man bought with his R 40 is 40x + 1

. Since 40x + 1

is

2 litres less than the amount of paint that the man had got two weeks earlier, we have

that40

x 40

x + 1= 2.

Now,

40

x 40

x + 1= 2 40x(x + 1)

x 40x(x + 1)

x + 1= 2x(x + 1)

40(x + 1) 40x = 2x(x + 1) 40x + 40 40x = 2x2 + 2x 2x2 + 2x 40 = 0 2(x2 + x 20) = 0 x2 + x 20 = 0 (x + 5)(x 4) = 0 x = 5 or x = 4.

Since x > 0, we must have that x = 4.

5. A man travels 180km from his farm to Pretoria in a loaded truck. On the return journey,he is able to travel 30km/h faster having unloaded his produce in Pretoria. He saves 1

hours on the return journey by travelling faster. How fast did he travel to Pretoria?

Soln: Distance =speed time.

Journey Distance Speed Time

From farm to Pretoria 180km x 180x

From Pretoria to farm 180km x + 30180

x + 30

Now we can form the time equation in hours:

180

x

=180

x + 30

+ 1, x > 0.

76


82/202

Now,

180

x=

180

x + 30+ 1 180x(x + 30)

x=

180x(x + 30)

x + 30+ x(x + 30)

180(x + 30) = 180x + x(x + 30)

180x + 5400 = 180x + x2 + 30x x2 + 30x 5400 = 0 (x 60)(x + 90) = 0 x = 60 or x = 90.

We discard the value x = 90 as the speed cannot be negative. Hence x = 60km/h.6. A tank has two pipes entering it. When operating together, the tank is filled in 40

mintures. Operating independently, one of the pipes fills the tank 60 minutes faster

than the other is able to on its own. Find the time taken by each pipe to fill the tankon its own.

Soln: Let x be the time (in minutes) taken by slower pipe to fill the tank on its own. Then

x 60 is the time taken by the faster pipe to fill the tank on its own. 1x

is the fraction

of the tank filled by the slower pipe in 1 minute and1

x 60 is the fraction of the tank

filled by the faster pipe in 1 minute.1

40is the fraction of the tank filled by both pipes

in 1 minute. Now we form an equation.

In 1 minute, we have 1

x+

1

x 60 =1

40.

Now,

1

x+

1

x 60 =1

40 40x(x 60)

x+

40x(x 60)x 60 =

40x(x 60)40

40(x 60) + 40x = x(x 60) 40x 2400 + 40x = x2 60x

x2

140x + 2400 = 0 (x 120)(x 20) = 0 x = 120 or x = 20.

We discard the value of x = 20 as this will make x 60 negative. Therefore x = 120minutes. That is, the slower pipe takes x = 120 minutes to fill the tank on its own while

the faster pipe takes x = 120 60 = 60 minutes to fill the tank on its own.

Exercise 4.17. 1. A man calculated that if his speed had been increased by 4km/h he

would have completed his 48km journey 1 hour sooner. Determine his speed.

77


83/202

2. One water pipe fills a swimming bath in x hours. A second pipe takes 3 hours longer

to fill the bath. The two pipes together fill the bath in two hours. How long will it take

each pipe to fill the swimming bath separately?

3. The lengths of three sides of a right-angled triangle are consecutive integers. Find the

lengths.

4. The length of a rectangle is three times its width. If the length is increased by 2cm

and the width by 1cm, the new rectangle has area 70cm2. Find the dimensions of the

rectangle.

5. Find two consecutive integers whose product is 56.

6. A man jogs for 10km and then walks for 10km. He jogs 212

km/hr faster than he walks.

The entire distance of 20km takes 6 hours. Find his jogging and walking speeds.

7. It takes a woman twice as long to do a job as a man. Working together, they take 40minutes to do the job. How long would it take them individually to complete the job?

8. A boat can travel 30km/h in still water. The boat takes one hour to travel 12km

upstream and back again. Find the speed at which the stream is flowing. (Leave your

answer in surd form.)

78


84/202

Chapter 5

Linear and Quadratic Functions

5.1 Concept of Function

The concept of a function is central in Mathematics. A function is used to represent a

relationship between object.

Definition 5.1. A function from a set A into a set B is a rule that assigns to each

element x in A exactly one element, denoted by f(x), in B. The numberf(x) is called

the value of f at x. The set A is calledthe domain of the function f. We shall denote

by Dom(f) the domain of the function f. Therange of f is the set of all possible values

of f(x) as x varies throughout the domain.

If f is a function with domain A, then itsgraph is the set of ordered pairs

{(x, f(x))|x A}.

'

&

$

%

'

&

$

%

A B

z

z

$$$$$$

$X

'

&

$

%

'

&

$

%

A B

E

E

E

In this course we consider functions for which the sets A and B are sets of real numbers.

79


85/202

Let f be a function from a set A into a set B. When deciding on the domain of a function

f, always bear in mind that

1. Division by zero is not defined. Any value of the variable x that makes the denominator

cannot be in the domain of f.

2. Square roots of negative numbers are not real numbers. Any value of the variable x

that makes the expression under the square root negative cannot be in the domain of

the function f.

How does one decide whether a given curve in the plane is the graph of a function or not?

Vertical line test: A curve in the plane is the graph of a function if and only if no vertical

line cuts that curve in more than one point.

Put differently: if a vertical line cuts the curve at two or more points, then that curve is not

a graph of a function.

Examples 5.2. 1. The graph of a function f is shown below.

(a) Find the values of f(1) and f(2).(b) What are the domain and range of f?

T

E...

...

...

..

.

.... . .. . .

1

3

y

x0...

...

...3

. . .. . .. . . 2......

...42...

.... . .. . .. . .1

Soln: (a) f(1) = 3 and f(2) = 1.(b) The domain of f is the interval [3;4] and the range of f is the interval [2;3].

2. Find the domain of each of the following functions.

(a) f(x) =1

(x 2)(x 3) (d) k(x) =

x2 4

(b) g(x) =

2x 1 (e) p(x) = 1x

+1

x2 9

(c) h(x) =

x

x 2 (f) q(x) =

3 xx + 1

Soln:

80


86/202

(a) The domain of f is the set of all real numbers except 2 and 3. That is,

Dom(f) = {x R | x = 2, x = 3} = (, 2) (2, 3) (3, +).

(b) The square root function is defined only for nonnegative numbers. Therefore, for

the function g to be defined, we require that 2x

1

0, and so x

1

2. Therefore

the domain of g is the set of all real numbers that are greater than or equal to1

2,

i.e.,

Dom(g) =

x R | x 1

2

=

1

2,

.

(c) The square root function is defined only for nonnegative numbers. Therefore, for

the function h to be defined, we require that x 0. But since h is a rationalfraction, we must also require that the denominator is not zero, i.e., x = 2.Therefore the domain of h is

Dom(h) = {x R | x 0 and x = 2} = [0, 2) (2, ).

(d) The square root function is defined only for nonnegative numbers. Therefore, for

the function k to be defined, we require that x2 4 0. Now,

x2 4 0 (x 2)(x + 2) 0 x 2 0 and x + 2 0 OR x 2 0 and x + 2 0 x 2 and x 2 OR x 2 and x 2

x

2 OR x

2.

It now follows that the domain of the function k is the set of all real numbers that

are greater than or equal to 2 or less than or equal to 2. That is,

Dom(k) = {x R | x 2 or x 2} = (, 2] [2, ).

(e) The function p is a sum of two rational fractions. For the function p to be defined,

we require that the denominators should not be zero; i.e., x = 0 and x2 9 = 0.Equivalently, x = 0 and x = 3. Therefore, the domain of the function p is theset of all real numbers x except 0,

3 and 3. That is,

Dom(p) = {x R | x = 0, x = 3, x = 3} = (, 3)(3, 0)(0, 3)(3, ).

(f) Note that x = 1. (Why?) For the square root function to be defined, we requirethat

3 xx + 1

0. Now,

3 xx + 1

0 3 x 0 and x + 1 > 0 OR 3 x 0 and x + 1 < 0 3 x and x > 1 OR 3 x and x < 1 1 < x 3,

81


87/202

since there is no real number that is simultaneously greater than 3 and less than

1. We noted in the beginning of the solution that x = 1. Therefore, thedomain of q consists of real numbers that are strictly bigger than 1 and less thanor equal to 3. That is,

Dom(q) = {x R

| 1 < x 3} = (1, 3].3. Determine whether the given curve is the graph of a function ofx or not.

T

E

y

x0

(a)

T

E0

y

x

(b)

T

E

y

x0

DDDDD

DDDDD

(c)

T

E0

y

x

(d)

T

E

&&&&&&

y

x0

(e)

T

E0

y

x

(f)82


88/202

Soln: (a) Yes (b) No (c) Yes (d) No (e) Yes (f) No.

4. Plot the graphs of the following functions f : X Y where X = {2, 1, 0, 1, 2} andY = Z.

(a) f(x) = 2x (b) f(x) = x + 1 (c) f(x) = 1 (d) f(x) = x2.

Soln:

T

E| | | |

y

x0

(a)

12 1 212

34

1

23

4

T

E| | | |

y

x0

(b)

12 1 212

34

1

23

4

T

E| | | |

y

x0

(c)

12 1 212

1

2

T

E| | | |

y

x0

(b)

12 1 21

1

2

3

4

83


89/202

5. Let X = {1, 2, 3, 4, 5, 6} and Y = Z. Plot the graph of the following functionsf : X Y

(a) f(x) =

1 ifx is even0 ifx is odd.

(b) f(x) = 1 ifx is divisible by 31 otherwise.Soln:

T

E

| | | | | | |

y

x0

(a)

1 1 2 3 4 5 612

1

2

T

E

| | | | | | |

y

x0

(b)

1 1 2 3 4 5 612

1

2

5.2 Properties of Functions

5.2.1 Increasing and Decreasing Function

Definition 5.3.

1. A function f is said to beincreasing on an interval[a, b] iff(x1) < f(x2) whenever

x1 < x2 in [a, b].

2. A function f is said to bedecreasing on an interval[a, b] iff(x1) > f(x2) whenever

x1 < x2 in [a, b].

3. A function f is said to be constant on an interval [a, b] if f(x1) = f(x2) whenever

x1 < x2 in [a, b].

Intuitively, a function f is increasing on an interval [a, b] if the graph of f goes up as one

moves from left to right in the interval [a, b]; and decreasing if its graph goes down as one

moves from left to right in the interval [a, b]. A function f is constant on the interval [a, b] if

it remains horizontal as one moves from left to right in the interval [a, b].

84


90/202

T

E

y

x0

Increasing function

T

E

y

x0

Decreasing function

T

E

y

x0

Constant function

5.2.2 Even and Odd Functions

Definition 5.4. 1. A function f is said to be even if f(x) = f(x) for every realnumber x in its domain.

2. A function f is said to be odd if f(x) = f(x) for every real number x in itsdomain.

The graph of an even function is symmetric with respect to the y-axis; the graph of an oddfunction is symmetric with respect to the origin.

T

E

y

x0

Graph of an even function issymmetric about the y-axis

T

E

y

x0

Graph of an odd function issymmetric about the origin

Examples 5.5. Determine whether each of the functions is even, odd or neither.

1. f(x) = x2 + x4

2. f(x) = x3 x

85


91/202

3. f(x) =

4 x2

4. f(x) = 3

x

5. f(x) = x4 + x.

Soln: 1. Since f(x) = (x)2 + (x)4 = x2 + x4 = f(x), the function f is even.

2. Since f(x) = (x)3 (x) = x3 + x = (x3 x) = f(x), the function f isodd.

3. Since f(x) = 4 (x)2 = 4 x2 = f(x), the function f is even.4. Since f(x) = 3x = 3x = f(x), the function f is odd.

5. Since f(x) = (x)4 + (x) = x4 x = (x4 + x) = f(x), f is not an oddfunction.

Similarly, since f(x) = (x)4 + (x) = x4 x = x4 + x = f(x), f is not an evenfunction.

It follows that the function f is neither odd nor even.

Note: The functions f(x) = cos x and f(x) = sec x are even, whereas the functions

f(x) = tan x, f(x) = cot x, f(x) = sin x and f(x) = cosec x are odd.

5.2.3 Zeroes of functions

Definition 5.6. Let f be a function from a set A into a set B. A zero of the function f

is a number x0 in A such that f(x0) = 0.

Examples 5.7. Find zeroes of the following functions.

1. f(x) = 3x + 1 (b) f(x) =

9 x2 (c) f(x) = x + 2x 1 (d) f(x) =

1

x + 1.

Soln: 1. Let f(x) = 3x + 1. We want to solve the equation f(x) = 0. Now

f(x) = 0 3x + 1 = 0 x = 13

.

It follows that x = 13

is the zero of the function f(x) = 3x + 1. Indeed,

f

1

3

= 3

1

3

+ 1 = 1 + 1 = 0.

86


92/202

2. Let f(x) =

9 x2. We want to solve the equation f(x) = 0. Now,

f(x) = 0

9 x2 = 0 9 x2 = 0 (3 x)(3 + x) x = 3.

Therefore x = 3 and x = 3 are zeroes of the function f(x) = 9 x2. Indeed,

f(3) = 9 (3)2 = 9 9 = 0 andf(3) =

9 (3)2 = 9 9 = 0.

3. Let f(x) =x + 2

x 1 . We want to solve the equation f(x) = 0. Now,

f(x) = 0 x + 2x 1 x + 2 = 0 x = 2.

Therefore x = 2 is the zero of the function f(x) = x + 2x 1 . Indeed,

f(2) = (2) + 2(2) 1 =

0

3 = 0.

4. Let f(x) =1

x + 1. If we try to solve the equation f(x) = 0, we get that 1 = 0, which

is absurd. Therefore the function f(x) =1

x + 1= 0 has no zeroes.

Exercise 5.8. 1. Iff(x) = 4x + 1, find f(0), f(2), f(2).

2. Ifg(x) = x2 x + 1, find g(2), g(k), g(x).

3. The domain of a function f is A = {1, 2, 3, 4, 5} and f(1) = 2, f(2) = 1, f(3) = 0,f(4) = 1, f(5) = 2. What is the range off? Draw an arrow diagram and a graph of f.

4. Find the domain of the following functions.

(a) f(x) = x2 x (d) k(x) =

x2 x

(b) g(x) =1

x2 + x 6 (e) (x) =3

x + 2

(c) h(x) = 4

3x + 4 (f) p(x) =

x 1

2x 5 .

5. Determine whether each of the following functions is even, odd or neither.

(a) f(x) = x7 + x (d) F(x) = sin xx

87


93/202

(b) g(x) = 2 3x4 (e) G(x) = x cos x(c) h(x) = 5x + 3x2

6. Find zeroes of each of the following functions.

(a) f(x) =x2 1

x(d) f(x) =

x + 1x

(b) f(x) =1

x2 4 (e) f(x) = x2 5x + 6

(c) f(x) =x

x2 + 1(f) f(x) =

2 x + x 2

7. Let f : X Y be a function, where X = {1, 2, 3, 4, 5, 6, 7, 8, 9} and Y = Z.Draw graphs of the following functions.

(a) f(x) =

x2

ifx is even

1 ifx is odd

(b) f(x) =

x + 1 ifx is divisible by 3

x 1 otherwise

(c) f(x) = 1 ifx is even

0 ifx is odd

5.3 Linear Function

In this section we discuss linear functions and their graphs.

Definition 5.9. A linear function in the variablex is any function f that can be writtenin the form

f(x) = ax + b, where a and b are real numbers.

If a = 0 then the linear function f is called a polynomial of first degree.

Here are some examples of linear functions:

f(x) = 2x + 1, g(x) = 6x, h(x) = 3, k(x) = 12

x 23

.

The graph of a linear function is a straight line. Graphing a linear function is quite easy.

88


94/202

The y-intercept of a straight line is the point where the graph of the line cuts the y-axis.

The y-intercept of the straight line y = ax + b is (0, b).

The x-intercept of a straight line is the point where the graph of the line cuts the x-axis.

The x-intercept of the straight line y = ax + b is

b

a, 0

.

Consider the straight line

y = ax + b, where a = 0.The number a is called the slope (or gradient) of the straight line y = ax + b.

If a > 0, i.e., the slope is positive, then the graph of the function f(x) = ax + b rises as one

moves from the left to the right.

If a < 0, i.e., the slope is negative, then the graph of the function f(x) = ax + b falls as one

moves from the left to the right.

5.3.1 Sketching a graph of a linear function

In this subsection we discuss two easy methods of sketching a graph of a linear function. We

shall call these the intercepts method and the slope-intercept method.

Suppose that we want to sketch the graph of

y = ax + b, where a = 0.

Method of Intercepts: Find the x-intercept by setting y = 0. The x

intercept is

b

a

, 0.Find the y-intercept by setting x = 0. The y-intercept is (0, b).Plot these two points on your graph and draw a straight line between them.

T

E

y

x0

(((

((((((((((

(

b

a

, 0)

(0, b)

Method of Intercepts

Slope-intercept Method The slope a of the straight line y = ax + b, where a = 0, is givenby

a =

vertical change

horizontal change =

change in y

change in x .

89


95/202

Find the y-intercept of the straight line. It is (0, b).

T

E

y

x0&&&&&&&

&&&&&&&

(0, b) 1

a

Slope-intercept Method

Examples 5.10. 1. Use the two methods discussed above to sketch the graphs of the

following linear functions.

(a) y = 2x + 3 (b) y = 13

x 1.

Soln: (a) Using the Method of Intercepts: For x intercept, set y = 0. The x-intercept is32

, 0

.

For the y-intercept, set x = 0. The y-intercept is 0, 3).

Plot these two points and join them by a straight line.

T

E

y

x0

eeeeeeeeeee

eee

(0, 3)

(32

, 0)

Using the Slope-intercept method: Note that the slope of the straight line y =

2x + 3 is a =

2 =

21

and its y-intercept is (0, 3).

T

E

y

x0

e

eeeeeeeeeeeee

(0, 3)1

2

90


96/202

(b) Using the Method of Intercepts: For the x-intercept, set y = 0. Then1

3x 1 = 0 x = 3. Therefore, the x-intercept is (3, 0).

For the y-intercept, set x = 0. Then y = 1. That is, the y-intercept is (0, 1).Plot these two points and join them by a straight line.

T

E

y

x0

(3, 0)

(0, 1)

Using the Slope-intercept method: The slope of the line y = 13

x 1 is a = 13

and

its y-intercept is (0, 1).T

E

y

x0

(0, 1)

3

1

2. On the same system of axes, sketch the graph of

(a) y = 2x + 3 (b) y = 2x (c) y = 2x 3.

Soln: Slope = 2 = 21

.

T

E

y

x0

eee

eeeeeeeeeeeeeee

eee

eee

eeeeeeeeeeeeeee

eee

eee

eeeeeeeeeeeeeee

eeee

y

=2x

y

=2x

3

y

=2x

+3

91


97/202

Since these three lines have the same slope of 2, they are parallel.

3. Let f(x) = 3x 2; x R.

(a) Find f(0), f(1), f(1) and indicate them on the graph of f.(b) Find the values of x for which

(i) f(x) = 0 (ii) f(x) > 0 (iii) f(x) < 0.

(c) Is f increasing or decreasing in R?

Soln: (a) f(0) = 2; f(1) = 3 1 2 = 1; f(1) = 3(1) 2 = 5.

T

E

y

x0

-

-

-

-

-

-

1

12345

-

1-

1

f(1)

f(0)

f(1)

(b) (i) f(x) = 0 3x 2 = 0 x =2

3 .

(ii) f(x) > 0 = 3x 2 > 0 = x > 23

.

(iii) f(x) < 0 = 3x 2 < 0 = x < 23

.

(c) f(x) = 3x 2 is increasing in R since its slope, a = 3, is positive.

4. Find the linear function f(x) = ax + b such that f(3) = 1 and f(2) = 4.

Soln:

1 = a(3) + b4 = a 2 + b = 3a + b = 1

2a + b = 4From these two equations, we have that a = 1 and b = 2. Therefore f(x) = x2.

5. The graph of the function f(x) = ax + b passes through the points (1, 4) and (1, 2).Find the function f.

Soln:

92


98/202

4 = a 1 + b

2 = a(1) + b=

a + b = 4

a + b = 2.From these two equations we have that a = 3 and b = 1. Therefore f(x) = 3x1.

6. Draw the graphs of the following functions.

(a) f(x) =

x + 1 if x 0

x + 1 if x < 0

(b) f(x) =

2x 2 if x < 2

2 if 2 x < 1

x + 3 if x 1Soln:

T

E

y

x0

dd

dd

ddd

1

y=

x+1

y=

x+1

(a)

T

E

y

x0

ddddddddd

eeeeee

y

=2x

2

y=

x+3

-1

-2

-

1-

2-

31--2

(b)

7. The relationship between the temperature measured in Celsuis (C) and Fahrenheit (F)

is given byC

5

=F 32

9

.

(a) Obtain C as a function of F.

(b) Obtain F as a function of C.

(c) What is

(i) 100 degrees F in terms of degrees C?

(ii) 32 degrees F in terms of degrees C?

(iii) 100 degrees C in terms of degrees F?

(iv) 30 degrees C in terms of degrees F?

93


99/202

Soln: (a) C =5

9F 160

9(Linear function of F).

(b) 9C = 5F 160 5F = 9C+ 160 F = 95

C+ 32.

(Linear function of C).

(c) (i) C = 59

F 1609

; C =5

9 100 160

9 37.8.

(ii) C =5

9 32 160

90= 0.

(iii) F =9

5C+ 32; F =

9

5 100 + 32 = 212.

(iv) F =9

5 30 + 32 = 86.

We can solve a sytem of two linear equations in two unknowns

(i) algebraically and (ii) graphically.

Two linear functions either (i) intersect at one point, (ii) are parallel (have the same slope) or

(iii) are coincident (they represent the same straight line).

Example 5.11. Solve the following systems of linear equations.

(a)

x + y = 2 . . . (1)

x y = 0 . . . (2)

(b)

x + y = 1 . . . (1)

x y = 1 . . . (2)

(c) 2x

y = 0 . . . (1)

2x + y = 1 . . . (2)

Soln: (a) Consider the system:

x + y = 2 . . . (1)

x y = 0 . . . (2)

94


100/202

Algebraic solution: Adding equations (1) and (2), we have that 2x = 2 and so

x = 1. Substituting the value x = 1 in equaton (2), we have that y = 1. Therefore

the solution of the given system if (1, 1).

Graphical solution:

T

E

y

x0

dd

dd

dd

dd

dd

dd

d

(1, 1)1

y=

x

1

...

...

...y=

x+2

(b) Consider the system: x + y = 1 . . . (1)

x y = 1 . . . (2)Algebraic solution: Equations (1) and (2) stand for the same line x y = 1.Therefore this system has infinitely many solutions.

Graphical solution:

T

E

y

x0

-

1

xy

=1

1

(x, x 1)

(c) Consider the system: 2x y = 0 . . . (1)

2x + y = 1 . . . (2)Algebraic solution: Adding equations (1) and (2), we have that 0

x + 0

y =

1 0 = 1, which is absurd. Therefore the system has no solution.

95


101/202

Graphical solution:

T

E

y

x0

y=2

x

y=2x

+1

Exercise 5.12. 1. Draw the graphs of the following functions.

(a) y = x 2 (b) y = 12x + 1

(c) y = 2x + 4 (d) y = 23

x + 2

(e) f(x) =

2x 1 if x 0

1 if x < 0

(f) f(x) = 3x + 3 if x 1

0 if 1 x < 13x + 3 if x < 1

(g) f(x) =

x + 4 if x 1

2x + 6 if 1 < x < 5

x 6 if x 52. Points A(1, 2) and B(1, 4) lie on the graph of y = ax + b. Determine whether point

C(2, 6) lies on this graph.

3. The graph of the linear function y = ax + b is parallel to the graph of y = 2x + 1 and

is passing through A(1, 5). Find the function.

4. Show that ifA(p, q) lies on the graph of y = x + 5 then p + q = 5.

5. For what values(s) of m is y = (2m 3)x + 1.

(a) increasing?

(b) decreasing?

(c) constant?

96


102/202

6. Let f(x) = 2x 5. Solve the following inequalities graphically.

(a) f(x) > 0.

(b) f(x) < 0.

(c) f(x)

1.

(d) f(x) < 3.

7. Let f(x) = x + 5 and g(x) = 2x + 10.

(a) Draw the graphs of f and g on the same system of axes.

(b) Use the graphs of f and g to find the values of x for which

(i) f(x) = g(x).

(ii) f(x) g(x).

8. Solve the following systems graphically.

(a)

x + 2y = 1

x y = 2(b)

3x + y = 2

x y = 49. There is a linear function which describes population growth in town. Assume that the

population changes in time t linearly. This means that the population at any time t is

given by P(t) = at + b, (t in years).

Given that the population size in 1990 was 30 000 people and in 2000 it was 32 000

people.

(a) Find the linear function P.

(b) What is the population size in 2004?

10. The monthly cost of driving a car depends on the number of miles driven. Jane found

that in May it cost her $380 to drive 480 miles and in June it cost her $460 to drive

800 miles.

(a) Express the monthly cost C as a function of the distance driven d, assuming that

a linear function gives a suitable model.

(b) Use part a) to predict the cost of driving 1500 miles per month.

(c) Draw the graph of the function.

(d

maths basics 2010

Documents