Mind Action Series Maths Grade 10

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Mathematics Textbook Grade 10 NCAPS (Revised - New Edition)
ISBN 13:
Print:
E-pub:
PDF:
9781776112098
9781776112104
9781776112111
Product Code:
MAT 216
Author:
M.D Phillips
J Basson
C Botha
J Odendaal
First Edition:
August 2016
PUBLISHERS
ALLCOPY PUBLISHERS
P.O. Box 963
Sanlamhof, 7532
Tel: (021) 945 4111, Fax: (021) 945 4118
Email: [email protected]
Website: www.allcopypublishers.co.za
i
MATHS TEXTBOOK
GRADE 10 NCAPS INFORMATION
(Revised - New Edition)
The AUTHORS
MARK DAVID PHILLIPS
B.SC, H Dip Ed, B.Ed (cum laude) University of the Witwatersrand.
Mark Phillips has over twenty-five years of teaching experience in both state and private schools and has a
track record of excellent matric results. He has presented teacher training and learner seminars for various
educational institutions including Excelearn, Isabelo, Kutlwanong Centre for Maths, Science and
Technology, Learning Channel, UJ and Sci-Bono. Mark is currently a TV presenter on the national
educational show Geleza Nathi, which is broadcast on SABC 1. He has travelled extensively in the United
Kingdom, Europe and America, gaining invaluable international educational experience. Based on his
experiences abroad, Mark has successfully adapted and included sound international educational
approaches into this South African textbook. The emphasis throughout the textbook is on understanding
the processes of Mathematics.
JURGENS BASSON
B.A (Mathematics and Psychology) HED RAU (UJ)
Jurgens Basson is a Mathematics consultant with more than 25 years experience and expertise in primary,
secondary and tertiary education. His passion is the teaching of Mathematics to teachers, students and
learners. Jurgens founded and started the RAU / Oracle School of Maths in 2002 in partnership with
Oracle SA. Due to the huge success of this program in the community, there are now several similar
programs running in the previously disadvantaged communities that are being sponsored by corporate
companies. During 2012, one of his projects sponsored by Anglo Thermal Coal in Mpumalanga was voted
Best Community Project of the Year. Jurgens had the privilege to train and up skill more than 15 000
teachers countrywide since 2006. He was also part of the CAPS panel that implemented the new syllabus
for Grade 10 - 12 learners. He successfully co-authored the Mind Action Series Mathematics textbooks,
which received one of the highest ratings from the Department of Education.
CONRAD BOTHA
Maths Hons, PGCE, B.Sc (Mathema cs and Psychology) Rand Afrikaans University.
Conrad Botha has been an educator in both state and private schools for the past eight years. He is
currently Head of Department at The King’s School West Rand and has also been extensively involved in
teaching Maths at The Maths Centre (Grades 4-12). The institution tutors both primary and high school
learners in Maths, Science as well as conducts examination training sessions for Grade 11 and 12
learners. The experience he obtained at The Maths Centre resulted in him gaining valuable knowledge and
teaching methodology, particularly in the implementation of the new curriculum. Conrad has also done
editing work for the Learning Channel.
JACO ODENDAAL
B.Sc (Mathematics and Applied Mathematics)
Jaco Odendaal has been an educator of Mathematics and Advanced Programme Mathematics for the past
ten years. He is currently the Head of Mathematics at Parktown Boys’ High School in Parktown,
Johannesburg. He has worked as a teacher trainer in many districts in Limpopo and Mpumalanga. Jaco
has a passion for Mathematics and the impact it can have in transforming our society. He has been
successful in teaching and tutoring Mathematics at all levels, from primary school to university level. He
has also worked with learners at both ends of the spectrum, from struggling to gifted learners. He is known
for his thorough, yet simple explanation of the foundational truths of Mathematics.
ii
MATHS TEXTBOOK
GRADE 10 NCAPS INFORMATION
(Revised - New Edition)
The CONCEPT
This revised edition of the Mind Action Series Mathematics Grade 10 is suitable for both DoE and IEB
schools and is completely CAPS aligned. It contains the following exciting new features:
•
The approach to Functions is more effective and includes more examples and exercises on graph
interpretation. The axes of symmetry for hyperbolas has also been included.
•
Euclidean Geometry contains more challenging questions. The reasons used for statements are
aligned with the requirements of the DoE.
•
Financial Mathematics contains exercises that have realistic interest rates and population growth
has been included.
•
Statistics is a major improvement and percentiles are discussed in detail. The examples and
exercises are more relevant to the modern world.
•
Probability is far more learner-friendly. Teaching Venn diagrams using the methods discussed will
lead to a greater understanding of the rules.
•
The chapter on Measurement is a major improvement on the original chapter.
•
The other topics (Trigonometry, Analytical Geometry, Number Patterns and Algebra) have also
been improved.
ENDORSEMENT
The original textbook was great, but this revised book is in a league of its own. The approaches used help
learners get to the point quickly and completing the syllabus will be a breeze. The layout of the book is
user-friendly and the exercises are not too little and not too much. The consolidation and extension
exercises are brilliant and are a must for the top kids. I am definitely going to switch to this new book. Well
done to the authors!
Mrs Judy Ennis (HOD Maths)
iii
MATHEMATICS
TEXTBOOK
GRADE 10 NCAPS
(Revised - New Edition)
CONTENTS
CHAPTER 1
ALGEBRAIC EXPRESSIONS
1
CHAPTER 2
EXPONENTS
32
CHAPTER 3
NUMBER PATTERNS
41
CHAPTER 4
EQUATIONS AND INEQUALITIES
46
CHAPTER 5
TRIGONOMETRY
62
CHAPTER 6
FUNCTIONS
103
CHAPTER 7
EUCLIDEAN GEOMETRY
163
CHAPTER 8
ANALYTICAL GEOMETRY
188
CHAPTER 9
FINANCIAL MATHEMATICS
206
CHAPTER 10
STATISTICS
219
CHAPTER 11
MEASUREMENT
244
CHAPTER 12
PROBABILITY
257
CHAPTER 13
SOLUTIONS TO EXERCISES
274
iv
CHAPTER 1
ALGEBRAIC EXPRESSIONS
REVISION OF THE REAL NUMBER SYSTEM
In Grade 9 you learnt about real numbers which consist of rational and irrational numbers.
Let’s briefly revise these numbers.
Rational numbers
a
where b ≠ 0 and
b
A rational number is a number that can be expressed in the form
where a and b are integers.
Rational numbers include all of the following numbers which can be expressed as common
fractions.
(a)
Integers, whole numbers and natural numbers
6
−2
0
For example: 6 =
−2=
0=
1
1
1
Mixed numbers
1 5
For example: 2 = where 5 and 2 are integers
2 2
Terminating decimals
125 1
For example: 0,125 =
=
1000 8
Recurring decimals
A recurring decimal has an infinite pattern. For example, 0,1 = 0,111111... and
  = 0,52525252........... are examples of recurring decimals.
0, 52
(b)
(c)
(d)
A recurring decimal like 0,111111.... can be expressed as a fraction in the form
a
.
b
Irrational numbers
Irrational numbers are non-terminating, non-recurring decimals. They cannot be expressed
as a ratio between integers. Examples include:
•
•
2, 6, 8
Square roots of numbers that are not perfect squares. For example:
Cube roots of numbers that are not perfect cubes. For example:
3
2 , 35 , 39
The following number line contains a few real numbers:
−2
−3
−1, 5
−1
−1
0
0
1
2
3
1
2
3
2
2
π
Not all calculations will produce real numbers. There are two such calculations:
•
Square roots of negative numbers do not produce real numbers.
−2; −3; −4; −π ; ..
These numbers exist, but don’t have a position on the number line. We call them nonreal numbers. This can be extended to any even root of a negative number.
•
Division by zero does not produce a real number.
There are no real numbers resulting from dividing by zero. Division by zero is
undefined.
Summary of the REAL Number System
1.
Natural numbers:
 = {1; 2;3; 4;5;...}
2.
Whole numbers:
 0 = {0;1; 2;3; 4;5;...}
3.
Integers:
 = {...; −3; −2; −1; 0;1; 2;3;...}
4.
Rational numbers:
a

 =  : a ∈  ; b ∈  ; b ≠ 0
b

- Whole numbers and integers
- Proper fractions
- Improper fractions and mixed numbers
- Terminating decimals
- Recurring decimals
5.
Irrational numbers: |
- Non-terminating, non-recurring decimals.
- Square roots of numbers that are not perfect squares, cube roots of numbers that
are not perfect cubes etc.
-π
6.
Real numbers: 
Any number on the number line. All rational and irrational numbers put together.
7.
Calculations that do NOT produce real numbers:
- Square roots of negative numbers
- Division by zero
EXAMPLE 1
State whether the following numbers are rational, irrational or neither:
(a)
0,25
−16
(e)
(b)
−2
16
(f)
(i)
0, 543215432154321...
(k)
3
−27
(l)
3
(c)
π+6
(g)
0
(h)
π
3
(n)
(j)
−9
(m)
2
(d)
10
5
0
0, 7931156480518346...
9
16
Solutions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
Rational (terminating decimal)
Rational (integer)
Irrational (the sum of an irrational number and a rational number is always irrational)
Irrational (10 is not a perfect square, therefore the square root of 10 is irrational)
Neither (16 is a perfect square but, a square root of a negative is non-real)
Rational (16 is a perfect square, therefore the square root of 16 is rational)
Rational (whole number)
Neither (division by 0 is undefined)
Rational (recurring decimal)
Irrational (non-terminating, non-recurring decimal)
Rational (this equals −3 and −3 is an integer and an integer is rational)
Irrational (real since cube roots of negatives are real, but, since 9 is not a perfect
cube, the cube root of −9 is irrational).
Irrational (any fraction of an irrational number is irrational)
Rational (9 and 16 are perfect squares and 169 = 43 which is a fraction)
EXAMPLE 2
Show that the following numbers are rational:
(a)
0,1
(b)

1, 75
Solutions
(a)
0,1 can be shown to be rational by expressing it as a common fraction.
Let x = 0,1111111...
Now what you need to do is multiply both sides by 10, 100, 1000 and so forth to
get the following equations:
x = 0,111111111...
(1)
10 x = 1,111111111...
(2)
100 x = 11,111111111...
(3)
Look for two equations where the decimals after the comma are the same and
then subtract the equations.
Two equations where the decimals are the same after the comma are (2) and (1).
Subtract (1) from (2) as follows:
10 x − x = 1,111111111... − 0,111111111...
∴ 9 x = 1, 000000000...
∴9x = 1
1
∴ x = which is a rational number.
9
You could have also used equation (2) and (3) or (1) and (3).
3
(b)
x = 1, 75757575...
10 x = 17,57575757...
(1)
(2)
100 x = 175, 75757575...
(3)
Carefully consider the decimals of each line. It should be clear that the decimals of
(2) and (1) are not the same. Two equations where the decimals are the same after
the comma are (3) and (1). Subtract (1) from (3).
∴ 99 x = 174, 00000..........
174 58
which is a rational number.
=
99 33
It may be necessary in other cases to continue multiplying by 10 until you have
established that the decimals to the right of the comma are equal.
∴x =
It is useful to take note of a few patterns regarding some recurring decimals and their
equivalent fractions.
1
2
3
  = 13 ; 0, 25
  = 25 ; 0, 76
  = 76 ; etc.
0,1 = ; 0, 2 = ; 0,3 = ; etc. and 0,13
9
9
9
99
99
99
EXERCISE 1
(a)
From the list of numbers: −3;
(1)
(4)
(b)
(d)
2;
9; 0; 2;
−4 , write down all the
whole numbers
irrational numbers
(3)
(6)
integers
real numbers
State whether each of the following numbers are rational, irrational or neither.
1
π
3
33
(1)
(2)
(3)
(4)
4,01345
121
2
(5)
(c)
natural numbers
rational numbers
3
;
4
(2)
(5)
−2
(6)
− 2
(7)
2
2
State why we may conclude that each number below is rational.
3
1
(1)
(2)
(3)
4
1
−3
4
8
(5)
(6)
0, 52
1, 212
Show that the following recurring decimals are rational.

0, 4
0, 21
0,14
(1)
(2)
(3)
 

0,124
0,124
−1,124
(6)
(7)
(5)
(8)
3
(4)
−5
(4)
(8)
19, 45
−2,35
27 + 1
EXAMPLE 3
Find a rational number between
5
3
and .
7
4
Solution
Start by getting the lowest common denominator of the fractions and rewrite them over this
5 20
3 21
denominator:
and
. Since we can’t find any twenty-eights between 20
=
=
7 28
4 28
twenty-eights and 21 twenty-eights we can try doubling the denominators (and therefore
20 40
21 42
41
also the numerators):
and
. Now we have a fraction, namely,
,
=
=
28 56
28 56
56
between the two fractions.
4
EXAMPLE 4
Between which two consecutive integers do the following irrational numbers lie?
12
(a)
(b)
− 12
(c)
3
(b)
9 < 12 < 16
20
(d)
−2π
Solutions
(a)
(c)
9 < 12 < 16
∴ 9 < 12 < 16
∴ 3 < 12 < 4
∴ 3 < 12 < 4
∴−3 > − 12 > −4
∴ 12 lies between 3 and 4.
∴ − 12 lies between − 4 and − 3.
8 < 20 < 27
(d)
−2π = −6, 2833185307...
∴ 3 8 < 3 20 < 3 27
−7 < −6, 283318... < −6
∴ 2 < 3 20 < 3
∴−7 < −2π < −6
∴ 3 20 lies between 2 and 3.
∴ −2π lies between − 7 and − 6
Rounding off numbers to certain decimal places
The rules for rounding off numbers to certain decimal places are as follows:
•
Count to the number of decimal places after the comma that you want to
round off to.
•
Look at the digit to the right of this decimal place.
 If it is lower than 5, drop it and all the digits to the right of it.
 If it is 5 or more than 5, then add one digit to the digit immediately to
the left of it and drop it and all the digits to the right of it.
 If necessary, keep or add zeros as place holders.
EXAMPLE 5
Round off the following numbers to the number of decimal places indicated:
(a)
4,31437 (2 decimal places):
(b)
4,31437 and the answer: 4,31
1,77777 (3 decimal places)
(c)
1,77777 and the answer: 1,778
365,1534 (1 decimal place)
(d)
365,1534 and the answer: 365,2
594,2 (2 decimal places)
594,200 and the answer: 594,20
(e)
12,07963 (3 decimal places )
12,07963 and the answer: 12,080
(f)
9,998 (1 decimal place)
9,998 and the anwer:
10,0
5
EXERCISE 2
(a)
Without using a calculator, determine between which two integers the following
irrational numbers lie. Then verify your answers by using a calculator.
3 45
50
29
− 54
(1)
(2)
(3)
(4)
5 30
(5)
(6)
π
(b)
Find a rational number between:
3
2
(2)
(1)
and
3,14 and π
5
3
(c)
Round off the following numbers to the number of decimal places indicated.
(1)
9,23584 (3 decimal places)
(2)
67,2436 (2 decimal places)
(3)
4,3768534 (4 decimal places)
(4)
17,247398 (5 decimal places)
(5)
79,9999 (3 decimal places)
(6)
34,27846 (4 decimal places)
3π (3 decimal places)
(7)
5,555555 (5 decimal places)
(8)
Representing real numbers
Sets and subsets of real numbers can be represented in a variety of ways:
Number lines, set builder notation and interval notation.
Set builder notation and number lines
Set builder notation is a useful way of representing sets and subsets of real numbers. When
a set is represented in set builder notation, first describe the set in words before representing
it on a number line. Set builder notation consists of three parts, namely:
{ x : − 4 < x < 3 ; x ∈ }
EXAMPLE 6
(a)
(b)
(c)
(d)
Integers greater than and including 3
{ x : x ≥ 3 ; x ∈ }
2
3
4
5
6
Integers less than 5
{x : x < 5 ; x ∈ }
1
2
3
4
5
Natural numbers less than 4
{x : x < 4 ; x ∈ }
0
1
2
3
4
0
1
2
Integers between − 2 and 3 excluding − 2 but including 3
{ x : −2 < x ≤ 3 ; x ∈ }
−1
6
3
(e)
Real numbers greater than or including 3
{ x : x ≥ 3 ; x ∈ }
3 is included
3
(f)
Real numbers less than 7
{ x : x < 7 ; x ∈ }
7 is not included
7
(g)
Real numbers between −5 and 4 excluding −5 but including 4]
{ x : −5 < x ≤ 4 ; x ∈ }
−5 is not included
4 is included
−5
4
Interval notation and number lines
Interval notation is another way of representing real numbers on a number line. Interval
notation may not be used to represent any subset of real numbers (natural numbers, whole
numbers, integers and rational numbers).
EXAMPLE 7
(a)
Real numbers between −2 and 4 excluding −2 but including 4
( −2 ; 4]
(b)
−2
Real numbers between −4 and 7 including −4 but excluding 7
[ −4 ; 7 )
(c)
−4
7
Real numbers greater than or including 1
[1; ∞ )
(d)
4
1
Real numbers less than but not including − 1
( −∞ ; − 1)
−1
EXERCISE 3
(a)
Represent the following sets on a number line:
(1)
(2)
{ x : −1 < x < 12 ; x ∈ }
{ x : x ≥ −4 ; x ∈ }
(3)
(4)
{ x : −3 ≤ x < 1 ; x ∈ }
{ x : x < 2 ; x ∈ }
(5)
(6)
{x : x > 4 12 ; x ∈ }
{ x : x ≤ 7 ; x ∈ }
(7)
{ x : x < 7 ; x ∈ }
(8)
7
{ x : x ≥ 0 ; x ∈ }
(b)
Write the following sets in set builder notation:
(1)
0
1
2
(3)
(c)
(d)
(2)
4
[ −6 ; ∞ )
(6)
−6
4
(4)
−7
Represent on a number line:
(1)
(2)
[ −2 ; 7 )
(5)
3
5
( −3 ; 10]
( −∞ ; 7 ]
(3)
(7)
[ −1 ; 4 ]
( −∞ ; 6 34 )
(4)
( −5 ; 5)
(8)
 3 ; 8


Write the following in interval notation:
(1)
(3)
(2)
−7
−2
9
11
(4)
−6
1
1
2
MULTIPLICATION OF ALGEBRAIC EXPRESSIONS
The distributive law (revision)
a ( b + c + d ) = ab + ac + ad
a ( b + c ) = ab + ac
The variable “a” is distributed to and multiplied with all the other terms in the brackets.
The product of two binomials (revision)
The FOIL method can be used to multiply two binomials. Here you must first multiply the
first terms in each bracket. Then you multiply the outer terms, then the inner terms and
finally the last terms.
F = Firsts
O = Outers
I = Inners
L = Lasts
OUTERS
FIRSTS
( a + b )( c + d ) = ac + ad + bc + bd
INNERS
LASTS
The product of a binomial and a trinomial
The method revised above can be extended to the product of a binomial and a trinomial.
( x + y )( a + b + c )
= ( x + y )a + ( x + y )b + ( x + y )c
= ax + ay + bx + by + cx + cy
( x + y )( a + b + c )
or
= ax + bx + cx + ay + by + cy
8
EXAMPLE 8
The examples below illustrate the methods of multiplying binomials and trinomials.
(a)
(c)
(e)
(g)
( x + 3)( x + 2)
(b)
(3 y − 1)(2 y + 4)
= x 2 + 2 x + 3x + 6
= 6 y 2 + 12 y − 2 y − 4
= x2 + 5x + 6
= 6 y 2 + 10 y − 4
(2 x − 4 y )( x − 3 y )
(d)
(2ab + 3)(ab − 5)
= 2 x 2 − 6 xy − 4 xy + 12 y 2
= 2a 2b 2 − 10ab + 3ab − 15
= 2 x 2 − 10 xy + 12 y 2
= 2a 2b 2 − 7 ab − 15
(2 x3 + 7 y )( x3 − 2 y )
(f)
( x − y )( x 2 − 3 xy + 2 y 2 )
= 2 x 6 − 4 x3 y + 7 x3 y − 14 y 2
= x3 − 3 x 2 y + 2 xy 2 − x 2 y + 3 xy 2 + 2 y 3
= 2 x 6 + 3 x3 y − 14 y 2
= x3 − 4 x 2 y + 5 xy 2 + 2 y 3
(8 − 3n 2 )(4 + n) − (n − 3)(n 2 + 3n + 9)
= 32 + 8n − 12n 2 − 3n3 − (n3 + 3n 2 + 9n − 3n 2 − 9n − 27)
= 32 + 8n − 12n 2 − 3n3 − (n3 − 27)
= 32 + 8n − 12n 2 − 3n3 − n3 + 27
= −4n3 − 12n 2 + 8n + 59
EXERCISE 4
(a)
Expand and simplify:
(1)
3 x( x + 3)
(4)
( x − 5)( x − 2)
(7)
(3x − 1)(2 x + 3)
(9)
(b)
(c)
(2)
(5)
(8)
(2 x 4 − 3 y 2 )(3x 4 + 2 y 2 )
Expand and simplify:
(1)
( x + 1)( x 2 + 2 x + 3)
−3a(3a3 − 6a 2 + a)
( x + 5)( x − 2)
(7 m − 2n)(3m + 4n)
(3)
(4)
( x + 5)( x + 2)
( x − 5)( x + 2)
(10)
(4 x 4 + 3 y5 )(2 x 4 − 4 y3 )
(2)
( x − 1)( x 2 − 2 x + 3)
(3)
(2 x + 4)( x 2 − 3x + 1)
(4)
(2 x − 4)( x 2 − 3x + 1)
(5)
(3x − y )(2 x 2 + 4 xy − y 2 )
(6)
(a + 2b)(4a 2 − 3ab + b2 )
(7)
(3x − 2 y )(9 x 2 + 6 xy + 4 y 2 )
(8)
(3x + 2 y )(9 x 2 − 6 xy + 4 y 2 )
(2)
(5 y + 1) 2 − (3 y + 4)(2 − 3 y )
Expand and simplify:
(1)
2 x (3 x − 4 y ) − (7 x 2 − 2 xy )
(3)
(2 x + y ) 2 − (3 x − 2 y ) 2 + ( x − 4 y )( x + 4 y )
(4)
x 6 + ( x 3 − 3 y )( x 3 + 3 y )
(5)
9
(3a + b)(3a − b)(2a + 5b)
Special products
We will now consider the following special products:
Products which lead to the difference of two squares
•
Squaring a binomial
•
Cubing a binomial
•
Products which lead to the difference of two squares (revision)
Consider the product ( x + y )( x − y ) . The product can be simplified as follows:
( x + y )( x − y )
= x 2 − xy + xy − y 2
= x2 − y2
In other words, the pattern is as follows:
(first term + last term)(first term − last term) = (first term) 2 − (last term) 2 or
(first term − last term)(first term + last term) = (first term) 2 − (last term) 2
EXAMPLE 9
Expand and simplify the following:
(a)
(3 x + 2 y )(3 x − 2 y )
= 9x2 − 4 y2
(b)
( a − 5b)( a + 5b)
= a 2 − 25b 2
(c)
(4 x 4 − 3 y 3 )(4 x 4 + 3 y 3 )
(d)
= 16 x8 − 9 y 6
( a − b + d )(a − b − d )
Alternatively we can substitute (a − b) = k
= [ (a − b) + d ][ (a − b) − d ]
∴[ (a − b) + d ][ (a − b) − d ]
= ( a − b) 2 − d 2
= ( k + d )(k − d )
= (a − b)( a − b) − d 2
= k2 − d2
= a 2 − 2ab + b 2 − d 2
= ( a − b) 2 − d 2
= a 2 − 2ab + b 2 − d 2
The alternative method makes it easier to recognise that the product leads to
the difference of two squares.
Squaring a binomial (revision)
Consider the squares of the following binomials:
(a + b) 2
= ( a + b )( a + b )
( a − b) 2
= (a − b)( a − b)
= a 2 + ab + ba + b 2
= a 2 − ab − ba + b 2
= a 2 + 2 ab + b 2
= a 2 − 2ab + b 2
Therefore:
( a + b) 2 = a 2 + 2ab + b 2
and
( a − b) 2 = a 2 − 2ab + b 2
In other words, the pattern is as follows:
(first term + last term)2 = (first term) 2 + 2(first term)(last term) + (last term) 2
(first term − last term) 2 = (first term) 2 − 2(first term)(last term) + (last term) 2
10
EXAMPLE 10
Expand and simplify the following:
(a)
( x + 4)2
(b)
( x − 4)2
(c)
(2 x + 4 y ) 2
(d)
(2 x − 4 y ) 2
(e)
( − 5 a − 2b ) 2
(f)
( a − 3b )3
Solutions
(a)
(c)
(e)
( x + 4)2
(b)
( x − 4) 2
= x 2 + 2( x)(+4) + (4)2
= x 2 + 2( x)(−4) + (−4) 2
= x 2 + 8 x + 16
= x 2 − 8 x + 16
(2 x + 4 y ) 2
(d)
(2 x − 4 y ) 2
= (2 x ) 2 + 2(2 x )(4 y ) + (4 y ) 2
= (2 x ) 2 + 2(2 x )( −4 y ) + ( −4 y ) 2
= 4 x 2 + 16 xy + 16 y 2
= 4 x 2 − 16 xy + 16 y 2
(−5a − 2b) 2
= (−5a ) 2 + 2(−5a )(−2b) + (−2b) 2
= 25a 2 + 20ab + 4b 2
(f)
This is an example where a binomial is cubed.
( a − 3b)3
= (a − 3b)1 ( a − 3b) 2
= (a − 3b)( a 2 − 6ab + 9b 2 )
= a 3 − 6a 2b + 9ab 2 − 3a 2b + 18ab 2 − 27b3
= a 3 − 9a 2b + 27 ab 2 − 27b3
EXERCISE 5
(a)
(b)
Expand and simplify:
(1)
( x + 7)( x − 7)
(2)
( x − 3)( x + 3)
(3)
(2 x − 1)(2 x + 1)
(5)
(3 x − 2 y )(3 x + 2 y )
(6)
(4a 3b + 3)(4a 3b − 3)
(4)
(9 x + 4)(9 x − 4)
(7)
(6 − 3 x 4 y )(6 + 3 x 4 y ) (8)
(9)
(1 − a 4 )(1 − a 2 )(1 + a 2 )
(3 x − 2 + y )(3 x − 2 − y )
Expand and simplify:
(1)
( x + 5) 2
(2)
( x − 5)2
(3)
(2a + 3) 2
(4)
(2a − 3)2
(5)
(a − 4b)2
(6)
(−a − 3b) 2
(7)
(−3a + 5b) 2
(8)
3( x − 3 y ) 2
(9)
[ 2(m − 4n)]2
(10)
( x3 − 3 y 6 ) 2
(11)
(2a + 3b)3
(12)
(2a − 3b)3
11
FACTORISATION OF ALGEBRAIC EXPRESSIONS
Factorisation is the process of writing a number or an expression as the product of its
factors. It can also be seen as the reverse procedure of the distributive law.
The following factorisation types will be revised and new types will be introduced:
•
Highest common factor (revision)
Difference of two squares (revision)
•
Quadratic trinomials of the form x 2 + bx + c (revision)
•
Quadratic trinomials of the form ax 2 + bx + c where a ≠ 0 (new type)
•
Grouping (revision)
•
Sum and difference of two cubes (new types)
•
Taking out the highest common factor
Consider the reverse procedure of the distributive law: ab + ac = a (b + c )
The highest common factor (a) has been “taken out” of the expression ab + ac and the
expression is said to be factorised as the product a ( b + c ) .
EXAMPLE 11
Factorise the following expressions:
(a)
15x2 + 9x5
The factors of 15 are 1 ; 3 ; 5 ; and 15 and the factors of 9 are 1 ; 3 ; and 9
Therefore the highest common factor between 15 and 9 is 3.
The factors of x2 are 1; x and x 2 and the factors x5 are 1 ; x ; x 2 ; x 3 ; x 4 and x 5
Therefore, the highest common factor between x2 and x5 is x 2 .
Therefore, the highest common factor between 15x2 and 9x5 is 3x2 .
We can now take out the highest common factor (HCF) as follows and thus factorise
the expression by writing it as a product of factors:
∴15 x 2 + 9 x5
= 3x 2 . 5 + 3x 2 . 3x3
(b)
(c)
= 3 x 2 (5 + 3 x3 )
Verify: 3 x 2 (5 + 3 x3 ) = 15 x 2 + 9 x5
15a4b6 − 3ab2
= 3ab2 × 5a3b4 − 3ab2 ×1
HCF = 3ab2
= 3ab2 (5a3b4 − 1)
Verify: 3ab2 (5a3b4 − 1) = 15a 4b6 − 3ab2
2a ( x + y ) − 3b( x + y )
For the above example the common factor is a common bracket.
∴ 2a ( x + y ) − 3b( x + y )
= ( x + y )(2a − 3b)
HCF = ( x + y )
12
(d)
( x − 1)7 − ( x − 1) 4
HCF = ( x − 1) 4
= ( x − 1) 4 × ( x − 1)3 − ( x − 1) 4 ×1
= ( x − 1) 4 ( x − 1)3 − 1
Alternatively we can substitute ( x − 1) = k
( x − 1)7 − ( x − 1) 4
= k7 − k4
HCF = k 4
= k 4 (k 3 − 1)
= ( x − 1) 4 ( x − 1)3 − 1
The next examples involve the sign-change rule discussed in Grade 9. Let’s revise this rule
by using the following examples.
(a)
3b − 2a
(b)
= + (3b − 2a )
= −( −3b + 2a )
= −(2a − 3b)
∴ 3b − 2a = −(2a − 3b)
−4 y + 7 x
(c)
= +(−4 y + 7 x)
= −(4 y − 7 x)
∴−4 y + 7 x = −(4 y − 7 x)
−5b − 2a
= + ( −5b − 2a )
= −(5b + 2a )
∴−5b − 2a = −(5b + 2a)
You can simplify this method by “taking out a negative” and changing signs. Whenever you
“take out a negative sign”, the signs of the terms inside the brackets will be different to the
signs of the the terms in the original expression. Consider the following examples using this
short-cut approach.
(a)
3 y − 9x
= −3(− y + 3 x )
= −3(3 x − y )
(b)
−3 y − 6 x
= −3( y + 2 x)
= −3(2 x + y )
In these examples, the HCF of 3 as well as a negative sign has been “taken out”.
EXAMPLE 12
Factorise:
(a)
2a (3 x − 2 y ) + 5b(2 y − 3 x )
(b)
Solutions
(a)
2a (3 x − 2 y ) + 5b(2 y − 3 x )
= 2a (3 x − 2 y ) − 5b(−2 y + 3 x)
= 2a (3 x − 2 y ) − 5b(3 x − 2 y )
= (3 x − 2 y )(2a − 5b)
(b)
3a (5 x + y ) − ( −5 x − y )
= 3a (5 x + y ) + (5 x + y )
= (5 x + y )(3a + 1)
13
3a (5 x + y ) − ( −5 x − y )
EXERCISE 6
(a)
(b)
Factorise:
(1)
6 x3 + 12 x
(2)
6 x3 + 4 x 2
(3)
5 x3 + 5 x
(4)
12 x3 − 18 x 2
(5)
3 x 2 − 9 y + 12 xy
(6)
8a 2b 2 − 64ab
(7)
16m 4 n8 − 8m3n7 + 36m 2 n3
Factorise:
(1)
x (a + b) + y (a + b)
(3)
3 p ( q + r ) − 4m( r + q )
(2)
(4)
k ( x + y) + p( x + y)
7 k ( m − 3n) − 3 p ( m − 3n)
(5)
( x − y )2 − 3( x − y )
(6)
( a + c ) 4 + ( a + c )5
(7)
(9)
(11)
( m − n) 6 + ( m − n) 2
7 x ( m − 3n ) + 4 y (3n − m )
2 x (3 p + q ) + 4 y ( − q − 3 p )
(8)
(10)
(12)
7 x ( m − 3n ) − 4 y (3n − m )
(13)
(15)
3
3
4 x(a − 2) + (2 − a )
(14)
( a − 3b) − c (3b − a ) + d (3b − a )
7 x(m + 3n) + 4 y ( −m − 3n)
( a − b) − ( pb − pa )
2 x2 (3a − b) − 12 x(b − 3a)
Difference of two squares (revision)
Consider the product ( a + b)( a − b) = a 2 − b 2 .
The reverse process a 2 − b 2 = ( a + b)( a − b) is called the factorisation of the difference of
two squares. Another way of seeing this type of factorisation is:
a 2 − b2 = ( a 2 + b2 )( a 2 − b2 ) = ( a + b)( a − b) where a > 0, b > 0
EXAMPLE 13
Factorise fully:
x2 − 25
(a)
(b)
= ( x2 + 25)( x2 − 25)
= ( x + 5)( x − 5)
(c)
64 x 4 − 49 y 2
= (8 x 2 + 7 y )(8 x 2 − 7 y )
2a8 − 2b8
= 2(a8 − b8 )
[always take out the HCF first]
= 2( a 4 + b 4 )( a 4 − b 4 )
= 2( a 4 + b 4 )( a 2 + b 2 )(a 2 − b 2 )
= 2( a 4 + b 4 )( a 2 + b 2 )(a + b)(a − b)
(d)
16 − ( x − y ) 2
[alternatively: Let ( x − y ) = k ]
= [ 4 + ( x − y ) ][ 4 − ( x − y ) ]
∴16 − ( x − y ) 2
= (4 + x − y )(4 − x + y )
= 16 − k 2
= (4 − k )(4 + k )
= [ 4 + ( x − y ) ][ 4 − ( x − y ) ]
= (4 + x − y )(4 − x + y )
14
EXERCISE 7
(a)
Factorise fully:
(1)
x 2 − 16
(4)
(b)
x2 − 1
(7)
100 x8 − y 6
Factorise fully:
(2)
x 2 − 36
(3)
9 x2 − 4
(5)
169 x 2 − 100
(6)
16a 2 − 121b 2
(8)
x4 − x2
(9)
a 4 − 16
(1)
n8 − 81
(2)
12 x 2 − 75 y 2
(3)
81a3 − 49ab 2
(4)
27 a3 − 3ab 2
(5)
4 x8 y 2 − 16 y 2
(6)
25 p 4 q − 100 p 4 q 3
(7)
( a + b) 2 − c 2
(8)
(2 x + y )2 − y 2
(9)
25a 2 − 16(a − b) 2
Quadratic trinomials (revision)
Consider the product ( x + a )( x + b)
By multiplying out, it is clear that this product will become:
( x + a )( x + b)
= x 2 + ax + bx + a × b
= x 2 + ( a + b) x + a × b
Take note that x 2 has a coefficient of 1.
So the expression x 2 + ( a + b) x + ab can be factorised as ( x + a )( x + b) .
For example, the trinomial x2 + 7 x + 12 can be factorised as follows:
Write the last term, 8, as the product of two numbers ( a × b ).
The options are: 1 × 12, 3 × 4 and 2 × 6
The middle term ( a + b ) is now obtained by adding the numbers of one of the above
options. The obvious choice will be the option 3× 4 because the sum of the numbers 3 and 4
gives 7. Therefore:
x 2 + 7 x + 12
= x 2 + (3 + 4) x + (3 × 4)
= ( x + 3)( x + 4)
So the strategy to factorising quadratic trinomials is as follows:
• Write down the last term as the product of two numbers.
• Find the two numbers (using the appropriate numbers from one of the products) which
gets the coefficient of the middle term by adding or subtracting.
• Check that when you multiply these numbers you get the last term.
EXAMPLE 14
Factorise:
(a)
x2 + 5x − 36
The last term can be written as the following products: 1× 36, 2 × 18, 4 × 9, 6 × 6
We now need to get +5 from one of the options above.
Using 4 × 9 will enable us to get +5 since
−4 + 9 = 5 which is the coefficient of the middle term and ( −4)( +9) = −36 which is
the last term.
Therefore: x 2 + 5 x − 36 = ( x − 4)( x + 9)
15
(b)
x 2 − 5x + 6
The last term can be written as the following products: 3 × 2, 1× 6
We now need to get the middle term −5 from one of the options above.
Try the option 3 × 2 . Clearly −3 − 2 = −5 which is the coefficient of the middle
term and −3 × −2 = +6 which is the last term.
Notice that the option 1 × 6 will not work because even though +1 − 6 = −5 is the
coefficient of the middle term, +1 × −6 = −6 , is not the last term.
Therefore: x 2 − 5 x + 6 = ( x − 3)( x − 2)
(c)
3a 2 − 21a − 24
Here it is necessary to first take out the highest common factor:
3a 2 − 21a − 24
= 3(a 2 − 7a − 8)
The last term of the trinomial in the brackets can be written as the following
products: 1 × 8, 4 × 2
The option 1 × 8 will work because: +1 − 8 = −7 which is the coefficient of the
middle term, and ( +1)( −8) = −8 which is the last term.
Therefore,
3a 2 − 21a − 24 = 3(a 2 − 7a − 8)
= 3(a + 1)(a − 8)
Take note:
• If the sign of the last term of a trinomial is negative, the signs in the brackets are
different (see Example 8a and 8c).
• If the sign of the last term of a trinomial is positive, the signs in the brackets are the
same i.e. both positive or both negative (see Example 8b)
EXERCISE 8
(a)
(b)
(c)
(d)
Factorise the following quadratic trinomials fully:
x 2 − 11x + 28
x 2 − 5 x − 84
(1)
(2)
(3)
x2 + 6x + 9
(4)
x 2 − 4 x − 12
(5)
a 2 − 9a + 20
(6)
a 2 − 11a − 12
(7)
a 2 − 12a + 35
(8)
a 2 − 2a − 48
(9)
m 2 + 5m + 6
(10)
m 2 − 5m − 6
(11)
m 2 + 5m − 6
(12)
m 2 − 5m + 6
(14)
x 2 − 7 x + 12
(15)
x 2 + 4 x − 12
(2)
3x 2 − 21x − 54
(3)
4a 2 + 12a − 40
(6)
90 − 45 x + 5 x 2
x 2 + 7 x + 12
(13)
Factorise fully:
2 x 2 − 2 x − 12
(1)
(4)
6a 2 + 24a − 30
(5)
x3 + 2 x 2 − 8 x
Explain why the trinomial x2 + 4 x + 5 cannot factorise.
Explain why the following factorisations are incorrect:
x 2 − 5 x + 6 = ( x − 6)( x + 1)
(1)
(2)
x 2 + 10 x − 24 = ( x + 6)( x − 4)
16
Factorising more advanced quadratic trinomials
The method to factorise these trinomials is a little more involved than with the previous
trinomials. A suggested method is as follows and will be discussed in the examples that
follow:
Step 1:
Write down the product options of the first and last terms.
Step 2:
Write the product options in a table format.
Step 3:
Select any product option from the first term and last term and write these
options using what is called the “cross method”. Then multiply diagonally.
Step 4:
Add and subtract the products in order to get the middle term
Step 5:
Obtain the factors by reading off horizontally.
EXAMPLE 15
Factorise the following fully:
(a)
21x 2 + 25x − 4
12 x 2 − 11x + 2
(b)
(c)
24 x 2 − 10 xy − y 2
Solutions
(a)
21x 2 + 25x − 4
Step 1
Write down the product options of the first and last terms:
21x 2 : 1x × 21x, 7 x × 3x
4:
Step 2
1 × 4, 2 × 2
Write the product options in a table format as follows:
First term
21x
7x
1x
3x
1
4
Last term
4
2
1
2
The product option 1 × 4 for the last term must also be written in
reverse order as 4 × 1 . This is not necessary for the product options for
the first term.
Step 3
Select any product option from the first term and last term and
write these options using what is called the “cross method” and
multiply diagonally:
3x
1
7x
3x
Step 4
4
28x
The strategy is to now get the middle term, +25x , from 3x and
28x using different signs (because the sign of the last term of the
trinomial is negative). Insert the signs as indicated below.
7x
1
−3x
3x
4
+28x
+25x
17
The factors are now obtained by reading off horizontally.
The first factor is (7 x − 1) . The negative sign for 1 comes from
−3x . The second factor is (3 x + 4) . The positive sign for 4 comes
from +28x .
Step 5
7x
1
− 3x
3x
4
+ 28x
+25x
∴ 21x 2 + 25 x − 4 = (7 x − 1)(3 x + 4)
This method involves trial and error and you need to keep trying different
options until you get ones that will work. For example, the following
option will not work:
Take note:
7x
4
12x
3x
1
7x
These two terms cannot
give the middle term.
Therefore, the options
selected are not helpful.
+25x
(b)
2
12 x − 11x + 2
The product options of the first and last terms:
12 x 2 :
2:
1x × 12 x, 4x × 3 x, 6 x × 2 x
1 × 2, 2 × 1
First term
1x
4x
6x
12x
3x
2x
Last term
1
2
2
1
4x
1
−3x
3x
2
−8x
−11x
The signs in the brackets must be the same because the sign of the last term of
the trinomial is positive.
∴12 x 2 − 11x + 2 = (4 x − 1)(3 x − 2)
(c)
24 x 2 − 10 xy − y 2
The product options of the first and last terms:
24 x 2 : 24 x× 1x, 8 x × 3x, 6 x × 4 x, 12 x × 2 x
y2 :
24x
1x
1y ×1y
First term
6x
12x
4x
2x
8x
3x
Last term
1y
1y
12x
1y
+2xy
2x
2y
−12xy
−10xy
The signs in the brackets must be different because the sign of the last term of
the trinomial is negative. The options that work are as follows:
∴ 24 x 2 − 10 xy − y 2 = (12 x + y )(2 x − y )
18
EXERCISE 9
(a)
(b)
Factorise fully:
(1)
3x 2 + 4 x + 1
2 x 2 − 3x + 1
(3)
12 x 2 − 7 x + 1
(4)
18 x 2 + 3x − 1
(5)
2 x2 − 5x − 3
(6)
5 x 2 + 14 x + 8
(7)
2 x2 − 7 x + 6
(8)
6 x 2 − 11x − 10
(9)
6 x 2 − 5 x − 21
(10)
20 x 2 + 24 x − 9
(11)
18 x 2 − 3x − 10
(12)
15 − x − 6 x 2
(2)
15 x 2 − 18 x + 3
(3)
x 2 − xy − 6 y 2
(5)
10m2 − 13mn − 3n2
(6)
12a 2 + 23ab + 10b 2
(2)
a3 − 6a2b + 9ab2
(3)
(5)
(a + b)2 + 4(a + b) − 32
Factorise fully:
(1)
4 x 2 + 10 x − 6
4 p 2 + 7 pq − 2q 2
(4)
(c)
(2)
Factorise fully:
(1)
10 x2 − 10 xy − 120 y 2
x4 + 5x2 y 2 + 6 y 4
(4)
x4 − 12 x2 + 32
Factorisation by grouping in pairs (revision)
EXAMPLE 16
Factorise the following expressions fully:
(a)
ax + 3a + 2 x + 6
(b)
2 x 2 + 4 xy − 3xy − 6 y 2
(c)
a 3 − 4a 2 − 4a + 16
(d)
6t 2 + 3t − 6 x − 12tx
Solutions
(a)
ax + 3a + 2 x + 6
= ( ax + 3a ) + ( +2 x + 6)
= ( ax + 3a ) + (2 x + 6)
= a ( x + 3) + 2( x + 3)
= ( x + 3)( a + 2)
(b)
[put brackets around the pairs separated by a + sign]
[ +2 x = 2 x ]
[factorise the pairs]
[take out the common bracket]
2 x 2 + 4 xy − 3xy − 6 y 2
= (2 x 2 + 4 xy ) + (−3xy − 6 y 2 ) [put brackets around the pairs separated by a + sign]
= (2 x 2 + 4 xy ) − (3xy + 6 y 2 )
= 2 x( x + 2 y ) − 3 y ( x + 2 y )
= ( x + 2 y )(2 x − 3 y )
(c)
[apply sign-change rule]
[factorise the pairs]
[take out the common bracket]
a 3 − 4a 2 − 4a + 16
Method 1
a 3 − 4a 2 − 4a + 16
= (a 3 − 4a 2 ) + (−4a + 16)
3
2
= (a − 4a ) − (4a − 16)
2
= a (a − 4) − 4(a − 4)
2
= (a − 4)(a − 4)
= (a − 4)(a + 2)(a − 2)
[put brackets around the pairs separated by a + sign]
[apply sign-change rule]
[factorise the pairs]
[take out the common bracket]
[factorise the difference of two squares]
19
Method 2
Group the first and third terms together and the second and fourth terms together.
a 3 − 4a 2 − 4a + 16
= a 3 − 4a − 4a 2 + 16
[put brackets around the pairs separated by a + sign]
= ( a 3 − 4a ) + ( −4a 2 + 16)
= ( a 3 − 4a ) − (4a 2 − 16)
2
2
= a ( a − 4) − 4( a − 4)
2
= (a − 4)( a − 4)
= ( a + 2)( a − 2)( a − 4)
(d)
[apply sign-change rule]
[factorise the pairs]
[take out the common bracket]
[factorise the difference of two squares]
6 p 2 + 12 pq − 8q − 4 p
= (6 p 2 + 12 pq ) + ( −8q − 4 p ) [put brackets around the pairs separated by a + sign]
= (6 p 2 + 12 pq ) − (8q + 4 p )
= 6 pq ( p + 2q ) − 4(2q + p )
= 6 pq ( p + 2q ) − 4( p + 2q )
= ( p + 2q )(6 pq − 4)
= ( p + 2q )2(3 pq − 2)
= 2( p + 2q )(3 pq − 2)
[apply sign-change rule]
[factorise the pairs]
[ 2 q + p = p + 2q ]
[take out the common bracket]
[factorise further]
EXERCISE 10
(a)
(b)
Factorise fully:
px + ky + px + ky
(1)
(2)
rx − ry + tx − ty
(3)
2 p − 2q + np − nq
(4)
a 2 − 2a + ab − 2b
(5)
(7)
(9)
4 x3 + 4 x 2 + 5 x + 5
bx + by − ax − ay
3ax − 3ay − x + y
(6)
(8)
(10)
bx − by − ax + ay
a − b − ac + bc
3ax + 3ay − x − y
(11)
(13)
(15)
x3 − 2 x 2 − 2 x + 4
2 x3 − 3x 2 − 6 x + 9
a − b + ab − 1
(12)
(14)
(16)
x3 + 2 x 2 − 2 x − 4
3x 4 − 3x 2 − 27 x 2 + 27
6a 2 px − 4ap 2 y − 6a 2 py + 4ap 2 x
Factorise the following:
(this is a challenge for you)
3
2
x 2 − (a + b) x + ab
6 x − 2 x − 54 x + 18
(1)
(2)
a 2 − b2 − a − b
(3)
x 2 − 9 − ( x − 3)(1 − 2 x)
(5)
a 2 − 4ab + 4b 2 − 16 x 2 (Hint: Group the four terms in the ratio 3 to 1)
(c)
(1)
(2)
Prove that (b − a ) 2 = ( a − b) 2
Hence, prove by means of factorisation, that
x( x − 3) 2 − 3(3 − x ) 2 = ( x − 3)3
(d)
Consider the expression:
x 2 ( x + 3) − ( x + 3)
Show how the method of factorising the expression is different from the method of
expanding and simplifying.
(4)
20
The sum and difference of two cubes
Consider the following products.
(a)
( x + y )( x 2 − xy + y 2 )
(c)
(e)
( x + 2)( x 2 − 2 x + 4)
(b)
= x 3 − x 2 y + xy 2 + x 2 y − xy 2 + y 3
= x3 − 2 x 2 + 4 x + 2 x 2 − 4 x + 8
= x3 + y 3
= x3 + 8
(2 x + 3)(4 x 2 − 6 x + 9)
(3 x + 1)(9 x 2 − 3 x + 1)
(d)
= 8 x3 − 12 x 2 + 18 x + 12 x 2 − 18 x + 27
= 27 x3 − 9 x 2 + 3 x + 9 x 2 − 3 x + 1
= 8 x3 + 27
( x − y )( x 2 + xy + y 2 )
= 27 x3 + 1
( x − 2)( x 2 + 2 x + 4)
(f)
= x 3 + x 2 y + xy 2 − x 2 y − xy 2 + y 3
= x3 + 2 x 2 + 4 x − 2 x 2 − 4 x − 8
= x3 − y 3
= x3 − 8
The answer for each product turned out to be either a sum or a difference of two cubes.
Therefore from the above products we noticed that:
x3 + y 3 = ( x + y )(x 2 − xy + y 2 ) of which we may conclude:
(first term) 3 + (last term)3 = (first term + last term)  (first) 2 − (first)(last) + (last) 2 
x3 − y 3 = ( x − y )(x 2 + xy + y 2 ) of which we may conclude:
(first term)3 − (last term) 3 = (first term − last term)  (first) 2 + (first)(last) + (last) 2 
EXAMPLE 17
Factorise the following:
(a)
8 − 27 b 3
Find the cube root of each of the terms:
3
8 = 2 and
3
27b 3 = 3b
∴ 8 − 27b3
= (23 − (3b)3 )
= (2 − 3b)  (2) 2 + (2)(3b) + (3b) 2 
= (2 − 3b)(4 + 6b + 9b 2 )
(b)
2 x 3 + 2000 y 6
= 2( x 3 + 1000 y 6 ) always take out the highest common factor first
= 2( x 3 + 1000 y 6 )
3
= 2( x 3 + (10 y 2 )3 )
x 3 = x and
3
1000 y 6 = 10 y 2
= 2( x + 10 y 2 )( x 2 − ( x)(10 y 2 ) + (10 y 2 ) 2
= 2( x + 10 y 2 )( x 2 − 10 xy 2 + 100 y 4 )
(c)
( x + y )3 − x 3
= [ ( x + y ) − x ]  ( x + y ) 2 + ( x + y )( x ) + ( x ) 2 
= [ y ]  x 2 + 2 xy + y 2 + x 2 + xy + x 2 
= y (3 x 2 + 3 xy + y 2 )
21
3
( x + y )3 = x + y and
3
x3 = x
EXERCISE 11
(a)
(b)
Factorise the following
(1)
27 x 3 − 1
(2)
8 x3 + 1
(4)
125 − 729 x 3
(5)
a 3b 3 −
(7)
8a 4 − 64a
(10)
8 − ( a − 1)3
(3)
64 x 3 − y 3
1
8
(6)
5 x 3 + 40
(8)
− x 3 − 27
(9)
(11)
x3 +
1
x3
(12)
1 3
x + 216
27
1
x3 − 3
x
Consider x 6 − 64
(1)
Factorise the above expression by using the difference of two cubes method
first.
(2)
Factorise the above expression by using the difference of two squares method
first.
The Golden Rules of Factorisation
Two terms
Three terms
Step 1:
Step 1:
Apply the sign-change rule
Apply the sign-change rule if
if necessary.
necessary.
Step 2:
Step 2:
Take out the HCF if it
Take out the HCF if it exists.
exists.
Step 3:
Factorise the trinomial.
Step 3:
Apply difference of two
squares or sum and
difference of two cubes if
possible.
Four terms
Step 1:
Group in pairs and put brackets
around each pair separated by
the + sign.
Step 2:
Apply the sign-change rule if
necessary.
Step 3:
Factorise the pairs.
Step 4:
Take out the common bracket.
Step 5:
Factorise further if needs be.
EXERCISE 12 (Revision of all factorisation methods)
Factorise the following:
(d)
(g)
(j)
2a2 − a − 15
8x3 − 27
4x3 − x − 4 x2 + 1
7 x 7 y14 − 14 x14 y 7
(b)
(e)
(h)
(k)
5a2 − 20a + 15
x 2 − 5x + 6
x2 − x − 2
16x2 −100
(c)
(f)
(i)
(l)
x2 − x4
6 x2 − 2 x − 20
16a3 − 2b3
1 − x3
(m)
12 x 2 − 18 xy + 6
(n)
12 x 2 − 18 xy + 6 y 2
(o)
6 y 2 − 3 y − 18
(p)
8 x 4 y8 − 4 x 2 y 4
(q)
4 x2 − 6 x − y 2 + 3 y
(r)
4 x 2 ( y − 1) + 36(1 − y)
(s)
25 − ( x − y )2
(t)
(a − b)2 − 3(b − a)
(u)
4ax2 − a 2 + x 2 + 4 x3
(a)
22
SIMPLIFICATION OF ALGEBRAIC FRACTIONS
Multiplication and division of algebraic fractions
EXAMPLE 18
Simplify the following expressions which have monomial (single term) denominators:
(a)
4 x2 + 8x
4 x2
(b)
12 x3 y + 3 xy
9 y2
Solutions
In these examples, always factorise the numerators before simplifying:
(a)
4 x2 + 8x
4 x2
(b)
12 x3 y + 3 xy
9 y2
=
4 x( x + 2)
4 x2
=
3 xy (4 x 2 + 1)
9 y2
=
x+2
x
=
x(4 x 2 + 1)
3y
EXAMPLE 19
Simplify the following expressions which have multi-term denominators:
(a)
x 2 − 12 x + 27
4 x 2 − 12 x
(c)
x3 − 8
2− x
(b)
x−2
x2 − 4
÷
x2 + 4 x − 5 x2 + 5x
(d)
6 x3 − 2 x 2 x 2 − 9
×
−3x − 9 2 − 6 x
Solutions
Whenever the numerator and denominator contains two or more terms, both the
numerator and denominator must be factorised. Only then can you cancel and simplify.
(a)
x 2 − 12 x + 27
4 x 2 − 12 x
( x − 9)( x − 3)
=
4 x( x − 3)
x −9
=
4x
(b)
x−2
x2 − 4
÷
x2 + 4 x − 5 x2 + 5x
x−2
x2 + 5x
"tip and times"
= 2
×
x + 4 x − 5 x2 − 4
x−2
x( x + 5)
=
×
( x + 5)( x − 1) ( x − 2)( x + 2)
x
=
( x − 1)( x + 2)
23
(c)
x3 − 8
2− x
x3 − 8
=
+ (2 − x)
=
6 x3 − 2 x 2 x 2 − 9
×
−3x − 9 2 − 6 x
6 x3 − 2 x 2
x2 − 9
=
×
−3( x + 3) −2(−1 + 3 x)
(d)
( x − 2)( x 2 + 2 x + 4)
−( x − 2)
= −( x 2 + 2 x + 4)
= − x2 − 2 x − 4
=
2 x 2 (3x − 1) ( x − 3)( x + 3)
×
−3( x + 3)
−2(3 x − 1)
=
x 2 ( x − 3)
3
EXERCISE 13
(a)
Simplify the following single algebraic fractions:
(1)
(4)
(7)
(10)
(13)
(b)
x2 − x
x
4m 2 + 4m
4m 2
x2 − 5x + 6
2x − 6
27 p3 − 8
2
27 p + 18 p + 12
10 − 2 x
x −5
(2)
4a 2 − 8a
4a
(3)
(5)
a3 − a2
a2
÷
4
8
(6)
(8)
(11)
(14)
x2 − 9
(9)
2 x2 − 5x − 3
9 − k2
(12)
2
k − 3k
x2 ( x + 2) − 4 ( x + 2)
( x + 2 )2
(15)
3 p2 − 6 p
6p
x2 − 4
x2 − 2 x
9 p3 − 81 p 2
6 p3
6 + 9k
9k 2 − 4
(3 − x)2
x2 − 9
Simplify the following algebraic fractions:
(1)
x 2 + x − 6 x 2 − 16
× 2
2x − 8
x − 2x
(2)
(3)
x 2 − 4 xy + 4 y 2 x − 2 y x 2 − 4
÷
×
x−2
2x − 4
4
(4)
(5)
2k 2 + k − 6 k 4 − 9
×
÷ ( 2k − 3)
k+2
2k 2 − 6
(6)
x2 + x x2 + 2 x + 1
÷
x2
x3
6 + k − k2
2k + 4
÷ ( k − 3) ÷ 2
3 + 3k
k −1
x+2
14 − 7 x
×
2
x + x − 2 3x − 6
Addition and subtraction of algebraic fractions
Whenever you add or subtract algebraic fractions, the first thing you need to do is to
determine the lowest common multiple of the denominators (LCD). This may have been
referred to as the LCM.
You then have to convert each fraction to an equivalent fraction so that each fraction’s
denominator is the same as the LCD.
The last thing to do is to write the numerators over the LCD. In the example below we will
establish the skill of finding an LCD for different kinds of denominators.
24
EXAMPLE 20
Determine the lowest common multiple of the denominator (LCD) for each of the following
sets of algebraic fractions.
(a)
1
1
1
; 3 ;
2
2 xy
4x
3x
For the numbers 4, 3 and 2 the lowest common multiple is 12.
For the variables x, x 2 and x 3 , the term with the highest exponent will be used in
the LCD. In this case, the term used will be x 3 .
Since there are no other terms in y, you will just use y in the LCD.
The LCD is 12 x 3 y .
(b)
(c)
(d)
1
1
Take note that x + 2 is a binomial whereas x is a single term.
;
x+2 x
Therefore both the denominators have to be accounted for in the LCD.
The LCD is x ( x + 2) .
1
1
x + 2 and x2 + 2 are binomials but are not identical.
; 2
x+2 x +2
Therefore both the denominators have to be accounted for in the LCD.
The LCD is ( x + 2)( x 2 + 2) .
1
1
;
x + 2 ( x + 2)2
( x + 2)2 is the term with the highest exponent for ( x + 2) .
The LCD: ( x + 2)2 or ( x + 2)( x + 2)
(e)
(f)
1
1
x 2 + 2 x can be factorised: x 2 + 2 x = x( x + 2)
; 2
x + 2 x + 2x
Therefore there is x to consider as a single term and ( x + 2) to consider as a
binomial. The LCD is x ( x + 2)
1
1
1
4 x + 4 can be factorized: 4 x + 4 = 4( x + 1)
;
;
4x + 4 4x x + 4
Therefore there are 4 and 4x to consider as single terms and ( x + 1) and ( x + 4) to
consider as binomials.
The LCD is 4 x ( x + 1)( x + 4)
EXAMPLE 21
Simplify the following:
1
x−2
−
2
4x y
3x3
(a)
1 +
(b)
2 
1

 x −  x + 
3 
3

(c)
3
2
− 2
x+3
x +3
25
Solutions
(a)
Insert brackets around a numerator that has more than one term.
1 +
1
( x − 2)
−
2
4x y
3x3
LCD = 12 x3 y
1 12 x3 y
1 3x
( x − 2) 4 y
= ×
+
×
−
×
3
2
1 12 x y
4y
4 x 3x
3x3
=
12 x3 y
3x
4 y ( x − 2)
+
−
3
3
12 x y
12 x y
12 x3 y
=
12 x3 y + 3x − 4 y ( x − 2)
12 x3 y
12 x3 y + 3x − 4 xy + 8 y
=
12 x3 y
(b)
Method 1
2 
1

 x −  x + 
3 
3

1
2
2
= x2 + x − x −
3
3
9
2
9 x + 3x − 6 x − 2
=
9
9 x 2 − 3x − 2
=
9
(c)
Method 2
2 
1

 x −  x + 
3 
3

 3x − 2   3x + 1 
=


 3  3 
(3 x − 2)(3 x + 1)
=
9
2
9 x − 3x − 2
=
9
find the LCD of each bracket
top × top and bottom × bottom
3
2
− 2
LCD = ( x + 3)( x 2 + 3)
x+3
x +3
3
( x 2 + 3)
2
( x + 3)
=
× 2
− 2
×
( x + 3) ( x + 3)
( x + 3) ( x + 3)
3( x 2 + 3)
2( x + 3)
=
−
2
( x + 3)( x + 3)
( x + 3)( x 2 + 3)
=
3( x 2 + 3) − 2( x + 3)
( x + 3)( x 2 + 3)
=
3x 2 + 9 − 2 x − 6
3x 2 − 2 x + 3
=
( x + 3)( x 2 + 3) ( x + 3)( x 2 + 3)
26
EXERCISE 14
(a)
Simplify the following:
(1)
2x
1
+
3
6
(4)
1+ a −
(7)
(10)
(12)
(b)
(c)
(2)
2a − 1
2
(5)
2
5
(8)
+
x
x +1
x +1
4
(11)
−
2
2x +1
(2 x + 1)
x−3
x−2
x +1
−
+
3
2
6
7
2
1
−
+
(3)
6x
9 xy
3x3
5
x+2
−
(6)
6 xy
2x
x+2
6
(9)
−
2
x+2
x +2
2
7x + 1
5 xy
+
−
+ 1
2
2 xy
4
4x y
2x
(13)
x −
3x − 2
3
2
+ 2
x
x
5
x +1
−
2x
x−2
x −3
x+3
−
x+3
x −3
Simplify the following and write your answers as single fractions:
(1)
1 
1

 x +  x − 
3
3

(4)
1 
1

 x +  x − 
x 
x

2
(2)
1 
1

 2 x + y   3x −
3 
2


y  (3)

1

x− 
4

(5)
1  2
1 

 x +   x − 1 + 2  (6)
x 
x 

 x 2
 + 
2 x
2
Simplify the following complex fractions:
1
9
x−
2+
x
x
(1)
(2)
3−
1
x −3
x2
Simplification of expressions involving the factorisation of denominators
EXAMPLE 22
Simplify the following:
(a)
x −1
2x +1
−
x + 4x
4x
x −1
2x +1
−
x( x + 4)
4x
( x − 1) 4
(2 x + 1) ( x + 4)
=
× −
×
x( x + 4) 4
4x
( x + 4)
4( x − 1)
(2 x + 1)( x + 4)
=
−
4 x( x + 4)
4 x( x + 4)
4( x − 1) − (2 x + 1)( x + 4)
=
4 x( x + 4)
2
=
4 x − 4 − (2 x 2 + 9 x + 4)
4 x ( x + 4)
=
4x − 4 − 2x2 − 9x − 4
4 x( x + 1)
=
−2 x 2 − 5 x − 8
4 x( x + 1)
[factorise the denominators first]
[LCD = 4 x( x + 4)]
27
(b)
5
3x
4
+
+
x − x −6 3− x 2+ x
5
3x
4
[factorise the denominators first]
=
+
+
( x − 3)( x + 2)
3− x
2+ x
[ x + 2 = 2 + x but x − 3 ≠ 3 − x and we know that 3 − x = −( x − 3)]
5
3x
4
=
+
+
( x − 3)( x + 2)
−( x − 3)
x+2
5
3x
4
[LCD = ( x − 3)( x + 2)]
=
−
+
( x − 3)( x + 2)
( x − 3)
( x + 2)
3x
( x + 2)
4
( x − 3)
5
−
×
+
×
=
( x − 3)( x + 2)
( x − 3) ( x + 2)
( x + 2) ( x − 3)
5 − 3x( x + 2) + 4( x − 3)
=
( x − 3)( x + 2)
2
=
5 − 3x 2 − 6 x + 4 x − 12
( x − 3)( x + 2)
=
−3 x 2 − 2 x − 7
( x − 3)( x + 2)
EXERCISE 15
(a)
(b)
Simplify the following as far as possible.
x
2
(1)
(2)
+
3− x
x−3
x +1
2x
(4)
(3)
+
x−3
9 − x2
x
5
(6)
(5)
−
2
5+ x
2 x + 11x + 5
3
2− x
(8)
(7)
−
6x + 6
6x
x+2
8x + 1
(9)
−
2
4x − 2x +1
8 x3 + 1
x
2
−
3+ x
x+3
3
x +1
−
2
4− x
x − 3x − 4
x
2 x − 24
− 2
4− x
x − 4x
x+4
x
x+2
−
+
2
2
x−2
x − 2x
(2 − x)
Mr Faulty was given the following question.
x
2
Simplify
−
3x + 1
3x
Mr Faulty answered it in the following manner:
x
2 ( +1)
…………..Line 1
∴
−
×
3x + 1
3 x ( +1)
x
3
…………..Line 2
=
−
3x + 1
3x + 1
x−3
…………..Line 3
=
3x + 1
(1)
(2)
What were the fundamental errors made in line 1 and line 2?
Show Mr Faulty the correct way of doing it.
28
Summary of the strategies for working with algebraic fractions
Multiplication and division
Addition and subtraction
1. When dividing remember to “tip and
1. Factorise the denominators
times” when required
2. Consider the sign-change rule
2. Factorise the numerator and
3. Find the LCD and convert each
denominator
fraction to an equivalent fraction
3. Consider the sign-change rule
with the same denominator (LCD)
4. Cancel and simplify
4. Write your answer as one fraction
CONSOLIDATION AND EXTENSION EXERCISE
(a)
State whether the following numbers are rational or irrational. If you consider the
number to be rational, provide an explanation of why it is rational.
6
5
36
(1)
(2)
(3)
−1
7
9
6
36
7
(4)
(5)
(6)
(7)
( 7)
(10)
3
2
8
(8)

0, 45
(11)
3
−8
`
1
4
(9)
6
(12)
225 + 400
(b)
Without using a calculator and by showing all of your workings, determine between
which two integers the following irrational numbers lie:
4
54
− 54
25
(2)
(3)
(1)
(c)
Write down three rational numbers between
(d)
Represent the following on a number line:
(1)
(2)
{ x : − 2 ≤ x < 5 ; x ∈ }
(3)
(e)
Simplify the following:
(1)
(2 x − 3 y )(3 x + 2 y )
(3)
(5)
(7)
(8)
(f)
[4 ; ∞)
2 and 10 .
{ x : x < 3 ; x ∈ }
(4)
(−
(2)
( x 2 + 1)( x 4 + 1)( x 2 − 1)
−3(2 x − 5)2
(4)
2
3 
4 
x(6 x − y ) − x  4 x − y  (6)
3
2 
9 
5 ; 25 
(9 x 2 + 12 xy + 16 y 2 )(3x − 4 y)
( a − 2b + 3)( a + 2b + 3)
(3 x 2 − 4 y 3 ) 2 − ( x 2 + 2 y 3 )( − x 2 + 2 y 3 )
(5a − 2b)
3
(9)
(2 x 4 + 3 x 4 )3
(2 x 4 .3 x 4 ) 2
Factorise the following:
(1)
3 x 2 − 5 x − 12
(2)
3 x 2 − 9 x − 12
(3)
12 − 16 x − 3 x 2
(4)
x3 − 64 x
(5)
x3 − 64
(6)
x 2 − 64
(7)
4 x 2 − 100
(8)
4 x 2 − 100 x
(9)
x 2 + 64
(10)
x 2 ( x − 3) − 4( x − 3)
(11)
x 2 ( x − 3) − 4(3 − x)
(12)
− x 2 + 64
(13)
3x3 − x 2 + 6 x − 2
(14)
3 x3 − x 2 − 6 x + 2
(15)
2 − 16 x 3
29
(g)
(h)
(i)
Simplify:
(1)
3x 2 + 6 x
6x
(3)
3x −
(5)

3 + 4x2 
 2 x −

2 x 

(7)
(2 − x)3
x3 − 8
(9)
x 4 − 16 4 x 2 + 16
÷
16 x
4 x2
1
x3 − 2
+
12 x
4x2
(2)
3 2 x + 3 17 x
−
−
4x
3y
x
(4)
4 x4 − 8x2
8 x3
(6)
3+
(8)
3x − 1
x
−
x − 5 x + 6 ( x − 3) 2
2
2x + 4
x−4
+
x − 2 2 x − x2
2
The number π is the ratio of the circumference of a circle to its diameter.
(1)
Use your calculator to determine π . Is the number π rational or irrational?
(2)
(3)

 is rational.
Show that the recurring decimal 3,142857
22
. Is this true?
It has been stated in the past that π =
7
Use the results in (1) and (2) to justify your answer.
(1)
Simplify: 9 x 2 − (3 x 2 − 2) 2
(2)
Factorise: 9 x 2 − (3 x 2 − 2) 2
Factorise:
x3 + 2 x 2 + 2 x + 1
(k)
Simplify:
2
x −1
2x
x+
x −1
(l)
(1)
Factorise:
(2)
Hence or otherwise factorise: (a + b)2 − 12(a + b) + 32
(j)
(m)
If x −
(1)
(n)
x−
p 2 − 12 p + 32
1
= 7, determine the value of:
x
1
1
x2 + 2
(2)
x3 − 3
x
x
If (a + b) 2 = 10 , ab = −2 and a > b , calculate the value of:
(1)
a 2 + b2
(2)
a −b
a+b
if a + b > 0, and a > b
a −b
(o)
If a 2 + b 2 = 10 and ab = 3 calculate the value of
(p)
(ax + b)2 = 4 x 2 + px + 9 determine the value of a, b, and p where a, b, p > 0
30
(q)
(r)
2011.(20122 − 9).2013
Calculate without the use of a calculator:
2015.(20122 − 1)
Factorise:
1 2
7
(1)
(2)
(3)
3x 2 − x + 1
x −x−4
2
2
1 2 1
3
x + x+
7
2
7
(s)
Simplify: ( x −2 − y −2 ) ÷ ( x −1 + y −1 )
(t)
If the product of two numbers is 40 and their sum 20. Show that the sum of their
1
reciprocals is equal to .
2
31
CHAPTER 2
EXPONENTS
REVISION OF THE EXPONENTIAL LAWS AND DEFINITIONS
In earlier grades you were introduced to the expression
a n = a × a × a × a × ... (n factors)
where a represents the base, n the exponent and a n the power.
an
Exponent
Power
Base
Here is a summary and examples of the exponential laws and definitions. The bases are
positive and the exponents are integers.
LAWS
Bases are variables
EXAMPLES
Bases are numerical
1.
am. an = am+n
2 x 4 . 3 x = 2 . 3 . x 4 . x1 = 6 x5
33. 33. 33 = 39 = 19 683
2(a)
am
= a m −n ( m > n )
n
a
y7
= y3
4
y
66
= 62 = 36
4
6
2(b)
am
1
= n−m ( m < n )
n
a
a
y4
1
=
y7 y3
64 1
1
= 2=
6
36
6
6
3.
(a m ) n = a m×n
( p 3 ) 2 = p 3× 2 = p 6
(55 )5 = 55 × 5 = 525
4.
( ab) m = a m b m
(4 x3 y ) 2 = 42 x3× 2 y 2 = 16 x 6 y 2
(26. 38 )3 = 218. 324
5.
am
a
=
 
bm
b
m
3
3
8d 3
 8d   2 d 
=
=

 

27e3
 12e   3e 
DEFINITIONS
Bases are variables
a0 = 1
2.
x −n =
3.
ax − n =
4.
( ax ) − n =
5.
1
= xn
−n
x
1
xn
y −7 =
a
xn
1
( ax ) n
3x −4 =
1
y7
(2 x ) −3 =
6 . 30 = 6 × 1 = 6
4 −2 =
3
x4
3

26 64
=
=

3
125
5

EXAMPLES
Bases are numerical
5x0 + (2 x)0 = 5 ×1 + 1 = 6
1.
 22

 5
1
1
=
2
16
4
4.2−3 = 4.
1
1
= 3
3
(2 x )
8x
1
= a5
−5
a
1 4 4 1
= = =
23 23 8 2
(4 × 2) −2 =
1
1
=
2
64
(4 × 2)
1
= 53 = 125
−3
5
32
6.
a
= ax n
−n
x
7
= 7 y6
−6
y
3
7.
1
xn
=
a
ax − n
1
p9
=
8
8 p −9
1
32 9
4
=
= =1
−2
5 5
5
5.3
8.
1
= ( ax ) n
−n
( ax )
1
= (2m) 4 = 16m 4
−4
(2m)
1
= (3 × 2)2 = 62 = 36
−2
(3 × 2)
a
 
b
9.
−n
b
= 
a
n
a − m bn
=
b−n a m
10.
 5x 


 4y 
−2
2
−2
2
16 y 2
 4y 
=
=

25 x 2
 5x 
2
 
7
a −5 . b −8
c6
=
c −6 . d 2 a5 . b8 . d 2
−2
2
49
1
7
=  =
= 12
4
4
2
2−3
32
9
9
=
=
=
−2
3
4 . 8 32
4.3
4.2
OTHER RULES
EXAMPLES
1368 = 1
(80 − 79)1000 = 11000 = 1
1n = 1×1×1×1× .........(n times) = 1
1.
= 2.32 = 2.9 = 18
( −3) 4 = 34 = 81
( − a ) n = a n if n is even
2.
( −1) 2016 = 12016 = 1
( −2 x 3 )6 = ( −2)6 . x18 = 26 x18 = 64 x18
( −3)5 = −35 = −243
( − a ) n = − a n if n is odd
3.
( −1) 2017 = −12017 = −1
( −2 x3 )5 = ( −2)5 . x15 = −25 x15 = −32 x15
EXERCISE 1 (REVISION OF GRADE 8 AND 9)
Simplify:
2 p3 × 8 p 2
(2)
6 x 2 y 3 . 6 xy 4
(3)
25. 23. 2
(4)
94. 98. 32
(5)
72. 72. 7
(6)
72 + 72 + 7
(7)
55. 55. 55
(8)
12 .1212
(9)
313. 312
(10)
88. 88
(11)
32.22
(12)
4.23
(13)
(3 x3 y 5 ) 2
(14)
2(2a 2 )3
(15)
(16)
(2 x 3 )3 × (3 x 2 ) 2
(17)
(25. 67 )3
(18)
3(2a 2b3 ) 


3 3 3
3(3 .2 )
(a) (1)
5
5
(19)
x
x3
(20)
x
x9
(21)
83
8
(22)
46
410
(23)
−12a14
−18a6
(24)
x 40 y16
x36 y 20
(25)
−6a 3b10c
12a8b 4c
(26)
(3x 2 y 3 )2
(3xy )(3xy 7 )
(27)
 3a 3 . 2a 5 


10
 12a b 
33
2
2
(b) (1)
2 p0
(2)
(2 p )0
(3)
2(3a0 )2
(4)
x−5
(5)
( a 2 ) −2
(6)
2−6
(7)
23. 2−5
(8)
11 y −3 .11 y 6 . y −3
(9)
4x−2
(10)
(4 x) −2
(11)
3a−3
(12)
(13)
4.2−2
(14)
(4.2) −2
(15)
(3a ) −3
1
x −4
1
2.3−2
1
(2 x ) −4
(16)
(19)
(22)
(25)
1
7 −2
1
(2.3) −2
5
 
 3
x −4
y −5
(17)
(20)
−3
2
3−2
1
2x −4
(18)
(21)
(23)
5−3
2−3
(24)
32
3−5
(26)
a 2b3c −5d 8
ab −4c −6 d 8
(27)
3 x −4 y −3
9 x −2 y −1
(29)
( x −2 y 4 )2
x 2 y −3
(30)
6 x −2 .
(33)
 x −4 . y 8 
 2 −3 
x .y 
(28)
3−4
5 . 3−5
(31)
2(3a 7 ) 2 . ab −7
3c −1 × 4a10b 9
(32)
2( a −2b2 ) −3 × ( ab) −6
(2b −6 )2
160
(2)
(49 − 48) 47
(3)
(2016 − 2015) 2014
(4)
(−1)60
(5)
(−1)61
(6)
(48 − 49)50
(7)
(2015 − 2016) 2017
(8)
( −2) 4
(9)
( −2)5
(10)
( − a )9
(11)
( − a )12
(12)
−( −2)6
(13)
−( −2)7
(14)
( −34 )( −33 )
(15)
(−3) 4 (−3)3
(16)
( −2 a 3 ) 4
(17)
( −2 a 3 ) 5
(18)
( − a 3b )30
(19)
( − a 3b )33
(20)
−( −3 xy ) 4
(21)
−(−3 xy )5
(22)
− ( −2 x 2 y 3 ) 4
(23)
− ( −2 x 2 y 3 ) 7
(24)
( −3 x 3 ) 2 ( −3 x 3 ) 3
(c) (1)
1
(2 x −3 ) 2
−2
Remember the difference between multiplication and addition:
[add the exponents of like bases]
(4 x 4 )(4 x 4 ) = 42. x8 = 16 x8
4 x4 + 4 x4 = 8x4
[add the coefficients but not the exponents]
(d) Simplify:
12 x3 − 3x3 × 2
(1)
(2)
2 x 3 . x 2 + 3 x. x 4
(3)
(5)
4 x5 . 4 x5 + 3x3 . x 6
(6)
3 x 3 . x − 4 x. x 3
( −2)( −2a 2b 2 )(−2a 3b3 )
(4)
( −2 x 2 y )( −3 xy 3 )
(7)
(5a 5 )(4a 5 ) − 15a. a 9 + 10b10
(8)
( −6 x3 . x 4 ) + (−2 x 5 )(− x 2 )
(9)
( −2 x 4 )(−3 y 2 ) − (−2 xy 2 )(−2 x 3 )
(10)
−3( −2a 3 ) 2 + 2( −2a 2 )3
(11)
2 x . 3 x 4 − ( −2 x 3 ) 2 + (−3 x 2 )( −2 x 3 )
34
EXAMPLE 1
Simplify the following:
16 2. (32 ) 4
27. 25. 81
(a)
(b)
(2 x )4 . 9 x
122 x
(c)
16 x −1
24 x. 32
(b)
(2 x )4 . 9 x
122 x
(c)
16 x −1
24 x. 32
Solutions
16 2. (32 ) 4
27. 25. 81
(a)
(24 ) x −1
24 x. 25
=
(24 )2 . 38
212. 34
=
24 x. (32 ) x
(22. 3)2 x
=
=
28. 38
212. 34
=
2 4 x. 32 x
2 4 x. 32 x
24 x − 4
= 4 x +5
2
=
34 81
=
24 16
= 24 x−4−(4 x −5)
=1
= 24 x − 4− 4 x +5
=2
EXERCISE 2
(a) Simplify the following leaving your answers in exponential form where necessary:
515
34. 35
28. 24
(2)
(3)
(1)
512. 56
32
25
(4)
25. 24
(23 )2
(5)
(34 )2
35. 32
(6)
5 . 55
(53 )3
(7)
(2 . 5)3
2 . 52. 53
(8)
(23 . 3)2
23. 35
(9)
(33 )3 . 73
32. 35. 49
(10)
(32 )3 . (24 )2
9 . 36. 64
(11)
363.163
62. 63. (27 )2
(12)
(112 )5 . (7 4 ) 2
(11)(11)9 . 493
(13)
817. 84
3 . (312 )2 . 47
(14)
1254. (52. 3) 2 . 3
154
(15)
(2−4 ) 4
8 . 2−20
(2)
(25 ) x
(3)
(3 x )3
(53. 35 ) x
(6)
9 x . 32 x
(b) Simplify the following:
(1)
2 x. 2 x
(4)
2 x. 2 3
(5)
(7)
81x. 27 2 x
(8)
(10)
7 x. 7 2 x
(7 2. 7) x
(11)
(13)
12 x. 3− x
2 . 4x
(14)
5x. 5x
25x
25 . 9x
3x. 3x. 5
32 x
4x
(9)
(12)
(15)
35
(2 x )3. 2 x
16 x
(2 x )3 . 27 x
8 x. (32 ) x . 3x
1
 
4
−3 x
.
1
16 x
(c) Simplify the following:
32 x +1
(1)
32 x
23 x +6
(4)
23 x + 4
7 2 x −1
(7)
49 x −1
(10)
5 x. 25 x −1
5 .125 x
(13)
50 x +1
2 x +1. 25 x + 2
(2)
(5)
(8)
2 x +5
2 x+2
64 x − 2
64 x − 4
9 x +1
(3)
(6)
(9)
32 x. 81
(11)
9a. 4a −1
32a −1. 22a
(14)
5 . 45 y
9 y. 5 y+2
5m+4
5m+5
4n
2 2 n −3
16 p −3
8 p−4
(12)
36 x + 2
6 2 x +5
(15)
18 x. 8 x −1
9 x +1. 42 x −1
EXPRESSIONS WITH RATIONAL EXPONENTS
Exponents need not always be restricted to integers. It is possible to work with powers that
include other rational exponents such as fractions. The next example works with fractional
exponents.
EXAMPLE 2
1
(4 x3 ) 2 . x
x −3
Simplify:
− 12
Solution
1
(4 x3 ) 2 . x
x −3
− 12
1
(22 x3 ) 2 . x
=
x −3
3
2x 2 . x
=
x −3
− 12
[write 4 to the base 2]
− 12
[multiply exponents]
3 −1
2x 2 2 2 x1
= −3 = −3 = 2 x1. x 3 = 2 x 4
x
x
[apply exponential laws and simplify]
EXERCISE 3
(a) Simplify:
1
(1)
27 3
(4)
(9 x 4 ) 2 . x −2
1
1
(2)
125 3
(5)
(16a 2 ) 4 . x 2
(2)
(64 x −6 ) 3
2 x3
1
1
1
(3)
64 6
(6)
(27 y 6 ) 3 . y −1
1
(b) Simplify:
1
(1)
(36 x 2 ) 2
3 x −1
1
(4)
(49 x 2 ) 4
1
(73 ) 2 . x
− 32
1
1
1
(5)
(3)
1
(9 x3 ) 2 . ( x 2 )−1
1
(3x 2 )2
36
(125m) 3
−1
25 . (m 3 )2
−1
(6)
1
( x3 ) 9 . ( x 3 ) 2
1
( x 6 )2
EXAMPLES INVOLVING FACTORISATION
EXAMPLE 3
Simplify:
3x +1 + 2 . 3x + 2
7 . 3x
(a)
(b)
9 x + 3x − 2
9x − 4
(challenge)
Solutions
(a)
3x +1 + 2 . 3x + 2
7 . 3x
=
3x. 31 + 2 . 3x. 32
7 . 3x
[apply the rule a m + n = a m . a n ]
=
3x (31 + 2 . 32 )
7 . 3x
[factorise numerator]
3x (3 + 2 . 9)
=
7 . 3x
=
3x (21)
7 . 3x
=3
(b)
[simplify]
9 x + 3x − 2
9x − 4
(32 ) x + 3x − 2
=
(32 ) x − 4
=
(3x )2 + 3x − 2
(3x )2 − 4
(3x + 2)(3x − 1)
= x
(3 + 2)(3x − 2)
=
(3x − 1)
(3x − 2)
x
[write 9 to base 3]
[apply the rule ( a m ) n = ( a n ) m ]
[factorise the numerator and denominator]
[simplify]
EXERCISE 4
(a) Simplify:
2 x + 2 + 2 x +3
(1)
12 . 2 x
(2)
3x +1 + 3x + 2
8 . 3x +1
(4)
5 x + 4 − 5 x +3
100 . 5 x +1
(7)
8 x. 2 x + 2 .16 x +1
11. 2 x +1
(3)
2 x + 2 − 2 x +1
2 x + 2 x+2
(5)
4 x + 3 . 22 x +1
7 . 22 x +1
(6)
(3x ) 2 − 9 x −1
9 x −1
(8)
12 x + 4 x . 3x +1
22 x + 4. 3x
(9)
2. 3x + 3x −2
5 . 2 x +1 − 7 . 3x −1
37
(b) Simplify:
9x −1
(1)
3x + 1
36 x − 36
(4)
6x + 6
9 x − 3x − 6
(7)
3x − 3
(2)
(5)
(8)
4x − 9
2x − 3
36 x − 6 x
6x −1
4x − 4 . 2x + 3
2x −1
(3)
(6)
(9)
25 x − 36
5x + 6
16 x − 49
4x + 7
25 x − 5 x +1 − 6
5x − 6
EXPONENTIAL EQUATIONS
In Grade 9 you learnt that in an exponential equation, the exponent is the unknown.
EXAMPLE 4
Solve the following equations:
(a)
4 . 25
x +3
=4
(b)
(0,5)
x−1
1
= 
4
x
Solutions
(a) 4 . 25 x +3 = 4
∴25x+3 = 1
2 x +3
∴ (5 )
=5
[divide both sides by 4]
∴52x+6 = 50
∴2x + 6 = 0
∴2x = −6
∴ x = −3
1
(b) (0,5) x−1 =  
4
1
∴ 
2
x −1
0
[write 25 to base 5 and 1 as 5 ]
0
[multiply exponents]
[equate exponents]
x
 1 
= 2 
2 
x
∴ (2−1 ) x −1 = (2−2 ) x
[apply the definition
∴ 2 − x +1 = 2 −2 x
[multiply exponents]
[equate exponents]
[solve]
∴− x + 1 = −2x
∴ x = −1
(c)
[write 0,5 as a fraction and 4 as 2 2 ]
1
= a−n ]
n
a
9x + 32x+1 = 36
∴ (32 ) x + 32 x. 3 = 36
[write 9 to base 3]
∴ 32 x + 32 x. 3 = 36
[apply the rule a m + n = a m . a n ]
∴ 32 x (1 + 3) = 36
[factorise]
2x
[simplify]
2x
[divide both sides by 4]
∴ 3 (4) = 36
∴3 = 9
∴32x = 32
∴2x = 2
∴x =1
[write 9 to base 3]
[equate exponents]
[solve]
38
(c)
9x + 32x+1 = 36
EXERCISE 5
(a) Solve the following equations:
(1)
(5)
(9)
2x = 1
74 x = 49
32( x−1) = 81
(2)
(6)
(10)
3x = 3
(13)
3x =
1
9
(14)
4x =
3.3x = 243
5.5x−5 = 5
x
1
16
(3)
(7)
(11)
3x = 27
(15)
5x =
(19)
1
3 
3
x
1
1
1
(18)   =
(17)   = 16
27
4
3
(b) Solve the following equations:
(1)
5 . 9 x−1 = 5
(4)
5 .125 x+3 =
(7)
(0, 2) x−2 = 0, 04
(10)
4x = 16 x−1
(2)
1214 x = 11
(16)
1
  =4
2
1
(20)
3
1 2x
. 2 =1
8
23 x−1 = 64
2.3x = 162
8 x. 2 = 128
x
1
125
x−1
7 . 49 x+ 2 = 49
4x = 16
(4)
(8)
(12)
=
(3)
x− 2
1
49 x
= 343
(6)
(0, 25) x = 0,125
(8)
8
2
=
 
27
3
x
0, 4 = 0, 064
(9)
(3 x +1 )3 = 9 x −3
(11)
3x ⋅ 9x−1 = 81
(12)
(0,5) x −1 = 4− x
(14)
(13) 812 x +1 = 27 x − 2
(c) Solve the following equations:
(1)
(2)
2 x +1 + 2 x + 2 = 24
8− x = 2. 4 x −1
(15)
4. 2 x = (0,5) x− 2
5x+1 − 2 . 5x = 75
(3)
3x + 3x + 3x = 33
1
25
7 x +1 + 14 . 7 x = 147
(4)
(5)
(5)
2 . 32 x +1 + 3 . 9 x = 243
EQUATIONS WITH RATIONAL EXPONENTS (FRACTIONS)
EXAMPLE 5
Solve the following equations:
1
x3 = 4
(a)
(b)
1
1
x 2 − 5x 4 + 6 = 0
Solutions
1
(a) x 3 = 4
1
∴ ( x 3 ) 3 = 43
∴ x = 81
1
[raise both sides to the reciprocal of the exponent]
1
(b) x 2 − 5x 4 + 6 = 0
1
1
1
1
∴ ( x 4 )2 − 5 x 4 + 6 = 0
∴ ( x 4 − 3)( x 4 − 2) = 0
1
∴ x4
= 3 or
1
∴ ( x 4 )4
1
x4
= 34 or
∴ x = 81
or
1
1
[change the expression x 2 to ( x 4 ) 2 ]
[factorise the trinomial]
=2
1
( x 4 ) 4 = 24
x = 16
39
EXERCISE 6
(a) Solve for x:
1
1
1
(1)
x2 = 5
(2)
x3 = 2
(3)
x4 = 1
(4)
x5 = 2
(5)
x3 = 4
(6)
x 2 = 27
(7)
x 2 = 32
(8)
x 6 = 128
(9)
3x5 = 729
1
5
2
7
3
(b) Solve for x:
1
1
1
1
1
1
1
1
1
1
(1)
x 2 − 7 x 4 + 12 = 0
(2)
x 2 − 8 x 4 + 16 = 0
(3)
x 2 − 8x 4 + 7 = 0
(4)
x3 − 6x 6 + 8 = 0
(5)
x 7 − 3x14 + 2 = 0
(6)
x − 9 x 2 + 18 = 0
1
CONSOLIDATION AND EXTENSION EXERCISE
(a) Simplify without using a calculator:
(1)
2 x −2 + (2 x ) −2
(2)
2 x 0 + (2 x )0
(3)
a−1 + b−1
(4)
(a + b) −1
(5)
(3011 − 3012)3013
(6)
(3x 4 )2 . 2( x 2 ) 4
(7)
(3 x 4 ) 2 + 2( x 2 ) 4
(8)
(3 x3 + 3 x3 )3
(9)
( − 4 x 2 y ) 2 . ( −4 x 2 y ) 3
(10)
 12 x −2 y 4 

−6 7 

 18 x y 
(11)
32 2. 253
100 . 84
(13)
9 x−1. 24 x+1
27 x. 8x
(16)
121x − 4
11x + 2
(b) Solve for x:
(1)
4 x = 0, 25
(4)
3
2 x 2 = 250
(c) Simplify:
(1)
4 x. 4 x. 4 x
(d) (1)
(2)
−2
(12)
1
1
16 2
−1
1
1
(14)
2 x+2 + 5 . 2 x + 2 x
5 . 2x
(17)
4 x − 3 . 2 x +1 − 27
2x + 3
(18)
50 x − 10 x
10 x − 2 x
(2)
9 . 27 x − 2 = 3
(3)
16 . 4 x +3 = 8 x − 2
(5)
x 2 − 8x 4 + 15 = 0
(6)
23x+2 + 8x+1 = 48
(2)
4x + 4x + 4x
(3)
410 + 2 . 410 + 3 . 410
1
1
What is one half of 224 ?
36
What is one sixth of 6 ?
(e) Show that:
88 + 412 + 166 + 224 = 226
(1)
(2)
2
8 3 . (2 −3 ) 3
330 is greater than 420
(f) If a = 1 + 2n and b = 1 + 2−n , show that b =
a
.
a −1
40
(15)
( x 4 ) 8 . ( x 2 )3 . ( x 3 ) −1
1
( x 3 )2
CHAPTER 3
NUMBER PATTERNS
REVISION OF LINEAR NUMBER PATTERNS (GRADE 9)
In Grade 9, you studied linear number patterns in which there is a constant difference between
consecutive terms. For example, consider the sequence 5 ; 7 ; 9 ; 11 ; ...................
T2 − T1 = 7 − 5 = 2
T3 − T2 = 9 − 7 = 2
T4 − T3 = 11 − 9 = 2
There is a constant difference of 2 between
T3
T2
T1
T4
T5
T7
T6
consecutive terms.
We call this constant difference d.
The sequence is generated by adding d
to each term. In this example, d = 2 .
It is easy to determine consecutive terms since you are adding 2 to each previous term.
Finding the 5th, 6th, 10th or even the 20th term is easy to do. However, if you are required to
determine the 100th term, this would be extremely time-consuming. We need to therefore find
a rule which helps us do this. This rule is called the general rule for the sequence.
One way to find this general rule is to link the position of the term to the constant difference
and work from there. Notice that in the sequence above:
T1 = 5 = 2(1) + 3 where 2 is the constant difference, 1 is the position of 5 (first term) and
3 is added to keep the actual term 5.
T2 = 7 = 2(2) + 3 where 2 is the constant difference, the 2 in brackets is the position of 7
(second term) and 3 is added to keep the actual term 7.
T3 = 9 = 2(3) + 3 where 2 is the constant difference, the 3 in brackets is the position of 9
(third term) and 3 is added to keep the actual term 9.
We can continue to generate terms of the sequence in this way.
T4 = 2(4) + 3 = 11
T5 = 2(5) + 3 = 13
T6 = 2(6) + 3 = 15
The general rule is Tn = 2n + 3 and this rule can help us to find other terms:
T10 = 2(10) + 3 = 23
T100 = 2(100) + 3 = 203
A table is useful for determining the general rule of a sequence with a constant difference.
Using the previous example, draw a table as follows:
The position of the term
The constant difference
multiplied by the position of
term
What to do to get the actual
term
The actual term in the
sequence
T1
T2
T3
T4
T5
T10
Tn
2(1)
2(2)
2(3)
2(4)
2(5)
2(10)
2( n)
+3
+3
+3
+3
+3
+3
+3
5
7
9
11
13
23
2( n ) + 3
The general rule using the letter n is Tn = 2n + 3 where 2 represents the constant difference
and n the position of the term in the sequence. It is important to note that the value of n is
always a natural number. We say that the nth term of the sequence is Tn = 2n + 3 .
This general rule can now be used to determine any term of the sequence.
T9 = 2(9) + 3 = 21
T30 = 2(30) + 3 = 63
For example:
Let’s revise another example to make sure that you fully understand these concepts.
41
EXAMPLE 1
6 ; 10 ; 14 ; 18 ; ......... is a given sequence.
(a) Determine the general rule (nth term).
(b) Calculate the 100th term.
(c) Which term of the sequence is equal to 242?
Solutions
(a) Draw a table. The constant difference is 4.
The position of the term
The constant difference
multiplied by the position
of term
What to do to get the actual
term
The actual term in the
sequence
T1
T2
T3
T4
Tn
4(1)
4(2)
4(3)
4(4)
4( n)
+2
+2
+2
+2
+2
6
10
14
18
4( n ) + 2
The nth term is Tn = 4n + 2
(b)
Tn = 4n + 2
∴ T100 = 4(100) + 2 = 402
(c)
The actual term in the sequence is 242 and we want to find its position.
Let Tn = 242 where n represents the position to be determined.
Tn = 4n + 2
∴ 242 = 4n + 2
∴ 242 − 2 = 4n
∴ 240 = 4n
∴ 60 = n
∴ T60 = 242 [242 is the 60th term in the sequence]
AN ALTERNATIVE METHOD FOR LINEAR NUMBER PATTERNS
The general rule for any linear number pattern takes the form Tn = bn + c , so let’s explore this
a little further. We can determine the first few terms using this rule.
Tn = bn + c
T1 = b(1) + c = b + c
b+c
T2 = b(2) + c = 2b + c
T3 = b(3) + c = 3b + c
2b + c
b
T4 = b(4) + c = 4b + c
The constant difference is b and the first term is b + c
b+c
2b + c
b
3b + c
b
4b + c
b
42
3b + c
b
4b + c
b
Now let’s determine the general term of the linear
pattern 5 ; 7 ; 9 ; 11 ; ................... using these
findings.
b=2
b+c = 5
∴2 + c = 5
∴c = 3
Tn = bn + c
b+c
b
(first term)
(substitute b = 2 into the equation b + c = 5 )
∴ Tn = 2n + 3
EXAMPLE 2
6 ; 10 ; 14 ; 18 ; ......... is a given sequence. Determine the general rule (nth term) and hence the
1000th term.
Solutions
b=4
Tn = bn + c
b+c = 6
∴4 + c = 6
∴c = 2
b+c
b
∴ Tn = 4n + 2
[Teacher’s note: The main reason for using Tn = bn + c is to avoid confusion in Grade 11
when quadratic patterns of the form Tn = an 2 + bn + c are dealt with].
Tn = 4n + 2
T1000 = 4(1000) + 2 = 4002
EXERCISE 1
(a) For each of the following sequences, determine the general rule (nth term) and hence
calculate the 100th term.
(1)
(2)
6 ; 9 ; 12 ; 15 ; ........
9 ; 13 ; 17 ; 21 ; ........ (3)
3 ; 8 ; 13 ; 18; ........
(5)
(4)
3 ; 7 ; 11 ; 15 ; ........
10 ; 16 ; 22 ; 28 ; ..... (6)
4 ; 11 ;18 ; 25 ; .......
(7)
(9)
5 ; 0 ; − 5 ; − 10 ; ....... (8)
0 ; − 3 ; − 6 ; .......
−6 ; − 11 ; − 16 ; .......
(b)
(c)
(d)
(e)
(10)
5 ; 1 ; − 3 ; − 7 ; ....
(11)
−5 ; − 11; − 17 ; ........ (12)
(13)
2 12 ; 4 12 ; 6 12 ; .....
(14)
1
4
;1; 74 ; ........
(15)
3 12 ; 4 ; 4 12 ; ........
0,5 ; 0, 7 ; 0,9 ; ....
(16) −13 ; − 7 ; − 1 ; .....
(17) 1 ; − 9 ; − 19 ; .......
(18) 13 ; 12 ; 11 ; 10 ; .....
4 ; 11 ; 18 ; 25 ; ......... is a given sequence.
(1)
Determine the 45th term.
(2)
Which term of the sequence is 627?
19 ; 16 ; 13 ; 10 ; ......... is a given sequence.
(1)
Determine the 65th term.
(2)
Which term of the sequence is −113 ?
Tn = 9n − 4 is the nth term of a linear number pattern (sequence).
(1)
Determine the first four terms of the sequence.
(2)
Which term is equal to 986?
Consider the number pattern: 4 × 7 ; 7 × 15 ; 10 × 23 ;13 × 31 ; ......
(1)
Determine the nth term.
(2)
Determine the 50th term
43
REVISION OF OTHER TYPES OF NUMBER PATTERNS (GRADE 9)
EXAMPLE 3
Determine the general term (nth term) and hence the 20th term of the following number
patterns:
(a)
(b)
3 ; 6 ; 12 ; 24 ; .........
3 ; 6 ; 11 ;18 ; .........
Solutions
[ 1 = 20 ]
(a) T1 = 3 = 3 × 1 = 3 × 20
T2 = 6 = 3 × 21
T3 = 12 = 3 × 2 2
T4 = 24 = 3 × 23
The nth term is Tn = 3 × 2 n −1
∴ T20 = 3 × 219 = 1 572 864
T2
T1
(b) Start by adding 3 to 3 to get 6. Then add 5
to 6 to get 11. Increase the number added
each time by 2 to get consecutive terms.
To get the nth term of this type of sequence,
work with the position squared and take it from there.
[square position 1 and add 2 to get 3]
T1 = 3 = (1) 2 + 2
T2 = 6 = (2) 2 + 2
[square position 2 and add 2 to get 6]
T3 = 11 = (3) 2 + 2
[square position 3 and add 2 to get 11]
T4 = 18 = (4) 2 + 2
[square position 4 and add 2 to get 18]
T3
T4
The nth term is Tn = n 2 + 2
Tn = n 2 + 2
∴ T20 = (20) 2 + 2 = 402
EXERCISE 2
(a) For each of the following number patterns, determine the general rule and hence the 10th
term.
(1)
(2)
(3)
2 ; 4 ; 8 ; 16 ; ........
1 ; 3 ; 9 ; 27 ; ........
4 ;12 ; 36 ; ........
(4)
32 ;16 ; 8 ; 4 ; .......
(5)
−2 ; − 6 ; − 18 ; ........
(6)
1
2
(7)
16 ; 4 ;1; 14 ........
(8)
1;1;1; 1
2 4 8 16
(9)
7 ........
28 ; 7 ; 74 ; 16
; ........
;1; 2 ; 4 ; ........
(b) For each of the following sequences, determine the nth term and hence the 100th term.
(1)
(2)
(3)
1 ; 4 ; 9 ; 16 ;.........
2 ; 5 ; 10 ; 17 ; .....
4 ; 7 ; 12 ; 19 ; .....
(4)
(5)
(6)
5 ; 8 ; 13 ; 20 ; ....
0 ; 3 ; 8 ; 15 ; ....
−1 ; 2 ; 7 ; 14 ; ....
(c) Determine the general term of the sequence:
44
1 3 9 27
; ; ;
; ............
5 8 13 20
CONSOLIDATION AND EXTENSION EXERCISE
(a) Consider the number pattern: 7 ;16 ; 25 ; 34 ; .......
(1)
Determine the nth term and hence the 300th term.
(2)
Determine which term of the number pattern equals 448.
(b) Consider the number pattern: −2 ; − 5 ; − 8 ; − 11 ; .......
(1)
Determine the nth term and hence the 145th term.
(2)
Determine which term of the number pattern equals −389 .
(c) Consider the diagram made up of black dots joined by thin black lines.
(1)
How many dots are there in figure 4?
(2)
How many lines are there in figure 4?
(3)
How many dots are there in figure 8?
(4)
How many lines are there in figure 8?
(5)
Determine the general rule to find the
number of dots in the nth figure.
(6)
How many dots are there in the 186th figure?
(7)
Which figure will contain 272 dots?
(8)
Determine the general rule to find the number of lines in the nth figure.
(9)
How many lines are there in the 900th figure?
(10) Which figure will contain 650 lines?
(d) Consider the following designs.
(1)
(2)
(3)
Write down the number of squares in design 1, 2, 3, 4, and 5.
Determine the number of squares in design n.
How many squares are there in design 20?
5 14
22 26 30
; ; 1;
;
;
;............
4 13
23 28 33
(1)
Determine the nth term.
(2)
Calculate the 20th term.
(f) Consider the number pattern: 1 ; 3 ; 1 ; 6 ; 1 ; 9 ; 1 ; 12 ; 1 ;.......
Determine the 999th and 1000th terms.
(g) Sipho wrote the name SWEET over and over again as follows:
SWEETSWEETSWEETSWEET.................................
(1)
What is the 23rd letter?
(2)
Find the 402nd letter.
(3)
The first W is in the second position, the second W is in the seventh position, the
third W is in the twelfth position, and so forth. Determine in what position is the
100th W?
(e) Consider the sequence: 2 ;
45
CHAPTER 4
EQUATIONS AND INEQUALITIES
LINEAR EQUATIONS (REVISION OF GRADE 9)
A linear equation is an equation of degree one with at most one solution.
EXAMPLE 1
Solve the following linear equations:
(a)
3 − 4(k + 2) = 3k + 16
(b)
4x −1
7x + 2
x
−
=
3
6
2
Solutions
(a)
(b)
3 − 4(k + 2) = 3k + 16
∴ 3 − 4k − 8 = 3k + 16
[use the distributive law]
∴−4k − 5 = 3k + 16
[collect like terms]
∴−4k − 3k = 16 + 5
[add 5 and subtract 3k from both sides]
∴−7k = 21
[collect like terms]
∴ k = −3
[divide both sides by −7 ]
To verify that this value is the solution, substitute into the left and right sides:
Left side = 3 − 4(( −3) + 2) = 7
Right side = 3 − 4(( −3) + 2) = 7
∴ Left side = Right side
∴ k = −3 is the solution.
4x −1
7x + 2
x
−
=
3
6
2
The equation contains fractions. Insert brackets around the numeratorors if they
are expressions of more than a single term. Determine the lowest common multiple
of the denominators (LCD).
(4 x − 1)
(7 x + 2)
x
−
=
3
6
2
LCD = 6
In Grade 9 two approaches were discussed. Method 1 is the preferred approach.
Method 1
Method 2
Multiply both sides by the LCD
(4 x − 1)
(7 x + 2)
x
×6 −
×6 = ×6
3
6
2
6(4 x − 1)
6(7 x + 2)
6x
∴
−
=
3
6
2
∴ 2(4 x − 1) − (7 x + 2) = 3x
Write every term as a fraction with
the common denominator
(4 x − 1) 2
(7 x + 2) 1
x 3
× −
× = ×
3
2
6
1
2 3
2(4 x − 1)
(7 x + 2)
3x
∴
−
=
6
6
6
∴ 2(4 x − 1) − (7 x + 2) = 3x
∴ 8x − 2 − 7 x − 2 = 3x
∴ x − 4 = 3x
∴ x − 3x = 4
∴ −2 x = 4
∴ x = −2
∴ 8x − 2 − 7 x − 2 = 3x
∴ x − 4 = 3x
∴ x − 3x = 4
∴ −2 x = 4
∴ x = −2
46
EXERCISE 1
(a)
Solve the following equations:
x + 12 = 6 − 3x
(2)
3( p + 2) = 2( p − 1)
(4)
7(m − 2) − 2(7 − 3m) + 2 = 0 (6)
(1)
(3)
(5)
(b)
Solve the following equations:
x x
− =2
(1)
(2)
4 3
y + 2 y −3
−
=5
(5)
(4)
3
6
x+3
− 2( x + 1) = 2 (8)
(7)
4
2 x − 5 = 3x + 3
4( x − 3) = 3 − 6( x − 2)
4(2 x − 7) − 8(5 − x) = 3(2 x + 4) − 5( x + 7)
3x
1
− x =1
(3)
4
2
2a − 1
−a +5=0
(6)
5
3
4
5
1
 2k +  − 4 =  k − 
2
3
4
5
x+5
=7
3
m+2 m−6 1
−
=
4
3
2
x+
QUADRATIC EQUATIONS (REVISION OF GRADE 9 AND EXTENSION)
A quadratic equation is a second degree equation with a standard form ax 2 + bx + c = 0 and
has at most two real solutions. From the standard form it is clear that one side of the
equation has to equal zero. When the right hand side is zero then the quadratic expression on
the left hand side has to be factorised. Then you apply the zero-factor law, which states that
if a.b = 0 then either a = 0 or b = 0 . This is due to the fact that if you multiply any number by
zero the answer will always be zero.
For example, the zero-factor law can be applied to the equation ( x + 3)( x − 4) = 0 as
follows:
( x + 3)( x − 4) = 0
∴ x + 3 = 0 or x − 4 = 0
∴ x = −3
or
x=4
EXAMPLE 2
Solve each of the following equations:
(a)
x 2 − 5 x − 14 = 0
(b)
5x + 2x2 = 3
(c)
2x2 = 4 x
(d)
8 x − 16 − x 2 = 0
(e)
(3 x − 1)( x + 2) = 0
(f )
(3 x − 1)( x + 2) = 10
Solutions
(a)
x 2 − 5 x − 14 = 0
∴ ( x − 7)( x + 2) = 0
∴ x − 7 = 0 or x + 2 = 0
∴ x = 7 or x = −2
(b)
5x + 2 x2 = 3
∴ 2x2 + 5x − 3 = 0
∴ (2 x − 1)( x + 3) = 0
∴ 2 x − 1 = 0 or x + 3 = 0
∴ 2x = 1
∴x =
[factorise]
[apply the zero-factor law]
[standard form]
[factorise]
[apply the zero-factor law]
1
or x = −3
2
47
(c)
2 x2 = 4 x
∴ 2x2 − 4 x = 0
∴ 2 x( x − 2) = 0
Alternatively:
First divide by 2:
∴ x2 = 2 x
[standard form]
[factorise]
∴ 2 x = 0 or x − 2 = 0
∴ x = 0 or x = 2
[apply the zero-factor law]
∴ x2 − 2 x = 0
∴ x ( x − 2) = 0
∴ x = 0 or x = 2
However, please take note that you may never divide both sides by the variable you are
solving for. The reason for this is that you will lose one of the solutions.
(d)
8 x − 16 − x 2 = 0
(e)
∴ 0 = x 2 − 8 x + 16
∴ ( x − 4)( x − 4) = 0
[factorise]
∴x−4 = 0
[no need to write both]
[standard form]
∴x = 4
(f)
(3 x − 1)( x + 2) = 0
∴ 3 x − 1 = 0 or x + 2 = 0
∴ 3x = 1
1
∴ x = or x = −2
3
(3x − 1)( x + 2) = 10
The right side is not zero so proceed as follows:
∴ 3x 2 + 6 x − x − 2 = 10
[expand the product]
∴ 3x 2 + 5 x − 12 = 0
∴ (3x − 4)( x + 3) = 0
[standard form]
[factorise]
∴ 3x = 4
4
∴x =
or x = −3
3
[apply the zero-factor law]
EXAMPLE 3
(Quadratic equations of the form x 2 = constant )
Solve each of the following equations:
(a)
(b)
x 2 = 36
3 x 2 = 48
( x + 5)2 = 9
( x + 5)2 = −9
(d)
(e)
(c)
x 2 + 25 = 0
Solutions
(a)
x 2 = 36
∴ x 2 − 36 = 0
[standard form]
x 2 = ± 36
∴ ( x − 6)( x + 6) = 0
(b)
Alternatively:
x 2 = 36
[factorise]
∴ x − 6 = 0 or x + 6 = 0
∴ x = 6 or x = −6
∴ x = ±6
∴ x = 6 or x = −6
3 x 2 = 48
∴ x 2 = 16
Alternatively:
∴ x 2 = 16
2
∴ x − 16 = 0
∴ ( x − 4)( x + 4) = 0
[divide by 3]
[standard form]
[factorise]
∴ x = 4 or x = −4
48
∴ x 2 = ± 16
∴ x = ±4
∴ x = 4 or x = −4
(c)
x 2 + 25 = 0
The expression on the left cannot be factorised into real factors.
Proceed as follows:
x 2 = −25
∴ x = ± − 25
Since −25 is non-real, the equation has no real solutions.
(d)
( x + 5)2 = 9
2
∴ x + 10 x + 25 = 9
∴ x 2 + 10 x + 16 = 0
∴ ( x + 2)( x + 8) = 0
∴ x = −2 or x = −8
(e)
[the right side is not zero]
[expand the left side]
[standard form]
[factorise]
[apply zero-factor law]
Alternatively:
( x + 5) 2 = 9
∴ ( x + 5) 2 = ± 9
∴ x + 5 = ±3
∴ x + 5 = −3 or x + 5 = 3
∴ x = −8 or x = −2
( x + 5)2 = −9
∴ x + 5 = ± −9
There are no real solutions
EXERCISE 2
(a)
(b)
(c)
Solve the following equations:
(1)
(2)
( x − 3)( x + 6) = 0
(4)
(2 x − 5)(3 x + 1) = 0 (5)
( x + 4)( x − 8) = 0
x + 4( x − 8) = 0
(3)
(6)
x ( x + 3) = 0
3 x (2 x + 5) = 0
Solve the following equations:
(1)
x2 = 9
(4)
x2 = 9 x
(7)
4 x 2 = 16
(10)
x( x − 5) + 6 = 0
(2)
(5)
(8)
(11)
x 2 = 49
9x 2 = x
4 x 2 = 16 x
3x 2 − 5 x − 2 = 0
(3)
(6)
(9)
(l2)
2 x 2 = 200
2 x2 = 5x
x2 − 2 x = 8
2 x 2 − x = 10
(13)
−2 x 2 − 14 x + 16 = 0
(14)
( m − 3)( m − 2) = 12
(15)
( x + 2)2 = 16
(16)
6(1 − x 2 ) = 5 x
(17)
( p − 7)( p + 3) = 24
(18)
x 2 = −4
(19)
(2 x + 1)2 = 4
(20)
( x + 3)2 = −16
(21)
9x2 + 4 = 0
(22)
5( x − 1)2 = 45
(23)
x = x2 − 2
(24)
− x2 + 9 = 0
(25)
(2 x − 1)2 = (2 x + 1)(2 x − 1)
Consider the equation 4x 2 = x
Sandy solved it in the following
way:
(26)
( x − 3)( x + 4) − (3x − 2)2 = x − 8 x( x − 1)
(d)
Consider the equation x 2 + 64 = 0
Andy solved it in the following way:
x 2 + 64 = 0
∴ ( x + 8)( x + 8) = 0
2
4x = x
4 x2 x
=
x
x
∴ 4x = 1
∴ x = −8
State his error(s) and then solve the
equation correctly.
∴
∴x =
1
4
State her error(s) and then solve the
equation correctly.
49
EQUATIONS WITH FRACTIONS (VARIABLES IN THE DENOMINATOR)
When solving equations with fractions (variables in the denominator) it is important to
remember that division by 0 is not permissible. You must always check that your solutions
do not make any one of the denominators 0. That is the reason why we must state
restrictions when solving equations with variables in the denominators.
EXAMPLE 4
Solve the following equations. Consider the solution(s) carefully as the solution(s) may
have been stated as a restriction.
(a)
(c)
16
4( x + 4)
=
x
x
2x
5x − 3
x
+
=
2
x−3
x+3
9− x
10 +
(b)
4
10
2
−
= 2
x−5
x
x − 5x
Solutions
(a)
In this example, there are variables in the denominator. Since the denominators can
never equal zero, it is important to note that x ≠ 0 .
LCD = x and the restriction is x ≠ 0
16
4( x + 4)
∴10 × x +
×x =
×x
[multiply every term by the LCD]
x
x
∴10 x + 16 = 4( x + 4)
∴10 x + 16 = 4 x + 16
[this is a linear equation]
∴6x = 0
∴x = 0
However, since x ≠ 0 , the equation does not have a solution.
(b)
With the addition and subtraction of algebraic fractions, you needed to factorise the
denominators first before finding the LCD. With equations you will do the same.
4
10
2
−
= 2
x −5
x
x − 5x
4
10
2
∴
−
=
x −5
x
x( x − 5)
∴ LCD = x( x − 5) and the restrictions are:
x ≠ 0 and x − 5 ≠ 0
x≠5
4
10
2
× x( x − 5) −
× x( x − 5) =
× x( x − 5) [multiply by the LCD]
x −3
x
x( x − 5)
∴ 4 x − 10( x − 5) = 2
∴
∴ 4 x − 10 x + 50 = 2
∴−6 x = 2 − 50
∴−6 x = −48
∴x = 8
This is a valid solution since it is not stated as a restriction.
50
(c)
2x
5x − 3
x
+
=
2
x−3
x+3
9− x
2x
5x − 3
x
∴
+
=
[change in sign: 9 − x 2 = −( x 2 − 9)]
2
x−3
x+3
−( x − 9)
2x
5x − 3
x
∴
+
=
[factorise]
x−3
x+3
−( x − 3)( x + 3)
x
2x
5x − 3
∴
−
=
x−3
x+3
( x − 3)( x + 3)
∴ LCD = ( x − 3)( x + 3) and the restrictions are: x ≠ 3 and x ≠ −3
x
2x
5x − 3
× ( x − 3)( x + 3) −
× ( x − 3)( x + 3) =
× ( x − 3)( x + 3)
x−3
x+3
( x − 3)( x + 3)
∴ 2 x( x + 3) − (5 x − 3) = x( x − 3)
∴
∴ 2 x2 + 6 x − 5 x + 3 = x 2 − 3x
[this is a quadratic equation]
∴ x2 + 4 x + 3 = 0
∴ ( x + 1)( x + 3) = 0
[standard form]
∴ x = −1 or x = −3
Since x ≠ −3, the solution is x = −1
Summary of the strategy for solving equations with fractions
1.
2.
3.
4.
Apply the sign-change rule if necessary and then factorise the denominators.
State the restrictions.
Multiply every term by the lowest common denominator (LCD).
Identify the equation as linear or quadratic and solve the equation.
EXERCISE 3
(a)
Solve the following equations. Remember to verify whether or not the solution to
the equation is viable by considering the restrictions.
6
3
1
3
1
2
5
=
−
−
=
+
(2)
(1)
x
2x
4
x
x
x
2
2
1
5
x−2
= −1
+
=
+ 1
(3)
(4)
2
3x
x
3x
7x
5
3
3
x+3
6
+
=
−
=
(5)
(6)
2
2
x−2
x+2
x
x +1
4− x
x +x
3
5
x − 24
2x + 1
5x
−
= 2
− 2
= 2
(7)
(8)
x−6
2x − 8
x
x − 10 x + 24
x − 4x
(b)
Solve the following equations:
18
x−3 =
(1)
x
x+6
2
−1
−
=
(3)
2
x−2
x+2
x −4
x
1
2
=
−
(5)
x−2
x −3
2− x
3x + 4
3x − 2
23x
=
− 2
(7)
x+6
x −3
x + 3x − 18
51
(2)
(4)
(6)
x−2
5
7
−
=
x −1
x+2
x −1
x2 − 1
= −2
x +1
3x − 4
4
=
4
3x − 4
SIMULTANEOUS LINEAR EQUATIONS
Consider the graphs of the straight lines with equations 3 x − 2 y = 8 and 4x + 2 y = 6 . Take
careful note of the point of intersection of these two graphs, which is (2 ; − 1) . How can we
determine this point algebraically?
3
2
1
−1 0
−1
1
2
3
−2
−3
−4
The sketching of straight line graphs will be revised in Chapter 6. These two lines intersect
at the point (2 ; − 1) or at x = 2 and y = −1 . This can be verified by substituting these values
into both equations as follows:
3x − 2 y = 8
4x + 2 y = 6
3(2) − 2( −1) = 8
4(2) + 2(−1) = 6
True statement
True statement
The values x = 2 and y = −1 are said to satisfy the equations simultaneously. We will now
discuss algebraic methods as an alternative method when solving two linear equations
simultaneously. There are two algebraic methods: Elimination and Substitution.
THE METHOD OF ELIMINATION
This method is the preferred method when solving two linear equations simultaneously.
EXAMPLE 5
Solve for x and y simultaneously:
3 x − 2 y = 8 and 4x + 2 y = 6
Solution
Method 1
(Eliminating y)
Label each equation as follows:
3 x − 2 y = 8.......... A
4x + 2 y = 6.......... B
By vertically adding these two equations you will eliminate one of the variables (y in this
case) and be able to solve for the other variable (x in this case):
3 x − 2 y = 8.......... A
4x + 2 y = 6.......... B
∴ (A + B) is
7 x + 0 y = 14
∴ 7 x = 14
∴x = 2
Add the like terms of equation A and B
52
Now solve for y if x = 2 . Therefore, substitute x = 2 into either A or B to get y:
If using equation A:
If using equation B:
If x = 2 then 3(2) − 2 y = 8
If x = 2 then 4(2) + 2 y = 6
∴6 − 2 y = 8
∴8 + 2 y = 6
∴ −2 y = 2
∴ 2 y = −2
∴ y = −1
∴ y = −1
Method 2
(Eliminating x)
3 x − 2 y = 8.......... A
4x + 2 y = 6.......... B
If we want the terms in x to be eliminated when added, the coefficients have to be additive
inverses (1 and − 1 or 2 and − 2 or 3 and − 3 etc.) . Let us therefore consider the lowest
common multiple (LCM) of 3 and 4 (the coefficients of x) in order to attain that. The LCM
is 12. We want the coefficient of x of equation A to be 12 and that of equation B to be −12 .
Multiply each term of equation A by 4
and call the new equation C:
Multiply each term of equation B by −3
and call the new equation D:
3x × 4 − 2 y × 4 = 8 × 4.......... (A × 4)
∴12 x − 8 y = 32......... C
4x × −3 + 2 y × −3 = 6 × −3.......... (B × −3)
−12 x − 6 y = −18.......... D
Now by vertically adding the equations C and D you will eliminate the x:
12 x − 8 y = 32............... C
−12 x − 6 y = −18.......... D
∴ (C + D) is
0 x − 14 y = 14
∴−14 x = 14
∴ y = −1
If using equation A:
If y = −1 then 3 x − 2(−1) = 8
∴ 3x + 2 = 8
∴ 3x = 6
∴x = 2
Add the like terms of equation C and D
If using equation B:
If y = −1 then 4 x + 2( −1) = 6
∴ 4x − 2 = 6
∴ 4x = 8
∴x = 2
EXERCISE 4
Solve for x and y by using the method of elimination:
(a)
(c)
(e)
x − y = 3 and 2 x + y = 9
x + 2 y = 5 and x − y = −1
2 x − 3 y = 10 and 4 x + 5 y = 42
(b)
(d)
(f)
53
x + y = −6 and 3 x + y = −10
3 x + 5 y = 8 and x − 2 y = −1
3 y − 4 x = 7 and 2x + 5 y = 16
THE METHOD OF SUBSTITUTION
The elimination method works mainly for two linear equations only, but the method of
substitution is a method you can use for other systems of equations as well. It may seem a
little longer than the elimination method but its process is much easier to understand.
EXAMPLE 6
Solve for x and y simultaneously:
3 x − 2 y = 8 and 4x + 2 y = 6
Solution
Method 1
(Making y the subject of one of the equations)
Label each equation as follows:
3 x − 2 y = 8.......... A
4x + 2 y = 6.......... B
Now pick either one of the equations and solve for one of the variables. Let’s solve for the
variable y in equation B as it will be the easier one because the resultant equation will have
no fractions.
4x + 2 y = 6
∴ 2 y = −4 x + 6
∴ y = −2 x + 3.........C
[divide both sides by 2]
Now replace the variable y in equation A with −2 x + 3 and solve for x:
3x − 2 y = 8
∴ 3 x − 2(−2 x + 3) = 8
∴ 3x + 4 x − 6 = 8
∴ 7 x = 14
∴x = 2
Now substitute x = 2 into either equation A, B or C. Equation C will always be the
easiest choice.
If x = 2 then y = −2(2) + 3 = −4 + 3 = −1
Method 2
(Making y the subject of both equations)
3 x − 2 y = 8.......... A
4x + 2 y = 6.......... B
Solve for the variable y in both equation A and B:
3x − 2 y = 8
4x + 2 y = 6
∴−2 y = −3x + 8
3x
∴ y = − 4........C
2
∴ 2 y = −4 x + 6
∴ y = −2 x + 3.........D
Now you equate equations C and D because both equations are equal to y.
54
3x
− 4 = − 2x + 3
2
∴ 3x − 8 = −4 x + 6
[multiply by the LCD of 2]
∴ 3x + 4 x = 8 + 6
∴ 7 x = 14
∴x = 2
Now substitute x = 2 into either A, B, C or D
Equation C or D will be the easiest. We will choose equation D.
∴ y = −2(2) + 3 = −4 + 3 = −1
Method 3
(Making x the subject of one of the equations)
3 x − 2 y = 8.......... A
4x + 2 y = 6.......... B
Choose either one of the equations and solve for x. Let’s use equation A:
3x − 2 y = 8
∴ 3x = 2 y + 8
2y 8
∴x =
+ .........C
[divide both sides by 3]
3 3
2y 8
+ and solve for y:
Now replace the variable x in equation B with
3 3
4x + 2 y = 6
 2y 8 
∴4
+  + 2y = 6
 3 3
8 y 32
∴ + + 2y = 6
3
3
∴ 8 y + 32 + 6 y = 18 [multiply by 3]
∴14 y = −14
∴ y = −1
∴ If y = −1 then x =
2( −1) 8 −2 8
+ =
+ =2
3
3 3 3
EXERCISE 5
(a)
Solve for x and y by using the method of substitution:
(1) x − y = 8 and 2 x + y = 10
(2)
3 x − y = 2 and 7 x − 2 y = 8
(4)
(3) 3 x + 5 y = 8 and x − 2 y = −1
7 x − 3 y = 41 and 3 x − y = 17
(b)
Solve for x and y using either the elimination or substitution method:
3 x + 2 y = 2 and 5 x − 2 y = −18
(1) x + y = 1 and x − 2 y = 1
(2)
(4)
(3) x + 4 y = 14 and 3 x + 2 y = 12
2 y − 3 x = 7 and 4 y − 5 x = 21
(5) 3 x + 2 y = 6 and 5 x + 3 y = 11
y−2
x −1
y−4
x −3
−
= 1 and
−
=1
(6)
3
2
5
4
(c)
At ABC shop, 3 hamburgers and 2 cooldrinks cost R89, whereas 2 hamburgers and
3 cooldrinks cost R73,50. Determine the cost of 1 hamburger.
55
LITERAL EQUATIONS
A literal equation is one in which letters of the alphabet are used as coefficients and
constants. These equations, usually referred to as formulae, are used a great deal in
Mathematics, Science and Technology. The aim is to solve the equation or formula for a
specific letter (or to make that letter the subject of the formula). This is a very useful skill in
Grade 12.
EXAMPLE 7
(a)
(b)
(c)
Make t the subject of the formula v = u + at .
The area of a circle is given by A = π r 2 . Make r the subject of the formula.
The surface area of a cylinder is given by A = 2πr ( h + r ) .
A − 2πr 2
.
Show that h =
2πr
Hence determine the value of h if the radius is 3 cm and the surface area is
86 cm 2 . Round off your answer to one decimal place.
(1)
(2)
Solutions
(a)
v = u + at
Subtract u from both sides:
∴ v − u = at
Divide both sides by a where a ≠ 0 :
v −u
∴
=t
a
(b)
A = πr 2
Divide both sides by π :
A
∴ = r2
π
Square root both sides:
∴
(c)
A
= r2
π
∴r =
A
(not ± as r is positive)
π
(1)
A = 2πr (h + r )
A
= h+r
2πr
A
∴
−r = h
2πr
∴
∴
(2)
A − 2πr 2
=h
2πr
[divide both sides by 2πr ]
.
[LCD is 2πr ]
86 − 2 π(3) 2
h=
= 1, 6 cm
2 π(3)
56
EXAMPLE 8
The variable you are solving for may sometimes be present in more that one term. For these
types of equations, you will need to factorise first before solving the equation.
(a)
Make d the subject of the formula if T = a + nd − d
(b)
Solve for x in terms of y if x 2 = 6 xy + 8 y 2
Solutions
(a)
T = a + nd − d
∴T − a = nd − d
∴T − a = d (n − 1) [factorise]
T −a
∴
=d
n −1
(b)
x 2 = 6 xy + 8 y 2
[quadratic equation]
∴ x 2 − 6 xy − 8 y 2 = 0
[standard form]
∴ ( x − 4 y )( x + 2 y ) = 0 [factorise]
∴ x − 4 y = 0 or x + 2 y = 0
∴ x = 4 y or x = −2 y
EXERCISE 6
Make a the subject of the formula v = u + at
Make s the subject of the formula v 2 = u 2 + 2as
(a)
(1)
(2)
(b)
Solve for x in the following equations:
ax + b = c
3ax = 2bx + c
(1)
(2)
(3)
(4)
(6)
(c)
4 x2 − y 2 = 0
(5)
x 2 − 5 xy + 6 y 2 = 0
ax 2 + bx = 0
18 y 2 = 8x 2
Make the variable that is written in brackets after the formula the subject of the
formula.
(1)
(3)
n
( a + L), ( a )
2
mv 2
F=
, (g)
gr
S=
(2)
9
F = C + 32, (C)
5
1 1 1
+ =
u v f
(d)
Make v the subject of the formula:
(e)
Make the variable that is written in brackets after the formula the subject of the
formula.
E = mc 2 , (c)
v 2 = u 2 + 2as, (u )
(2)
(1)
(3)
F=
mv 2
, (v )
gr
57
LINEAR INEQUALITIES
Consider the following true statement:
−5 < 3
If we add or subtract any value to both sides, the statement above will still be true:
−5 + 2 < 3 + 2
−5 − 2 < 3 − 2
∴ −3 < 5
∴ −7 < 1
But, if we multiply (or divide) both sides by −1 , the statement will become 5 < −3 , which
is clearly now false. However, if the direction of the inequality sign was reversed, then the
statement would become true, i.e. 5 > −3 .
Therefore the following rules are always applicable when working with inequalities:
•
Change the direction of the inequality sign whenever you multiply or divide by a
negative number.
•
Do not change the direction of the inequality sign if you multiply or divide by a
positive number.
•
Do not change the direction of the inequality sign if you add or subtract by a
number or expression.
EXAMPLE 9
Solve the following inequalities and then represent their solutions on a number line:
(a)
2 x > −6
(b)
(c)
4(2 x − 1) > 5 x + 2
(d)
−2 x > −6
2x −1
≥ 2x +1
3
Solutions
(a)
2 x −6
>
2
2
∴ x > −3
(b)
−3
[All real numbers larger than −3 ]
[The inequality remains the same]
(c)
4(2 x − 1) > 5 x + 2
3
[All real numbers smaller than 3]
[The inequality is reversed]
(d)
∴8x − 4 > 5x + 2
∴8x − 5x > 2 + 4
∴ 3x > 6
∴x > 2
−2 x −6
>
−2 −2
∴x < 3
2
[All real numbers larger than 2]
2x −1
≥ 2x +1
3
2x −1
∴
× 3 ≥ (2 x + 1) × 3
3
∴ 2x −1 ≥ 6x + 3
∴ 2x − 6x ≥ 3 + 1
∴−4 x ≥ 4
∴ x ≤ −1
−1
[All real numbers smaller than and
including −1 ]
58
EXAMPLE 10
Solve the following compound inequalities and then represent the solutions on a number
line:
1
−2 < 3 − x ≤ 5
−3 < 3x − 1 < 5
(a)
(b)
2
Solutions
(a)
The inequality −3 < 3x − 1 < 5 is read as follows:
3x − 1 is smaller than 5 AND larger than −3 which implies that both the inequality
statements must be true.
2
x>−
−3 < 3x − 1 and 3x − 1 < 5
3
x
<
2
∴−3x < 2 and 3x < 6
2
− <x<2
2
∴ x > − and x < 2
3
3
2
2
−2
−
∴ <x<2
3
3
An alternative and relatively easier approach is the following:
∴−3 < 3 x − 1 < 5
∴−3 + 1 < 3 x − 1 + 1 < 5 + 1
[add 1 to all terms in order to get x by itself]
∴−2 < 3 x < 6
[divide all three terms by 3]
−2
∴ <x<2
[inequality signs remain the same]
3
−
2
3
2
[All real numbers greater than −
(b)
2
and smaller than 2]
3
1
−2 < 3 − x ≤ 5
2
1
∴−5 < − x ≤ 2 [subtract 3 to all terms]
2
∴10 > x ≥ −4
[multiply all terms by − 2]
[Note that the inequality signs had to be reversed]
Rewrite the inequality as follows:
∴−4 ≤ x < 10
59
−4
10
EXERCISE 7
(a)
(b)
(c)
Solve the following inequalities and represent the solutions on a number line.
x + 18 ≤ 9 − 2 x
x − 3 < 2x + 5
(1)
(2)
(4)
(3)
5( x − 1) > 7( x − 1)
4( x − 3) − 2( x − 1) ≥ 0
Solve the following inequalities and represent the solutions on a number line.
y +5
x x
3x
1
− x ≤1
+ y ≤1
− >1
(1)
(2)
(3)
4
2
3
3 2
3y + 2 y − 6
−
>0
(4)
(5)
( x + 3)( x − 4) ≤ ( x − 3)( x + 4)
4
3
Solve the following inequalities and represent the solutions on a number line.
−2 ≤ x − 1 < 5
−3 < x + 2 < 4
(2)
(1)
−5 ≤ 2 x + 1 ≤ 5
−8 < 3x − 2 < 4
(4)
(3)
−9 < 1 − 5x ≤ 21
−2 ≤ 2 − x < 2
(6)
(5)
CONSOLIDATION AND EXTENSION EXERCISE
(a)
Solve for x:
2 x = 5x + 3
(1)
(4)
4 x 2 = 16
(7)
x − 2(2 x + 3) = 0
3( x − 2)
2x − 5
−
=
(9)
4
2
3
x2 + 6
=2 −
(11)
2
2x
x +x
(13)
(b)
(2)
(5)
(8)
2 x2 = 5x + 3
− x 2 + 3 x = −10
( x − 2)(2 x + 3) = −6
4x
(10)
3
x +1
x2 + 7
−
=
2x
3x
6x
(12)
6 x − 3 ( x + 4) < 5x
(14)
2 (3x − 5)
4x −1
−
≤ x+2
5
3
−3 < 7 − 2 x ≤ 11
Solve the following equations
3x = 9
(2)
(1)
3x = 9
−2 < 2 −
(3)
Solve for x:
(d)
(1)
Solve: p +
(2)
Hence solve for x if: x 2 − 2 x +
(f)
(g)
(h)
3 x 2 = 27
4 x 2 = 16 x
( x − 2)(2 x + 3) = 0
(4)
3x 2 = 27 x
2x
< 10
3
(c)
(e)
(3)
(6)
24
= 11
p
24
= 11
x − 2x
2
and
x − 3y = 7
Given: 4 x − 2 y = 8
Solve for x and y simultaneously using the:
(1)
elimination method
(2)
substitution method
(3)
graphical method (sketch the two lines and find the point at which they
intersect.
Solve for x and y using the method of your choice.
5x − 3 y = 10 and 2 y − 3x = 12
Solve for x and represent your answer on a number line: 9 x ≤ 6 x − 10 < 14
Solve for x in the equation (2 x − 1)( y + 2) = 0 if:
y=4
(1)
(2)
y = −2
60
(i)
(j)
(k)
(l)
(m)
(n)
(o)
1
= 5, x > 1
x
1
1
1
x−
(1)
(2)
(3)
x2 + 2
x3 + 3
x
x
x
1
x−
x =1
Given:
1
1+
x
(1)
For which values of x is this equation undefined.
(2)
Solve the equation.
3
+4
Consider the equation of a hyperbolic function: y =
x−2
(1)
For what value of x will the equation be undefined?
(2)
By making x the subject of the equation, show that the equation is also not
defined for y = 4 .
There were 600 spectators at a tennis match. Of these 144 were children and there
were twice as many men as women. Determine how many women were there.
Tickets to the DJ Sbu concert cost R200 and R300. A total of 250 tickets were sold.
The total amount taken for the show was R55 000. Determine how many of each
ticket were sold.
The length of a rectangular field is 3 metres more than its breadth. The area of the
field is 70 m 2 . Calculate the length of the field.
The area of a room in a house has the following measurements as indicated on the
diagram below. If the total area of the room is 43 m 2 , calculate the value of x.
Given: x +
x+
3
2
2x
2m
8m
61
CHAPTER 5
TRIGONOMETRY
INTRODUCTION TO TRIGONOMETRY
Trigonometry can be found in many fields such as navigation, surveying, engineering and
architecture. It is the study of the relationships between the angles and the lengths of the
sides of triangles. The word “Trigonometry” is derived from two Greek words: ‘trigon’
which means triangle and ‘metron’ which means a measure and therefore trigonometry
means literally the measurement of a triangle. Angles in Trigonometry are usually indicated
by means of Greek letters: θ = theta, β = beta, α = alpha, φ = phi
Right-angled triangles
These triangles are fundamental to the study of trigonometry. In Grade 8 you were
introduced to the theorem of Pythagoras which describes the relationship between the three
sides of a right-angled triangle.
From the Theorem of Pythagoras:
c2 = a 2 + b2
a 2 = c2 − b2
b2 = c2 − a 2
Furthermore, we will label the three sides as adjacent, opposite and hypotenuse. The
longest side (opposite the right-angle) is called the hypotenuse. The opposite and adjacent
sides are dependent on which angle is used as the reference point.
Opposite to B
Hy
po
ten
us
e
Adjacent to A
Adjacent to B
Hy
po
ten
us
e
Opposite to A
TRIGONOMETRIC RATIOS
There are three special ratios in the sudy of Trigonometry, namely the sine, cosine and
tangent ratios. When these ratios are applied to a right-angled triangle, they define the
relationships between its sides and angles.
The triangles sketched on the next page are all similar because their corresponding angles
are equal and their corresponding sides are in proportion. Each of the triangles has
angles 30°, 60° and 90°. The lengths of the sides are also indicated.
62
30°
Opposite 60°
Adjacent to 30°
30°
30°
60°
60°
60°
Adjacent to 60°
Opposite 30°
With respect to the angle 60° in the three triangles, the ratio
opposite side
will be equal.
hypotenuse
opposite side 4,33
=
= 0,866
hypotenuse
5
opposite side 6,928
In ΔDEF:
=
= 0,866
hypotenuse
8
opposite side 3, 464
In ΔGHI:
=
= 0,866
hypotenuse
4
In ΔABC:
This will be true for any triangle with angles 30°, 60° and 90°. The constant 0,866 is called
the sine of 60°.
We can write this as follows:
sin 60° = 0,866
In general, we can define the sine of an angle as follows:
length of the side opposite angle θ
length of the hypotenuse
opposite
∴ sin θ =
hypotenuse
sine of an angle θ =
Notice that in all three triangles:
sin 30° =
opposite
2,5 4 2 1
=
= = =
hypotenuse
5
8 4 2
63
With respect to the angle 60° in the three triangles the ratio
adjacent side
will be equal.
hypotenuse
adjacent side 2,5
=
= 0,5
hypotenuse
5
adjacent side 4
In ΔDEF:
= = 0,5
hypotenuse 8
adjacent side 2
In ΔGHI:
= = 0,5
hypotenuse 4
In ΔABC:
This will be true for any triangle with angles 30°, 60° and 90°. The constant 0,5 is called the
cosine of 60°.
We can write it as follows:
cos 60° = 0, 5 .
In general, we can define the cosine of an angle as follows:
length of the side adjacent to angle θ
length of the hypotenuse
adjacent
∴ cos θ =
hypotenuse
cosine of an angle θ =
Notice that in all three triangles:
cos 30° =
adjacent
4,33 6,928 3, 464
=
=
=
= 0,866
hypotenuse
5
8
4
With respect to the angle 60° in the three triangles the ratio
opposite side
will be equal.
adjacent side
opposite side 4,33
=
= 1, 732
adjacent side 1,5
opposite side 6,928
In ΔDEF:
=
= 1, 732
adjacent side
4
opposite side 3, 464
In ΔGHI:
=
= 1, 732
adjacent side
2
In ΔABC:
This will be true for any triangle with angles 30°, 60° and 90°. The constant 1,732 is called
the tangent of 60°.
We can write it as follows: tan 60° = 1, 732
In general, we can define the tangent of an angle as follows:
length of the side opposite angle θ
length of the side adjacent to angle θ
opposite
∴ tan θ =
adjacent
tangent of an angle θ =
64
Notice that in all three triangles:
tan 30° =
opposite 2,5
4
2
=
=
=
= 0,577 (rounded off to 3 decimal places)
adjacent 4,33 6,928 3, 464
We can therefore conclude that the trigonometric ratios of any two similar triangles will be
the same.
Summary of the trigonometric ratios
sin θ =
opposite
hypotenuse
cos θ =
adjacent
hypotenuse
opposite
adjacent
tan θ =
EXAMPLE 1
ˆ = β and Fˆ = θ.
In ΔDEF , DE = 5, EF = 12, Eˆ = 90°, D
(a)
Determine the length of the hypotenuse DF.
(b)
Write the value of sin θ , cos θ and tan θ .
(c)
Write the value of sin β , cos β and tan β .
D
β
E
Solutions
(a)
(b)
F
opp 5
=
hyp 13
adj 12
cos θ =
=
hyp 13
opp 5
tan θ =
=
adj 12
sin θ =
(c)
E
opp 12
=
hyp 13
adj 5
=
cos β =
hyp 13
opp 12
=
tan β =
adj 5
sin β =
P
(a)
(b)
tan Q
cos V
Solutions
PR 2 = 152 − 92
β
Opposite θ
Adjacent to β
EXAMPLE 2
(a)
θ
D
DF2 = 52 + 122 [Pythagoras]
∴ DF2 = 169
∴ DF = 13 cm
Determine:
12 cm
S
25 cm
T
15 cm
(b)
R
SV 2 = 252 − 7 2
∴ PR 2 = 144
∴ SV 2 = 576
∴ PR = 144 = 12 cm
opp 12 4
∴ tan Q =
=
=
adj 9 3
∴ SV = 576 = 24 cm
adj 24
∴ cos V =
=
hyp 25
65
F
V
7 cm
9 cm
Q
θ
12 cm
Opposite β
Adjacent to θ
EXERCISE 1
(a)
Redraw the triangles below and indicate which sides are opposite, adjacent and
hypotenuse with respect to θ .
θ
(b)
(c)
(d)
(e)
θ
State the following in terms of g, k and m:
(1)
sin K
(2)
cos K
(3)
(5)
tan K
cos G
(4)
(6)
sin G
tan G
State the following in terms of p, q and r:
(1)
sin θ
(2)
cos θ
(3)
(5)
tan θ
cos α
(4)
(6)
sin α
tan α
θ
α
F
ΔDEF is shown with given dimensions.
(1)
Determine the length of EF.
(2)
Write the value of sin θ , cos θ and tan θ
B
17 cm
5k cm
C
E
15cm
θ
E
D
O
4k cm
P
N
Determine the value of:
(f)
(1)
sin N
(1)
Determine cos L and cos B
(2)
What do you notice?
(3)
What can you deduce about
the two triangles and why?
(2)
tan C
P
24cm
L
N
12 cm
E
16cm
30cm
K
66
B
CALCULATING THE TRIGONOMETRIC RATIOS OF A GIVEN ANGLE
The sine, cosine and tangent ratio can be calculated with the use of a calculator which
must be on the DEG (degree) mode.
Consider the triangles we used to discover the sine, cosine and tangent ratios on page 64.
adj
We discovered that cos 60° =
= 0, 5 and it is possible to use a calculator to do this
hyp
calculation as well as for the other trigonometric ratios of 60°.
On your calculator, press the button “cos” and then type in “60” (some calculators will
expect you to close the brackets first). Then press “ = ” and you will get 0,5. Now verify the
ratios of sin 60° = 0,866 and tan 60° = 1, 703 rounded to three decimal places.
EXAMPLE 3
Evaluate the following trigonometric ratios rounded off to two decimal places where
necessary.
(a)
cos 35°
(b)
tan 36°
(c)
sin 83°
(d)
sin 43°
(e)
cos14°
(f)
tan 45°
(g)
5cos 60°
(h)
sin12°
12
(i)
1
tan 36°
3
( j)
20,35
sin 38°
(k)
sin(35° + 75°) = 0,94
(l)
sin 35° + sin 75° = 1, 54
Solutions
(a)
cos(35°) = 0,82
(b)
tan(36°) = 0, 73
(c)
sin(83°) = 0, 99
(d)
sin(43°) = 0, 68
(e)
cos(14°) = 0, 97
(f)
tan(45°) = 1
(g)
5 cos(60°) = 2, 5
(h)
sin(12°)
= 0, 02
12
(i)
1
tan(36°) = 0, 24
3
(j)
20,35
= 33,05
sin(38°)
(k)
sin(35° + 75°) = 0, 94
(l)
sin 35° + sin 75° = 1, 54
Take note: sin(35° + 75°) ≠ sin 35° + sin 75°
(An important result)
EXAMPLE 4
Determine the decimal value of the following if A = 23,8° and B = 18,1°
(Round off your answers to one decimal place)
(a)
sin(A + B)
(b)
tan 2B
(c)
(b)
tan(2(18,1°))
cos2 (2A − 10°)
Solutions
(a)
sin(23,8° + 18,1°)
(c)
cos 2 (2(23,8°) − 10°)
= sin(41,9°)
= tan(36, 2)
= cos 2 (37, 6°)
= 0, 667... ≈ 0, 7
= 0, 7318... ≈ 0, 7
= (cos(37, 6°))2
= 0, 627... ≈ 0, 6
67
EXERCISE 2
(a)
(b)
Calculate with the use of a calculator the following rounded off to two decimal
places where appropriate.
(1)
cos 23°
(2)
sin 68°
(3)
tan 64°
(4)
cos34°
(5)
2cos 45°
(6)
5sin 65°
(7)
7 tan 58°
(8)
−4sin 30°
(9)
1 sin 70°
3
(10)
sin 60°
20
(11)
(12)
tan(54° + 25°)
(13)
tan 54° + tan 25°
(14)
(15)
sin 60° cos 70°
tan 46° sin 30°
cos 24°
24
40
−25sin 45°
Calculate with the use of a calculator the value of the following:
2
[ sin 2 θ can be written as (sin θ)2 , eg. sin 2 23° = ( sin(23°) ) = 0,1526... ]
(1)
(3)
sin 2 61° + cos2 61°
sin 2 42° + cos2 42°
(2)
(4)
sin 2 49° + cos2 49°
cos2 30° + sin 2 30°
What possible conclusion can you make from the above calculations?
(c)
Determine the decimal value of the following if A = 35° and B = 52°
(Round off your answers to two decimal places)
(1)
(4)
cos (A + B)
1
3 tan A
3
(2)
cos A + cos B
(3)
(5)
2sin 2 (2A − B)
(6)
3sin 2B
cos 3A + sin B
SOLVING SIMPLE TRIGONOMETRIC EQUATIONS
From the previous example and exercise, we saw that the trigonometric ratio of a given
angle can easily be found by using a calculator. The reverse process is also possible. This
process is finding the size of an angle when given its sine, cosine or tangent ratio.
Consider the trigonometric equation cos θ = 0, 5 . We want to find the size of the angle that
will result in the ratio 0,5.
In order to do this, we will make use of the cos−1 function on the calculator. This function
can be found above the “cos” button. In order to use this function, you have to press the
“SHIFT” button on your calculator.
Therefore, the sequence of buttons to press on most of the calculators will be as follows:
•
Press the SHIFT button followed by the “cos” button. The cos−1 will appear on
your calculator screen.
•
Enter the ratio value (in this case 0.5)
•
Press the “=” button (some calculators expect you to close the brackets first)
Some calculators may have the button INV or 2nd F instead of SHIFT.
Always ensure that your calculator is on the degree mode [DEG]
68
EXAMPLE 5
(a)
Determine the size of the acute angle θ in each of the following trigonometric
equations. Round your answers off to one decimal place where necessary.
(1)
(b)
cos θ = 0, 5
(2)
tan θ = 4,123
(3)
Solve the following equations. Round your answers off to one decimal place where
necessary. All angles are acute.
(1)
2sin θ = 1,124
(2)
sin 2θ = 0, 435
(3)
1
tan 2 x = 3
2
(4)
1 + 2 cos( x + 10°) = 2,356
(1)
cos θ = 0,5
press [shift] [cos] then 0,5
as it appears on the calculator screen
(2)
∴θ = cos −1 (0,5)
∴θ = 60°
tan θ = 4,123
Solutions
(a)
−1
(3)
(1)
press [shift] [tan] then 4,123
∴θ = tan (4,123)
∴θ = 76, 4°
as it appears on the calculator screen
sin θ = 0, 706
press [shift] [sin] then 0,706
−1
(b)
sin θ = 0, 707
∴θ = sin (0, 706)
∴θ = 44,9°
as it appears on the calculator screen
2sin θ = 1,124
∴ sin θ = 0,562
sin θ has been multiplied by 2
isolate sin θ by dividing by 2
∴θ = sin −1 (0,562)
∴θ = 34, 2°
(2)
(3)
sin(2θ) = 0, 435
insert the brackets
∴ 2θ = sin −1 (0, 435)
∴ 2θ = 25, 78529...°
∴θ = 12,9°
determine θ
divide by 2 and then determine θ
1
tan 2 x = 3
2
1
∴ 2 × tan 2 x = 3 × 2
2
∴ tan 2 x = 6
LCD: 2
isolate tan 2 x
∴ 2 x = tan −1 (6)
∴ 2 x = 80,537...°
∴ x = 40,3°
69
(4)
1 + 2 cos( x + 10°) = 2, 356
∴ 2 cos( x + 10°) = 1,356
∴ cos( x + 10°) = 0, 678
∴ x + 10° = cos −1 (0, 678)
∴ x + 10° = 47, 3124...°
∴ x = 37, 3°
EXERCISE 3
(a)
Determine the size of the acute angle θ in each of the following, rounding your
answers off to two decimal places where necessary.
(1)
(2)
(3)
sin θ = 0,866
cos θ = 0,866
tan θ = 1, 703
sin θ = 1
cos θ = 1
tan θ = 1
(5)
(6)
(4)
(b)
Solve the following equations by finding the size of the acute angle θ in each case
rounding your answers off correct to one decimal place where necessary.
(1)
(2)
(3)
cos 3θ = 0,33
3cos θ = 0,33
sin 4θ = 0,888
tan 4θ = 4
4 tan 4θ = 4
(5)
(6)
(4)
4 sin θ = 0,888
(c)
Determine the value of the acute angle x in each of the following equations.
(1)
(2)
sin( x − 20°) = 0, 678
3cos( x + 30°) = 2,121
(4)
(3)
2 tan(2 x − 10°) = 3, 4641
2 tan 2 x − 10 = 3, 4641
(d)
Michael attempted to solve the equation 2 cos(4θ + 20°) = 0,8 . Identify what errors
were made by Michael in lines 1 and 2 and then provide the correct solution.
2 cos(4θ + 20°) = 0,8
Line 1
∴ cos(2θ + 10°) = 0, 4
Line 2
Line 3
Line 4
∴θ = 2 cos −1 (0, 4) + 10°
∴θ = 2(66, 421...°) + 10°
∴θ = 142,8°
TRIGONOMETRIC RATIOS OF THE SPECIAL ANGLES 30°, 45° AND 60°
Before discussing the trigonometric ratios of special angles, it is essential for you to
understand the following concepts involving surds.
Addition and subtraction of surds
Consider the following:
(a)
2 6 +5 6 = 7 6
(add the like surds)
(b)
7 3+3 7 − 5 7 = 7 3 − 2 7
(subtract the like surds)
Note that 7 3 − 2 7 cannot be simplified any further because the surds are unlike.
Multiplication of surds
The following rules are useful when multiplying or dividing surds:
●
●
a × b = ab for a > 0 ; b > 0
( a )2 = a . a = a2 = a for a > 0
70
Consider the following:
(a)
5× 7 = 5 7
(b)
3 5 × 2 = 3 5 × 2 = 3 10
(c)
( 3) 2 = 3
(d)
(5 2) 2 = (5) 2 ( 2) 2 = 25.2 = 50
(e)
2× 5
2 5
2
=
=−
3
−3 × 5 −3 5
(f)
2+
2 2
2 2 2 2 +2
= ×
+
=
2 1
2
2
2
The angles 30° , 45° and 60° are referred to as special angles. The reason for them being
called special angles is because the trigonometric ratios of these angles can be evaluated
without using a calculator.
The following two triangles can be used to determine trigonometric ratios of special angles.
Triangle A
Triangle B
30° 30°
45°
2
3
45°
60°
60°
For 45° we will sketch a right-angled triangle with sides 1 and 1 and then find the third side
using Pythagoras. For 30° and 60°, we will sketch an equilateral triangle with side-lengths of
2 units. The dotted line AD is a perpendicular bisector of side BD and divides  into two
equal angles ( 30° ). We can find the length of the AD using Pythagoras.
EXAMPLE 6
Evaluate the following without using a calculator. Use the above triangles to assist you.
(a)
cos 45°
(b)
cos 60°
(c)
tan 45°
(d)
sin 30°
Solutions
(a)
cos 45° =
adj
1
=
(Triangle A)
hyp
2
(b)
cos 60° =
adj 1
= (Triangle B)
hyp 2
(c)
tan 45° =
opp 1
= = 1 (Triangle A)
adj 1
(d)
sin 30° =
opp 1
= (Triangle B)
hyp 2
71
EXAMPLE 7
Calculate the following without using a calculator:
(a)
sin 30° + cos 60°
(b)
tan 60° − cos 30°
(c)
sin 60°
tan 30°
(d)
cos2 45°
(b)
tan 60° − cos30°
Solutions
(a)
(c)
sin 30° + cos 60°
1 1
+
2 2
=1
3
3
−
1
2
2 3
3
=
−
2
2
2 3− 3
3
=
=
2
2
=
1
sin 30°
= 2
1
tan 30°
3
cos 2 45° = (cos 45°) 2
(d)
 1 
=

 2
1
=
2
1
3
3
= ×
=
2 1
2
EXAMPLE 8
2
Solve the equation 2 cos x − 3 = 0 without using a calculator where x is an acute angle.
Solution
2 cos x − 3 = 0
30° 30°
∴ 2 cos x = 3
3
3
60°
2
3 is adjacent to 30° and 2 is the hypotenuse
∴ cos x =
∴ x = 30° since cos 30° =
60°
opp
3
=
hyp 2
EXERCISE 4
(a)
Evaluate the following without the use of a calculator:
(1)
sin 30° + cos 2 45°
(2)
cos 30° + tan 60°
(3)
(4)
sin 2 45° − cos 2 30°
(5)
tan 60° + sin 60°
(6)
72
tan 30°
tan 60°
cos 30°
tan 30°
(b)
(c)
(7)
cos 30°.tan 30° − tan 45°
(8)
sin 45°.cos 45°
tan 60°.tan 30°
(9)
2 cos 45° − 3 tan 30°
(10)
4 cos 2 30° + tan 30°.sin 60°
(1)
Using the fact that 1 =
(2)
Hence show that:
a
where a > 0 , show that
a
(i)
cos 45° =
2
2
(ii)
tan 30° =
(iii)
tan 60° =
3
3
(iv)
sin 60° =
1
2
=
2
2
3
3
3
2 3
Without using a calculator, solve the following equations, where the angles are acute.
(1)
(4)
sin x =
1
2
2 cos x = 1
1
2
(2)
cos x =
(5)
tan x = 3
(6)
(3)
2sin x = 3
3 tan x − 1 = 0
(7)
tan x = 1
(8)
sin 2 x = 0,5
(9)
cos (3 x − 15°) = 0, 5
(10)
2 cos 2 x = 2
(11)
3 tan x = 3
(12)
1
2
sin 4 x =
4
8
FINDING SIDES AND ANGLES USING TRIGONOMETRIC RATIOS
EXAMPLE 9
(Finding the length of a side)
F
ˆ = 90° and N
ˆ = 28° .
In ΔFUN , FN = 10, U
Calculate the length of FU, rounded off to one decimal place.
.
Solution
N
Let N be your point of reference (given angle).
Side FU is opposite 28° and side FN is the hypotenuse.
You now need to create an equation involving
opp
and the angle 28°.
the ratio
hyp
opp
= sin 28°
hyp
FU
∴
= sin 28°
10
multiply by LCD
∴ FU = 10sin 28°
∴ FU = 4, 7 units
73
10
28°
U
EXAMPLE 10
.
(Finding the length of a side)
Consider the triangle sketched alongside.
Calculate the length of ON correct to one decimal
place.
C
57°
12,1 cm
Solution
Let O be your point of reference (given angle).
You want side ON, which is adjacent to 57°.
You have side NC, which is opposite to 57°.
N
You now need to create an equation involving the ratio
opp
= tan 57°
adj
12,1
∴
= tan 57°
ON
∴12,1 = ON tan 57°
12,1
∴
= ON
tan 57°
∴ ON = 7,9 cm
EXAMPLE 11
opp
and the angle 57°:
adj
ˆ = 33°
Alternatively: C
ˆ opp = tan 33°
From C:
adj
ON
∴
= tan 33°
12,1
∴ ON = 12,1tan 33°
∴ ON = 7,9 cm
(int ∠ s of Δ )
(Finding an angle)
Calculate the size of θ correct to one decimal place in each case.
(a)
P
10 m
θ
U
(b)
F
13cm
16 m
A
N
θ
Solutions
(a)
We have side PU, which is adjacent to θ.
Side PN is the hypotenuse.
Now form an equation involving the ratio
adj
hyp
10
∴ cos θ =
16
∴ cos θ = 0, 625
cos θ =
∴θ = 51,3°
74
adj
and angle θ:
hyp
11, 2cm
N
O
(b)
We have side AF, which is adjacent to θ .
Now form an equation involving the ratio
opp
adj
11, 2
∴ tan θ =
13
∴ tan θ = 0,8615384615
Side FN is opposite θ .
opp
and angle θ :
adj
tan θ =
(don't round off)
∴θ = 40, 7°
EXERCISE 5
(Round answers off to one decimal place in this exercise)
(a)
Q
Calculate the length of PQ in Δ PQR .
R
3
42°
C
P
(b)
(1)
Calculate the length of AB.
(2)
Calculate the length of BC.
67°
5,6
B
A
(c)
Calculate:
(1)
the size of θ .
(2)
the the length of AC.
θ
(d)
Calculate:
(1)
the size of α .
(2)
the size of θ .
α
θ
(e)
Calculate:
(1)
the length of AC.
(2)
the length of AB.
67°
75
(f)
In the diagram, BD ⊥ AC .
Using the information provided,
calculate the length of AC.
(g)
In ΔABC , CD ⊥ AB , Â = θ , B̂ = 40° ,
AD = 15 cm and DB = 16 cm .
Calculate the size of θ .
48°
42°
40°
(h)
θ
Using the information provided on the given
diagram, calculate the length of BC.
14°
36°
20°
(i)
(j)
In the given diagram, ΔABC is right-angled at C.
3
ˆ ∈ ( 0° ; 90° ) .
and A
It is given that AC = 4 units, tan A =
2
(1)
Determine the length of BC without solving for  .
(2)
Calculate the size of B̂ .
(3)
Determine the length of AB.
In ΔPQR, QS ⊥ PR, QS = h units, PQ = m units, QR = n units.
(1)
Express sin P in terms of h and m.
(2)
Express sin R in terms of h and n.
(3)
Hence show that m sin P = n sin R .
(4)
Now use the result in (3) to calculate
the size of P̂ if it is given that
m = 40 cm, n = 30 cm and Rˆ = 80° .
76
ANGLES OF ELEVATION AND DEPRESSION
β
θ
θ is the angle of elevation of F from N.
β is the angle of depression of N from F.
[Note that N̂ = β since alt ∠ s are equal]
EXAMPLE 12
The angle of depression of a boat on the ocean from the top of a cliff is 55° . The boat is 70
metres from the foot of the cliff.
(a)
What is the angle of elevation of the top of the cliff from the boat?
(b)
Calculate the height of the cliff.
55°
Solutions
(a)
The angle of elevation of the top of the cliff from the boat is 55° , i.e. B̂ = 55 ° .
(b)
We can calculate the height of the cliff as follows:
55°
h
= tan 55°
70 m
∴ h = (70 m) tan 55°
∴ h = 100 m
T
h
B
77
55°
70 m
C
EXERCISE 6
(Round your answers off to one decimal place in this exercise)
(a)
(b)
The Cape Town cable car takes tourists
to the top of Table Mountain. The cable
is 1,2 kilometers in length and makes
an angle of 40° with the ground.
Calculate the height (h) of the mountain.
An architectural design of the front of a
house is given below. The length of the
house is to be 10 metres. An exterior
stairway leading to the roof is to form an
angle of elevation of 30° with ground level.
The slanted part of the roof must be 7 metres
in length.
(1)
Calculate the height of the vertical
wall (DE).
(2)
Calculate the size of θ , the angle of
elevation of the top of the roof (A)
from the ceiling BCD.
(3)
(c)
(d)
40°
A
roof
B
F
θ
C
10 m
30°
Front view of a house
Calculate the length of the beam AC.
In a soccer World Cup, a player kicked the
ball from a distance of 11 metres from the
goalposts (4 metres high) in order to score
a goal for his team. The shortest distance
travelled by the ball is in a straight line.
The angle formed by the pathway of the ball
and the ground is represented by θ .
θ
(1)
Calculate the largest angle θ for which the player will possibly score a goal.
(2)
Will the player score a goal if the angle θ is 22° ? Explain.
Treasure hunters in a boat, at point A, detect a
treasure chest at the bottom of the ocean (C) at
an angle of depression of 13° from the boat to
the treasure chest. They then sail for 80 metres
so that they are directly above the treasure chest
at point B. In order to determine the amount of
oxygen they will need when diving for the treasure,
they must first calculate the depth of the treasure
(BC). Calculate the depth of the treasure for the
treasure hunters.
78
13°
D
E
ANGLES IN THE CARTESIAN PLANE
In this section, we will extend the trigonometric definitions to include angles in the interval
[0° ; 360°] .
Consider a circle with centre O(0 ; 0) and radius r with R( x ; y ) any point on the circle. θ is
the angle measured anti-clockwise from the positive side of the x-axis to the radius OR,
which is referred to as the terminal arm and θ is said to be in standard position.
Note that for every point R( x ; y ) on the circumference of the circle, x2 + y 2 = r 2
For each point R( x ; y ) on the terminal arm of θ , the following trigonometric functions of θ
are defined:
R(x ; y )
opp y
=
hyp r
adj x
=
cos θ =
hyp r
opp y
=
tan θ =
adj x
sin θ =
θ
The Cartesian plane is divided into four quadrants allowing for angles in the interval
(0° ; 360°)
Angles in the first quadrant will lie in the interval (0° ; 90°)
Angles in the second quadrant will lie in the interval (90° ; 180°)
Angles in the third quadrant will lie in the interval (180° ; 270°)
Angles in the fourth quadrant will lie in the interval (270° ; 360°)
90°
0°
180°
360°
270°
79
Consider the following table:
Quad
θ
r
x
y
1
θ∈ (0° ; 90°)
+
+
+
2
3
θ ∈ (90° ;180°)
θ∈ (180° ; 270°)
−
+
−
+
+
−
sin θ
y
r
+
=+
+
cos θ
x
r
+
=+
+
tan θ
y
x
+
=+
+
+
=+
+
−
=−
+
+
=−
−
−
=−
+
−
=−
+
−
=+
−
Diagram
x+
y+
θ
x−
y+
θ
θ
x−
y−
4
θ ∈ (270° ; 360°)
+
+
−
−
=−
+
+
=+
+
−
=−
+
θ
x+
y−
Conclusion
•
•
•
●
All trigonometric functions are positive in the first quadrant.
sin θ is positive in the second quadrant and tan θ and cos θ are negative.
tan θ is positive in the third quadrant and sin θ and cos θ are negative.
cos θ is positive in the fourth quadrant and sin θ and tanθ are negative.
sin θ +
All +
tan θ +
cos θ +
Here is a useful way of remembering this rule of signs in the quadrants:
Quadrant 1
All
Quadrant 2
singers (sin)
Quadrant 3
take (tan)
80
Quadrant 4
cough sweets (cos)
EXERCISE 7
In which quadrant does the terminal arm of the angle θ lie if:
(a)
(c)
sin θ > 0 and cos θ > 0
tan θ > 0 and cos θ < 0
(b)
(d)
sin θ < 0 and cos θ < 0
tan θ < 0 and cos θ < 0
(e)
sin θ < 0 and θ∈ [90°;270°]
(f)
cos θ < 0 and 0° < θ < 180°
EXAMPLE 13
4
and θ∈ [ 0°;90°] , calculate without the use of a calculator and with the aid of a
5
diagram the value of tan 2 θ .
If cos θ =
Solution
4 x
=
5 r
x 2 + y 2 = r 2 .......Pythagoras
(4 ; y )
cos θ =
Given :
and
5
θ
∴ (4) 2 + y 2 = (5) 2
x=4
2
∴16 + y = 25
∴ y2 − 9 = 0
∴ ( y + 3)( y − 3) = 0
∴ y = ±3
But y is positive in Quadrant 1
∴y =3
2
2
9
 y 3
∴ tan 2 θ =   =   =
 x   4  16
y
4
y=3
r =5
Quicker method to solve for y:
y2 − 9 = 0
∴ y2 = 9
∴y =± 9
∴ y = ±3
EXAMPLE 14
If 13sin θ = −5 and θ∈ [90°; 270°] calculate without the use of a calculator and with the aid
of a diagram the value of cos θ + sin θ .
Solution
13sin θ = −5
5 −5 y
∴ sin θ = − =
=
(r is always positive)
13 13 r
sin θ is negative and therefore the terminal arm
will lie in the third or fourth quadrant.
But with θ∈ [90°; 270°] , the terminal arm will lie
in the third quadrant.
81
x θ
−5
x = −12
( x ; − 5)
r = 13
y = −5
x 2 + y 2 = r 2 .......Pythagoras
Quicker method:
∴ x 2 + (−5) 2 = (13) 2
x 2 − 144 = 0
∴ x 2 + 25 = 169
∴ x 2 = 144
∴ x 2 − 144 = 0
∴ x = ± 144
∴ x = ±12
∴ ( x + 12)( x − 12) = 0
∴ x = ±12
But x is negative in Quadrant 3
∴ x = −12
cos θ + sin θ
 −12 
 −5 
=
 +  
 13 
 13 
4
−17
=
= −1
13
13
EXERCISE 8
(a)
(1)
(b)
3
and 0° ≤ θ ≤ 90° , determine by means of a diagram:
5
2 tan θ
cos 2 θ
(2)
If sin θ =
5
and sin θ > 0 , determine by means of a diagram:
12
13cos θ
cos2 θ + sin 2 θ
(2)
If tan θ =
(1)
(c)
If 5cos A + 3 = 0 and 180° < A < 360° , determine by means of a diagram:
sin A
tan 2 A
(2)
(1)
cos A
(d)
If 8 tan θ + 15 = 0 and θ ∈ [90° ; 270°] , determine by means of a diagram:
(1)
sin θ + cos θ
(2)
34sin θ − 17 cos θ
(e)
If 13cos θ − 5 = 0 and 180° ≤ θ ≤ 360° , determine by means of a diagram:
sin 2 θ + cos2 θ
25 tan 2 θ
(1)
(2)
(f)
If 4 tan B − 3 = 0 and cos B < 0 , determine by means of a diagram:
(1)
(2)
(sin B + cos B) 2
25(sin B − cos B) 2
(g)
If 2sin θ + 1 = 0 and 90° < θ < 270° calculate without the use of a calculator and with
the aid of a diagram the value of the following:
(2)
(1)
4 cos 2 θ
81tan 2 θ
(h)
(i)
12
and P(b ; 18)
5
Determine the value of b without using a calculator.
In the diagram alongside tan θ =
If tan θ =
y
P(b ; 18)
θ
a
where θ ∈ [ 0° ; 90°] , determine sin 2 θ by means of a diagram.
b
82
x
RECIPROCALS OF THE TRIGONOMETRIC RATIOS
Note to the educator:
According to CAPS, the reciprocals of the trigonometric ratios are to be taught and
examined in Grade 10 only. Should you be pressurized for time, simply introduce the
reciprocals and let the learners study this topic on their own for interest.
The reciprocal of a number is a number which when multiplied by the original number gives
an answer of 1. Here are some examples:
1
1
since 6 × = 1
6
6
2
3
2 3
The number has a reciprocal of since × = 1
3 2
3
2
The number 6 has a reciprocal of
We can define reciprocal funtions for the sine, cosine and tangent ratios.
The cosecant function (cosec θ ) is the reciprocal of sin θ
1
hyp
cosec θ =
=
sin θ opp
The secant function sec θ is the reciprocal of cos θ
1
hyp
sec θ =
=
cos θ adj
The cotangent function cot θ is the reciprocal of tan θ
1
adj
cot θ =
=
tan θ opp
EXAMPLE 15
Consider the following:
(a)
sin N =
n
k
and cosec N =
k
n
(b)
cos N =
m
k
and sec N =
k
m
(c)
tan N =
n
m
and cot N =
m
n
EXAMPLE 16
(a)
Evaluate the following without using a calculator:
(1)
(b)
cosec 45°
(2)
sec 30°. tan 60°
Evaluate the following rounded off to two decimal places:
(1)
cot 26°
(2)
sec36° + cot 63°
83
Solutions
(a)
(1)
cosec 45°
2
= =2
1
(2)
sec 30°. cot 60°
2 1
2
=
=
.
3 3 3
(b)
(1)
cot 26°
1
=
= 2, 05
tan 26°
(2)
sec36° + cosec 63°
1
1
=
+
= 2,36
cos 36° sin 63°
EXAMPLE 17
Without using a calculator, solve the equation cosec θ = 2 if θ is an acute angle.
Solution
cosec θ = 2
2
2
1
2 is the hypotenuse and 1 is opposite 45°
hyp
2
=
∴ x = 45° since cosec 45° =
opp 1
∴ cosec θ =
45°
45°
Alternatively:
2
cosecθ =
1
1
∴ sin θ =
2
2 is the hypotenuse and 1 is opposite 45° .
opp
1
=
∴ x = 45° since sin 45° =
hyp
2
EXAMPLE 18
Solve the equation sec θ = 1, 5 rounded off to two decimal places if θ is an acute angle.
Solution
sec θ = 1,5
1
1,5
2
∴ cos θ =
3
2
∴ θ = cos −1   = 48,19°
3
∴ cos θ =
84
EXERCISE 9
(a)
Refer to the diagram alongside to answer the following
questions.
State the following:
(1)
(3)
(5)
(b)
(d)
(e)
cot A
cot C
sec C
sin θ
sec θ
tan α
(2)
(4)
(6)
α
cot θ
cosec α
cos α
θ
Evaluate the following without using a calculator:
(1)
cosec 60°
(2)
sec 60°
(3)
cot 45°
(4)
cot 60°
(5)
sec 2 45°
(6)
2cosec2 45°
(7)
cot 30°. tan 30°
(8)
cot 30° + tan 30°
(9)
cosec 30°. cosec 45°
Evaluate the following rounded off to two decimal places:
(1)
cosec 43°
(2)
sec 78°
(3)
cot 64°
(4)
cosec 94° + sec 35°
(5)
3cot 2 57°
(6)
( cosec 25° )3
Without using a calculator, solve the following equations if the angles are acute.
(1)
(f)
(2)
(4)
(6)
Refer to the diagram alongside and then
answer the questions below.
State the following:
(1)
(3)
(5)
(c)
sec A
cosec A
cosec C
3 cosec x = 2
(2)
sec x = 2
(3)
cot x = 3
Solve the following equations rounded off to two decimal places if the angles are
acute.
(1)
cosec θ = 1, 2
(2)
2sec x = 9
85
(3)
cot 2 x = 2
TRIGONOMETRIC FUNCTIONS
(It is advisable to first do Chapter 6 on Functions before doing this topic)
THE GRAPHS OF THE BASIC SINE AND COSINE FUNCTIONS
In this section, we will discuss the graphs of the basic functions y = sin θ and y = cos θ and
then introduce the functions y = a sin θ + q and y = a cos θ + q .
Consider y = sin θ
The following table contains specific values of θ and the corresponding y-values.
For example, if θ = 30° , then y = sin 30° = 0,5 .
θ
0°
0
θ
210°
−0, 5
y = sin θ
y = sin θ
30°
0,5
45°
0,7
225°
−0, 7
60°
0,9
240°
−0, 9
90°
1
270°
−1
120°
0,9
300°
−0, 9
135°
0,7
150°
0,5
180°
0
315°
−0, 7
330°
−0, 5
360°
0
We can represent the values of θ on the horizontal axis and the values of y on the vertical
axis and then draw the graph of y = sin θ .
y
°
°
°
°
°
°
°
°
°
°
°
°
θ
The graph of y = sin θ has the following characteristics:
(a)
The maximum value is 1 and the minimum value is −1 .
(b)
The range is y ∈ [−1;1]
(c)
The amplitude of a graph is defined to be
1
[distance between max and min value].
2
1
For the graph of y = sin θ , the amplitude is [1 − (−1)] = 1
2
86
If we increase the values of θ , the graph of y = sin θ will repeat its basic shape over
360° intervals. Let’s draw the graph for the interval θ ∈ [0° ; 720°] .
Please note that we are dealing with angles that are greater than 360° . These types of
angles will be explained in more detail in Grade 11.
(d)
y
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
°
We say that the period of the graph of y = sin θ is 360° . The graph of y = sin θ is
therefore cyclical in nature and repeats its basic shape every 360° .
Consider y = cos θ
The following table contains specific values of θ and the corresponding y-values.
θ
y = cos θ
θ
y = cos θ
0°
1
210°
−0,9
30°
0,9
45°
0,7
225°
−0, 7
60°
0,5
240°
−0,5
90°
0
270°
0
120°
−0,5
300°
0,5
135°
−0, 7
150°
−0, 9
180°
315°
0, 7
330°
0,9
360°
1
−1
We can represent the values of θ on the horizontal axis and the values of y on the vertical
axis and then draw the graph of y = cosθ .
1,2
1
0,8
0,6
0,4
0,2
−0,2
30 °
60 °
90 °
120 °
150°
210°
180°
−0,4
−0,6
−0,8
−1
−1,2
87
240°
270°
300°
330°
360°
θ
°
θ
The graph of y = cos θ has the following characteristics:
(a)
The maximum value is 1 and the minimum value is −1 .
(b)
The range is y ∈ [−1 ; 1]
(c)
The amplitude of a graph is defined to be
(d)
If we increase the values of θ , the graph of y = cos θ will repeat its basic shape
over 360° intervals. Let’s draw the graph for the interval θ ∈ [0° ; 720°] .
1
[distance between max and min value].
2
1
For the graph of y = cos θ , the amplitude is [1 − (−1)] = 1
2
y
1,2
1
0,8
0,6
0,4
0,2
30° 60° 90 ° 120° 150° 180° 210° 240 ° 270° 300° 330° 360 ° 390 °420 ° 450° 480° 510° 540° 570° 600° 630° 660° 690° 720°
- 0,2
- 0,4
- 0,6
- 0,8
-1
- 1,2
We say that the period of the graph of y = cos θ is 360° . The graph of y = cos θ is
therefore cyclical in nature and repeats its basic shape every 360° .
SINE AND COSINE GRAPHS INVOLVING CHANGES IN AMPLITUDE
EXAMPLE 19
Sketch the graphs of y = 2 sin θ and y = 4 cos θ for θ ∈ [0° 360°] .
Solution
Select a few values for θ , calculate the corresponding y-values and then draw the graphs.
For θ = 0° y = 2 sin 0° = 0
y = 2 sin θ
For θ = 90° y = 2 sin 90° = 2
For θ = 180° y = 2 sin180° = 0
For θ = 270° y = 2 sin 270° = −2
For θ = 360° y = 2 sin 360° = 0
For θ = 0° y = 4 cos 0° = 4
y = 4 cos θ
For θ = 90° y = 4 cos 90° = 0
For θ = 180° y = 4 cos180° = −4
For θ = 270° y = 4 cos 270° = 0
For θ = 360° y = 4 cos 360° = 4
88
θ
y
3
y = 2sin θ
2
1
y = sin θ
90°
180
270 °
°
360 °
θ
-1
-2
-3
Notice that the graph of y = 2 sin θ is a vertical stretch of the basic graph y = sin θ by a
factor of 2.
The maximum value is 2 and the minimum value is −2 and the range is y ∈ [−2 ; 2] .
1
The amplitude of the graph of y = 2 sin θ is [2 − (−2)] = 2 and the period is 360° .
2
This vertical stretch of the graph of y = sin θ is called an amplitude shift.
The number 2 in the equation y = 2 sin θ tells us what the amplitude of the graph is.
y
°
°
°
°
θ
y = cos θ
y = 4cos θ
Notice that the graph of y = 4 cos θ is a vertical stretch of the basic graph y = cos θ by a
factor of 4.
The maximum value is 4 and the minimum value is −4 and the range is y ∈ [−4 ; 4] .
1
The amplitude of the graph of y = 2 sin θ is [2 − (−2)] = 2 and the period is 360° .
2
This vertical stretch of the graph of y = cos θ is called an amplitude shift.
The number 4 in the equation y = 4 cos θ tells us what the amplitude of the graph is.
89
EXAMPLE 20
Sketch the graph of y = −3sin θ for the interval θ ∈ [0° 360°] .
Solution
As with the graphs of other functions (See Chapter 6), the negative sign indicates a
reflection in the horizontal axis.
All you need to do is first draw the basic graph of y = sin θ , stretch this graph vertically by a
factor of 3 and then reflect this graph in the horizontal axis to obtain the graph of
y = −3sin θ .
y
44
y = 3sin θ
3
2
y = sin θ
1
90 °
180°
270°
360°
θ
-1
-2
-3
y = −3sin θ
-4
Note:
1
[3 − (−3)] = 3 .
2
Therefore, in the equation y = −3sin θ , the number 3 tells us that the amplitude is 3 and the
The amplitude of the graph of y = −3sin θ is:
negative sign indicates a reflection in the horizontal axis.
EXERCISE 10
(a)
Given: y = 3sin θ and y = −2 sin θ
(1)
Sketch the graphs on the same set of axes for θ ∈ [0° ; 360°] .
(2)
Write down the maximum and minimum values for each graph.
(3)
Write down the range, amplitude and period for each graph.
(b)
Given: y = 2 cos θ and y = −3cos θ
(1)
Sketch the graphs on the same set of axes for θ ∈ [0° ; 360°] .
(2)
Write down the maximum and minimum values for each graph.
(3)
Write down the range, amplitude and period for each graph.
(c)
Given:
(1)
(2)
(3)
1
y = cos θ and y = − sin θ
2
Sketch the graphs on the same set of axes for θ ∈ [0° ; 360°] .
Write down the maximum and minimum values for each graph.
Write down the range, amplitude and period for each graph.
90
SINE AND COSINE GRAPHS INVOLVING VERTICAL SHIFTS
EXAMPLE 21
Sketch the graph of y = sin θ + 1 and y = − cos θ − 1 for θ ∈ [0° 360°] .
Solution
The graph of y = sin θ + 1 is the graph of y = sin θ shifted 1 unit up. The graph is shown
below.
y
3
y = sin θ + 1
2
1
y = sin θ
90 °
180 °
270°
360°
θ
-1
-2
The maximum value is 2 and the minimum value is 0. The range is y ∈ [0 ; 2] .
1
The amplitude is [2 − 0] = 1 and the period is 360° .
2
The graph of y = − cos θ − 1 is the graph of y = cos θ reflected in the x-axis and then shifted
1 unit down. The graph is shown below.
y
2
1
y = cos θ
90 °
-1
-2
180 °
270 °
y = − cos θ
y = − cos θ − 1
-3
The maximum value is 0 and the minimum value is −2 . The range is y ∈ [−2 ; 0] .
1
The amplitude is [0 − ( −2)] = 1 and the period is 360° .
2
91
360 °
θ
EXERCISE 11
(a)
Given: y = sin θ + 2 and y = cos θ − 1
(1)
Sketch the graphs on the same set of axes for θ ∈ [0° ; 360°] .
(2)
Write down the maximum and minimum values for each graph.
(3)
Write down the range, amplitude and period for each graph.
(b)
Given: y = − cos θ + 3 and y = − sin θ − 2
(1)
Sketch the graphs on the same set of axes for θ ∈ [0° ; 360°] .
(2)
Write down the maximum and minimum values for each graph.
(3)
Write down the range, amplitude and period for each graph.
(c)
Given: y = 2 sin θ + 4 and y = −3cos θ − 1
(1)
Sketch the graphs on the same set of axes for θ ∈ [0° ; 270°] .
(2)
Write down the maximum and minimum values for each graph.
(3)
Write down the range, amplitude and period for each graph.
THE GRAPH OF THE TANGENT FUNCTION
Consider y = tan θ
The following table contains specific values of θ and the corresponding y-values.
θ
tan θ
0°
0
30°
0,6
45°
1
60°
1,7
89°
57,2
90°
error
91°
−57, 2
120°
−1, 7
135°
θ
tan θ
210°
0,6
225°
1
240°
1,7
269°
57,2
270°
error
271°
error
300°
315°
330°
−1, 7
−1
−0, 6
−1
150°
−0, 6
180°
0
360°
0
We can represent the values of θ on the horizontal axis and the values of y on the vertical
axis and then draw the graph of y = tan θ .
y
4
θ = 90°
y = tan θ
θ = 270°
3
2
1
.
(45° ;1)
45°4
90°
.
135°3
−1
.
(225° ;1)
8
180°
225°2
(135° ; − 1)
−2
270° 7
.
315° 1
6
360°
(315° ; − 1)
−3
−4
The graph of y = tan θ has the following characteristics:
(a)
The graph has no maximum value, no minimum value and hence no amplitude.
(b)
The range is y ∈ ( −∞ ; ∞ )
92
θ
(c)
The points (45° ;1) , (135° ; − 1) , (225° ;1) and (315° ; − 1) may be referred to as the
critical points on the basic graph y = tan θ . The points are useful when sketching
tan graphs involving vertical stretches, reflections in the x-axis and vertical shifts.
(d)
As the values of θ approach 90° from the left, the values of y tend towards +∞ .
At 90° , the y-value is undefined. This means that the curve moves upwards and
never cuts or touches the line θ = 90° . As the values of θ approach 90° from the
right, the values of y tend towards −∞ . The curve moves downwards and never cuts
or touches the line θ = 90° .
An asymptote is a vertical line that a graph approaches but never touches. Therefore,
the line x = 90° is an asymptote of the graph of y = tan θ .
All of this applies to 270° . The line θ = 270° is therefore also an asymptote.
(e)
If we start at −90° and then increase the values of θ , it is possible to get an idea of
how the graph of y = tan θ repeats its basic shape over 180° intervals. Let’s draw
the graph for the interval θ ∈ (−90° ; 810°) .
Please note that we are dealing with angles that are negative. These types of
angles will be explained in more detail in Grade 11.
y
θ = 90°
5
θ = 270°
θ = 450°
θ = 630°
θ = 810°
44
33
22
11
180°
−90°
4
4
90°
0
3
8
360°
2270° 7
1
11
6
720°
540°
0450° 5
9
4
8630°
7
2
6
810°
22
3
−4
5
We say that the period of the graph of y = tan θ is 180° . The graph of y = tan θ is
therefore cyclical in nature and repeats its basic shape every 180° .
The graph cuts the horizontal axis at θ = 0° + k.180° where k represents integer
values. In other words, every 180° starting from 0° in a negative and positive `
direction.
For example, the graph cuts the horizontal axis at:
θ = 0° + (1).180° = 180°
θ = 0° + (3).180° = 540°
θ = 0° + (2).180° = 360°
θ = 0° + (4).180° = 720°
There are asymptotes at θ = 90° + k.180° where k represents integer values.
For example, the graph has asymptotes at:
θ = 90° + (−1).180° = −90°
θ = 90° + (1).180° = 270°
θ = 90° + (0).180° = 90°
θ = 90° + (2).180° = 450°
93
θ
EXAMPLE 22
Sketch the graph of y = −2 tan θ for the interval θ ∈ [0° ; 270°) .
Solution
First draw the graph of y = 2 tan θ by vertically stretching the graph of y = tan θ by a factor
of 2. The y-values of the critical points of y = tan θ are multiplied by 2 to form the
following new points on the graph of y = 2 tan θ :
(45° ; 2) , (135° ; − 2) , (225° ; 2)
The graph of y = −2 tan θ is then formed by reflecting y = 2 tan θ in the x-axis.
The signs of the y-values of the critical points of y = 2 tan θ now become:
(45° ; − 2) , (135° ; 2) , (225° ; − 2)
The asymptotes of this graph remain the same as well as the x-intercepts.
Now plot the critical points, x-intercepts and asymptotes for y = −2 tan θ and restrict the
graph in the interval θ ∈ [0° ; 270°) . The graph is shown below.
θ = 90°
θ = 270°
y
5
4
3
2
1
−1
−2
−3
y = 2 tan θ
(45° ; 2)
.
45°
y = −2 tan θ
(225° ; 2)
y = tan θ
90°
.
135°
180°
.
225°
270°
(135° ; − 2)
−4
−5
The period of y = −2 tan θ is 180° and the asymptotes are θ = 90° and θ = 270° .
EXERCISE 12
(a)
Sketch the graph of y = − tan θ for the interval θ ∈ [0° ; 360°]
(b)
Sketch the graph of y = 3 tan θ for the interval θ ∈ [0° ; 360°]
(c)
Sketch the graph of y = − 12 tan θ for the interval θ ∈ [0° ; 270°]
(d)
Sketch the graph of y = tan θ + 1 for the interval θ ∈ [0° ; 360°]
(e)
Sketch the graph of y = tan θ − 2 for the interval θ ∈ [90° ; 360°]
(f)
Sketch the graph of y = −2 tan θ − 1 for the interval θ ∈ [0° ; 360°]
(g)
Given: y = 2 tan θ and y = −3sin θ − 3
(1)
Sketch the graphs on the same set of axes for θ ∈ [0° ; 270°] .
(2)
Write down the period for each graph.
94
θ
Summary of the trigonometric functions
For graphs of the form y = a sin x + q and y = a cos x + q :
The value of a (ignoring negative signs) represents the vertical stretch of the graph as well
as the amplitude.
If a is negative, then there is a reflection in the x-axis.
The value of q represents a vertical shift of the graph y = sin x or y = cos x up or down.
The period of these graphs is 360° .
For graphs of the form y = a tan x + q :
The value of a (ignoring negative signs) represents a vertical stretch of the graph y = tan x
from the x-axis. The critical points for y = a tan x are
(45° ; a ) , (135° ; − a ) , (225° ; a ) and (315° ; − a )
If a is negative, then there is a reflection in the x-axis.
The value of q represents a vertical shift of the graph y = tan x up or down.
The equations of the asymptotes are x = 90° + k .180° where k represents integer values.
The x-intercepts are at the points (0° + k .180° ; 0) where k represents integer values.
The period of these graphs is 180° .
FINDING THE EQUATION OF A GIVEN TRIGONOMETRIC GRAPH
EXAMPLE 23
In the diagram below, the graphs of y = f ( x ) = a sin x + q and y = g ( x) = m cos x + n are
shown for the domain x ∈ [0° ; 360°] .
y
.
(270° ; 3)
3
f
2
.
.
(360° ;1)
1
.
90°
−1
180°
(90° ; − 1)
−2
g
−3
270°
360°
x
(360° ; − 1)
.
(180° ; − 3)
Note: The variable x in the equations may also be used to represent the angles.
(a)
Write down the amplitude and range of f.
(b)
Write down the amplitude and range of g.
(c)
Determine the values of a and q.
(d)
Determine the values of m and n.
Solutions
(a)
For f :
Amplitude: 2
Range:
y ∈ [ −1 ; 3]
(b)
For g:
Amplitude: 1
Range:
y ∈ [ −3 ; − 1]
95
(c)
Substitute two points on the graph into the equation y = a sin x + q and solve
simultaneously.
(90° ; − 1) :
−1 = a sin 90° + q
∴ −1 = a (1) + q
∴ −1 = a + q
a + q = −1
−a + q = 3
(A)
(B)
∴ 2q = 2
∴q = 1
(A) + (B)
(270° ; 3) :
3 = a sin 270° + q
∴ 3 = a ( −1) + q
∴ 3 = −a + q
∴ a + 1 = −1
∴ a = −2
Here is a quicker method:
Use the following formulae to determine the positive value of a and the value of q in
the equations of y = a sin x + q or y = a cos x + q :
1
a = [max − min]
2
1
q = [max + min]
2
1
(this is the amplitude)
a = [3 − (−1)] = 2
2
The given sine graph involves a reflection in the x-axis.
∴ a = −2
1
q = [3 + ( −1)] = 1
2
Notice the transformations of y = sin x into y = −2 sin x + 1
y
3
2
1
−1
−2
−3
(d)
y = −2sin x + 1
y = 2sin x
y = sin x
90°
180°
270°
360°
x
y = −2sin x
Substitute two points on the graph into the equation y = m cos x + n and solve
simultaneously:
(180° ; − 3) :
−3 = m cos180° + n
(360° ; − 1) :
−1 = m cos360° + n
∴ −3 = m( −1) + n
∴ −1 = m (1) + n
∴−3 = −m + n
∴−1 = m + n
96
−m + n = −3 (A)
m + n = −1
(B)
∴ 2n = −4
(A) + (B)
∴ n = −2
∴ m + (−2) = −1
∴m = 1
Alternatively:
1
a = [max − min]
2
1
q = [max + min]
2
1
a = [−1 − ( −3)] = 1
2
1
q = [−1 + (−3)] = −2
2
Notice the transformations of y = cos x into y = cos x − 2
y
y = cos x
1
90°
360°
270°
180°
x
−1
−2
y = cos x − 2
−3
EXERCISE 13
(a)
In the diagram below, the graphs of y = f ( x) = a cos x + q and y = g ( x ) = m sin x + n
are shown for the domain x ∈ [0° ; 360°] .
.
y
2
y = a cos x + q
1
90°
−1
−2
.
(90° ; − 2)
(360° ; 2)
.
(180° ; 0)
270°
y = m sin x + n
(1)
(2)
(3)
Write down the amplitude and range of f.
Write down the amplitude and range of g.
Determine the values of a and q.
(4)
Determine the values of m and n.
97
360°
x
(b)
In the diagram below, the graphs of y = f ( x) = a sin x + q and y = g ( x) = m cos x + n
are shown for the domain x ∈ [0° ; 360°] .
.
y
(180° ; 3)
3
g
2
1
.
−2
−3
.
x
360°
(360° ; − 1)
270°
180°
90°
−1
(360° ; − 2)
f
(90° ; −3)
(c)
(1)
Write down the range, amplitude and period of f.
(2)
Write down the range, amplitude and period of g.
(3)
Determine the values of a and q
(4)
Determine the values of m and n.
In the diagram below, the graphs of two trigonometric functions are shown.
.
y
(180° ; 4)
4
3
2
1
−1
−2
−3
−4
(45° ; 2)
45°
90°
180°
135°
225°
270°
x
(135° ; −2)
(1)
Determine the equations of the two graphs.
(2)
Write down the range, amplitude and period of each graph, where possible.
The following exercise includes graph interpretation and the graphical interpretation of
inequalities dealt with in Chapter 6.
98
EXERCISE 14
(a)
The diagram below represents the graphs of f ( x ) = 2 cos x and g ( x ) = − cos x for
the interval x ∈ [0° ; 360°] .
y
A
F
C
f
x
E
B
g
D
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(b)
Sketch the graphs of y = f ( x ) = − tan x and y = g ( x ) = tan x + 2 for x ∈ [ 0°; 270°]
on the same set of axes and then answer the questions that follow.
(1)
(2)
(3)
(4)
(c)
Determine the lengths of OA and OB.
Determine the length of CD.
Determine the length of EF if OE = 315° .
Determine graphically the values of x ∈ [0° ; 360°] for which:
(ii)
(i)
f ( x) = 0
g ( x) = 0
(iv)
(iii)
f ( x) = 2
g ( x ) = −1
(vi)
(v)
f ( x) < 0
g ( x) ≤ 0
(viiii) g ( x ) > f ( x )
(vii)
f ( x) ≥ g ( x)
Calculate the value of x for which f ( x ) = 1 in the interval x ∈ [0° ; 90°] .
For which values of x is the graph of f ( x ) = 2 cos x increasing?
For which values of x is the graph of g ( x ) = − cos x decreasing?
For which values of k will the graph of y = − cos x + k not cut the x-axis?
For which values of t will the graph of y = 2 cos x + t touch the x-axis in one
point only?
Calculate f (45°) − g (45°)
For which values of x is g ( x ) = f ( x ) ?
For which values of x is g ( x ) ≥ f ( x ) ?
For which values of x g ( x ) − f ( x) = 2 ?
Sketch the graphs of y = f ( x ) = sin x and y = g ( x ) = cos x + 1 for x ∈ [ 0° ; 360°]
on the same set of axes and then answer the questions that follow.
(1)
(2)
(3)
(4)
(5)
(6)
For which values of x is
For which values of x is
For which values of x is
For which values of x is
For which values of x is
For which values of x is
f ( x) = g ( x) ?
g ( x) ≥ f ( x) ?
f ( x) ≤ 0 ?
g ( x) > 0 ?
g ( x) − f ( x) = 2 ?
f ( x ). g ( x ) ≥ 0 ?
99
CONSOLIDATION AND EXTENSION EXERCISE
(a)
If A = 38, 6° and B = 141, 4° , determine of the following to two decimal places:
1 
2 tan 2 B
(1)
(2)
(3)
sec A
cos 2A − sin  B 
2 
(4)
2 cosec A + cot 3B
(b)
Calculate θ in each case where 0° ≤ θ ≤ 90° (rounded off to two decimal places):
(c)
(d)
(1)
sin 2θ = 0, 4
(2)
2 sin θ = 0, 4
(4)
θ
sin = 0, 4
2
(5)
cosec θ =
(f)
(1)
tan 3θ = 2
(2)
3 tan θ = 2
(3)
4 cos θ − 1 = 0
(4)
tan(θ − 13°)
= 6,5
2
(5)
sec θ = 2
(6)
3cot θ = 1
If A = 25,34° and B = 134, 66° , determine the following to two decimal places:
sin 2A
2
(2)
cos 2 (A + B)
(3)
2
tan A − 2cos B
3
Calculate the following without the use of a calculator:
(1)
sin 2 45° − cos 60°
(2)
tan 30°.tan 60° − cos 30°.sin 60°
(3)
(sin 30°)tan 45°
(4)
tan 45°. sin 60°
tan 30°(1 − sin 30°)
(5)
cosec 30°
(6)
cot 60°. sin 60°
Without using a calculator, solve the following equations for θ ∈ (0°;90°) :
(1)
(4)
(7)
(g)
5
2
sin θ
= 0, 4
2
Calculate θ in each case where θ ∈ (0°;90°)
(1)
(e)
(3)
sin θ = 0,5
3 tan θ = 1
cosec θ = 2
(2)
tan θ = 1
(5)
2cos θ = 2 (6)
(8)
(3)
2 cos θ − 3 = 0
2sin θ = 3
3 cot θ = 1
Two right-angled triangles, ΔGHK and ΔPQR are given.
Calculate, rounded off to two decimal places:
(1)
the length of PQ and PR
(2)
the value of K̂
42°
100
(h)
Calculate the value of x in each case.
(1)
(2)
38°
63°
(3)
(i)
(j)
(4)
In the accompanying figure AB represents a lamp
pole with height 25 m. Two cables from the top of
the pole, are anchored at points C and D. From A,
the angles of depression of C and D in the same
horizontal line as B are 75° and 67° .
Calculate the distance (CD) between the two
anchor points.
67°
75°
In the diagram below, Δ PQR is right-angled at Q.
15
and
It is given that PR = 34 cm, tan P =
8
P̂ ∈ (0° ; 90°) .
(1)
(k)
Determine the length of QR without
solving for P̂ .
(2)
Determine the length of QR by first
solving for P̂ .
5
In ΔEAY , ES ⊥ YA , tan θ =
and ES = 7, 5 cm
12
(1)
Calculate the lengths of EA and YA
without solving θ .
(2)
Calculate the length of YS by first solving θ .
101
θ
(l)
(m)
A handy man attempts to reach the roof of a hall
with a ladder 5 metres in length. Unfortunately,
the ladder is too short and a new ladder will be
required. Suppose that the length of the ladder needed
to reach the top has to be double the distance from the
foot of the ladder to the wall. Also, the angle between
his current ladder and the ground will need to be
equal to the angle between the two ladders.
2x
5m
θ
θ
(1)
Calculate the value of θ
(2)
Hence, or otherwise, determine what the length of the ladder should be to get
the handy man to the roof.
x
On the Cartesian Plane below O is the origin. Q lies on the x-axis and P is the
point (1 ; 3)
y
P(1 ; 3)
(1)
Calculate θ
(2)
Determine the length of OQ.
θ
(n)
x
A rectangular slab is placed against a wall
as shown in the diagram. It has a length
of 3 m and a width of 1 m. It is inclined at
an angle of 27° to the ground.
Calculate the distance (h) of the slab’s
highest point above the ground.
27°
102
CHAPTER 6
FUNCTIONS
In this chapter, you will revise the graphs of linear functions (straight lines). Then you will
explore the graphs of quadratic functions (parabolas), hyperbolic functions (hyperbolas) and
exponential functions (exponential graphs).
LINEAR FUNCTIONS (SKETCHING STRAIGHT LINE GRAPHS)
Consider the graph of y = x
We can select a few input values (x-values) and hence determine the corresponding output
values (y-values). These values will be represented in a table.
x
y
−1
−1
0
0
1
1
The graph of this line is obtained by plotting the points on the Cartesian plane and drawing a
solid line through the points.
y=x
2
1
(1;1)
(0 ; 0)
−2 −1
(−1;1) −1
1
2
−2
This graph is referred to as the “mother” graph of straight lines and based on this graph, we
can generate different types of straight lines depending on the value of a and q in the general
equation of a line, which is y = ax + q .
Investigation of the effect of the value of a
In Grade 9, you learnt about the gradient and steepness of a straight line. Let’s briefly
summarise these concepts.
The gradient of a line represents the ratio of the change of the y-values with respect to the xvalues. In other words, gradient tells us the direction (or slope) of the line.
We will revise the concept of gradient later on in this chapter. However, the focus now will be
on the steepness of a line, which is the way that the line slants upwards or downwards from
left to right. We will compare the steepness of different lines to the mother graph and show
how the mother graph is transformed by changing the value of a (the coefficient of x).
103
Let’s investigate this by comparing the graphs of the following straight lines:
A:
y=x
B:
y=
1
x
2
y = 2x
C:
y = 3x
D:
Lines A, B, C and D pass through the origin since for all of the given graphs, the y-value is 0
if x = 0 .
To sketch the graphs of these lines, select one x-value and then determine the corresponding
y-value. For all four graphs, choose x = 1 .
A:
y = (1) = 1
Line A passes through the points (0 ;0) and (1;1)
B:
1
1
y = (1) =
2
2
 1
Line B passes through the points (0 ;0) and 1; 
 2
C:
y = 2(1) = 2
Line C passes through the points (0 ;0) and (1; 2)
D:
y = 3(1) = 3
Line D passes through the points (0 ;0) and (1; 3)
We will now plot the points and then draw the lines on the same set of axes.
y = 3x
3
y = 2x
(1; 3)
y = 1x
2
(1; 2)
1
(1;1)
(0 ; 0)
−3
−2
−1
y=
1
x
2
(1 ; 12 )
1
2
3
−1
−2
−3
Notice that line A (the mother graph) is closer to the y-axis than line B. We say that line A is
steeper than line B. The coefficient of x in the equation of line A is greater than the coefficient
of x in line B ( 1 > 12 ). Line C is steeper line A ( 2 > 1 ) and line D is steeper than line C ( 3 > 2 ).
Line D is the steepest of all the lines.
104
Let’s now consider what happens if the value of the coefficient of x is negative.
(a)
Consider the graph of y = − x
We can select a few input values (x-values) and hence determine the corresponding
output values (y-values). These values will be represented in a table.
−1
1
x
y
1
−1
0
0
The graph of this line is obtained by plotting the points on the Cartesian plane and
drawing a solid line through the points.
y = −x
2
(−1;1)
−2
1
(0 ; 0)
1
−1
−1
2
(1; − 1)
−2
If you now compare the mother graph y = 1x to the graph of y = −1x , it is interesting
to note that the graph of y = − x is the reflection of the mother graph in the x-axis. The
negative sign therefore causes a reflection in the x-axis.
(b)
1
Consider the graph of y = − x
2
1
affects the steepness of the line. The mother
2
graph y = x transforms into a line that is not as steep as the mother graph line.
We already know that the number
As already seen, the negative sign causes a reflection in the x-axis.
To draw this graph involves transforming the mother graph into the less steep line
y=
1
x and then reflecting this newly-formed graph about the x-axis.
2
To sketch the graph of this line, select one x-value, determine the corresponding yvalue, plot the point formed and the y-intercept (0 ;0)
Choose x = 1
1
1
y = − (1) = −
2
2
105
The line passes through the point (1; − 12 )
Now plot this point and the y-intercept and draw the graph of the line.
y = 1x
2
1
y=
1
x
2
(1;1)
(1; 12 )
−3
−2
−1
1
−1
2
3
(1; − 12 )
1
y=− x
2
−2
1
x and then this new line
2
1
is reflected in the x-axis to form the graph of the line y = − x .
2
Notice that the mother graph y = x is transformed into y =
Conclusion:
The value of a in the equation y = ax + q (ignoring negative signs), determines the
steepness of the line (closeness to the y-axis). The larger the value of a, the steeper the line.
A negative sign will cause a reflection in the x-axis.
Investigation of the effect of the value of q
(a)
Consider the graphs of the following:
y = x + 3 and y = x − 2
For y = x + 3 :
If x = 0 then y = 0 + 3 = 3
If x = 1 then y = 1 + 3 = 4
The line passes through the points (0 ; 3) and (1; 4)
For y = x − 2 :
If x = 0 then y = 0 − 2 = −2
If x = 1 then y = 1 − 2 = −1
The line passes through the points (0 ; − 2) and (1 ; − 1)
106
y = x+3
(1; 4)
4
(0 ; 3) 3
y=x
2
y = x−2
1
(1;1)
(0 ; 0)
−3 −2 −1
(−1;1) −1
1
2
(1; − 1)
−2
(0 ; − 2)
It should be clear to you from the graphs that y = x + 3 is the mother graph y = x
shifted 3 units up and y = x − 2 is the mother graph y = x shifted 2 units down.
Also, the y-intercept of y = x + 3 is 3 and the y-intercept of y = x − 2 is −2 .
(b)
y = − x + 1 and y = − x − 2
Consider the graphs of the following:
We will first draw the graph of y = − x which is the reflection of the mother graph
y = x in the x-axis and then based on this graph, we will form y = − x + 1 by
shifting y = − x one unit up and y = − x − 2 by shifting y = − x two units down.
y = −x +1
y = −x
y=x
3
2
y = −x − 2
−3 −2
1
−1
1
2
−1
−2
−3
It should be clear to you from the graphs that y = − x + 1 is the line y = − x
shifted 1 unit up and y = − x − 2 is the line y = − x shifted 2 units down.
Also, the y-intercept of y = − x + 1 is 1 and the y-intercept of y = − x − 2 is −2 .
107
Conclusion:
The value of q in the equation y = ax + q determines the shift of the graph of y = ax up or
down. It also represents the y-intercept of the graph of y = ax + q .
EXAMPLE 1
Draw a neat sketch graph of y = −2 x + 4
Solution
First draw the graph of y = 2 x , reflect this graph in the x-axis to form y = −2 x and then shift
y = −2 x four units up to form y = −2 x + 4 .
The line y = 2 x cuts the y-axis at 0.
y = −2 x + 4
y = 2x
4 (0 ; 4)
y = −2 x
Now choose x = 1
∴ y = 2(1) = 2
3
Plot the point (1; 2) and draw the line y = 2 x
2
(1; 2)
1
Now reflect y = 2 x in the x-axis to form
(0 ; 0)
y = −2 x . The point (1; 2) transforms into
−2
the point (1; − 2) .
−1
1
2
−1
−2
Then shift y = −2 x four units up.
(1; − 2)
Draw the graph of y = −2 x + 4 .
Alternatively, you can make use of the dual-intercept method that you studied in Grade 9.
This method involves determining the intercepts with the axes algebraically. Let’s use this
method for the line y = −2 x + 4
y-intercept:
Let x = 0
x-intercept:
y = −2(0) + 4 = 4
Let y = 0
0 = −2 x + 4
∴ 2x = 4
∴x = 2
The coordinates of the y-intercept are (0 ; 4) and the coordinates of the x-intercept are (2 ; 0).
You would now plot these points and draw the straight line.
108
Revision of horizontal and vertical lines
In Grade 9 you learnt that horizontal lines have the general equation y = n where n is any real
number. Vertical lines have the general equation x = n where n is any real number.
For example, the graphs of the lines y = 2 and x = 1 are shown below.
(−2 ; 2)
(1; 2)
2
y=2
1
−1
−2
2
1
(1; 2)
1
2
−2
−1
1
−1
−1
−2
−2
2
(1; − 1)
x =1
For the line y = 2 , the y-values will be constant but the x-values will vary.
For the line x = 1 , the x-values will be constant but the y-values will vary.
EXERCISE 1
(a)
(b)
(c)
Given:
y=x
y=
3
x
2
y=
1
x
4
(1)
Which of the three lines is the steepest? Explain
(2)
Sketch the three graphs on the same set of axes.
Given:
y = −x
3
y=− x
2
1
y=− x
4
(1)
Which of the three lines is the steepest? Explain
(2)
Sketch the three graphs on the same set of axes.
Given:
y = −x + 4
y = 3x − 6
(1)
Describe the transformation of y = x into the graph of y = − x + 4
(2)
Describe the transformation of y = x into the graph of y = 3x − 6
(3)
Sketch the graphs of these two functions on the same set of axes using
transformations or the dual-intercept method.
109
(d)
Match the equations on the left to the graphs on the right.
(1)
y = −2 x
(2)
y=
(3)
y = −x − 2
(4)
y = 4x
(5)
y=
(6)
y = −4 x
(7)
x=4
(8)
y=4
1
x+2
4
1
x
4
QUADRATIC FUNCTIONS (SKETCHING PARABOLAS)
Consider the graph of y = x 2
We can select a few input values (x-values) and hence determine the corresponding output
values (y-values). These values will be represented in a table.
x
y
−2
4
−1
1
0
0
1
1
2
4
The graph of y = x 2 is obtained by plotting the points on the Cartesian plane and drawing a
curve through the points.
Notice that:
●
all output values are positive
●
the graph is not linear but rather
a curve referred to as the graph
of a parabola
y = x2
(−2 ; 4)
(2 ; 4)
4
3
2
(−1;1)
1
−2
0
−1
−1
110
(1;1)
1
2
This graph is referred to as the “mother” graph of parabolas and based on these graphs, we
can generate different types of parabolas depending on the value of a and q in the general
equation of a parabola, which is y = ax 2 + q .
Investigation of the effect of the value of a
Let’s investigate the effect of value of a on the shape of different parabolas by comparing the
graphs of the following parabolas:
A:
y = x2
y=
B:
1 2
x
2
y = 2x2
C:
D:
y = 3x2
Parabolas A, B, C and D pass through the origin. To sketch the graphs of these parabolas,
select one x-value and then determine the corresponding y-value. For all four graphs, choose
x =1.
A:
y = (1) 2 = 1
Parabola A passes through the points (0 ;0) and (1;1)
B:
1
1
y = (1)2 =
2
2
 1
Parabola B passes through the points (0 ;0) and 1; 
 2
C:
y = 2(1) 2 = 2
Parabola C passes through the points (0 ;0) and (1; 2)
D:
y = 3(1) 2 = 3
Parabola D passes through the points (0 ;0) and (1; 3)
The points are plotted and the parabolas have been drawn on the same set of axes.
3
(1; 3)
2
(1; 2)
1
(1;1)
(1 ; 12 )
−2
−1 (0 ; 0)
1
2
Notice that the arms of the mother graph parabola (A) are closer to the y-axis than those of B.
The arms of parabola C are closer to the y-axis than those of A. The arms of parabola D are
closer to the y-axis than those of C. The value of the coefficient of x affects the shape of the
parabola (or what is called its vertical stretch). The greater the value of this number, the
closer the arms of the parabola will be to the y-axis. Also note that the coefficient of x2 for
each parabola is positive and the graphs are concave up (happy!).
111
Let’s now consider what happens if the value of the coefficient of x is negative.
(a)
Consider the graph of y = − x 2
We can select a few input values (x-values) and hence determine the corresponding
output values (y-values). These values will be represented in a table.
−2
x
y
−4
−1
−1
0
0
2
−4
1
−1
The graph of y = − x 2 is obtained by plotting the points on the Cartesian plane and
drawing a curve through the points.
y = x2
4
3
2
1
−2 −1 0
(−1; −1) −1
2
1
(1; − 1)
−2
−3
(−2 ; − 4)
(2 ; − 4)
−4
y = − x2
If you now compare the mother graph y = 1x 2 to the graph of y = −1x 2 , it is interesting
to note that the graph of y = − x 2 is the reflection of the mother graph in the x-axis.
The negative sign therefore causes a reflection in the x-axis.
(b)
1
Consider the graph of y = − x 2
2
1
affects the shape of the curve. The mother graph
2
y = x 2 transforms into a parabola with arms that arms that are less close to the y-axis.
We already know that the number
To sketch the graph of this parabola, select one x-value, determine the corresponding
y-value, plot the point forms and the y-intercept (0 ; 0) .
Choose x = 1
1
1
∴ y = − (1) 2 = −
2
2
112
1

The parabola passes through the point 1; − 
2

Now plot this point and the y-intercept and draw the graph of the parabola.
y = x2
y=
2
1
1 2
x
2
(1;1)
(1 ; 12 )
−2
−1
1
−1
2
(1; − 12 )
−2
1
y = − x2
2
−3
1 2
x and then this graph was
2
1
reflected in the x-axis to form the graph of the parabola y = − x 2
2
Notice that the mother graph y = x 2 transformed into y =
Conclusion:
The value of a in the equation y = ax 2 + q (ignoring negative signs), determines the
closeness of the arms of the parabola to the y-axis. The larger the value of a , the closer
the
arms are to the y-axis. A negative sign will cause a reflection in the x-axis.
If a > 0 , then the parabola is concave up (happy face!)
If a < 0 , then the parabola is concave down (sad face!)
The value of a is sometimes referred to as the vertical stretch factor
113
Investigation of the effect of the value of q
Consider the following graphs:
A:
y = x2 −1
y = x2 + 2
B:
We will first draw the mother graph y = x 2 and then based on this graph, we will draw
parabola A and B.
The graph of A is the graph of y = x 2 shifted 1 unit down.
The graph of B is the graph of y = x 2 shifted 2 units up.
y = x2 + 2
5
4
3
y = x2 − 1
(1; 3)
2
1
(−1; 0)
−2 −1
(1;1)
(1; 0)
1
2
0
−1 (0 ; − 1)
Conclusion:
The value of q in the equation y = ax 2 + q determines the shift of the graph of y = ax 2 up
or down. It also represents the y-intercept of the graph of y = ax 2 + q .
EXAMPLE 2
y = −2 x 2 + 8 and y = −2 x 2 − 2
Given:
(a)
Sketch the graphs of y = −2 x 2 + 8 and y = −2 x 2 − 2 on the same set of axes.
(b)
For these graphs, determine algebraically the coordinates of the intercepts with the
axes.
Solutions
(a)
Reflect the graph of y = 2 x 2 in the x-axis to form y = −2 x 2 and then shift y = −2 x 2
eight units up to form y = −2 x 2 + 8 . Then shift the graph of y = −2 x 2 two units
down to form y = −2 x 2 − 2 .
114
For y = −2 x 2 , choose x = 1
∴ y = −2(1) 2 = −2
y = −2 x 2 passes through the origin (0 ; 0) and the point (1; − 2)
8
y = −2 x 2 + 8
6
4
2
−4 −2
−2
2
4
(1; − 2)
−4
−6
y = −2 x 2
−8
(b)
y = −2 x 2 − 2
Consider the graph y = −2 x 2 + 8 :
y-intercept:
Let x = 0
Let y = 0
x-intercept:
∴ y = −2(0) 2 + 8 = 8
∴0 = −2x2 + 8
(0 ; 8)
∴ 2 x2 − 8 = 0
∴ x2 − 4 = 0
∴ ( x + 2)( x − 2) = 0
∴ x = −2 or x = 2
(−2 ; 0)
(2 ; 0)
Consider the graph y = −2 x 2 − 2 :
y-intercept: Let x = 0
∴ y = −2(0) 2 − 2 = −2
(0 ; − 2)
x-intercept:
The graph doesn’t cut the x-axis
There are no x-intercepts
Also notice that the equation 0 = −2 x 2 − 2
has no real solutions:
0 = −2 x 2 − 2
∴ 2 x 2 = −2
∴ x 2 = −1
∴ x = ± −1 which is non-real
115
EXERCISE 2
(a)
(b)
(c)
(d)
(e)
(f)
Given:
y = x2
y = 3x2
y = 4x2
(1)
Which parabola has arms that are closest to the y-axis?
(2)
Sketch the graphs of these parabolas on the same set of axes.
(3)
Are the parabolas concave up or down? Explain
Given:
y = − x2
1
y = − x2
2
1
y = − x2
4
(1)
Which parabola has arms that are closest to the y-axis?
(2)
Sketch the graphs of these parabolas on the same set of axes.
(3)
Are the parabolas concave up or down? Explain
Given:
y=
1 2
x
4
y=
3 2
x
2
y = −4 x 2
(1)
Which parabola has arms that are closest to the y-axis?
(2)
Sketch the graphs of these parabolas on the same set of axes.
(3)
Are the parabolas concave up or down? Explain
Given:
y = x 2 − 4 and y = x 2 + 3
(1)
Sketch the graphs of y = x 2 − 4 and y = x 2 + 1 on the same set of axes.
(2)
For these graphs, determine algebraically the coordinates of the intercepts with
the axes.
Given:
y = −4 x 2 + 4 and y = − x 2 − 2
(1)
Sketch the graphs of y = −4 x 2 + 4 and y = − x 2 − 2 on the same set of axes.
(2)
For these graphs, determine algebraically the coordinates of the intercepts with
the axes.
Given:
(1)
(2)
1
y = − x2 + 8
2
1
Sketch the graph of y = − x 2 + 8
2
Determine algebraically the coordinates of the intercepts of this parabola with
the axes.
116
(g)
Match the equations on the left to the graphs on the right.
1 2
x
4
(1)
y=
(2)
1
y = − x2
3
(3)
1
y = − x2
5
(4)
y = −5 x 2 −
(5)
y = x2 −1
(6)
y = 2x2
1
2
HYPERBOLIC FUNCTIONS (SKETCHING HYPERBOLAS)
Consider the graph of y =
1
x
We can select a few input values (x-values) and hence determine the corresponding output
values (y-values). These values will be represented in a table.
x
−2
−1
− 12
0
1
2
1
2
y
− 12
−1
−2
1
0
2
1
1
2
Notice from the table that if x = 0 then y =
1
which is undefined.
0
Since the y-value is undefined at x = 0 , this means that the graph has no y-intercept.
For the graph to have an x-intercept we let y = 0 and solve for x.
1
∴0 =
x
There is no value of x which satisfies this equation. Hence the graph has no x-intercept.
1
The graph of y = is obtained by plotting the points on the Cartesian plane and drawing a
x
curve through the points. From the above discussion, it should be clear that the graph has no
intercepts with the axes.
The graph gets closer and closer to the axes but never actually cuts them.
117
An asymptote is a horizontal or vertical line that a graph approaches but never touches.
The vertical line x = 0 (lying on the y-axis) is called the vertical asymptote of the graph.
The horizontal line y = 0 (lying on the x-axis) is called the horizontal asymptote of the
graph.
3
x=0
y=
1
x
( 12 ; 2)
2
(1;1)
1
−3
−2
−1
1
(−2 ; − 2 )
(−1; − 1)
0
1
(2 ; 12 )
2
3
y=0
−1
(− 12 ; − 2) −2
−3
This graph is referred to as the “mother” graph of hyperbolas and based on this graph, we can
generate different types of hyperbolas depending on the value of a and q in the general
a
equation of a hyperbola, which is y = + q .
x
Investigation of the effect of the value of a
Let’s investigate the effect of value of a on the shape of different hyperbolas by comparing
the graphs of the following hyperbolas:
A:
y=
1
x
B:
y=
2
x
C:
y=
4
x
The graph of A is the mother graph. In order to sketch the graph of B and C, we can select a
few input values (x-values) and hence determine the corresponding output values (y-values)
for each graph.
118
y=
For B:
x
y
−2
−1
y=
For C:
x
y
2
x
−4
−1
−1
−2
2
−1
−4
4
2
1
1
4
x
4
1
1
Let’s now plot the points and draw the graphs of A, B and C.
y=
4
y=
4
x
(1; 4)
2
x
3
−4
(−4 ; − 1)
−3
−2
−1
2
(1; 2)
1
1
y=
x
0
(4 ;1)
(2 ; 1)
1
2
3
4
−1
(−2 ; − 1)
(−1; − 2)
−2
−3
(−1; − 4)
−4
Notice that as the number in the numerator (a) gets larger, the branches of the hyperbolas
are stretched vertically away from the x-axis. The branches of the graph of y =
stretched further away from the x-axis when compared to the graph of y =
119
2
.
x
4
are
x
Let’s now consider what happens if the value of the number in the numerator is negative.
−1
Consider the graph of y =
x
As with lines and parabolas, the negative sign indicates a reflection in the x-axis.
x=0
3
y=
(−2 ;
−3
−1
x (− 12 ; 2)
1)
2
(−1;1)
−2
−1
2
1
0
−1
−2
2
1
(2 ; −
(1; − 1)
3
1)
2
y=0
( 12 ; − 2)
−3
As with the mother graph y =
1
−1
gets closer and closer to the axes but
, the graph of y =
x
x
never actually cuts them.
The vertical line x = 0 (lying on the y-axis) is the vertical asymptote of the graph.
The horizontal line (lying on the x-axis) is the horizontal asymptote of the graph.
Conclusion:
a
+ q (ignoring negative signs), determines the vertical
x
stretch of the branches of the hyperbola from the x-axis. The larger the value of a , the
The value of a in the equation y =
further the stretch away from the axes. A negative sign will cause a reflection in the x-axis.
120
Investigation of the effect of the value of q
2
+1
x
1
2
The branches of the mother graph y = stretch vertically to form the graph of y = and
x
x
2
then the newly-formed graph is shifted 1 unit up to form the graph of y = + 1 .
x
2
Let’s draw the graph of y = and then shift it up 1 unit and see what effect this shifting has.
x
2
2
y=
= −1
(−2 ; − 1) lies on y =
For x = −2
−2
x
Consider the following graph: y =
For x = −1
y=
2
= −2
−1
(−1; − 2) lies on y =
For x = 1
y=
2
=2
1
(1; 2) lies on y =
2
x
For x = 2
y=
2
=1
2
(2 ;1) lies on y =
2
x
(1; 3)
3
2
2
x
(2 ; 2)
y=
(1 ; 2)
2
+1
x
y =1
1
(2 ;1)
y=
( −2 ; 0)
−4
−2
−3
−1
0
1
2
3
4
( −1 ; − 1)
−1
( −2 ; − 1)
( −1; − 2)
−2
−3
Notice that the graph of y =
2
+ 1 cuts the x-axis at (−2 ; 0) .
x
The horizontal asymptote shifts 1 unit up and has the equation y = 1 .
The constant in the equation y =
2
+ 1 therefore represents the horizontal asymptote.
x
The vertical asymptote is still the line x = 0 (lying on the x-axis).
121
2
x
Conclusion:
a
a
+ q determines the shift of the graph of y = up or
x
x
a
down. It also represents the horizontal asymptote of the graph of y = + q .
x
The value of q in the equation y =
EXAMPLE 3
3
y = − −1
x
Given:
[Note: −
3
−3
]
is the same as
x
x
(a)
Determine algebraically the coordinates of the x-intercept for this graph.
(b)
Describe the different transformations of y =
(c)
3
3
to y = − − 1 .
x
x
3
Sketch the graph of y = − − 1 on a set of axes, clearly showing the asymptotes and
x
x-intercept.
Solutions
(a)
(b)
Let y = 0
−3
0=
−1
x
∴ 0 = −3 − x
∴ x = −3
( −3 ; 0)
3
3
y = − − 1 is the graph of y = reflected in the x-axis and then shifted 1 unit down.
x
x
y=
3
x
(Start with y =
y=
−3
x
(Reflect y =
y=
−3
−1
x
(Shift y =
122
3
)
x
3
in the x-axis)
x
−3
one unit down)
x
(c)
First draw the horizontal asymptote y = −1 on a set of axes.
Then plot the x-intercept (−3 ; 0)
After this, select one negative x-value and one positive x-value. Find the
3
corresponding y-values by substituting these x-values into y = − − 1
x
−3
(−1; 2) lies on one branch
y=
−1 = 3 −1 = 2
For x = −1
−1
−3
(1; − 4) lies on the other branch
y=
− 1 = −3 − 1 = −4
For x = 1
1
We already know the shape from (b) and we can now draw the graph as follows:
3
y = − −1
x
4
3
2
( −1; 2)
1
( −3 ; 0)
−4
−3
−2
−1
0
1
2
3
4
y = −1
−1
−2
−3
−4
123
(1; − 4)
EXERCISE 3
(a)
(b)
(c)
(d)
(e)
Given:
y=
1
x
and
y=
5
x
(1)
Which graph has branches that have the furthest stretch away from the x-axis?
Explain.
(2)
Sketch the graphs on the same set of axes.
(3)
Now sketch the graph of y = −
Given:
y=
5
on the same set of axes.
x
2
+2
x
(1)
Write down the equations of the vertical and horizontal asymptotes.
(2)
Determine the coordinates of the x-intercept.
(3)
Sketch the graph on a set of axes.
Given:
4
y = − −1
x
(1)
Write down the equations of the vertical and horizontal asymptotes.
(2)
Determine the coordinates of the x-intercept.
(3)
Sketch the graph on a set of axes.
3
+ 2 on a set of axes. Indicate the coordinates of the
x
x-intercept as well as the asymptotes.
Sketch the graph of y =
Match the equations on the left to the graphs on the right.
(1)
(2)
(3)
(4)
2
x
4
y=
x
y=
3
x
3
y = − +1
x
y=−
124
EXPONENTIAL FUNCTIONS (SKETCHING EXPONENTIAL GRAPHS)
1
Consider y = 2 x and y =  
2
x
A set of x-values {−1; 0 ;1} has been selected and the corresponding y-values have been
calculated in each case. The graphs are shown on the right.
y = 2x
y = 2x
x
−1
1
2
y
0
1
1
2
2
1
(−1;
1)
2
The x-intercept is calculated by letting y = 0 :
0 = 2x
1
2
(1; 2)
(0 ;1)
−1
But there are no real values of x for which
2x = 0 and therefore there is no x-intercept.
You might like to check this by substituting a
few negative and positive values of x as well as 0 into the equation.
1
−1
The graph will therefore not cut the x-axis. In fact, the graph has a horizontal asymptote
lying on the x-axis with equation y = 0 . The y-axis is not a vertical asymptote.
1
y= 
2
x
x
x
−1
0
y
2
1
1
y = 
2
(−1; 2)
1
1
2
2
1 (0 ;1)
1
2
The x-intercept is calculated by letting y = 0 :
1
0= 
2
x
−1
(1 ; 12 )
1
−1
But there are no real values of x for which
x
1
  = 0 and therefore there is no x-intercept.
2
You might like to check this by substituting a few negative and positive values of x as well as
0 into the equation.
The graph will therefore not cut the x-axis. In fact, the graph has a horizontal asymptote
lying on the x-axis with equation y = 0 . The y-axis is not a vertical asymptote.
125
x
1
The graphs of y = 2 and y =   are called the “mother graphs” of the exponential
2
functions and based on these graphs, we can generate different types of exponential graphs,
x
depending on the value of a, b and q in the general equation of an exponential function, which
is y = a. b x + q where 0 < b < 1 or b > 1 . These restrictions on b will be explained later in the
chapter.
Investigation of the effect of the value of b
Let’s investigate the effect of b on the shape of different exponential graphs by comparing the
following graphs.
(a)
We will first consider graphs where b > 1 .
A:
y = 2x
y = 3x
B:
C:
y = 4x
For all of these graphs, the y-intercept is (0 ;1) since b0 = 1
Select one other x-value and determine the corresponding y-value.
For A:
Choose x = 1
y = 21 = 2
(1; 2) lies on the graph of A
For B:
Choose x = 1
y = 31 = 3
(1; 3) lies on the graph of B
For C:
Choose x = 1
y = 41 = 4
(1; 4) lies on the graph of C
The graphs are sketched below:
x
4
y=
x
3
y=
x
2
y=
4 (1; 4)
3
(1 3)
2
(1 ; 2)
1
−2
−1
0
1
2
Notice that if b > 1 , all of the exponential graphs move upwards as the x-values
increase. Also as the value of b increases in value, the steeper the graph becomes.
C is steeper than B since 4 > 3 and B is steeper than A since 3 > 2 .
126
(b)
Now let’s discuss exponential graphs where 0 < b < 1 .
Consider the following graphs:
D:
1
y = 
2
x
1
y= 
3
E:
x
F:
1
y = 
4
x
For all of these graphs, the y-intercept is (0 ;1) since b0 = 1
Select one other x-value and determine the corresponding y-value. We will choose
negative x-values to avoid fractions.
Choose x = −1
1
y= 
2
For E: Choose x = −1
1
y = 
3
Choose x = −1
1
y= 
4
For D:
For F:
−1
=2
(−1; 2) lies on the graph of D
=3
(−1; 3) lies on the graph of E
=4
(−1; 4) lies on the graph of F
−1
−1
The graphs are sketched below:
y=
y=
(1
2 )
x
(1
3 )
x
y=
(1
4 )
x
( −1; 4)
4
( −1 3)
3
( −1; 2)
2
1
−2
−1
1
0
2
Notice that if 0 < b < 1 , all of the exponential graphs move downwards as the x-values
increase. Notice that as the value of the base b decreases in value, the steeper the
graph becomes. F is steeper than E since
1
4
< 13 and E is steeper than D since
The value of b therefore affects the shape of the exponential graph.
127
1
3
< 12 .
Investigation of the effect of the value of a
Consider the following exponential graphs:
G:
y=2
x
H:
y = 2.2
x
I:
y = 3.2
x
J:
1
y = 2.  
2
We will calculate the y-intercept and one other point for each graph.
For G:
For H:
For I:
y-intercept
y = 20 = 1
Let x = 0
1
(0 ;1)
Select x = 1
y=2 =2
(1; 2)
y-intercept
Let x = 0
y = 2.20 = 2
1
Select x = 1
y = 2.2 = 4
(1; 4)
y-intercept
Let x = 0
y = 3.20 = 3
Select x = 1
y = 3.21 = 6
(1; 6)
y-intercept
Let x = 0
1
y = 2.   = 2
2
Select x = 1
1
y = 2.   = 1
2
(0 ; 2)
(0 ; 3)
0
For J:
(0 ; 2)
1
(1;1)
The graphs have been drawn below on the same set of axes.
6 (1 ; 6)
5
4
(1 ; 4)
3
−2
−1
2
(1 ; 2)
1
(1;1)
0
1
2
Notice that the value of a causes a vertical stretch of the mother graphs.
Also note that except for graphs H and J, the graphs have different y-intercepts.
128
x
As with lines, parabolas and hyperbolas, the negative sign indicates a reflection in the x-axis.
x
1
The graphs of y = −2 and y = −   are reflections of the two mother graphs in the x-axis.
2
x
1
y = 
2
x
y = 2x
Note:
In the expression −2x , the value of b
is 2 and not −2 .
Remember that −2 x = −(2) x and
(0 ;1)
−1
(−1; − 12 )
(−1; − 2)
1
y = − 
2
−2 x ≠ (−2) x for all values of x
(not true for even values of x).
For example, if x = 2 , then:
1
− 12
−22 = −4 and (−2)2 = 4
(0 ; − 1)
∴−22 ≠ (−2)2
(1; − 2)
−2
x
y = −2 x
Investigation of the effect of the value of q
As with other graphs discussed thus far, the graph of y = a. b x + q is the graph of y = a. b x
shifted up or down by q units.
For example, y = 2 x + 1 is the graph of y = 2 x shifted 1 unit up. The horizontal asymptote is
indicated by the constant in the equation y = 2 x + 1 . The equation of the asymptote is y = 1 .
y = 2x + 1
4
y = 2x
3
(1 ; 3)
2
(1 ; 2)
y =1
1
−2
−1
0
129
1
2
Conclusion:
The value of b in the equation y = a. b x + q determines the shape and steepness.
If b > 1 , then the graph moves upwards from left to right as the
x-values increase.
As the value of b increases, the graphs get steeper.
If 0 < b < 1 , then the graph moves downwards from left to right
as the x-values increase.
As the value of b decreases, the graphs get steeper.
The value of a determines the vertical stretch of the mother graph.
The value of q represents the shift of the graph of y = a. b x up or down. It also indicates
where the horizontal asymptote cuts the y-axis.
A negative sign will cause a reflection in the x-axis.
EXAMPLE 4
x
1
y = −3   + 1
3
Given:
(a)
Determine algebraically the coordinates of the intercepts with the axes for this graph.
(b)
Write down the equation of the horizontal asymptote.
(c)
1
1
Describe the different transformations of y =   to y = −3   + 1 .
3
3
x
x
x
(d)
1
Sketch the graph of y = −3   + 1 on a set of axes, clearly showing the horizontal
3
asymptote and intercepts with the axes.
Solutions
(a)
y-intercept:
Let x = 0
x-intercept:
1
y = −3   + 1 = −3 + 1 = −2
3
Let y = 0
0
(0 ; − 2)
x
1
0 = −3   + 1
3
x
1
∴3  = 1
3
x
1
1
∴  =
3
3
∴x =1
(b)
(1; 0)
Horizontal asymptote is y = 1
130
x
(c)
1
y =   undergoes a vertical stretch to form the
3
3
x
1
graph of y = 3   with a y-intercept of 3.
3
x
1
y = 3   is then reflected in the x-axis to form
3
x
1
y = −3   with a y-intercept of −3 .
3
−3
x
1
y = −3   is then shifted 1 unit up to form the graph
3
y =1
x
1
of y = −3   + 1 with a y-intercept of −2 , x-intercept
3
of 1 and a horizontal asymptote y = 1 .
(d)
1
−2
First draw the horizontal asymptote.
y =1
Then determine the intercepts with the
1
axes (if they exist).
(1; 0)
If there is no x-intercept, select one
−1
negative x-value and one positive
1
x-value and determine the corresponding
y-values. Plot these points and draw
−2 (0 ; − 2)
the graph.
Check the shape by considering the
transformations of the graph from the
x
1
y = −3   + 1
3
mother graph.
EXERCISE 4
(a)
(b)
Given:
y = 2x
y = 4x
y = 6x
(1)
Which graph is the steepest? Explain.
(2)
Write down the equation of the horizontal asymptote for the three graphs.
(3)
Sketch the graphs on the same set of axes.
1
y = 
2
x
1
y = 
4
x
1
y = 
6
x
(a)
Given:
(1)
Which graph is the steepest? Explain.
(2)
Write down the equation of the horizontal asymptote for the three graphs.
(3)
Sketch the graphs on the same set of axes.
131
(c)
(d)
Given:
y = 2.2
x
y = 4.2
1
y = 3 
2
x
x
(1)
Determine the coordinates of the y-intercept for each graph.
(2)
Write down the equation of the horizontal asymptote for all three graphs.
(3)
Sketch the graphs on the same set of axes indicating the y-intercept and one
other point.
Given:
1
y = − 
2
x
1
y = −3  
2
x
(1)
Determine the coordinates of the y-intercept for each graph.
(2)
Write down the equation of the horizontal asymptote for both graphs.
(3)
Sketch the graphs on the same set of axes indicating the y-intercept and
one other point.
x
(4)
(e)
Given:
1
1
Explain the transformations of y =   into the graph of y = −3  
2
2
y = 2x − 2
x
y = 2x + 1
(1)
Determine the intercepts of y = 2 x − 2 with the axes.
(2)
Write down the equation of the horizontal asymptote of y = 2 x − 2 .
(3)
Sketch the graph of y = 2 x − 2 on a set of axes.
(4)
Explain the transformation of y = 2 x into the graph of y = 2 x − 2
(5)
Sketch the graph of y = 2 x + 1 on the same set of axes.
(6)
Explain algebraically why the graph of y = 2 x + 1 does not cut the x-axis.
x
(f)
Given:
1
y =  −4
4
(1)
Determine the intercepts of this graph with the axes.
(2)
Write down the equation of the horizontal asymptote
(3)
Sketch the graph on a set of axes.
(4)
1
1
Explain the transformation of y =   into the graph of y =   − 4
4
4
x
132
x
(g)
On different axes, draw neat sketch graphs of the following exponential graphs.
Indicate the coordinates of intercepts with the axes as well as the horizontal
asymptotes.
(1)
y = 2.4 x − 2
(3)
1
y = 2  − 2
4
(2)
y = 2.4 x + 2
(4)
1
y = 2  + 2
4
x
(h)
x
Match the equations on the left to the graphs on the right.
(1)
y = 4x
(2)
y = 2x
(3)
1
y = 
2
(4)
1
y = − 
2
(5)
1
y= 
3
(6)
1
y =  −4
2
(7)
1
y = 4 
2
x
x
x
x
x
For interest
Here is a short explanation as to why we restrict the base b to 0 < b < 1 or b > 1 for
exponential graphs. The values b = 0 , b = 1 and all negative values of b are excluded.
If b = 0 , then y = 0 x = 0 which is a horizontal line and not an exponential graph.
If b = 1 , then y = 1x = 1 which is a horizontal line and not an exponential graph.
If b is negative, and you raise b to a rational power, you may not get a real number.
1
For example, if b = −2 , then (−2) 2 will not be possible to calculate since your calculator
will state Math ERROR.
1
In Grade 11 you will learn that (−2) 2 = −2 which is a non-real number.
133
Here is a short summary of the main features of the four functions discussed, the different
types of functions, how to recognise each function and the methods required to sketch the
graphs of these functions.
Linear functions
( y = ax + q )
The value of a determines the steepness of the line (ignoring negative signs). The greater
the value of a, the steeper the line. a < 0 means a reflection in the x-axis.
The value of q is the vertical shift up or down.
All of the different types of lines are shown below:
a>0
q=0
a>0
q>0
a<0
q>0
a>0
q<0
a<0
q<0
a<0
How to sketch a straight line:
Plot (0 ; 0) and one other point and draw the graph of y = ax
If q = 0 :
If q ≠ 0 :
Draw the graph of y = ax and shift it up or down. Find the x-intercept.
Alternatively, use the dual-intercept method to sketch the graph.
Quadratic functions
( y = ax 2 + q )
(x has been squared)
The value of a determines the vertical stretch (ignoring negative signs). The greater the
value of a, the closer the arms are to the y-axis. a < 0 means a reflection in the x-axis.
The value of q is the vertical shift up or down.
All of the different types of parabolas are shown below:
a>0
q=0
a>0
q>0
a<0
q=0
a<0
q<0
a>0
q<0
a<0
q>0
How to sketch a parabola:
Plot (0 ; 0) and one other point and draw the graph of y = ax 2
If q = 0 :
Draw the graph of y = ax 2 and shift it up or down. Indicate the y-intercept.
Determine the x-intercepts, if they exist.
a
Hyperbolic functions
( y = +q)
(x is in the denominator)
x
Ignoring negative signs, a determines the vertical stretch of the branches away from the
axes. The greater the value of a, the greater the stretch away from the axes.
If q ≠ 0 :
a>0
a < 0 means a reflection in the x-axis
a<0
134
The value of q is the vertical shift up or down.
The hyperbola has two asymptotes:
y = q is the equation of the horizontal asymptote
x = 0 is the equation of the vertical asymptote
All of the different types of hyperbolas are shown below:
a>0
q=0
a<0
q=0
a<0
q<0
a>0
q<0
a<0
q>0
a>0
q>0
How to sketch a hyperbola:
Select one negative x-value and one positive x-value and determine the
If q = 0 :
corresponding y-values. Plot these points and draw the two branches.
The asymptotes lie on the axes (vertical: x = 0 ; horizontal: y = 0 )
If q ≠ 0 :
Draw the horizontal asymptote y = q . The vertical asymptote is x = 0 .
Determine the x-intercept by letting y = 0 and solving for x.
Then select one negative x-value and one positive x-value and determine
the corresponding y-values. Plot these points and draw the two branches.
Exponential functions
( y = a .b x + q )
(x is the exponent)
The value of b affects the shape and steepness of the graph:
b >1
0 < b <1
For b > 1 :
As the value of b increases, the steeper the graph.
For 0 < b < 1 As the value of b decreases, the steeper the graph.
The value of a causes a vertical stretch of the mother graph and the y-intercept is affected.
a < 0 means a reflection in the x-axis.
The value of q is the vertical shift up or down.
The exponential graph has one horizontal asymptote: y = q
All of the different types of exponential graphs are shown below:
a>0
b >1
q=0
a>0
0 < b <1
q=0
a<0
b >1
q=0
a<0
0 < b <1
q=0
a>0
b >1
q<0
a>0
b >1
q>0
a>0
0 < b <1
q<0
a>0
0 < b <1
q>0
135
y
a<0
b >1
q<0
x
y
y
x
a<0
0 < b <1
q>0
x
a<0
b >1
q>0
y
a<0
0 < b <1
q<0
x
How to sketch an exponential graph:
If q = 0 :
Determine the y-intercept (let x = 0 ).
Then select one negative x-value and one positive x-value and determine
the corresponding y-values. Plot these points and draw the graph.
The horizontal asymptote will be the line y = 0 .
If q ≠ 0 :
Draw the horizontal asymptote y = q
Determine the y-intercept (let x = 0 ).
Determine the x-intercept (let y = 0 ).
If the graph has no x-intercept, select one negative x-value and one positive
x-value and determine the corresponding y-values. Plot these points and
draw the graph.
The following exercise is a mixed exercise in which you will be required to identify and
sketch the graphs of all functions studied thus far. Use the above summary to guide you.
EXERCISE 5
(a)
(b)
Identify the type of graph and then draw a neat sketch of the graph:
(1)
y = 2x
(2)
y=
2
x
(3)
y = 2x2
(4)
y=2
(5)
y = 2x
(6)
y = x2 + 2
(9)
1
y = −   +1
2
(12)
y = 2.2 x − 2
(7)
y = −2 x + 2
(8)
2
y = − +1
x
(10)
y = −2 x − 2
(11)
x=2
2
x
Sketch the graphs of the following functions on different axes.
You will need to re-work the equations algebraically into the standard equations for
the functions you have studied.
(3)
y=
x
3
(6)
y=
3− x
x
y = (3 − x )(3 + x )
(9)
y = 3− x + 3
y = 2 x+1 + 2 x − 3
(12)
y=
(1)
x = 3y
(2)
xy = 3
(4)
y=
x2
3
(5)
y=
(7)
3x + 3
y=
3
(8)
(10)
y = 3(3x − 1)
(11)
136
3− x
3
22 x − 4
2x + 2
FUNCTIONAL NOTATION
A function may be represented by means of functional notation.
Consider the function f ( x ) = 3x
The symbol f ( x) is used to represent the value of the output given an input value.
In other words, the y-values corresponding to the x-values are given by f ( x ) , i.e. y = f ( x ) .
For example, if x = 4 , then the corresponding y-value is obtained by substituting x = 4 into
3x . For x = 4 , the y-value is f (4) = 3(4) = 12 .
The brackets in the symbol f (4) do not mean f multiplied by 4, but rather the y-value when
x = 4 . Also, f ( x ) = 3x is read as “ f of x is equal to 3x ”.
We can also use other letters to name functions. For example, g ( x), h( x) and p( x) may be
used.
EXAMPLE 5
If f ( x) = 3 x 2 − 1 , determine the value of:
(a)
f (2)
(b)
f ( −3)
(c)
(d)
f (3x )
(e)
3 f ( x) + 1
(f)
Solutions
(a)
f ( x ) = 3x 2 − 1
(b)
f (2) = 3(2) 2 − 1
= 3(4) − 1
f ( x ) = 3x 2 − 1
f (−3) = 3(−3) 2 − 1
= 3(9) − 1
= 26
= 11
(c)
f ( x ) = 3x 2 − 1
(d)
f ( x ) = 3x 2 − 1
f (3 x ) = 3(3 x ) 2 − 1
f (a) = 3(a) 2 − 1
= 3(9 x 2 ) − 1
= 3a 2 − 1
= 27 x 2 − 1
(e)
f ( x ) = 3x 2 − 1
(f)
f ( x) = 3x 2 − 1
∴ 3 f ( x) = 3(3 x 2 − 1)
∴ 2 = 3x 2 − 1
∴ 3 f ( x) = 9 x 2 − 3
∴ 0 = 3x 2 − 3
∴ 3 f ( x) + 1 = 9 x 2 − 3 + 1
∴ 0 = x2 − 1
∴ 0 = ( x + 1)( x − 1)
∴ x = −1 or x = 1
∴ 3 f ( x) + 1 = 9 x 2 − 2
137
f (a)
x if f ( x) = 2
EXERCISE 6
(a)
(b)
(c)
(d)
If f ( x ) = 2 x 2 − x + 1 , determine the value of:
(1)
f (1)
(2)
f (−1)
(3)
f (2)
(4)
f (−2)
(5)
1
f 
2
(6)
 1
f − 
 2
(7)
f (a)
(8)
f (2 x)
(9)
2 f ( x)
(10)
f ( x) + 2
(11)
f (− x)
(12)
f ( x − 1)
(13)
2 f ( x − 1) − 3
(14)
f ( x + h)
(15)
f ( x + h) − f ( x )
If g ( x ) = x 2 − 5 , determine the value(s) of:
(1)
x if g ( x) = 4
(2)
x if g ( x) = 20
(3)
x if g ( x) = 1 − 5 x
(4)
x if g ( x) = 6 x − 14
Given:
g ( x) = −2 x − 4
(1)
Write down an expression for g ( p − 5)
(2)
If g (2 p) = 10 , determine the value of p.
Describe the transformation from f to g if:
(1)
f ( x) = x 2
and
g ( x) = 3 f ( x)
(2)
f ( x) = 2 x
and
g ( x) = − f ( x)
(3)
f ( x) =
2
x
and
g ( x) = f ( x) + 2
(4)
f ( x ) = x 2 + 5 and
g ( x) = x 2 + 1
(5)
f ( x) = 2 x 2
g ( x) = 8 x 2
(6)
f ( x) =
and
6
+ 1 and
x
6
g ( x) = − − 1
x
138
FURTHER CHARACTERISTICS OF THE FOUR FUNCTIONS
Domain and range
In Grade 8 and 9, you learnt that a function is a rule which when applied to a given set of
input values, produces a set of output values. For example, suppose that the rule is y = 2 x .
When applied to a set of input x-values x ∈ {−2 ; − 1; 0 ;1; 2; 3} , the output y-values can be
obtained by substituting the given x-values into the equation (rule) y = 2 x . The output values
are therefore y ∈ {−4 ; − 2 ; 0 ; 2 ; 4; 6}
The domain is simply all of the inputs (x-values) that are used.
The range is the set of outputs (y-values) obtained from the input values.
EXAMPLE 6
Determine the domain and range for the following functions:
(a)
There are four x-values in the domain:
2
(1; 2)
1 (0 ;1)
There are four y-values in the range:
(−1; 0)
−2
(−2 ; − 1)
1
−1
x ∈ {−2 ; − 1 ; 0 ; 1}
2
y ∈ {−1; 0 ; 1; 2}
For each point on the graph, there is a
−1
y-value corresponding to an x-value.
−2
(b)
There are an infinite number of x-values in
2
(1; 2)
excluding 1). The domain is written as:
1
−2
(−2 ; − 1)
−1
1
−1
−2
the domain from −2 to 1 (including −2 but
2
x ∈ [ −2 ; 1) or −2 ≤ x < 1
There are an infinite number y-values in
the range from −1 to 2 (including −1 but
excluding 2). The range is written as:
y ∈ [ −1; 2 ) or −1 ≤ y < 2
You can use the vertical ruler method to obtain the domain and range:
For domain: Keep the edge of the ruler vertical and slide it across the graph from
left to right. Where the edge starts cutting the graph, the domain starts
(as read off from the x-axis). Where it stops cutting the graph the
domain ends.
For range:
Keep the edge of the ruler horizontal and slide it across the graph from
bottom to top. Where the edge starts cutting the graph, the range starts
(as read off from the y-axis). Where it stops cutting the graph the range
ends.
139
(c)
There are an infinite number x-values in
2
the domain (as indicated by the arrows).
1
We write the domain as follows:
x ∈ (−∞ ; ∞) or x ∈ 
−2
−1
1
2
−1
There are an infinite number y-values in
the range (as indicated by the arrows).
−2
We write the range as follows:
y ∈ (−∞ ; ∞) or y ∈ 
(d)
There are an infinite number x-values in
9
the domain (as indicated by the arrows).
We write the domain as follows:
x ∈ (−∞ ; ∞) or x ∈ 
There are an infinite number y-values in
the range from 9 downwards. The range is
written as:
y ∈ ( −∞ ; 9] or y ≤ 9
(e)
There are an infinite number x-values in
the domain (as indicated by the arrows).
We write the domain as follows:
x ∈ (−∞ ; ∞) or x ∈ 
2
There are an infinite number y-values in
the range from 2 upwards (excluding 2).
The range is written as:
y ∈ ( 2 ; ∞ ) or y > 2
EXERCISE 7
State the domain and range of the following functions:
(a)
(b)
(c)
(−3 ; 2)
0
−3
(1; − 4)
140
(d)
(e)
(f)
2
2
1
Increasing and decreasing functions
For the graph of a given function, if the y-values increase as the x-values increase, then the
graph is said to be increasing. If the y-values decrease as the x-values increase, then the graph
is said to be decreasing. In short, a graph is increasing if it moves upwards when moving
from left to right and decreasing if it moves downwards when moving from left to right.
For example, the graph on the right is increasing
for all values of x < 0 . As we move from left to
right, the graph moves upwards. The graph is
decreasing for all values of x > 0 . As we move
from left to right, the graph moves downwards.
Note to the educator:
Increasing and decreasing depends on how we
define these concepts. For school purposes, we
will define increasing and decreasing in terms
of the gradient of the curve (positive or negative).
At x = 0 , the gradient is 0 and therefore we
exclude 0 in the solutions.
x<0
x>0
THE LINEAR FUNCTION
We will now briefly revise lines of the form ax + by = c , the gradient of a line and points of
intersection of lines.
EXAMPLE 7
In the diagram, two lines are drawn:
3x + 2 y = 6 and x − 4 y = 16
The first line cuts the y-axis at A and the x-axis at B.
The two lines intersect at C. Determine:
(a)
the coordinates of A and B.
(b)
the gradient of 3x + 2 y = 6
(c)
the gradient and y-intercept of x − 4 y = 16
(d)
the coordinates of C
(e)
the values of x for which the lines are
increasing or decreasing.
3x + 2 y = 6
x − 4 y = 16
141
Solutions
(a)
We can use the dual-intercept method:
x-intercept: Let y = 0
y-intercept:
∴ 3 x + 2(0) = 6
∴ 3x = 6
∴x = 2
B(2 ; 0)
(b)
In Grade 9, you learnt that the gradient of a line between any two points on the line is
change in y -values
given by the ratio:
change in x-values
We can determine the gradient by referring to the diagram or by re-writing the
equation of the line in the form y = ax + q , bearing in mind that the value of a
represents the gradient of the line. Between the points (0 ; 3) and (2 ; 0) ,
the y-values decrease ( −3 ) as the x-values increase ( +2 ).
−3
The gradient is therefore
(0 ; 3)
2
We can also determine the gradient as follows:
−3
3x + 2 y = 6
∴ 2 y = −3x + 6
−3
∴y =
x + 3 [The coefficient of x is the gradient]
2
Let x = 0
∴ 3(0) + 2 y = 6
∴2y = 6
∴y =3
A(0 ; 3)
(2 ; 0)
+2
(c)
x − 4 y = 16
∴−4 y = − x + 16
1
∴y = x−4
4
1
The gradient is and the y-intercept is −4
4
(d)
In order to determine the coordinates of C, we need to solve simultaneous equations:
3x + 2 y = 6
x − 4 y = 16
(1)
(2)
6 x + 4 y = 12
x − 4 y = 16
(1) ×2
(2)
∴ 7 x = 28
∴x = 4
Add like terms and constants
∴ 4 − 4 y = 16
∴−4 y = 12
Substitute x = 4 into (2) to get the corresponding y-value
∴ y = −3
The coordinates of C are C(4 ; − 3)
(e)
3x + 2 y = 6 decreases for all real values of x.
x − 4 y = 16 increases for all real values of x.
142
Finding the equation of a linear function
EXAMPLE 8
(a)
Determine the equation of the following line in the form f ( x) = ax + q .
Solution
The y-intercept is 3.
Therefore q = 3 .
∴ y = ax + 3
Substitute the point (8 ; − 1) to get a:
−1 = a (8) + 3
−1 = 8a + 3
−8a = 4
1
a=−
2
1
Therefore the equation is f ( x) = − x + 3
2
(b)
3
(8 ; − 1)
Determine the equation of the following line in the form g ( x) = ax + q .
Solution
Method 1
The y-intercept is 4.
Therefore q = 4 .
∴ y = ax + 4
Substitute the point (−2 ; 0) to get a:
0 = a(−2) + 4
4
−2
0 = −2a + 4
2a = 4
a=2
Therefore the equation is g ( x) = 2 x + 4
Method 2
+4
=2
+2
The y-intercept is 4
+2
The gradient is
+4
Therefore, the equation is g ( x) = 2 x + 4
143
−2
4
(c)
Determine the equation of the following line in the form f ( x) = ax + q .
Solution
Method 1
(−1; 3)
Substitute the two points into y = ax + q and
solve simultaneous equations:
For (−1; 3) : 3 = a(−1) + q
∴3 + a = q
−1 = a(3) + q
For (3 ; − 1)
∴−1 − 3a = q
∴ 3 + a = −1 − 3a
(both equal q)
∴ 4a = −4
∴ a = −1
∴ q = 3 + ( −1) = 2
(3 ; − 1)
Therefore the equation is f ( x) = − x + 2
Method 2
The gradient is a =
−4
= −1
+4
(−1; 3)
∴ y = −1x + q
Substitute either of the two points:
into the equation y = − x + q to get q:
(−1; 3) :
3 = −(−1) + q
∴q = 2
or
(3 ; − 1) :
−1 = −(3) + q
∴q = 2
The y-intercept is 2
−4
+4
Therefore the equation is f ( x) = − x + 2
Method 3
Use the Analytical Geometry formula to find the gradient (see Chapter 8)
−1 − 3 −4
Gradient =
=
= −1
3 − (−1) 4
Therefore a = −1 .
∴ y = −1x + q
Substitute one of the points into the equation y = − x + q to get q:
(−1; 3) :
3 = −(−1) + q
∴q = 2
The y-intercept is 2
Therefore, the equation is f ( x) = − x + 2
144
(3 ; − 1)
EXERCISE 8
(a)
(b)
(c)
In the diagram, line 4 x − 2 y = 8 cuts the axes at A and B. Line x + y = −1 cuts the
axes at C and D. The two lines intersect at E.
Determine:
4x − 2 y = 8
(1)
the coordinates of A and B.
(2)
the coordinates of C and D.
(3)
the gradient of 4 x − 2 y = 8
(4)
the gradient of x + y = −1
(5)
the coordinates of E
(6)
the values of x for which the lines are
increasing or decreasing.
Given:
x + y = −1
3 x − y = 4 and 2 x − y = 5
(1)
On the same set of axes, draw neat sketch graphs of the two functions.
(2)
Determine the coordinates of the point of intersection.
Determine the equations of the following lines in the form f ( x) = ax + q :
(1)
(2)
(3)
6
(−7 ; 2)
3
−3
−5
(−4 ; − 1)
(4)
(5)
(6)
(2 ; 4)
−4
−6
(d)
(−2 ; 7)
(2 ; 5)
(−3 ; − 1)
(1)
Determine the equation of the line passing through the point (0 ; − 1) and
parallel to the x-axis. Do you remember what the gradient of this line is?
(2)
Determine the equation of the line passing through the point (−1; 0) and
parallel to the y-axis. Do you remember what the gradient of this line is?
145
THE QUADRATIC FUNCTION
EXAMPLE 9
In the diagram, the graphs of f ( x) = 4 − x 2 ,
2
g ( x) = x 2 + 6
2
g ( x ) = x + 6 and h( x) = 3 x are shown.
The graph of f cuts the axes at A, C and D.
The graph of g cuts the y-axis at B. Determine:
(a)
the coordinates of A, B, C and D
(b)
the coordinates of E and F.
(c)
the values of x for which f is increasing
(d)
the values of x for which g is decreasing
(e)
the maximum or minimum values of f and g
(f)
the turning points of f and g
(g)
the equation of the axis of symmetry for f and g
h( x ) = 3 x 2
f ( x) = 4 − x 2
Solutions
(a)
The y-intercept of f is (0 ; 4) and the y-intercept of g is (0 ; 6)
A(0 ; 4) and B(0 ; 6)
Let y = 0
0 = − x2 + 4
∴ x2 − 4 = 0
∴ ( x + 2)( x − 2) = 0
∴ x = −2 or x = 2
C(−2 ; 0)
D(2 ; 0)
(b)
E and F are the points of intersection between f and h
h( x ) = f ( x )
(y-values are equal at the points of intersection)
∴ 3x 2 = 4 − x 2
∴ 4x2 − 4 = 0
∴ x2 −1 = 0
∴ ( x + 1)( x − 1) = 0
∴ x = −1 or x = 1
Now determine the corresponding y-values by substituting into either of the two
equations:
f (−1) = 3(−1) 2 = 3
f (1) = 3(1) 2 = 3
E(−1; 3)
F(1 ; 3)
(c)
f increases for all x < 0
(d)
g decreases for all x < 0
146
(e)
The maximum value refers to the largest y-value on a graph and the minimum value
refers to the smallest y-value on a graph.
The maximum value of f is 4 and the minimum value of g is 6
(f)
A turning point on a graph is the point at which the graph changes from increasing to
decreasing or decreasing to increasing.
The turning point of f is (0 ; 4)
The turning point of g is (0 ; 6)
(g)
The axis of symmetry of a parabola is a vertical line that divides the parabola into two
congruent halves. The y-axis is the axis of symmetry for both graphs. The equation is
x =0.
Finding the equation of a quadratic function
EXAMPLE 10
(a)
Determine the equation of the following graph in the form f ( x) = ax 2 + q .
Solution
The y–intercept is 3
∴q = 3
(2 ; 7)
∴ y = ax 2 + 3
Substitute the point (2 ; 7)
3
2
into y = ax + 3 to get the value of a
∴ 7 = a (2) 2 + 3
∴ 7 − 3 = 4a
∴ 4 = 4a
∴a = 1
∴ f ( x ) = 1x 2 + 3
(b)
Determine the equation of the following graph in the form f ( x) = ax 2 + q .
Solution
The x-intercept form of the equation of
a parabola can be used:
y = a( x − x1 )( x − x2 )
where x1 and x2 represent the x-intercepts.
∴ y = a( x − (−2))( x − 2)
∴ y = a( x + 2)( x − 2)
Substitute (1; 6) to get the value of a
6 = a (1 + 2)(1 − 2)
∴ 6 = a (3)(−1)
(1; 6)
−2
2
Notice that the x-intercept
form of y = −2 x 2 + 8 is:
∴ 6 = −3a
y = −2 x 2 + 8
∴ a = −2
∴ y = −2( x + 2)( x − 2)
= −2( x 2 − 4)
= −2( x + 2)( x − 2)
= −2( x − (−2))( x − 2)
∴ y = −2( x 2 − 4)
∴ f ( x) = −2 x 2 + 8
147
EXERCISE 9
(a)
(b)
In the diagram, the graph of f ( x) = 2 x 2 − 2
is shown. The graph of f cuts the axes at
A, B and C. Line h is parallel to the x-axis and
passes through A. Determine:
(1)
the coordinates of A, B and C
(2)
the values of x for which f is increasing
(3)
the values of x for which f is decreasing
(4)
the minimum value of f
(5)
the turning point of f
(6)
the equation of the axis of symmetry of f
(7)
the domain and range of f
(8)
the equation of h and the value of its gradient
(9)
the domain and range of h
f ( x) = 2 x 2 − 2
In the diagram, the graphs of f ( x ) = − x 2 + 9
g ( x ) = − x 2 , h and p are shown. The graph of
f cuts the axes at A, B and C. Line p cuts the
x-axis at B and is parallel to the y-axis.
Determine:
(c)
f ( x) = − x 2 + 9
(1)
the coordinates of A, B, and C.
(2)
the values of x for which f is increasing
(3)
the values of x for which g is decreasing
(4)
the maximum value of f and g
(5)
the turning points of f and g
(6)
the equation of the axis of symmetry for f
(7)
the domain and range of f and g
(8)
the equation of h if h( x) = − g ( x) + 12 . Describe the transformations.
(9)
the equation of p and the value of its gradient.
(10)
the domain and range of p
Given:
g ( x) = − x 2
f ( x ) = x 2 − 9 and g ( x ) = − x 2 − 1
(1)
Determine the coordinates of A, B, C and D
(2)
Determine the coordinates of E and F.
f ( x) = x 2 − 9
g ( x) = − x 2 − 1
148
(d)
Determine the equation of each of the following parabolas in the form f ( x) = ax 2 + q.
(1)
(−1; 6)
(2)
(3)
−1
(1; − 2)
(4)
(5)
(6)
25
(1;15)
−2
−4
3
−3
(−2 ; − 5)
2
4
2
5
−5
−8
THE HYPERBOLIC FUNCTION
An interesting characteristic of the hyperbola is that it has two axes of symmetry.
y = −x
y=x
y=
y = x+q
y = −x + q
a
x
y=
q
a
+q
x
y=q
Each axis of symmetry has a gradient of 1 or −1 and a y-intercept of q.
EXAMPLE 11
Given: g ( x) =
3
+2
x
Determine:
y=
(a)
the equations of the asymptotes
(b)
the equations of the axes of symmetry
(c)
the values of x for which g is decreasing
(d)
the domain and range of g
149
3
+2
x
Solutions
(a)
Horizontal asymptote is y = 2
(b)
y = − x + 2 and y = x + 2
(c)
The graph of g is decreasing on the interval x ∈ ( −∞ ; 0) as well as the interval
x ∈ (0 ; ∞) .
(d)
Domain of g: x ∈  x ≠ 0
y∈ y ≠ 2
Range of g:
Vertical asymptote is x = 0
Finding the equation of a hyperbola
EXAMPLE 12
Determine the equation of the following graph in the form f ( x) =
a
+q.
x
Solution
The horizontal asymptote is y = −3
∴ q = −3
a
∴ y = −3
x
Substitute the point (3 ; − 4) :
a
−4 = − 3
3
a
∴−1 =
3
∴−3 = a
−3
∴ f ( x) =
−3
x
y = −3
(3 ; − 4)
EXERCISE 10
(a)
The line y = x + 4 is an axis of symmetry of the
−2
graph of f ( x) =
+ q . The graph of f cuts the
x
x-axis at A.
(1)
Write down the value of q
(2)
Write down the equation of f
(3)
State the domain and range of f
(4)
For which values of x is f increasing?
(5)
Write down the equations of the asymptotes
(6)
Write down the equation of the other axis of symmetry.
150
y = x+4
(b)
Determine the equation of each of the following hyperbolas in the form f ( x) =
(1)
(2)
a
+ q.
x
(3)
(−3 ; 5)
(4 ; 3)
y=2
(−2 ; − 6)
(4)
(5)
(6)
3
−4
−2
(−8 ; − 8)
(1; − 2)
(7)
(2 ; − 6)
(8)
(9)
y = x −3
−3
−4
(−1; − 8)
y = −x +1
(10)
y = x−2
(11)
( 14
;
(2 ; 4)
5
)
4
−1
−1
1
1
2
(2 ; − 6)
(c)
For each function below, state the domain and range and determine the equation:
(1)
y
(2)
(3)
(−4 ; 2)
x
y = −3
(−2 ; − 7)
151
y = x+5
(2 ; − 4)
THE EXPONENTIAL FUNCTION
Finding the equation of an exponential graph
EXAMPLE 13
(a)
Determine the equation of the given
graph in the form f ( x ) = a .2 x + q
−4
Solution
The horizontal asymptote is y = −8
−8
x
∴ y = a .2 − 8
Substitute (0 ; − 4) :
−4 = a .20 − 8
∴−4 = a − 8
∴a = 4
∴ f ( x ) = 4.2 x − 8
(b)
Determine the value of b and q if the
equation of the given graph is
g ( x ) = −b x + q
3
−1
Solution
The horizontal asymptote is y = 3
∴ y = −b x + 3
Substitute (−1; 0) :
0 = −b −1 + 3
1
∴0 = − + 3
b
∴ 0 = −1 + 3b
∴−3b = −1
∴b =
1
3
x
1
∴ g ( x) = −   + 3
3
1
b=
and q = 3
3
(c)
Determine the equation of the given
graph in the form y = a . b x + q
(2 ;1)
Solution
1
3
The graph passes though the origin.
y = a .b x −
− 13
152
Substitute the point (0 ; 0) :
1
0 = a . b0 −
3
1
∴0 = a −
3
1
∴a =
3
1
1
∴ y = . bx −
3
3
Substitute the point (2 ;1) :
1
1
1 = .b2 −
3
3
2
∴3 = b −1
(2 ;1)
− 13
∴ 0 = b2 − 4
∴ 0 = (b + 2)(b − 2)
∴ b = −2 or b = 2
Choose b = 2 since b ≠ −2
1
1
∴ y = .2 x −
3
3
EXERCISE 11
(a)
Determine the equation of the given
graph in the form f ( x ) = a .2 x + q
−3
−4
(b)
Determine the equation of the given
graph in the form g ( x ) = b x + q
(−1; 2)
−1
(c)
Determine the equation of the given
graph in the form h( x ) = −b x + q
3
1
(d)
Determine the equation of the given
graph in the form f ( x ) = a . b x + q
3
153
(2 ;12)
(e)
Determine the equation of the given
graph in the form f ( x ) = a . b x + q
18
16
2
(f)
Determine the equation of the given
graph in the form g ( x ) = a . b x + q
(−2 ; 60)
−4
GRAPH INTERPRETATION
This topic involves determining the lengths of line segments using the different functions
you have studied thus far. It also discusses the graphical interpretation of inequalities,
which is so important for matric.
Determining the length of line segments using graphs
The following information is extremely important for determining the length of line segments:
The length of a line segment is always positive.
xA = 1 and OA = 1 unit
xB = −1 and OB = 1 unit
1
−1
yC = 1 and OC = 1 unit
yD = −1 and OD = 1 unit
●
●
1
−1
To determine a length OA along the y-axis, let x = 0
The y-value will be the length of OA.
To determine a length OB along the x-axis, let y = 0
The positive x-value will be the length of OB.
The point of intersection (B) is obtained by equating the
equations of the functions ( f ( x) = g ( x) ) and solving for
x and hence for y. The value of x will give the horizontal
length OA and the value of y will give the vertical length
OB.
154

Let x = 0
Let y = 0 



●

●
A vertical length between two graphs can be
calculated using the formula:
ytop graph − ybottom graph

 AB = yA − yB

(substitute the x-value into this formula to
get the required length)






x
A horizontal length between two graphs can be
calculated using the formula:
xright end point − yleft end point
CD = xD − xC
Using graphs to solve inequalities
The diagram below shows the graph of a parabola and a line. A and B are the points of
intersection of the two graphs. Both graphs cut the x-axis at −1 and the graph of the parabola
cuts the x-axis at 1.
g ( x) = − x − 1
f ( x) = − x 2 + 1
−1
1
2
We can use the graphs to solve the following inequalities graphically:
(a)
For which values of x is f ( x) > 0 ?
We are required to determine the x-values for which the y-values of f are positive.
Where the parabola lies above the x-axis will be where the y-values are positive.
The solution lies between x-values of −1 and 1.
∴−1 < x < 1 or we can write x ∈ (−1;1)
(b)
For which values of x is f ( x) ≤ 0 ?
We are required to determine the x-values for which the y-values of f are negative.
Where the parabola lies below the x-axis will be where the y-values are negative.
∴ x ≤ −1 or x ≥ 1 or we can write x ∈ (−∞ ; − 1] ∪ [1; ∞)
(c)
For which values of x is g ( x) > 0 ?
We are required to determine the x-values for which the y-values of g are positive.
Where the line lies above the x-axis will be where the y-values are positive.
∴ x < −1 or we can write x ∈ (−∞ ;1)
155
(d)
For which values of x is f ( x) ≥ g ( x) ?
We are required to determine the values of x for which the y-values of f are greater
than or equal to the y-values of g. This is where the graph of f is above the graph of g
(between A and B).
∴−1 ≤ x ≤ 2 or we can write x ∈ [−1; 2]
(e)
For which values of x is f ( x) < g ( x) ?
We are required to determine the values of x for which the y-values of f are smaller
than the y-values of g. This is where the graph of f is below the graph of g
∴ x < −1 or x > 2 or we can write x ∈ (−∞ ; − 1) ∪ (2 ; ∞)
EXAMPLE 14
Two lines cut the y-axis at A.
f ( x) = − x + 3
Determine:
(a)
the length of OA, OB, OC and BC
(b)
the length of OD, FE, OF and DE
(c)
the length of OJ, GH, OG and JH
(d)
the values of x for which f ( x) ≥ 0
(e)
the values of x for which g ( x) < 0
(f)
the values of x for which f ( x) ≤ g ( x)
g ( x) = 3x + 3
4
−3
Solutions
(a)
(b)
A(0 ; 3)
∴ OA = 3 units
For xB
0 = −x + 3
∴x = 3
∴ OB = 3 units
For xC
0 = 3x + 3
∴−3x = 3
∴ x = −1
∴ OC = 1 unit
BC = yB − yC
∴ BC = 3 − (−1)
∴ BC = 4 units
OD = 4 units
∴ FE = 4 units (opp sides rectangle)
Substitute x = 4 into y = − x + 3
y = −4 + 3 = −1
∴ yE = yF = −1
∴ OF = 1 units and DE = 1 unit
156
(c)
OJ = 3 units
∴ GH = 3 units (opp sides rectangle)
Substitute y = −3 into y = 3x + 3
−3 = 3x + 3
∴−3x = 6
∴ x = −2
∴ OG = 2 units and JH = 2 units
(d)
f ( x) ≥ 0
∴ x ≤ 3 or write x ∈ (−∞ ; 3]
(e)
g ( x) < 0
∴ x < −1
(f)
or write x ∈ (−∞ ; − 1)
f ( x) ≤ g ( x)
∴ x ≥ 0 or write x ∈ [0 ; ∞)
EXAMPLE 15
The diagram below shows the graphs of f ( x) = 3 x 2 − 12 , g ( x) = 3x − 6 and h( x) = −3x + 6 .
The graphs of f , g and h share a common x-intercept (D). The graph of f and g intersect
at D and F. The diagram is not drawn to scale.
f ( x) = 3x 2 − 12
h( x) = −3x + 6
g ( x) = 3 x − 6
Determine:
(a)
the length of AB and CD.
(b)
the length of OE and EF.
(c)
the length of GH if OG = 3 units
(d)
the length of OI if IJ = 6 units
(e)
the length of KL if OR = 1 unit
(f)
the length of ON if MP = 18 units
(g)
the values of x for which f ( x) ≥ 0
(h)
the values of x for which f ( x) < g ( x)
157
Solutions
(a)
AB = yA − yB = −6 − (−12) = 6 units
Let y = 0 in y = 3 x 2 − 12
0 = 3 x 2 − 12
∴ 0 = x2 − 4
∴ 0 = ( x + 2)( x − 2)
∴ x = −2 or x = 2
CD = xD − xC = 2 − (−2) = 4 units.
(b)
Determine the coordinates of E, the point of intersection of f and g
3 x 2 − 12 = 3 x − 6
∴ 3x 2 − 3x − 6 = 0
∴ x2 − x − 2 = 0
∴ ( x + 1)( x − 2) = 0
∴ x = −1 or x = 2
At E, x = −1
y = 3(−1) − 6 = −9
The coordinates of E are (−1; − 9)
∴ OE = 1 unit and EF = 9 units
(c)
OG = 3 which means that x = −3 at G.
Substitute x = −3 into g ( x) = 3x − 6 to get y.
g (−3) = 3(−3) − 6 = −15
∴ GH = 15 units
(d)
IJ = 6 which means that y = −6 at J.
Substitute y = −6 into h( x) = −3x + 6 to get x
−6 = −3 x + 6
∴ 3 x = 12
∴x = 4
∴ OI = 4 units
(e)
OR = 1 which means that x = 1
KL = yK − yL = (3 x − 6) − (3 x 2 − 12)
Substitute x = 1
∴ KL = (3(1) − 6) − (3(1) 2 − 12) = 6 units
(f)
MP = yM − yP = (3x − 6) − (−3x + 6) = 3x − 6 + 3x − 6 = 6 x − 12
∴18 = 6 x − 12 (since MP = 18 )
∴−6 x = −30
∴x = 5
∴ ON = 5 units
(g)
f ( x) ≥ 0 for x ≤ −2 or x ≥ 2
or
x ∈ (−∞ ; − 2] ∪ [2 ; ∞)
(h)
f ( x) < g ( x) for −1 < x < 2
or
x ∈ (−1; 2)
158
EXERCISE 12
(a)
(b)
Two lines f ( x) = − x + 4 and g ( x) = x − 2
intersect at R. By using the information on
the diagram, determine:
(1)
the length of OA, OP, OB and OC
(2)
the length of AP and BC
(3)
the length of OD, EF, OF and DE
(4)
the length of OK, GH, OG and KH
(5)
the length of RS and OS
(6)
the values of x for which f ( x) ≥ 0
(7)
the values of x for which g ( x) < 0
(8)
the values of x for which f ( x) < g ( x)
g ( x) = x − 2
−2
−1
The diagram shows the line f ( x) = x + 5
Determine:
(c)
f ( x) = − x + 4
f ( x) = x + 5
(1)
the length of OP and OQ
(2)
the length of AB if OB = 2 units
(3)
the length of DC if OD = 7 units
(4)
the length of OF if EF = 3 units
(5)
the length of OH if HG = 3 units
(6)
the values of x for which f ( x) < 0
Two lines f ( x) = − x + 4 and g ( x) = x + 2
intersect at E.
g ( x) = x + 2
Determine:
(1)
the length of AB, CD and DF
(2)
the length of PQ if OR = 3 units
(3)
the length of OU if ST = 8 units
(4)
the length of GH if HK = 1 12
(5)
the values of x for which f ( x) > 0
(6)
the values of x for which g ( x) ≤ 0
(7)
the values of x for which g ( x) < f ( x)
159
f ( x) = − x + 4
(d)
The diagram shows the graphs of f ( x ) = x 2 − 9
and g ( x) = 3 − x intersecting at E and D.
The graph of f cuts the axes at B, C and D.
f ( x) = x 2 − 9
Determine:
(e)
(1)
the length of AB, CT and CD
(2)
the length of OF and EF
(3)
the length of GH if OH = 1 unit
(4)
the length of OV if VW = 8 units
(5)
the length of JL if OK = 1 unit
(6)
the length of OQ if PR = 8 units
(7)
the values of x for which f ( x) > 0
(8)
the values of x for which f ( x) ≥ g ( x)
The graph of f ( x) =
g ( x) = 3 − x
2
− 2 and g ( x) = 2 is shown.
x
Determine:
g ( x) = 2
(1)
the length of OA.
(2)
the length of BC if OB = 4 units
(3)
the length of OD and OE
(4)
the coordinates of F
(5)
the values of x for which f ( x) ≥ 0
(6)
the values of x for which f ( x) < 0
(7)
the values of x for which f ( x) > −2
(8)
the values of x for which f ( x) > g ( x)
f ( x) =
2
−2
x
x
(f)
1
The graph of f ( x ) =   − 2 is shown.
2
Determine:
(1)
the length of OA and CD.
(2)
the values of x for which f ( x) > 0
(3)
the values of x for which f ( x) ≤ 0
(4)
the values of x for which f ( x) > −2
(5)
the values of x for which f ( x) > −1
(6)
the values of x for which f ( x) ≤ −1
160
x
1
f ( x) =   − 2
2
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(b)
(c)
Given:
f ( x ) = x 2 − 1 and g ( x) = x − 1
(1)
Sketch the graph of f and g on the same set of axes.
(2)
Determine the domain and range of f.
(3)
If h( x) = f ( x) + 3 , write down the coordinates of the turning point of h.
(4)
Determine the maximum value of p if p( x) = −h( x) .
(5)
Sketch the graphs of h and p on the same set of axes.
(6)
Determine the y-intercept of the graph of y = − g ( x) .
(7)
Determine the x-intercept of the graph of y = g ( x + 2) .
(1)
3
x
Sketch the graph of h if h( x) = 3 f ( x) − 1 and describe the transformations.
(2)
Sketch the graph of p if p( x) = − g ( x) + 1 and describe the transformations.
(3)
Write down the range of h.
(4)
Write down the domain of p.
Given:
f ( x) = 3x and g ( x) =
In the diagram below, the graphs of f and g are shown.
The graphs intersect at (1; 3) . Determine:
(1)
the equation of f
(2)
the coordinates of the turning point of f.
(3)
the equation of the axis of symmetry of f.
(4)
the equation of the horizontal asymptote of g
(5)
the equation of g and the coordinates of A
(6)
the domain and range of f and g
(7)
the values of x for which f is decreasing?
(1; 3)
x
(d)
1
The graph of f ( x) =   + q passes through
2
the origin. Determine:
(1)
the value of q
(2)
the length of AB of OA = 2 units
(3)
the length of OC if CD = 78 units
(4)
the length of OT if MN = 2 units
(5)
the equation of g , the image when the graph
of f is translated downwards so that the image
of M lies on the x-axis.
161
f
(e)
(f)
The graph of f ( x) = mx + c and g ( x ) = ax 2 + b
intersect at D and on the x-axis at A.
H and T lie on line f . Determine:
(1)
the equation of f
(2)
the coordinates of A and hence C
(3)
the equation of g
(4)
the length of AC and OB
(5)
the length of DE and DF
(6)
the length of ST
(7)
the length of OM if QR =
(8)
the values of x for which g ( x ) ≥ 0
(9)
the values of x for which g ( x ) < f ( x )
g
H(−1; 4)
T(4 ; − 1)
45
4
f
G(1; − 8)
a
+6
x
cut the x-axis at A. The graph of f passes through
The graph of f ( x) = a.b x + q and g ( x ) =
the point ( −2 ; − 12) and cuts the y-axis at 4.
Determine:
(1)
the range of f
(2)
the equation of f
(3)
the equation of g
(4)
the equation of h if the graph of f is
reflected in the x-axis
(5)
(−2 ; − 12)
the values of x for which f is increasing
*(g) Given:
f ( x) = −3 x 2 + 12 and g ( x ) = −3 x + 6
Determine:
(1)
the values of x for which f ( x ) ≥ 0
(2)
the values of x for which f ( x ) > g ( x )
(3)
the values of x for which f ( x ) ≥ 9
(4)
the values of x for which f ( x ). g ( x ) < 0
(5)
the values of x for which f ( x ). g ( x ) ≥ 0
(6)
the possible values of t if the graph of
f ( x ) = −3 x 2 + 12 − t
(i) does not cut or touch the x-axis.
(ii) cuts the x-axis in two distinct points.
162
f ( x) = −3x 2 + 12
g ( x) = −3x + 6
CHAPTER 7
EUCLIDEAN GEOMETRY
Revision of lines, angles, triangles and polygons
Line property 1
Adjacent angles on a straight line are supplementary.
ˆ +B
ˆ = 180° or
• If ABC is a straight line, then B
1
2
ˆ +B
ˆ = 180° then ABC is a straight line
• If B
1
2
Line property 2
If two lines AB and CD cut each other (intersect) at E,
then the vertically opposite angles are equal.
Eˆ 1 = Eˆ 3 and Eˆ 2 = Eˆ 4 .
Line property 3
The angles around a point add up to 360° .
ˆ +B
ˆ +B
ˆ +B
ˆ = 360°
B
1
2
3
4
Corresponding angles
Corresponding angles lie either both above or both below the parallel lines and on the same
side as the transversal. They are the angles in matching corners and are equal. Always look
out for the F shape.
Alternate angles
Alternate angles lie on opposite sides of the transversal and between the parallel lines. They
are equal in size. Always look out for the Z or N shape.
163
Co-interior angles
Co-interior angles lie on the same side of the transversal between the parallel lines.
These angles are supplementary. Always look out for the U shape.
ˆ +H
ˆ = 180°
G
1
1
ˆ +H
ˆ = 180°
G
1
1
ˆ +H
ˆ = 180°
G
1
1
ˆ +H
ˆ = 180°
G
1
1
Acute-angled triangles
Right-angled triangles
All three interior angles are smaller
than 90° (acute).
The largest interior angle is equal to 90° .
The other two angles are acute.
From Pythagoras:
AB2 = AC2 + BC2
AC2 = AB2 − BC2
BC 2 = AB2 − AC2
Obtuse-angled triangles
The largest interior angle is greater than 90° .
The other two angles are acute.
Scalene triangle
Isosceles triangle
No sides are equal.
Two sides are equal and the angles opposite
the equal sides are equal.
We can say that:
ˆ (sides opp = ∠ s )
AB = AC if B̂ = C
ˆ if AB = AC
B̂ = C
( ∠ s opp equal sides)
Equilateral triangle
Properties of triangles
Three sides are equal and the
interior angles are equal to 60° .
60°
60°
60°
ˆ +B
ˆ = 180°
ˆ +C
A
2
ˆC = A
ˆ +B
ˆ
1
164
(sum of the ∠ s of Δ)
(ext ∠ of Δ )
Congruency of triangles (four conditions)
Condition 1
Two triangles are congruent if three
sides of one triangle are equal in
length to the three sides of the other
triangle.
≡
≡
Condition 2
Two triangles are congruent if two
sides and the included angle are
equal to two sides and the included
angle of the other triangle.
Condition 3
Two triangles are congruent if two
angles and one side of a triangle are
equal to two angles and a
corresponding side of the other
triangle.
Condition 4
Two right-angled triangles are congruent
if the hypotenuse and a side of the one
triangle is equal to the hypotenuse and a
side of the other triangle.
Similar Triangles
If two triangles are similar (equiangular), then their corresponding sides are in the same
proportion.
AB BC AC
If ΔABC|||ΔDEF , then
=
=
DE EF DF
EXERCISE 1
(Revision)
200°
(a)
In the diagram, ABCG||FD, ABCG ⊥ GH
ˆ = 50° .
ˆ = 200° and CGD
The reflex angle BFE
(1)
Calculate the size of all angles
indicated by small letters.
(2)
Show that DF ⊥ FE .
2a − 50°
50°
(b)
In ΔABC , EF||BC. BA is produced to D.
(1)
Calculate a and hence show that
AE = AF .
(2)
Calculate, with reasons, the value of
b, c, d and e.
165
5a − 40°
2a
80°
(c)
ˆ = 28° , ABD
ˆ = 48°
In the diagram, CD||EF, DEF
ˆ = 160° . Prove that AB||CD.
and BDE
160°
48°
x
28°
(d)
ˆ = 82° ,
In the diagram, PU||QT, T̂ = 42° , RQS
ˆ = y , UPT
ˆ = x + 40° .
ˆ = x and QPT
PQT
(1)
Prove that PT||QS.
(2)
Calculate y.
x + 40°
42°
82°
(e)
A
ABCD is a parallelogram, ΔABE is equilateral
and ΔDEC is isosceles.
(1)
Show that AE ⊥ ED
(2)
Show that ED bisects D̂
1 2 3
B
(f)
In the diagram, AB = 7 cm , AD = 12 cm ,
CE = 8 cm and DE = 17 cm .
Calculate the length of BC.
(g)
ABCD is a kite in which
AB = AD and BC = CD .
Prove that:
ΔABC ≡ ΔADC
(1)
ˆ =D
ˆ?
(2)
Why is B
(h)
AB||DE and DC = CB .
Prove that:
AC = CE
(1)
(2)
AB = DE
1
2
1 2
A
B
1
2
D
(i)
Prove that ΔABC ≡ ΔADC using two
different conditions of congruency.
(j)
In the figure below, sides PR
and QS of triangles PQR and SQR
intersect at T. PQ = SR and P̂ = Sˆ = 90° .
Prove that ΔPQR ≡ ΔSRQ .
166
C
E
C
E
D
(k)
ˆ .
In the diagram, O is the centre of the circle. BO bisects AOC
Prove that:
(1)
OB bisects AC at B.
(2)
OB ⊥ AC
(l)
In the diagram, O is the centre of the circle passing through
A, B and C. AB = AC and AO ⊥ BC.
Prove that:
ˆ =O
ˆ
O
(1)
1
2
(2)
Δ ABC is a right-angled triangle.
(m)
Show that the following triangles
are similar.
(1)
(n)
If ΔABC|||ΔDEC ,
calculate x and y.
E
(2)
A
y
B
5
15
3
1
3
12
C
x
D
QUADRILATERALS
A polygon is a two-dimensional figure with three or more straight sides. A quadrilateral is a
polygon with four straight sides.
90°
90°
167
Summary of the properties of quadrilaterals (Revision of Grade 8 and 9)
Quadrilateral




45°
45°
45°
45°
45°
45°
45
45°
Diagonals
The diagonals of a
parallelogram
bisect each other.
The diagonals of a
rhombus bisect
each other at right
angles. The
diagonals bisect
the vertex angles.
The diagonals of a
rectangle bisect
each other and are
equal in length.
The diagonals of a
square bisect each
other at right
angles and are
equal in length.
The diagonals
bisect the vertex
angles.
The diagonals of a
trapezium
intersect but don’t
bisect each other.
They lie between
parallel lines and
therefore the
alternate angles
are equal.
The diagonals are
perpendicular and
one diagonal
bisects the other.
One of the
diagonals bisects
the vertex angles.
168
Angles
The opposite
angles of a
parallelogram are
equal. The interior
angles add up to
360° .
The opposite
angles of a
rhombus are
equal. The interior
angles add up to
360° .
The interior angles
of a rectangle are
equal to 90° . The
interior angles add
up to 360° .
Sides
The opposite sides
of a parallelogram
are parallel and
equal.
The interior angles
of a square are
equal to 90° . The
interior angles add
up to 360° .
The opposite sides
of a square are
parallel and all
sides are equal.
The opposite sides
of a rhombus are
parallel and all
sides are equal.
The opposite sides
of a rectangle are
parallel and equal.
The interior angles One pair of
opposite sides are
add up to 360° .
parallel.
One pair of
opposite angles
are equal. The
interior angles add
up to 360° .
Two pairs of
adjacent sides are
equal.
The concept of a theorem and its converse
A theorem is a statement that can be demonstrated to be true by accepted mathematical
operations and arguments. In general, a theorem is an embodiment of some general principle
that makes it part of a larger theory. The process of showing a theorem to be correct is called
a proof. In Grade 10, we introduce the concept of a theorem and its proof.
A converse of a theorem is a statement formed by interchanging what is given in a theorem
and what is to be proved. For example, the isosceles triangle theorem states that if two sides
of a triangle are equal then the two base angles are equal. In the converse, the given (that two
sides are equal) and what is to be proved (that two angles are equal) are swapped, so the
converse is the statement that if two angles of a triangle are equal then two sides are equal.
In Grade 10, we will discuss and then prove three theorems relating to parallelograms.
PARALLELOGRAMS
(b)
If the opposite
angles of a
quadrilateral are
equal, then the
quadrilateral is a
parallelogram.
(opp ∠ s of quad
equal)
Theorem 2
The diagonals of a
parallelogram bisect each
other.
(diags of parm bisect)
then AD = BC , AB = DC ,
ˆ =D
ˆ
ˆ =C
ˆ and B
A
>
>
If ABCD is a parallelogram
(AD||BC and AB||DC)
If AD = BC and AB = DC
then ABCD is a parm
A
>
>
D
C
B
ˆ =C
ˆ and B
ˆ =D
ˆ
If A
A
then ABCD is a parm
D
C
B
If ABCD is a parallelogram
then AE = EC and BE = ED
A
B
169
D
E
>
Theorem 1 (Converse)
(a)
If the opposite sides
of a quadrilateral are
equal, then the
quadrilateral is a
parallelogram.
(opp sides of quad
equal)
Diagrams
>
Statement of theorem
Theorem 1
The opposite sides and
angles of a parallelogram
are equal.
(opp sides of parm equal)
(opp ∠ s of parm equal)
C
Theorem 2 (Converse)
If the diagonals of
quadrilateral bisect each
other, then the quadrilateral
is a parallelogram.
(diags of quad bisect)
If AE = EC and BE = ED
A
D
E
C
B
Theorem 3
If one pair of opposite sides
of a quadrilateral are equal
and parallel, then the
quadrilateral is a
parallelogram.
(one pair opp sides equal
and ||)
then ABCD is a parm
then ABCD is a parm
If AD||BC and AD = BC
The following is information for use in the examples and exercises on parallelograms which
follow. The proofs of theorems (excluding the converses) are presented after the basic
exercises. If ABCD is a parallelogram, you may assume the following:
=
AD||BC ; AB||DC
AD = BC ; AB = DC
=
AE = EC ; BE = ED
ˆ ;C
ˆ =A
ˆ
ˆ =B
ˆ ;D
ˆ =B
ˆ ; Cˆ = A
D
1
2
2
1
1
2
2
1
ˆ
ˆ
ˆ =D
ˆ
A =C;B
EXAMPLE 1
DELM is a parallelogram.
(a)
Calculate the value of x and hence the sizes
of the interior angles.
(b)
If DE = 2DM and ML = 10 cm , determine
the length of the other sides of DELM.
Solutions
(a)
Statement
2 x + x = 180°
D
x
2x
M
10 cm
Reason
co-int ∠ s ; DM||EL
∴ 3 x = 180°
∴ x = 60°
∴ Ê = 60° and M̂ = 60°
opp ∠ s of parm equal
D̂ = 2(60 °) = 120 °
(b)
L̂ = 120°
opp ∠ s of parm equal
DE = 10 cm
opp sides of parm
∴ DM = 5 cm
DE = 2DM
∴ EL = 5 cm
opp sides of parm
170
L
E
EXAMPLE 2
In the diagram, PQRS is a parallelogram. PT ⊥ TQR ,
PT = 5 cm , TQ = 2 cm and PS = 10 cm .
Calculate the length of VR.
=
=
Solution
Statement
QR = 10 cm
Reason
opp sides of parm equal
TQ = 2 cm
given
∴ TR = 12 cm
PT = 5 cm
2
2
given
PR = (5) + (12)
ˆ = 90°
Pythagoras ; PTQ
2
PR 2 = 169
∴ PR = 13 cm
PV = VR
diags of parm bisect
∴ VR = 6,5 cm
Note: In order to prove that a quadrilateral is a parallelogram, you will need to prove at least
one of the following:
AD||BC and AB||DC
Opp sides ||
A
D
AD = BC and AB = DC
Opp sides =
AE = EC and BE = ED
Diagonals bisect
ˆ =C
ˆ and B
ˆ =D
ˆ
A
Opp angles =
E
B
C
AB||DC and AB = DC
One pair opp sides = and ||
AD||BC and AD = BC
One pair opp sides = and ||
EXAMPLE 3
70°
70°
In trapezium ABCD, AD||BC
ˆ =D
ˆ = 70° and EC = DC .
with A
Prove that ABCE is a parallelogram.
Solution
Statement
Reason
Ê 2 = 70°
∠s opp = sides
Ĉ1 = 70°
alt ∠ s ; AD||BC
171
ˆ =C
ˆ
∴A
1
Ê1 = 110°
adj ∠ s on a line
B̂ = 110°
ˆ
∴ Eˆ = B
co-int ∠ s ; AD||BC
Therefore, ABCE is a parallelogram
opp ∠ s of quad equal
1
EXAMPLE 4
Diagonals AC and BD of parallelogram ABCD intersect at M. AP = QC and AC = 600 mm ,
AB = 500 mm and AP = 150 mm . Prove that PBQD is a parallelogram.
Statement
Reason
AM = MC
diagonals of a parm
But AC = 600 mm
given
∴ AM = MC = 300 mm
AP = QC = 150 mm
given
PM = MQ = 150 mm
Also BM = MD
∴ PM = MQ and BM = MD
diagonals of parm
∴PBQD is a parallelogram
diags of quad bisect
EXERCISE 2
(a)
Determine the sizes of the interior angles of
parallelogram ABCD.
(b)
PQRS is a parallelogram with P̂ = 60° and PQ = PS .
ˆ ,Q
ˆ and Ŝ .
Calculate the sizes of R̂ , Sˆ , Q
1
(c)
1
2
2 x − 30°
2 x + 10°
60°
2
KLMN is a parallelogram. Calculate the size of
the interior angles.
5 x − 12°
3x + 18°
(d)
In Δ ABC , Â = 80° and Ĉ = 35° . Calculate the interior
angles of parallelogram MENB.
172
(e)
In parallelogram ABCD, AB = AD
and Ĉ = 110° . Calculate the size of all
interior angles.
(f)
Δ ABC is an equilateral triangle.
Determine the interior angles of
parallelogram LMCN.
(g)
A
ABCD is a parallelogram. AM bisects  .
AB = AM . Ĉ = 120° . Calculate the sizes
of all interior angles.
(h)
In parallelogram ABCD, AB = BE = DE .
Calculate the size D̂1 if D̂1 = x and
D
1 2
1 2
B
A
1
2
2 1
x
Â1 = 28°
1
B
(i)
(j)
(l)
2 3
C
E
In parallelogram ABCD, AB = 50 cm and
E is a point on AD such that AB = AE and
CD = DE . Determine:
(1)
DE
(2)
the perimeter of ABCD.
P
In parallelogram PQRS, Q̂ = 114° ,
1
PT bisects P̂ and TS bisects Ŝ .
Prove that PT ⊥ ST .
(k)
C
D
M
1 2
1
Q
B
ΔABD and Δ BCD are two isosceles
ˆ = 30° .
triangles. Ĉ = 75° and ADB
Prove that ABCD is a parallelogram.
2
S
T
R
C
1
2
2
A
1
D
In quadrilateral LMNP, Ê1 = 62° , P̂1 = 68° ,
P̂2 = 56° , FP = FN and LE = LM .
Prove that:
(1
LP||MN
(2)
LMNP is a parallelogram.
(m)
In quadrilateral ABCD, AB = 5 cm , BC = 10 cm ,
FD = 3 cm , BE = FD and AE = FC .
5 cm
AE ⊥ BC and CF ⊥ AD
Prove that ABCD is a parallelogram.
B
173
F 3 cm D
A
E
10 cm
C
(n)
A
Δ ABC is an equilateral triangle.
D, E and F are the midpoints of the sides
of the triangle and DE||BFC .
Prove that DECF is a parallelogram.
3
B
(o)
In the diagram, BCDF, EDCF and ABCF
are parallelograms. BC = 4 units and
CD = 6 units . Prove that ABDE is a
parallelogram.
PROOFS OF THEOREMS
THEOREM 1
The opposite sides and angles of a parallelogram are equal.
Required to prove:
ˆ =C
ˆ ;B
ˆ =D
ˆ
AB = CD ; AD = BC ; A
Proof
Draw parallelogram ABCD and join the diagonals AC and BD.
In Δ ABC and ΔCDA :
ˆ = Cˆ
(a)
alt ∠ s ; AB||DC
A
2
1
ˆ = Cˆ
(b)
alt ∠ s ; AD||BC
A
1
2
common side
AC = AC
SAA
∴ Δ ABC ≡ Δ CDA
∴ AB = CD and AD = BC
(c)
ˆ =D
ˆ
Also B
Similarly, it can be proved that Δ ABD ≡ Δ CDB
ˆ =C
ˆ
∴A
THEOREM 2
The diagonals of a parallelogram bisect each other.
Required to prove:
AE = EC and BE = ED
Proof
Draw parallelogram ABCD and join the
diagonals AC and BD.
In Δ ABE and Δ CDE :
ˆ = Cˆ
(a)
alt ∠ s ; AB||DC
A
2
1
ˆ =D
ˆ
(b)
alt ∠ s ; AB||DC
B
1
2
opp sides of a parm
(c)
AB = DC
SAA
∴ Δ ABE ≡ Δ CDE
∴ BE = ED and AE = EC
174
1 E
2 3
D 1
2
1
2
F
3
C
THEOREM 3
If one pair of opposite sides of a quadrilateral are equal
and parallel, then the quadrilateral is a parallelogram.
Required to prove: ABCD is a parallelogram
Proof:
In Δ ABC and ΔCDA :
ˆ =C
ˆ
A
alt ∠s ; AD||BC
(a)
1
2
(b)
common side
AC = AC
(c)
given
AD = BC
SAS
∴ Δ ABC ≡ Δ CDA
ˆ =C
ˆ
∴A
2
1
∴ AB||DC
alt ∠s =
∴ABCD is a parallelogram since the opposite sides are parallel
EXAMPLE 5
ABCD is a parallelogram with ED = BF .
Prove that BEFD is a parallelogram.
Solution
Statement
Reason
AD||BC
∴ ED||BF
opp sides of parm parallel
AED and BFC are straight lines
and ED = BF
given
∴ BEFD is a parm
one pair of opp sides equal and parallel
EXERCISE 3
(a)
Parallelograms ABCD and ABDE
are given with DF = DB .
Prove that BCFE is a parallelogram.
(b)
ABCD is a parallelogram. DM = BP and
DC = BN . Prove that:
(1)
APNM is a parallelogram.
PN = MC
(2)
(c)
PQRS is a parallelogram. PR and QS
intersect at T. QT = RM and SM = PT .
Prove that:
(1)
RTSM is a parallelogram.
QR = RN [Hint: Prove ΔPQR ≡ ΔSRN ]
(2)
175
ABCD is a parallelogram with AE = FC .
Prove that BEDF is a parallelogram.
(e)
BCDE and ABCG are parallelograms.
Prove that ABGE is a parallelogram.
(f)
AC and DB are diagonals of quadrilateral
ABCD. AO = OC and BO = OD . Prove:
(1)
Δ AOD ≡ Δ COB
(2)
Δ AOB ≡ Δ COD
(3)
ABCD is a parallelogram in two
different ways.
=
(d)
=
RECTANGLES, RHOMBUSES, SQUARES, TRAPEZIUMS AND KITES
The following is information for use in the exercise which follows. This exercise involves the
properties of rectangles, rhombuses, squares, trapeziums and kites. Familiarise yourself with
the properties of these quadrilaterals before attempting the exercise.
RECTANGLE
If ABCD is a rectangle, you may assume the following properties:
2
=
A
1
D
1
2
E
1
1
2
=
B
2
AD||BC ; AB||DC
AD = BC ; AB = DC
AE = EC = BE = ED
ˆ ;C
ˆ =A
ˆ
ˆ =B
ˆ ;D
ˆ =B
ˆ ; Cˆ = A
D
1
2
2
1
1
2
2
1
ˆ =C
ˆ =B
ˆ =D
ˆ = 90°
A
C
In order to prove that a quadrilateral is a rectangle, you will need to prove one of the
following:
(a)
The quadrilateral is a parallelogram with at least one interior angle equal to 90° .
(b)
The diagonals of the quadrilateral are equal in length and bisect each other.
RHOMBUS
If ABCD is a rhombus, you may assume the following properties:
A
1
2 1
2
E
2
2
>
>
B
=
1
=
1
D
1
C
AD||BC ; AB||DC
AD = BC = AB = DC
AE = EC ; BE = ED
ˆ =D
ˆ =B
ˆ =B
ˆ
D
1
2
1
2
ˆ =A
ˆ =C
ˆ =C
ˆ ;A
ˆ =C
ˆ ;B
ˆ =D
ˆ
A
1
2
1
2
Ê1 = 90° ; AC ⊥ BD
176
In order to prove that a quadrilateral is a rhombus, you will need to prove one of the
following:
(a)
The quadrilateral is a parallelogram with a pair of adjacent sides equal.
(b)
The quadrilateral is a parallelogram in which the diagonals bisect at right angles.
SQUARE
If ABCD is a square, you may assume the following properties:
>
1 45° >
45° 1 2
2
1
45°
E
B
=
=
1
45°
=
>
45°
D
>
=
A
45°
1
2 45° 45° 2
>>
C
AD||BC ; AB||DC
AD = BC = AB = DC
AE = EC = BE = ED
ˆ =A
ˆ =C
ˆ =C
ˆ = 45°
ˆ =D
ˆ =B
ˆ =B
ˆ =A
D
1
2
1
2
1
2
1
2
ˆ =C
ˆ =B
ˆ =D
ˆ = 90°
A
Ê1 = 90° ; AC ⊥ BD
In order to prove that a quadrilateral is a square, you will need to prove one of the following:
(a)
The quadrilateral is a parallelogram with an interior right angle and a pair of adjacent
sides equal.
(b)
The quadrilateral is a rhombus with an interior right angle.
(c)
The quadrilateral is a rhombus with equal diagonals.
TRAPEZIUM
If ABCD is a trapezium, you may assume the
following properties:
AD||BC
ˆ =C
ˆ ; D
ˆ =B
ˆ
A
2
2
1
2
In order to prove that a quadrilateral is a trapezium, you will need to prove that AD||BC.
KITE
If ABCD is a kite, you may assume the following properties:
AB = AD
BC = DC
BE = ED
ˆ =A
ˆ
A
1
2
ˆ =C
ˆ
C
1
2
ˆB = D
ˆ
Ê 2 = 90°
AC ⊥ BD
In order to prove that a quadrilateral is a kite, you will need to prove that the pairs of adjacent
sides are equal in length.
177
EXAMPLE 6
ABCD is a parallelogram.
ˆ and
BH bisects ABC
ˆ .
HC bisects BCD
ˆ = 60°. F̂ = 120° , BH||GC
ABC
and BG||HC. AD is produced to E such
that AB = DE = 30 cm . BC is produced
to F. Prove that:
(a)
BGCH is a rectangle.
(b)
DCFE is a rhombus.
Solutions
90°
30°
60°
60°
30°
Statement
(a)
Reason
BCGH is a parallelogram
ˆ = 60°
ABC
BH||GC and BG||HC ; given
ˆ =B
ˆ = 30°
∴B
1
2
ˆ
BH bisects ABC
co-int ∠ s ; AB||DC
ˆ = 120°
BCD
ˆ =C
ˆ = 60°
∴C
given
∴ Ĥ 2 = 90°
ˆ
HC bisects BCD
int ∠ s of Δ
∴BGCH is a rectangle
BGCH is a parm with an interior ∠ = 90°
F̂ = 120°
Cˆ + Cˆ = 120°
given
1
(b)
120°
1
2
2
proved
∴ F̂ = Cˆ 1 + Cˆ 2
∴ DC||EF
corr ∠s =
AD||BC
ABCD is a parallelogram
ADE and BCE are straight lines
∴ DE||CF
given
∴DCFE is a parallelogram
opp sides ||
DC = 30 cm
opp sides of parm
DC = DE = 30 cm
∴DCFE is a rhombus
DCFE is a parm with adjacent sides equal
178
EXERCISE 4
(a)
PQRS is a rhombus with Ŝ2 = 35° .
Calculate the size of all other interior
angles.
35°
(b)
Diagonals AC and BD intersect at E.
ABCD is a rectangle with AC = 10 cm
and BC = 8 cm . D̂ 2 = 20° . Calculate the
following:
ˆ ;A
ˆ ;B
ˆ ;B
ˆ ; Cˆ ; Cˆ ; D
ˆ , AD, AE and AB.
A
1
2
1
2
1
2
1
(c)
ˆ = 55° .
ABCD is a square. AEB
Calculate F̂1 .
(d)
In rhombus PQRS, PQ = 26 cm and
QS = 48 cm . Calculate the length of
PR.
(e)
In rectangle ABCD, AB = 3 x and
BC = 4 x . Find the length of AC
and BD in terms of x.
20°
3x
4x
(f)
The diagonals of parallelogram
LMNP intersect at T. LT = LM
ˆ = 120° .
and MTN
Prove that LMNP is a rectangle.
(g)
ABCD is a parallelogram. B̂1 = 40°
A
1
and Ĉ1 = 50° . Prove that ABCD
is a rhombus.
2
1
1
40°
1
B
(h)
ABCD is a square. DE = DA
and DF = DC . Prove that ACEF
is a square.
179
2
2
4
E
D
2
3
50°
2
1
C
(i)
In parallelogram PQRS, NR bisects
ˆ . SN||RT
ˆ and NS bisects PSR
SRQ
and NR||ST.
Prove that STRN is a rectangle.
(j)
ABCD is a trapezium with AD||BC.
AB = AD and BD = BC . Ĉ = 80° .
Determine the unknown angles.
A
d
B
(k)
ˆ =C
ˆ ).
ABCDE is an isosceles trapezium ( A
BD = CD and Ĉ = 60° .
Prove that:
(1)
ABDE is a parallelogram.
(2)
ΔBCD is equilateral.
ABCD is an isosceles trapezium
with  = x , BC = BP and AB = DB .
Prove that:
(1)
(2)
P̂ = x
Δ ABD ≡ Δ PDB
(m)
ABCD is a kite. The diagonals intersect at E.
BD = 30 cm , AD = 17 cm and
DC = 25 cm .
Determine:
(1)
AE
(2)
AC
(3)
B̂1 if Â1 = 20°
(n)
Circle centre M intersects circle
centre N at C and D. Prove that:
(1)
MDNC is a kite.
ˆ = MDN
ˆ by using
MCN
(2)
congruency.
180
a
b
A
80°
C
B
1
E
(l)
D
c
e
2
60°
2
1
D
C
MORE ON POLYGONS
A regular polygon is a polygon in which all the sides are equal in length and all the interior
angles are equal in size. Equilateral triangles and squares are regular polygons since their
sides and angles are equal. Pentagons, hexagons and octagons can be regular polygons if their
interior angles and sides are equal. It is important to note a rhombus is not a regular polygon
since its interior angles are not all equal. A rectangle is not a regular polygon since its sides
are not equal in length. Polygons that are not regular are called irregular polygons.
The formula for calculating the sum of the interior angles of a polygon of n sides is given by
the formula: 180°( n − 2)
180°( n − 2)
The size of an interior angle of a regular polygon is given by the formula:
n
Here are the regular polygons.
Polygon
3 sides
60°
60°
Regular polygon
Equilateral
Triangle
Interior angles
The sum of the interior angles:
180°(3 − 2) = 180°
The size of an interior angle:
180°(3 − 2)
= 60°
3
Square
The sum of the interior angles:
180°(4 − 2) = 360°
The size of an interior angle:
180°(4 − 2)
= 90°
4
60°
4 sides
5 sides
Pentagon
The sum of the interior angles:
180°(5 − 2) = 540° .
The size of an interior angle:
180°(5 − 2)
= 108°
5
The five triangles in the pentagon
are congruent isosceles triangles.
The five angles at the centre each
equal 72° since 72°× 5 = 360° .
The base angles of each triangle
all equal 54° .
Hexagon
The sum of the interior angles:
180°(6 − 2) = 720° .
The size of an interior angle:
180°(6 − 2)
= 120°
6
The six triangles in the hexagon
are congruent equilateral triangles.
The six angles at the centre each
equal 60° since 60°× 6 = 360° .
108°
108°
108°
108° 108°
54° 54°
54°
54°
72° 72°
54°
54°
72° 72°
72°
54°
54°
54° 54°
6 sides
120° 120°
120°
120°
120° 120°
60° 60°
60°
60°
60°
60° 60° 60° 60°
60° 60° 60° 60°
60°
60°
60°
60° 60°
181
8 sides
Octagon
135° 135°
135°
135°
135°
135°
135° 135°
67,5° 67,5°
67,5°
67,5°
67,5°
67,5°
45°
67,5°
45°
45°
67,5°
45°
45°
67,5°
45°
45°
45°
67,5°
67,5°
67,5°
The sum of the interior angles:
180°(8 − 2) = 1 080° .
The size of an interior angle:
180°(8 − 2)
= 135°
8
The eight triangles in the octagon
are congruent isosceles triangles.
The eight angles at the centre each
equal 45° since 45°× 8 = 360° .
The base angles of each triangle
all equal 67,5° .
67,5°
67,5°
67,5° 67,5°
Note:
Scalene and isosceles triangles, rectangles, kites and trapeziums are irregular polygons since
their sides are not equal in length.
Rhombuses are irregular polygons since their interior angles are not equal.
EXERCISE 5
(a)
ABCD is a polygon with four sides.
(1)
Calculate the value of x.
(2)
Hence show that ABCD is a
trapezium.
8x + 2
4x + 2
x−2
5x − 2
A
(b)
ABCD is a pentagon made up of five equal
sides and five equal interior angles.
Calculate the size of θ, β and α .
B
α
β
θ
2
E
D
C
(c)
1
In polygon ABCDE, Ĉ = 90° , BC = CD
and ABDE is a parallelogram.
Use TWO different methods to
determine the value of x.
60°
(d)
Using the information provided on
the diagram, determine β .
β
θ
113°
θ
182
132°
THE MIDPOINT THEOREM IN A TRIANGLE
The Midpoint Theorem can be stated in the following two ways:
If AD = DB and AE = EC ,
1
then DE||BC and DE = BC
2
If AD = DB and DE||BC,
1
then AE = EC and DE = BC .
2
EXAMPLE 7
In ΔABC , AD = DB and AE = EC . DE is produced to F.
DB||FC and BC = 32 mm .
(a)
Prove that DBCF is a parallelogram.
(b)
Calculate the length of DE.
Solutions
Statement
(a)
(b)
Reason
AD = DB and AE = EC
∴ DE||BC
given
But DEF is a straight line
∴ DF||BC
DE is produced to F
But DB||FC
given
∴DBCF is a parallelogram
opposite sides parallel
BC = 32 mm
given
∴ DE = 16 mm
midpoint theorem
midpoint theorem
183
EXERCISE 6
(a)
In ΔACD , AB = BC , GE = 15 cm ,
AF = FE = ED .
Calculate the length of CE.
(b)
In ΔABC , AE = EB and EF||BC.
In ΔACD , FG||CD.
Prove that AG = GD .
(c)
In ΔDEF , DS = SE , EU = EF and ST||EF.
Prove that SEUT is a parallelogram.
(d)
PQRS is a kite. A and B are the midpoints
of PQ and PS respectively.
QD = DR and SC = CR . Let AB = x
Prove that ABCD is a parallelogram
CONSOLIDATION AND EXTENSION EXERCISE
(a)
In Δ PQR , PQ = PR and STRE is a
ˆ.
parallelogram. Q̂ = x and P̂ = 2Q
Calculate the sizes of the angles of
STRE.
(b)
ABCD is a parallelogram. FD = DC
and DE = 2DO .
DO = x . Prove that BCEF is a
parallelogram.
184
(c)
ABCD is a parallelogram. BE ⊥ AC
and DF ⊥ AC . Prove that EBFD is
a parallelogram.
(d)
PQRS is a square. The diagonals intersect
at E. PA = BS . Prove that ΔAEB is an
isosceles triangle.
(e)
ΔPQT is inscribed in a circle. AO||QR,
PA = AQ and PB = BT .
Prove that:
(1)
AB||QT
(2)
O is the centre of the circle if
PR is a diameter.
(3)
BORT is a trapezium.
(f)
FDCE is a parallelogram. CE is produced
to A such that CE = EA and CD = DB .
Prove that ΔBDF ≡ ΔFEA
(g)
ABCD is a rhombus. Diagonals intersect
at E. EF = FA and EG = GA . Prove that
AGEF is a rhombus.
185
(h)
PQRS is a parallelogram. PQ = PE ,
ˆ = x.
QE = QR , ER = SR and PQE
ˆ .
Calculate the size of QER
(i)
3y
ˆ = 3y
ABCD is a rhombus. DEC
ˆ .
and Ĉ = y . Prove that EC bisects ACD
y
(j)
LMNP is a square. LE = EM and
MF = FN . L̂1 = x . Prove that:
ΔLMF ≡ ΔPLE
(a)
Ĝ1 = 90°
(b)
(k)
ˆ = x and FB = BC .
AFCE is a parallelogram. AB||DC and AF||BD. F̂ = FAB
Prove that:
(l)
(1)
ABCD is a rhombus
(2)
Â1 = 90° − x
ABCD is a trapezium, ABGH is a square,
ˆ .
AG||DJ and B̂2 = G
1
Prove that:
(1)
AEFD is a rectangle
(2)
BGJC is a trapezium
186
(m)
(n)
(o)
PSRQ is a square.
Diagonal PR and line SVT
intersect at V and Ŝ2 = 40°
(1)
Calculate V̂1
(2)
Prove that
PR
= 2
PS
40°
ABCD is a parallelogram with
AE = ED and AD = a .
GE = EF = BG = r and AB = b .
1
DF =
4r 2 − a 2
2
(1)
Prove that ABCD is a rectangle.
(2)
Express AG in terms of b and r.
(3)
Prove that r =
a 2 + 4b 2
8b
In the diagram, BCDE is a rhombus.
The diagonals intersect at O and
BD is produced to A. AE is joined.
AE ⊥ ED
Prove that:
(p)
(1)
AE 2 = a 2 + 2ab + d 2
(2)
AE 2 = a 2 + 4ab + 4b 2 + d 2
ABCD is a parallelogram.
GF = FH = x and AD = DE .
ADE is a straight line.
Prove that:
(1)
DECB is a parallelogram.
(2)
AG = 2GF
(3)
1
FH = AF
3
187
CHAPTER 8
ANALYTICAL GEOMETRY
Analytical Geometry is the study of Geometry, using the Cartesian plane. It is an algebraic
approach to the study of Geometry. In this chapter, we will address the following concepts:
• The distance between two points (length of a line segment).
• The midpoint of a line segment.
• The gradient of a line.
THE DISTANCE BETWEEN TWO POINTS (LENGTH OF A LINE SEGMENT)
Suppose that we wish to calculate the length of line segment AB, with endpoints A(−2 ; − 2)
and B(3 ; 2).
Consider the diagram below. The movement from A to B has been indicated. A right-angled
triangle ABC is formed, with lengths AC = 4 (4 units up) and CB = 5 (5 units right).
y
C
The theorem of Pythgoras can now be used to calculate
the length of AB:
5 units right B(3 ; 2)
∴ AB2 = BC2 + AC 2
4 units up
∴ AB2 = (horizontal movement) 2 + (vertical movement)2
∴ AB2 = 52 + 42 = 41
x
∴ AB = 41 ≈ 6, 40
A( − 2 ; − 2)
We can generalise this concept to create what is known as the “Distance Formula”.
Consider any two points A( xA ; yA ) and B( xB ; yB ).
y
C(xA ; yB )
From the diagram alongside:
BC = horizontal movement = xB − xA
B(xB ; yB )
AC = vertical movement = yB − yA
∴ AB2 = BC2 + AC2 Pythagoras
∴ AB2 =(xB − xA )2 + ( yB − yA )2
x
∴ AB = ( xB − xA )2 + ( yB − yA )2
A(xA ; yA )
The formula to calculate the length of a line segment between points A and B,
with A( xA ; yA ) and B( xB ; yB ), is:
AB2 = ( xB − xA ) 2 + ( yB − yA ) 2
or
AB = ( xB − xA ) 2 + ( yB − yA )2
188
EXAMPLE 1
Calculate the lengths of line segments
AB and CD in the given diagram.
B(5 ; 7)
C( − 5 ; 6)
A(2 ; 3)
D( − 2 ;1)
Solutions
(a)
AB2 = ( xB − xA )2 + ( y B − yA )2
(b)
CD2 = ( xD − xC )2 + ( y D − yC )2
AB2 = (5 − 2) 2 + (7 − 3) 2
CD 2 = (−2 − (−5)) 2 + (1 − 6) 2
AB2 = (3) 2 + (4) 2
CD 2 = (3)2 + (−5) 2
AB2 = 25
CD 2 = 34
AB = 25 = 5 units
CD = 34 ≈ 5,83 units
APPLICATIONS OF THE DISTANCE FORMULA
EXAMPLE 2
In the diagram, the vertices of ΔABC
are A(2 ; 3) , B(5 ; 7) and C(−2 ; 6).
Show that ΔABC is an isosceles triangle.
(a)
(b)
Calculate the perimeter of ΔABC
correct to one decimal place.
B(5 ; 7)
C( − 2 ; 6)
A(2 ; 3)
Solutions
(a)
(b)
We can show that ΔABC is an isosceles triangle by proving two sides equal. From the
diagram above, the obvious choice is to prove that AB = AC .
AB2 = ( xB − xA ) 2 + ( yB − yA ) 2
AC2 = ( xC − xA ) 2 + ( yC − yA ) 2
AB2 = (5 − 2) 2 + (7 − 3) 2
AC2 = (−2 − 2) 2 + (6 − 3) 2
AB2 = 25
AC2 = 25
AB = 25 = 5 units
AC = 25 = 5 units
∴ AB = AC
The perimeter of ΔABC is the sum of its three sides:
BC2 = ( xC − xB ) 2 + ( yC − yB )2
BC2 = (−2 − 5)2 + (6 − 7)2
BC2 = 50
BC = 50
Perimeter = AB + AC + BC
∴ Perimeter = 5 + 5 + 50 = 17,071... ≈ 17,1 units
189
EXAMPLE 3
Calculate the possible values of k, if the distance between A and B is 5 units, where A(2 ; 5)
and B(−1; k ).
Solution
AB2 = ( xB − xA ) 2 + ( yB − yA ) 2 , with AB = 5
∴ 52 = ( −1 − 2) 2 + ( k − 5) 2
[substitute into the distance formula]
∴ 25 = 9 + k 2 − 10k + 25
[quadratic equation]
∴ 0 = k 2 − 10k + 9
∴ 0 = ( k − 9)( k − 1)
[solve by getting one side = 0]
∴ k = 9 or k = 1
EXAMPLE 4
(a)
Show that the point Q( − 6 ;1) is equidistant from the points P( − 4 ; 5) and R( − 2 ; 3).
(b)
The point T(x ;1) is equidistant from the points A( − 2 ; − 1) and N(1; 2) . Determine
the value of x.
Solutions
(a)
We are required to show that QP = QR.
QP 2 = ( xP − xQ )2 + ( yP − yQ )2
QR 2 = ( xR − xQ ) 2 + ( yR − yQ ) 2
QP 2 = ( −4 − (−6)) 2 + (5 − 1) 2
QR 2 = (−2 − (−6)) 2 + (3 − 1) 2
QP 2 = 20
QR 2 = 20
QP = 20
QR = 20
∴ QP = QR and Q is therefore equidistant from P and R.
(b)
It is given that TA = TN . Therefore TA2 = TN2 .
∴ ( xA − xT )2 + ( yA − yT )2 = ( xN − xT )2 + ( yN − yT )2
∴ (−2 − x)2 + (−1 − 1)2 = (1 − x)2 + (2 − 1)2
∴ 4 + 4 x + x2 + 4 = 1 − 2 x + x2 + 1
∴ 6 x = −6
∴ x = −1
EXERCISE 1
(a)
D( − 1; 4)
Calculate the lengths of the line segments
in the given diagram.
C( − 5 ;1)
A(2 ; 3)
F(2 ;1)
B(5 ;1)
G(1; − 1)
E( − 5 ; − 2)
190
H(5 ; − 4)
(b)
In the given diagram, two triangles
have been drawn.
(1)
Show that ΔABC is an
isosceles triangle.
(2)
Determine the perimeter
of ΔABD.
C( − 5 ; 7)
A(2 ; 6)
B( − 2 ; 3)
D(5 ; 0)
(c)
Show that C(2 ; 3) is equidistant from the points A(3 ; 6) and B( −1 ; 4) .
(d)
C is the point (1; − 2). The point D lies in the second quadrant and has coordinates
( x ; 5). If the length of CD is
(e)
53 units, determine the value of x.
Given the points P( −3 ; 2), Q(2 ; 7) and R( −10 ; y ). Determine the values of y if P is
equidistant from Q and R.
THE MIDPOINT OF A LINE SEGMENT
The midpoint of a line segment is the halfway mark on the line segment. Consider the
numbers 1 and 7 for example. Halfway between 1 and 7 is 4. How do we get to 4? One way is
to take the difference between 1 and 7 (which is 6), half that (which is 3), and then add this 3
to the 1 to get to 4. (You could also subtract 3 from 7 to get to 4).
6 units
The diagram on the right illustrates this approach.
3 units
3 units
1
4
7
7 +1 8
= =4
2
2
A quicker approach is to add the two end values
and then divide the answer by 2.
1
4
7
This is basically working out the average of the two end values.
If we apply this concept to a line segment joining two points on the Cartesian plane, we can
easily find the midpoint of the line segment by calculating the average of the x-values and the
average of the y-values.
M(xM ; yM ) = M ( average of the x-values ; average of the y-values )
 x + x y + yB 
∴ M( xM ; yM ) = M  A B ; A

2
 2

M(xM ; yM )
A(xA ; yA )
The formula to calculate the midpoint of a line segment between points A and B,
 x + x y + yB 
M( xM ; yM ) = M  A B ; A
with A( xA ; yA ) and B( xB ; yB ), is:

2
 2

191
B(xB ; yB )
EXAMPLE 5
Determine the coordinates of M, if M is the midpoint of line segment AB, where A(2 ;1) and
B(8 ; 3) .
B(8 ; 3)
M
Solution
 x + xB yA + yB 
M A
;

2 
 2
 2 + 8 1+ 3 
= M
;

2 
 2
= M(5 ; 2)
A(2 ;1)
EXAMPLE 6
Calculate the midpoints of line segments AB and CD in the given sketch.
Solutions
Midpoint of AB is M:
 x + xB yA + yB 
M( xM ; yM ) = M  A
;

2
 2

B(5 ; 7)
C( − 5 ; 6)
M
N
 1+ 5 3 + 7 
;
∴M 

2 
 2
∴ M(3 ; 5)
A(1; 3)
D( − 1; 2)
Midpoint of CD is N:
y + yD 
x +x
N( xN ; yN ) = N  C D ; C

2
 2

 −5 + ( −1) 6 + 2 
;
∴N

2
2 

∴ N(−3 ; 4)
APPLICATIONS OF THE MIDPOINT FORMULA
EXAMPLE 7
Determine the values of x and y if M(5 ; 2) is the midpoint of the line segment joining the
points A( x ;1) and B(8 ; y) .
Solution
xA + xB
2
x +8
∴5 =
2
∴10 = x + 8
∴x = 2
xM =
M(5 ; 2)
yA + yB
2
1+ y
∴2 =
2
∴4 = 1+ y
∴y =3
A(x ;1)
yM =
192
B(8 ; y)
EXAMPLE 8
y
The following sketch shows parallelogram ABCD,
with P the point where the diagonals intersect.
D
A( − 1; 3)
(a)
Determine the coordinates of P, giving a reason.
(b)
Determine the coordinates of D.
P
C(2 ;1)
x
B( − 4 ; − 1)
Solutions
(a)
(b)
P is the midpoint of both AC and BD, because the diagonals of a parallelogram bisect
each other.
 x + xC yA + yC 
P( xP ; yP ) = P  A
;

2
2


 −1 + 2 3 + 1 
∴P
;

2 
 2
1 
∴P ; 2
2 
 x + xD yB + yD 
P( xP ; yP ) = P  B
;

2
2


x +x
y + yD
∴ xP = B D and yP = B
2
2
−1 + yD
1 −4 + xD
and 2 =
∴ =
2
2
2
4 = −1 + yD
∴1 = −4 + xD and
∴ xD = 5
and
yD = 5
∴ D(5 ; 5)
EXERCISE 2
(a)
y
Determine the midpoints of the given
line segments.
Use the midpoint formula.
D( − 1; 3)
A(2 ; 3)
F(2 ;1)
C( − 5 ;1)
B(6 ;1)
x
G(1; − 1)
E( − 5 ; − 2)
H(5 ; − 4)
(b)
A circle has diameter AB with endpoints A( − 1; 4) and B(5 ; − 2).
(1)
Determine the centre of the circle.
(2)
Determine the radius of the circle.
193
(c)
Answer the following questions. You may want to draw a diagram to help you
visualise the scenario.
(1)
If M(−3 ; 2) is the midpoint of the line segment joining the points
A( x ;1) and B( −1 ; y ) , calculate the values of x and y.
(2)
If M(−1; 7) is the midpoint of the line segment joining the points
A( x ; 6) and B(2 ; y ) , calculate the values of x and y.
(3)
If M(−1; − 5) is the midpoint of the line segment joining the points
A( x ; y ) and B(−6 ; − 3) , calculate the values of x and y.
(d)
Given below is rhombus FINE. Q is the point where the diagonals intersect.
Determine the coordinates of F.
y
E(3 ; 4)
F
Q
x
N(8 ; − 1)
I(1; − 2)
(e)
A( − 2 ; 3), B(x ; y ), C(1; 4) and D( − 1; 2) are the vertices of a quadrilateral. Find
B(x ; y ) if ABCD is a parallelogram.
THE GRADIENT OF A LINE SEGMENT
Gradient (or slope) measures the steepness and direction of a line. A line can either slant up
(gradient is positive), slant down (gradient is negative), be horizontal (gradient is zero) or be
vertical (gradient is undefined). The symbol used for gradient is m.
In Grade 9 we calculated the gradient of a line using the concept of “rise over run”. In other
change in y -values
vertical movement
words: Gradient =
.
=
change in x -values horizontal movement
This can easily be translated into a formula. For any two points A( xA ; yA ) and B( xB ; yB ) :
Vertical movement = yB − yA and Horizontal movement = xB − xA .
y
ver ical movement
horizontal movement B(x ; y )
B
B
x
A(xA ; yA )
194
A formula to calculate the gradient of a line joining two points A and B, with
y − yA
A( xA ; yA ) and B( xB ; yB ), is:
mAB = B
xB − xA
EXAMPLE 9
Calculate the gradients of the following lines using the formula for gradient.
y
D( − 1; 3)
A(2 ; 3)
F(2 ;1)
C( − 5 ;1)
B(6 ;1)
x
G(1; − 1)
E( − 5 ; − 2)
H(5 ; − 4)
Solutions
yB − yA 1 − 3 −2
1
=
=
=−
xB − xA 6 − 2 4
2
y −y
3 −1
2 1
mCD = D C =
= =
xD − xC −1 − (−5) 4 2
y −y
1 − (−2) 3
mEF = F E =
=
xF − xE 2 − (−5) 7
y − yG −4 − (−1) −3
3
mGH = H
=
=
=−
xH − xG
5 −1
4
4
mAB =
(slopes down from left to right)
(slopes up from left to right)
(slopes up from left to right)
(slopes down from left to right)
GRADIENTS OF HORIZONTAL AND VERTICAL LINES
Between any two points on a horizontal line there is no vertical movement (the vertical
movement is zero). Thers is only a horizontal movement.
The gradient of a horizontal line is always zero.
∴ gradient horizontal line =
change in y values
0
=
= 0.
change in x values horizontal movement
Between any two points on a vertical line there is no horizontal movement (the horizontal
movement is zero). There is only a vertical movement.
The gradient of a vertical line is always undefined.
change in y values vertical movement
∴ gradient vertical line =
=
which is undefined.
change in x values
0
195
APPLICATIONS OF GRADIENT
Parallel lines
Parallel lines slope in the exactly the same direction and will therefore never intersect.
Differently stated: Lines that are parallel have equal gradients.
y
B(3 ; 5)
4
D(5 ; 2)
4
3
3
x
A( − 5 ; − 1)
C( − 3 ; − 4)
For any pair of parallel lines AB and CD:
mAB = mCD
EXAMPLE 10
Given are the points A( − 1; 5), B( −2 ; 3), C(9 ;10) and D(5 ; 2). Show that AB||CD.
Solution
yB − yA
3−5
−2
−2
=
=
=
=2
xB − xA −2 − (−1) −2 + 1 −1
∴ mAB = mCD
∴ AB||CD
mAB =
mCD =
yD − yC 2 − 10 −8
=
=
=2
xD − xC
5 − 9 −4
Collinear points
y
Points are said to be collinear when they
lie on the same line. Refer to the diagram.
A, B and C lie on the same line and are
therefore collinear. This implies that the
gradients between each pair of points
are the same.
6
C(4 ; 3)
4
3
2
A( − 5 ; − 3)
196
x
.
6
B( − 2 ; − 1)
9
When points A, B and C are collinear:
mAB = mAC = mBC
In other words: mAB = mAC and mAB = mBC and mAC = mBC
EXAMPLE 11
Show that the points A, B and C are collinear if the coordinates of the points are:
A(2 ; − 2), B(1;1) and C( − 1; 7) .
Solution
We will consider the gradients of AB and BC, but any other pair could have been used.
y − yB 7 − 1
y − yA 1 − (−2) 3
6
=
=
= −3
=
=
= −3
mAB = B
mBC = C
and
−1
xB − xA
xC − xB −1 − 1 −2
1− 2
∴ mAB = mBC
Therefore A, B and C are collinear.
D( − 4 ; 6)
Perpendicular lines
Perpendicular lines intersect at a 90°
angle. The gradients of perpendicular
lines have a particular property.
Consider the diagram on the right.
4
Firstly:
The gradients of the lines have opposite
3
signs. AB has a positive gradient whereas
CD has a negative gradient.
Secondly:
The gradients (ignoring signs) are reciprocals
A( − 5 ; − 1)
of one another. In other words, the horizontal
movement of AB is the vertical movement of CD and vice versa.
y
B(3 ; 5)
3
4
C(2 ; − 2)
This can be summarised by the following relationship: The product of the gradients of
AB and CD will equal −1 when AB is perpendicular to CD.
For any pair of perpendicular lines AB and CD: mAB × mCD = −1
EXAMPLE 12
Given are the points A(3 ; − 3), B(6 ; − 7), C(−5 ; 0) and D( −1; 3).
Show that AB is perpendicular to CD.
Solution
mAB =
yB − yA −7 − (−3) −7 + 3 −4
=
=
=
xB − xA
6−3
3
3
mCD =
yD − yC
3−0
3
3
=
=
=
xD − xC −1 − (−5) −1 + 5 4
∴ mAB × mCD =
−4 3
× = −1
3 4
∴ AB ⊥ CD
197
x
y
EXAMPLE 13
A
In the diagram, ABCD is a rhombus.
F is the point where the diagonals intersect.
(a)
Determine the value of q.
(b)
Calculate the area of ΔCFD .
(c)
What is the area of the rhombus?
x
F(2 ; − 1)
C(3 ; − 3)
D( − 4 ; q )
Solutions
(a)
B
AC ⊥ BD because the diagonals of a rhombus are perpendicular to each other.
∴ mAC × mBD = −1 which implies that mFC × mFD = −1
∴ mFC =
yC − yF −3 − (−1)
y − yF q − (−1) q + 1
=
= −2 and mFD = D
=
=
−4 − 2
−6
xC − xF
xD − xF
3− 2
∴ But mFD =
1
2
q +1 1
=
−6 2
∴ 2q + 2 = −6
∴ q = −4
1
(−2 × = −1)
2
∴
(b)
(c)
cross multiplication
1
1
Area of ΔCFD = b × h = (FD)(FC)
2
2
2
2
FD = ( xD − xF ) + ( yD − yF ) 2
FC 2 = ( xC − xF ) 2 + ( yC − yF ) 2
FD 2 = (−4 − 2) 2 + (−4 − (−1)) 2
FC 2 = (3 − 2) 2 + ( −3 − ( −1)) 2
FD 2 = 45
FC2 = 5
FD = 45
FC = 5
1
1
∴ Area of ΔCFD = (FD)(FC) = ( 45)( 5) = 7,5 units 2
2
2
Area of Rhombus = 4 × 7,5 = 30 units 2
EXERCISE 3
(a)
Calculate the gradients of the lines joining the following points.
A(1; 3) and B(5 ; 7)
(2)
A(1; 3) and B(−5 ; − 7)
(1)
(3)
A(−1 ; − 3) and B(−5 ; − 7)
(4)
A(−1; 3) and B(5 ; − 7)
(b)
Calculate the gradients of the line
segments in the given diagram.
y D(1; 4)
A(2 ; 3)
E( − 1;1)
G( − 2 ; − 1)
C( − 6 ; − 1)
x
H(5 ; − 1)
F( − 1; − 3)
198
B(4 ; − 3)
(c)
Determine whether line segments AB and CD are parallel, perpendicular or neither, in
each of the following cases.
A(−1; − 3), B(2 ;1) and C(4 ; − 1), D(7 ; 3) .
(1)
(2)
A(1; − 3), B(2 ;1) and C(4 ; − 1), D(7 ; 3)
(3)
A(1; − 3), B(2 ;1) and C( −3 ; 1), D(1; 0)
(d)
(1)
(2)
Line segment AB is parallel to line segment CD. A( −5 ; − 1) and
B( −3 ; a ) are points on AB. C(−4 ; − 3) and D(−1; 3) are points on CD.
Calculate the value of a.
Line segment AB is perpendicular to line segment CD. A(−5 ; 2) and
B(b ; − 1) are points on AB. C(−4 ; − 3) and D(−1; 3) are points on CD.
Calculate the value of b.
(e)
Show that points F, R and N are collinear if F(3 ; 2), R(4 ; − 2) and N(7 ; − 14).
(f)
Calculate the value(s) of x if R ( x ; 4), U( x − 1; x + 4) and N(0 ;13), are collinear.
(g)
A(3 ; 4), B( − 1; 7), C(x ; − 1) and D(1; 8) are points on the Cartesian plane.
Calculate the value of x in each case if:
AB ⊥ CD
(1)
AB||CD
(2)
(3)
B, C and D are collinear
(h)
Determine the x-intercept A, and the
y-intercept B of lines LI and VE respectively.
y
L(−6 ; 5)
E(1; 7)
B
A
I(8 ; − 2)
V(−4 ; − 3)
(i)
B
Rhombus ABCD is drawn with
A( − 1; − 2) and C(3 ; 4).
Determine the gradient of BD.
C(3 ; 4)
A( − 1; − 2)
(j)
In the diagram below, a circle with centre
P is drawn. A, B and C are points on
C( − 2 ; 7)
the circle, with AC the diameter.
(1)
Determine the length of the radius.
(2)
Determine the coordinates
of P.
(3)
Show that AB ⊥ BC .
(4)
Hence determine the area
B(−1; 2)
of ΔABC.
199
x
D
y
.
P(a ; b)
A(4 ; 3)
x
MORE ON QUADRILATERALS
We will now discuss some of the most effective ways to show that a quadrilateral is a
parallelogram, rhombus, rectangle or square, using the methods of Analytical Geometry.
Proving that a quadrilateral is a parallelogram
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
To demonstrate this on a Cartesian plane we will show that the diagonals share the same
midpoint.
EXAMPLE 14
y
I(5 ; 5)
Quadrilateral TIME is drawn on the Cartesian plane.
Prove that it is a parallelogram.
Solution
Midpoint of IE:
T(0 ; 2)
 xE + xI yE + yI 
;


2 
 2
 −2 + 5 −3 + 5 
;
=

2 
 2
P
x
M(3 ; 0)
3 
=  ;1
2 
 x + x y + yT 
Midpoint of TM:  M T ; M

2 
 2
E( − 2 ; − 3)
Take note:
Proving that both pairs of opposite
sides are parallel, OR
that both pairs of opposite sides are
equal, OR
that one pair of opposite sides are
parallel and equal could have also
been used.
 3+ 0 0 + 2 
=
;

2 
 2
3 
=  ;1
2 
3 
∴ P  ;1 is the midpoint of both diagonals.
2 
Quadrilateral TIME is a parallelogram, because its diagonals bisect each other.
Proving that a quadrilateral is a rhombus
A rhombus can be defined as a parallelogram in which the adjacent sides are equal, OR as a
parallogram in which the diagonals are perpendicular (i.e. a parallelogram in which the
diagonals bisect at a 90° angle). The latter will be quicker to prove and will be the approach
that we will use.
EXAMPLE 15
H(4 ; 0), E( −1 ; 2), A( −3 ; 7) and R(2 ; 5) are the vertices
of quadrilateral HEAR.
(a)
Prove that quadrilateral HEAR is a parallelogram.
(b)
Prove that quadrilateral HEAR is a rhombus.
200
A( − 3 ; 7)
y
R(2 ; 5)
M
E( − 1; 2)
H(4 ; 0)
x
Solutions
(a)
We are required to prove that diagonals HA and ER share the same midpoint.
 x + x y + yH 
 xE + xR yE + yR 
Midpoint of HA:  A H ; A
;
 Midpoint of ER: 

2 
2 
 2
 2
 −3 + 4 7 + 0 
 −1 + 2 2 + 5 
;
;
=
=


2 
2 
 2
 2
1 7
= ; 
2 2
1 7
= ; 
2 2
1 7
∴ M  ;  is the midpoint of both diagonals.
2 2
Quadrilateral HEAR is a parallelogram because its diagonals bisect.
(b)
We are required to prove that the diagonals bisect at a 90° angle. We have already
proven that they bisect. We need to prove that HA ⊥ ER , in other words prove that
mHA × mER = −1.
y − yH
y − yE
0−7
5−2
=
= −1 and mER = R
=
=1
mHA = A
xA − xH 4 − (−3)
xR − xE 2 − (−1)
∴ mHA × mER = (−1) × (1) = −1
∴ HA ⊥ ER
Quadrilateral HEAR is a rhombus because its diagonals bisect at 90°.
Proving that a quadrilateral is a rectangle
A rectangle can be defined as a parallelogram of which the diagonals are equal, OR as a
parallelogram with one interior angle equal to 90°. The latter will be quicker to prove and will
be the approach that we shall use.
EXAMPLE 16
y
A(4 ; 9)
Quadrilateral LAND is shown in the given diagram.
(a)
Prove that quadrilateral LAND is a
parallelogram.
(b)
Prove that quadrilateral LAND is a
rectangle.
N(6 ; 6)
M
L( − 5 ; 3)
Solutions
(a)
D( − 3 ; 0)
We are required to prove that diagonals
NL and DA share the same midpoint.
 x + x y + yL 
Midpoint of NL:  N L ; N

2 
 2
 6 + (−5) 6 + 3 
;
=

2 
 2
 x + x y + yA 
Midpoint of DA:  D A ; D

2 
 2
 −3 + 4 0 + 9 
;
=

2 
 2
1 9
= ; 
2 2
1 9
= ; 
2 2
201
x
1 9
∴ M  ;  is the midpoint of both diagonals.
2 2
Quadrilateral LAND is a parallelogram, because its diagonals bisect each other.
(b)
We are required to prove that one interior angle of parallelogram LAND is 90°. We
need to prove that LA ⊥ LD , in other words prove that mLA × mLD = −1.
y − yL
y − yL
9−3
2
0−3
3
mLA = A
=
=
and mLD = D
=
=−
xA − xL 4 − ( −5) 3
xD − xL −3 − ( −5)
2
2  3
∴ mLA × mLD =   ×  −  = −1.
3  2
∴ LA ⊥ LD
Quadrilateral LAND is a rectangle because it is a parallelogram with one interior angle
equal to 90°.
Proving that a quadrilateral is a square, trapezium or kite
Square
Several approaches can be used to prove that a quadrilateral is a square. We will discuss two
approaches.
First approach:
A square is a rhombus with one interior angle 90°.
First prove that the quadrilateral is a rhombus and then prove one
interior angle 90°.
Second approach:
A square is a rectangle of which the adjacent sides are equal.
First prove that the quadrilateral is a rectangle, and then prove one pair
of adjacent sides equal.
Trapezium
Simply prove that one pair of opposite sides are parallel.
Kite
Simply prove that two pairs of adjacent sides are equal OR prove that one diagonal is bisected
by the other at 90°.
EXERCISE 4
(a)
Quadrilateral DEFG is formed by the points D( − 5 ; 3), E(3 ; 5), F(2 ;1) and
G( − 6 ; − 1) . Show that DEFG is a parallelogram.
(b)
Given: A( − 4 ; 3), B(3 ; 4), C(8 ; − 1) and D(1; − 2) . Show that ABCD is a rhombus.
(Hint: First show that it is a parallelogram)
(c)
Given: A(0 ; − 3), B(4 ; 0), C( − 2 ; 8) and D( − 6 ; 5) . Show that ABCD is a rectangle.
(d)
M( − 3 ; 2), N(3 ; 6), O(9 ; − 2) and P(3 ; − 6) are the points of quadrilateral MNOP.
Show that:
(1) MNOP is a parallelogram.
(2)
MNOP is not a rectangle.
(e)
A( − 2 ; 3), B(x ; y ), C(1; 4) and D( − 1; 2) are the vertices of a quadrilateral. Determine
B(x ; y ) if ACDB is a parallelogram. Drawing a diagram will be useful.
202
(f)
In the following sketch ABCD is a parallelogram.
The diagonals AC and BD intersect in P( − 1,5 ; 3).
D is a point on the x-axis. The gradient of BD is 6.
(1)
Determine the coordinates of C and D.
(2)
Use Analytical techniques to determine
whether or not ABCD is a rhombus.
B
C
P
A(3 ; 2)
x
D
CONSOLIDATION AND EXTENSION EXERCISE
(a)
A(4 ; 3) and B(10 ; 5) are two points on a Cartesian plane.
(1)
Calculate the length of AB. Round off your answer to one decimal place.
(2)
Determine the coordinates of M, the midpoint of AB.
(3)
Determine the coordinates of P if B is the midpoint of AP.
(b)
ΔABC has coordinates A( −4 ; 2), B(1 ; 2) and C( −1 ; 6)
(1)
Determine the perimeter of ΔABC
(2)
What kind of triangle is ΔABC?
(3)
Explain why ΔABC cannot be right-angled. Show all workings.
(c)
Points F( −3 ; − 4), A(1 ; b) and N(3 ; 5) are collinear. Determine the value of b.
(d)
Refer to the diagram alongside.
(1)
Determine the gradient of AC.
(2)
Determine the length of BC
(one decimal place).
(3)
Determine the coordinates of M,
the midpoint of line AB.
(4)
Determine the coordinates of D.
(5)
Show that BD ⊥ AC .
C(1,5 ; 5)
B(4 ; 0)
A( −3 ; − 4)
(e)
(f)
Refer to figure alongside. RU doesn’t necessarily
go through the origin. U is the midpoint of AE
and RU ⊥ AE .
(1)
Determine the coordinates of U.
(2)
Determine the gradient of RU.
(3)
Show that the origin lies on
the line RU.
E( − 10 ; − 5)
(4)
Calculate the area of ΔARE .
On the Cartesian plane below M( a ; 1) is
the midpoint of line AB with A( − 2 ; 4)
and B(5 ; k ) . Point D lies on the x-axis.
Determine the values of a and k.
Determine the coordinates of D.
Show all calculations.
203
U
A(−2 ; 4)
y
R(8 ; − 6)
M(a ;1)
The length of MD is 21,25 .
This graph is not sketched according to scale.
(1)
(2)
A(2 ;11)
21, 25
D
x
B(5 ; k )
(g)
E( − 7 ;15)
The sketch alongside is a circle with centre V
through points S, A and E.
(1)
Determine the length of the radius
of the circle.
(2)
Determine the coordinates of V.
S( − 15 ; 3)
V
(3)
Show that SE ⊥ SA .
y
.
x
A(3 ; − 9)
(h)
Use analytical methods to show that
PQRS is a parallelogram if P( −3 ; 2) ,
Q(3 ; 6) , R(10 ; − 1) and S(4 ; − 5) .
Q(3 ; 6)
P( − 3 ; 2)
R(10 ; − 1)
S(4 ; − 5)
(i)
L( − 1 ; − 1) , M( − 2 ; 4) , N(x ; y ) and P(4 ; 0) are the vertices of parallelogram
LMNP.
(1)
Determine the coordinates of N.
(2)
Show that MP ⊥ LN and state what type of quadrilateral LMNP is
other than a parallelogram.
(3)
Show that LMNP is a square.
(j)
SRQP is a parallelogram. S lies on the y-axis.
Determine the value of b and the coordinates of S.
y
R(6 ; 8)
Q(b ;1)
x
P(−3 ; − 1)
(k)
In the diagram below, two circles centres M(2 ; 2) and C respectively are drawn such
that they are tangential (touching) at P. B is a point on the larger circle such that MBC
is a right-angled triangle. If the radius of the smaller circle is 2 with MB = 4 units and
BC = r .
Determine:
(1)
the co-ordinates of B.
(2)
the co-ordinates of C in terms of r.
(3)
the length of the radius of the larger circle.
204
(l)
yB − yA
was used.
xB − xA
In Chapter 6, you determined the equation of a line given the y-intercept and another
point on the line. The coefficient of x in the equation y = ax + q represents the
gradient
or slope of the line and q represents the y-intercept. Use this information to determine
the equation of each of the following lines.
In this Chapter, the gradient of a line using the formula
(1)
(2)
B(1; 4)
A( −7 ; 2)
B( −2 ; − 3)
A( −4 ; − 1)
205
CHAPTER 9
FINANCIAL MATHEMATICS
In any financial situation in the real-world, the interest rate is the percentage charged, or paid,
for the use of money. It is charged when money is being borrowed, and paid when it is being
loaned. The function of banks is to grant loans or hold deposits. Banks convince people to
make deposits by paying interest to them. They are paying depositors for the right to use their
money.
They then use that money to grant loans. The interest rate for borrowers is higher than the
interest rate for depositors. Banks want to charge as much interest as possible on loans, and
pay as little as possible on deposits, so they can be more profitable.
There are two main types of interest rates that you were introduced to in previous grades:
simple interest and compound interest.
If interest is calculated using only the original amount of money saved or borrowed, then it is
called simple interest. Simple interest is used for short-term loans (hire-purchase accounts)
and investments.
If interest is calculated on the original sum plus interest already earned, then it is called
compound interest. Compound interest is used with long-term loans and investments. Saving
money over a long term-period (five or more years) with a compound interest rate well above
the inflation rate, will help you to accumulate wealth. Compounding can be your friend if you
are saving money over a long-term period. It can also be your worst enemy if you are paying
back a bank loan over a long-term period.
Let’s revise simple and compound interest in the following examples.
EXAMPLE 1
R5 000 is invested in a bank. Calculate the accumulated amount after three years if the
interest rate is:
(a) 3,75% per annum simple interest
(b)
3,75% per annum compound interest
Solutions
(a)
After 1 year:
A = 5 000 + 3, 75% of 5 000
A = 5 000 + 0, 0375 × 5000
∴ A = 5 000 + 187,50
∴ A = R5 187,50
After 2 years:
A = 5 187,50 + 0, 0375 × 5 000
∴ A = 5 187,50 + 187,50
∴ A = R5 375
After 3 years:
A = 5 375 + 0, 0375 × 5 000
∴ A = 5 375 + 187,50
∴ A = R5 562,50
(b)
After 1 year:
A = 5 000 + 3, 75% of 5 000
A = 5 000 + 0, 0375 × 5 000
∴ A = 5 000 + 187,50
∴ A = R5 187,50
After 2 years:
A = 5 187,50 + 0, 0375 × 5 187,50
∴ A = 5 187,50 + 194,53
∴ A = R5 382,03
After 3 years:
A = 5 382,03 + 0, 0375 × 5 382,03
∴ A = 5 382,03 + 201,83
∴ A = R5 583,86
Notice that after 1 year, the amount accumulated is the same for both the simple and
compound interest rates. After this, the amount accumulated using the compound interest rate
will always be higher than the amount accumulated using the simple interest rate.
206
This method of calculation is somewhat tedious. Let’s revise the formulae for simple and
compound interest rate calculations that you dealt with in Grade 9.
Simple interest formulae
Compound interest formulae
rn 

A = P 1 +
 or A = P(1 + in)
 100 
r 

A = P 1 +

 100 
where:
A = accumulated amount
n = number of years
r
i=
= interest rate as a decimal
100
n
or
A = P(1 + i ) n
P = original amount borrowed or invested
r = interest rate as a percentage
Note: r = i × 100
Let’s now use these formulae to verify the answers in Example 1 and focus on some further
examples.
EXAMPLE 2
(Calculating the value of A)
R5 000 is invested in a bank. Calculate, using the appropriate formulae, the accumulated
amount after three years and the total interest received, if the interest rate is:
(a)
3,75% per annum simple interest
(b)
3,75% per annum compound interest
Solutions
A=?
(a)
P = 5 000
i=
A = P(1 + in)
∴ A = 5 000(1 + 0, 0375 × 3)
3, 75
= 0, 0375
100
(b)
A = P(1 + i ) n
∴ A = 5 000(1 + 0, 0375)3
∴ A = R5 562,50
Interest received over 3 years
= R5 562,50 − R5000
= R562,50
 R562,50

= R187,50 each year 

3

EXAMPLE 3
n=3
∴ A = R5 583,86
Interest received over 3 years
= R5 583,86 − R5 000
= R583,86
(Calculating the value of P)
Five years ago, a certain amount of money was invested in a bank. The value of the
investment is currently R200 000. Calculate the original amount invested (P) if the interest
rate was:
(a) 5% per annum simple interest
(b)
5% per annum compound interest
Solutions
A = 200 000
(a)
P=?
A = P(1 + in)
i=
5
= 0, 05
100
(b)
n=5
A = P(1 + i ) n
∴ 200 000 = P(1 + 0, 05 × 5)
∴ 200 000 = P(1 + 0, 05)5
∴ 200 000 = P(1,25)
200 000
∴
=P
(1, 25)
∴ P = R160 000
∴ 200 000 = P(1, 05)5
200 000
=P
(1, 05)5
∴ P = R156 705,23
∴
207
Alternative method for (b):
200 000 = P(1, 05)5
200 000
∴
=P
(1, 05)5
∴ 200 000(1,05) −5 = P
[apply the exponent rule
∴ P = 200 000(1,05) −5
∴ P = R156 705,23
1
= a−m ]
am
The compound interest formula can therefore be written in the form P = A(1 + i ) − n . This new
formula can be used to calculate P when given A, i and n. With this formula, the interest is
removed from A to get back to P.
In the previous example, you can calculate P directly as follows:
P = 200 000(1 + 0,05)−5
∴ P = R156 705,23
EXAMPLE 4
(Calculating the value of P using the new formula)
Rachel has just opened a small business and takes out a loan to provide for the initial start-up
costs. She agrees to repay the loan four years later by means of a payment of R1 200 000. The
bank charges her an interest rate of 18% per annum compounded annually. What was the
amount of money she originally borrowed?
Solution
A = 1 200 000
P=?
i=
18
= 0,18
100
n=4
P = A(1 + i ) − n
∴ P = 1 200 000(1 + 0,18) −4
∴ P = R618 946,65
EXAMPLE 5
(Calculating the value of n)
How long would it take for an investment of R40 000 to increase by R5 000 if the interest rate
is 4,5% per annum simple interest?
Solution
A = 45 000
P = 40 000
i=
A = P(1 + in)
∴ 45 000 = 40 000(1 + 0, 045 × n)
45 000
∴
= 1 + 0, 045n
40 000
∴1,125 = 1 + 0, 045n
∴1,125 − 1 = 0, 045n
4,5
= 0, 045
100
n=?
Note:
Calculating the value of n in the compound
interest formula is not part of the Grade 10
syllabus since this requires the use of
logarithms which are studied in Grade 12.
∴ 0,125 = 0, 045n
0,125
∴n =
= 2, 7777..... years
0, 045
This is the same as 2 years and approximately 9 months
208
[0, 7777.... × 12 = 9,3333333....]
EXAMPLE 6
(Calculating the value of i)
Joseph invests R18 000 and it grows to R25 000 over a period of two years. Calculate the
interest rate to one decimal place if the investment earned:
(a) simple interest
(b)
compound interest
Solutions
A = 25 000
(a)
P = 18 000
A = P(1 + in)
∴ 25 000 = 18 000(1 + i × 2)
∴ 25 000 = 18 000(1 + 2i )
25 000
= 1 + 2i
18 000
25
∴ − 1 = 2i
18
7
∴ = 2i
18
7 1
1
∴ × = 2i ×
18 2
2
7
∴ =i
36
∴ i = 0,194444...
∴
r =?
(b)
n=2
A = P(1 + i ) n
∴ 25 000 = 18 000(1 + i ) 2
25 000
∴
= (1 + i ) 2
18 000
25
∴ = (1 + i ) 2
18
∴
25
= 1+ i
18
25
−1
18
∴ i = 0,178511302....
∴ r = 0,178511302.... ×100
∴i =
∴ r = 17,9%
∴ r = 0,194444... × 100
∴ r = 19, 4%
EXERCISE 1
(a)
(b)
(c)
(d)
Trevor invests an amount of R20 000 in a bank. Calculate, using appropriate formulae,
the accumulated amount after six years and the total interest he will receive, if the
interest rate is:
(1)
4,5% per annum simple interest
(2)
4,5% per annum compound interest
Refilwe wants to purchase a stove costing R12 000. She wants to pay back this amount
with interest in two years’ time. The interest rate is 24% per annum simple interest.
(1)
Calculate the amount that she will repay in two years’ time.
(2)
If she wants to pay the loan off in monthly payments over the two-year period,
what will her monthly payments be?
Nceba invested R500 000 in the share market. He managed to secure an average
compound interest rate of 14% per annum during the first two years.
(1)
Calculate the value of his investment at the end of the two-year period.
(2)
During the next three years, he managed to secure an average compound
interest rate of 12% per annum. What was his investment then worth at the end
of the next three years?
Seven years ago, a certain amount of money was invested in a bank. The value of the
investment is currently R350 000. Calculate the original amount invested (P) if the
interest rate was:
(1)
3,25% per annum simple interest
(2)
3,25% per annum compound interest
209
(e)
(f)
(g)
(h)
(i)
What amount must be invested now in order to receive R650 000 in five years’
time if the compound interest rate is 6% per annum?
How long would it take for an investment of R50 000 to increase to R65 000 if the
interest rate is 5,5% per annum simple interest?
Calculate how long it would take for an investment of R9 000 to double if the simple
interest rate is 11% per annum.
Paul invests R35 000 and it accumulates to R55 000 over a period of five years.
(1)
What simple interest rate would you need to secure?
(2)
What compound interest rate would you need to secure?
Janet invests R80 000 and it accumulates to R90 000 over a period of two years.
(1)
What simple interest rate will she need to receive in order to achieve this?
(2)
What compound interest rate will she need to receive in order to achieve this?
HIRE-PURCHASE AGREEMENTS
A Hire Purchase Agreement (HP) is a short-term loan. Household appliances and furniture are
often bought on HP. The buyer signs an agreement with the seller to pay a specified amount
per month. The interest paid on a hire purchase loan is simple interest and it is calculated on
the full value of the loan over the repayment period. Normally a deposit is paid initially and
the balance is paid over a short time period. The buyer will be required to pay the total interest
charged on the loan even if the loan can be paid off in a shorter time period.
EXAMPLE 7
Vanessa buys a laptop costing R16 000. She pays a 10% deposit and then takes out a twenty
four month hire-purchase loan on the balance. The interest rate charged on the loan is 22% per
annum simple interest. Calculate her monthly payments and what she will actually pay for the
computer.
Solution
Deposit = 10% of 16 000
= 0,1× 16 000 = R1 600
Balance = 16 000 −1 600 = R14 400 (P)
Interest is charged on R14 400 for a period of 24 months or 2 years (n).
HP loan amount (A):
A = P(1 + in)
∴ A = 14 400(1 + 0, 22 × 2)
∴A = R20 736
Monthly payments:
R20 736
∴
= R864
24
She will pay R20 736 plus the deposit of
R1 600 for the laptop, which is a total amount of R22 336.
210
EXAMPLE 8
Mike buys a mobile phone costing R8 000 on HP, pays a deposit of R800, and then pays 36
monthly payments of R344. Calculate the simple interest rate.
Solution
Balance after deposit has been paid is:
R8 000 − R800 = R7 200
(P)
Amount repaid over 3 years:
R344 × 36 = R12 384
(A)
Simple interest rate:
12 384 = 7 200(1 + i × 3)
12 384
∴
= 1 + 3i
7 200
12 384
∴
− 1 = 3i
7 200
18
∴ = 3i
25
18 1
1
∴ × = 3i ×
25 3
3
∴ i = 0, 24
∴ r = 24%
EXERCISE 2
(a)
(b)
(c)
(d)
(e)
(f)
David buys a computer which costs R17 000. He takes out a 36-month hire purchase
agreement. He pays a deposit of 10% and the interest charged on the balance is 14%
per annum simple interest. What will his monthly payments be and what will he
actually pay for the laptop?
Patricia wants to buy a furniture suite for R38 000. She decides to take out a hirepurchase loan involving equal monthly payments over five years. The deposit is 20%
and the simple interest rate charged per annum is 15%. Calculate:
(1) how much must be paid each month
(2) the amount of interest paid
(3) the actual amount paid for the furniture suite
Shaun buys a smartphone on HP which costs R11 799,90. He will have to pay R639
per month for 24 months. No deposit will be required. Calculate the simple interest
rate.
A tablet with WiFi costs R10 579. Mark buys a tablet on HP and agrees to pay a
deposit of R1 500 and 36 monthly payments of R350. Calculate the total simple
interest paid and the rate of simple interest.
A hire purchase contract for a sound system requires James to pay a deposit of R2 000
and to then make six monthly payments of R3 375. If the price of the sound system is
R20 000, calculate the total simple interest paid and the rate of simple interest.
Belinda buys a flat screen plasma television costing R20 000 on hire purchase. She
traded in an old television for R3 000 and paid a deposit of R 2000. The balance was
paid by means of monthly instalments of R900 over two years. Calculate the total
simple interest paid and the rate of simple interest.
211
(g)
(h)
Jeremy wants to buy a motorbike. He can only afford to pay R3 000 per month. He
wants to take out a hire purchase loan over 24 months at an interest rate of 12% per
annum simple interest. Calculate the price of the computer that he can afford to buy.
Ayanda wants to buy a new car and can afford to pay R4 899 per month. A car
dealership offers her a payment plan over 72 months at a simple interest rate of 10,5%
per annum. What is the price of the car she can afford to buy?
INFLATION IN THE ECONOMY
Inflation is the steady increase in the prices of goods and services over time throughout the
economy. It is measured by defining a ‘shopping basket’ of goods and services used by a
typical South African household and then keeping track of the cost of the basket. The
consumer price index, or CPI, is the cost of a shopping basket. For example, in one year, the
cost of the basket may rise by 4%. This increase of 4% in the CPI is referred to as the inflation
rate. The rate of inflation is therefore the percentage of money you’ll need more every year to
buy the same things you were able to buy the previous year. Inflation eats away at the value of
people’s money. With the same amount of money, fewer goods can be bought than before.
Inflation is caused by a wide range of factors. A few of them are:
An increase in the demand for goods due to shortage, which leads to prices going up. If there
are 100 consumers wanting to buy a product, but only 90 products are available, the price of
the product will increase.
Increases in production charges due to such factors as exchange rates, oil and petrol prices
will cause the producer to pass on the higher prices to the consumer in order to stay in
business. A higher petrol price has both an impact on the general price level, as it not only
impacts on the price we pay for petrol, but also on the whole range of goods and services that
are subject to transport costs.
EXAMPLE 9
In January 1985, the average home cost about R72 000. Assuming an average annual inflation
rate of 8,6% from 1985 up to 2015, which is a period of thirty years, what did the same house
cost in January 2015?
Solution
A = P(1 + i ) n
∴ A = 72 000(1 + 0, 086)30
∴ A = R855 514,26
EXAMPLE 10
The average salary of a computer programmer in South Africa in 1995 was R4 500. Assuming
an annual average rate of inflation of 6,1%, would a salary of R9 000 have the same buying
power in 2015?
Solution
A = P(1 + i ) n
∴ A = 4 500(1 + 0, 061) 20
∴ A = R14 706,87
A salary of R9 000 would be way below the buying power of a salary of R14 706,87.
212
EXERCISE 3
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Arnold paid R2 599,99 for a car sound system in 1997. Assuming an average inflation
rate of 7% per annum, what did he pay for a sound system with the equivalent value in
2014?
The average price of a hamburger and milkshake in 1991 was R6,60. Assuming an
average inflation rate of 8% per annum, what did hamburger and milkshake cost in
2006?
The average salary of a domestic worker in South Africa in the year 2000 was R2 000.
Assuming an annual average rate of inflation of 5,7%, would a salary of R3 000 have
the same buying power in 2015?
University fees for a student stuying a bachelor’s degree are, on average, about
R34 000 per year. Assuming an average inflation rate of 5% per annum, what will the
fees be in twenty years’ time?
Forty years ago, John deposited R5 000 in a bank paying him 3% per annum
compound interest. The average inflation rate over the forty years was 6%.
(1) How much money will he have saved after forty years?
(2) Calculate the buying power of R5 000 after forty years.
(3) Comment on the value of John’s savings after forty years.
If salaries double every seven years, what will the rate of inflation be?
Suppose that a person earns a monthly salary of R20 000. What would this salary have
been 30 years ago, assuming an annual average rate of inflation of 4,5%?
Suppose that a cold drink costs R4,50 now but will cost double in eight years’ time.
What will the average inflation rate be?
EXCHANGE RATES
There are different money systems in different countries. Currency is the term used to
describe the particular money system of a country. Here are the currencies of a few countries.
Country
South Africa
United States of America
United Kingdom
Several European countries
Currency used
Rand
US dollar
British Pound
Euro
Symbol for the currency
R
$
£
€
In every country, in order to purchase goods and services, you would need to use their
currency. In the USA, you would not be able to use rands to pay for things. This is because
the USA uses the dollar. Therefore, you would need to convert rands to dollars before you can
spend money in America. Because the rand is much weaker than the American dollar or
British pound, you will need to exchange a lot more rands for dollars and even more for
pounds. For example, say you want to buy an item online costing one pound in England. You
would therefore need about R18 to buy that item because the pound is much stronger than the
rand.
EXAMPLE 11
Sean wants to buy the latest DJ equipment, which has been advertised in a US catalogue for
$4 000. He wants to order and pay for the equipment online. The current rand/dollar exchange
rate is R12,56 to the US dollar. Calculate the cost in rands of the DJ equipment.
Solution
$1 = R12, 56
∴$1 × 4 000 = R12,56 × 4 000
213
∴$4 000 = R50 240
The DJ equipment will cost R50 240.
EXAMPLE 12
Simone is on a trip to the UK to visit her mom. The current rand/pound exchange rate is
R18,50 to the British pound. She has R40 000 to spend in the UK. How many pounds does
she have to spend?
Solution
£1 = R18,50
£1
R18,50
∴
=
18,50 18,50
£1
∴
= R1
18,50
£1
∴
× 40 000 = R1× 40 000
18,50
∴ £2162,16 = R40 000
Judy has £2 162,16 to spend in London.
EXERCISE 4
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
The latest Playstation game costs $645 in New York. What would it cost in South
Africa if the rand/dollar exchange rate is R12,25 to the US dollar?
Conrad wants to buy a fitness book costing £30 in London. How much will it cost
him in rands? The rand/pound exchange rate is R17 to the British pound.
Mark wants to buy Japanese sweets costing 50 yen (¥). If the rand/yen exchange rate is
one rand to 17,76 yen, calculate the cost of the sweets in rands.
Jill is visiting a friend in California for a week. She has R3 000 to spend and will
exchange the money for US dollars. How many dollars will she have to spend if the
rand/dollar exchange rate is R11,28 to the US dollar?
Nathan wants to purchase electric drums from a music dealership in England. The
drums cost £4 000 and Nathan has saved R75 000. The rand/pound exchange rate is
R18,24 per one pound. Will he have enough money, in rands, to buy the drums? Show
your reasoning.
One yen costs R0,22. How much yen can you spend in Japan if you have R4 500
available to you?
A married couple living in England, having saved £4 000, decided to have a holiday in
South Africa. The airfare cost £1 100 and they converted the balance of their money
into rand which they intended to spend in South Africa. They actually spent R23 000
in South Africa and converted their remaining rands back into pounds when they
returned to England. The exchange rate on arrival in South Africa was R18,24 to the
pound and R18,46 to the pound when they departed from South Africa.
(1) How much, in rands, had the couple planned to spend in South Africa?
(2) How many pounds did they take back to England?
A certain watch costs €350 in Germany or £245 in England. Which price is better for
the South African buyer if the exchange rates are R11,24 to the euro and R18,06 to the
pound?
A South African teacher works in England for two years. He saves £400 every month.
How much money in rand would he save in this time if the average exchange rate
during the two years is R18,54 to the pound?
214
(j)
Brenda won a competition where she can fly to three international destinations free of
charge with spending money. The destinations she chose were Germany, Japan and
England. For Germany, she was allocated €9 000. For Japan, she was allocated ¥30
000. For England, she was allocated £2 500. Use the exchange rates in the table on the
next page to calculate the total amount she had been allocated in rands.
Exchange Rates
Japan (¥)
R0,29
Germany (€)
R10,46
England (£)
R17,12
POPULATION GROWTH
Population growth is the increase in the number of individuals in a population. The population
growth rate is the rate at which the number of individuals in a population increases in a given
time period as a fraction of the initial population. It measures how the size of the population is
changing over time.
In the first few years, a population grows exponentially in the same way that money grows
through compounding. As time progresses, any population of living creatures is constrained
by the availablility of food, water, land or other important resources. Populations therefore
don’t always grow exponentially. In this section, we will focus on the early years of
exponential growth of populations. There are other models of population growth that are
studied at university which take into consideration changing population growth rates due to
the constraining factors mentioned.
The formula for calculating exponential growth of a population is similar to the compound
interest formula:
n
r 

Pfuture = Ppresent 1 +
 where:
 100 
Ppresent = present size of the population
Pfuture = future size of the population
r = average population growth rate expressed as a percentage
n = the number of years
EXAMPLE 13
The mid-year population in South Africa in 2014 was 54 002 200. Calculate the size of the
population in five years’ time if the average population growth rate is 1,56%.
Solution
Pfuture
r 

= Ppresent 1 +

 100 
∴ Pfuture
n
 1,56 
= 54 002 200  1 +

 100 
5
∴ Pfuture = 58 347 857,54
The population size will be approximately 58 347 857 people in five years’ time.
215
EXERCISE 5
(a)
(b)
(c)
(d)
(e)
(f)
The mid-year population in South Africa in 2008 was 48 910 000. Calculate the size of
the population in two years’ time if the average population growth rate was 1,34%.
The number of people in the USA as at June 2014 was estimated at 318 857 056. If the
average population growth rate was 1,42%, calculate the estimated population size of
the USA in June 2017.
The number of black rhinos in Africa during 2012 was estimated at 5 487. If the
average population growth rate of black rhinos is 4,9% per annum, calculate how
many rhinos were there in Africa in 2007.
The world population during the year 2015 was estimated to be 7 320 248 940. If the
average annual exponential growth rate was 1,14%, what was the population in the
year 2000?
A family of 6 mice can multiply into a family of 60 mice in 3 months.
(1) Calculate the estimated monthly growth rate for the mice population.
(2) If not controlled, how many mice will there possibly be in one year if the initial
population is 6 mice?
You are studying the population growth of a species of frog. In a pond constructed for
the frogs, you start off with 50 frogs and notice that after 10 months, the number of
frogs has increased to 61. What is the average monthly growth rate?
SITUATIONS INVOLVING CHANGING INTEREST RATES
EXAMPLE 14
R8 000 is deposited into a savings account. The interest rate for the first three years is 3% per
annum compounded annually. Thereafter, the interest rate changes to 4% per annum
compounded annually. Calculate the value of the investment at the end of the eighth year.
Solution
Time lines are useful when dealing with interest rate changes.
T0
T1
T2
T3
T4
Method 1
A = 8 000(1 + 0,03)3
T6
T7
T8
Method 2
(at T3 )
∴ A = 8 741,816
A = 8 741,816(1 + 0,04)5
∴ A = R10 635,76
T5
(at T8 )
Instead of actually calculating the value
of 8 000(1, 06)3 , you can do both
calculations in one line as follows:
A = 8 000(1, 03)3 . (1, 04)5
∴ A = R10 635,76
216
EXAMPLE 15
A certain amount of money was invested ten years ago and is now worth R200 000. The
interest rate during the first 6 years was 3,5% per annum compounded annually and for the
remaining four years, the interest rate was 4,6% per annum compounded annually. How much
money was invested ten years ago?
Solution
T0
T1
T2
T3
T4
T5
Method 1
P = 200 000(1 + 0,046)−4
T7
T8
T9
T10
Method 2
(at T6 )
∴ P = 167 071,8407
P = 167 071,8407(1 + 0,035)−6
∴ P = R135 913,05
T6
(at T0 )
Instead of actually calculating the value
of 200 000(1, 046) −4 , you can do both
calculations in one line as follows:
P = 200 000(1, 046)−4 . (1, 035)−6
∴ P = R135 913,05
EXERCISE 6
(a)
(b)
(c)
(d)
(e)
(f)
R24 000 is deposited into a bank account. The interest rate for the first seven years is
2,25% per annum compounded annually. Thereafter, the interest rate changes to 3,5%
per annum compounded annually. Calculate the value of the investment at the end of
the thirteenth year.
Portia invests R500 000 in the share market. Over a fifteen-year period, the average
interest rate was 6% per annum compounded annually for the first eight years and 7%
per annum compounded annually for the remaining seven years. How much money
will she have saved at the end of the fifteen-year period?
David invested R50 000 for nine years. The interest rate changed three times during
the nine-year period. At the start, the interest rate was 4,35% per annum compounded
annually. After two years, it increased by 1% and four years later, it decreased by 2%.
How much was David’s investment worth at the end of the investment period?
A certain amount was invested in the share market seven years ago and is now worth
R1 300 000. The average interest rate during the first five years was 9% per annum
compounded annually and for the remaining two years, 10% per annum compounded
annually. How much money was originally invested?
Sibongile wants to travel to New York in five years’ time. She estimates that, by then,
the cost of the trip will be R250 000. Suppose that the bank will give her a fixed
compound interest rate of 5% for the first year and 6% per annum for the remaining
four years, how much will she need to invest now in order to save R250 000?
Thembi invested R6 000 at the beginning of the year 2010. The interest rate then was
2% per annum compounded annually. At the beginning of the following year, the
interest rate increased to 2,5%. One year later, the interest rate increased by 0,8%. One
year after this, the interest rate dropped by 2%. How much money did Thembi save at
the beginning of the year 2015?
217
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
Sam invests R28 000 for two years ands earns 4,7% per annum compound interest.
(1)
Calculate the accumulated value at the end of the two-year period.
(2)
He leaves the money in the account for a further two years. The interest rate
changes to 6% per annum simple interest. How much money will he then have
saved?
What amount must Tanisha invest now if she wishes to accumulate R1 000 000 in ten
years’ time given that the interest rate is:
(1)
13% per annum simple interest?
(2)
13% per annum compound interest?
Theresa believes that she has the potential to double her money in three years by
investing in the share market. If she deposits R60 000 into a share-market investment
account,
(1) what simple interest rate will be required in order for her to do this?
(2) what compound interest rate will be required in order for her to do this?
Manhatten Island is said to have been bought from the Indians in 1626 for $24. If,
instead of making this purchase, the buyer had put the money in a savings account
drawing compound interest at 6% per annum, what would that account be worth in the
year 2015?
On his trip to New York, Robert booked into a hotel for three nights. The cost per
night was $450. How much did Robert spend in rands for the three nights if the
exchange rate at that time was $1 = R12,12 .
On his trip to London, Mark decides to buy a tablet from a computer store. The tablet
costs R10 000 in South Africa. How much is this in pounds if the exchange rate on the
day is £1 = R17, 30 ?
Ernest wants to buy a car costing R240 000. He wants to pay cash for this model of car
but can only do this in one and a half years’ time. If the inflation rate is running at 6%
per annum, calculate the cost of this car in one and a half years’ time.
A population of Canadian swans doubles every seven years. Calculate the average
annual growth rate of this swan population.
One pair of German cockroaches can theoretically produce enough offspring in one
year to carpet the floors of the average home to a depth of 1 metre per year. Reasearch
has found that one female cockroach give birth to 400 000 offspring in one year!
What is the annual growth rate for this population?
[http://www.stephentvedten.com/27_Roaches.pdf]
An investment of R9 000 earns 6% per annum compounded annually for a period of
four years. Thereafter, the interest rate changes to 7% per annum compounded
annually for a further two years. Calculate the future value of the investment at the end
of the six-year period.
Peter invests a certain sum of money for five years at 12% per annum compounded
annnually for the first two years and 14% per annum compounded annually for the
remaining term. The money grows to an amount of R65 000 at the end of the five-year
period. How much did Peter originally invest?
R6000 is deposited into a savings account. Four years later, R7000 is added to the
savings. The interest rate for the first three years is 8% per annum compounded
annually. Thereafter, the interest rate changes to 9% per annum compounded annually.
Calculate the value of the savings at the end of the seventh year.
Christo invests R12 000 in a savings account in order to save up for an overseas trip in
five years’ time. The interest rate for the five-year period is 11% per annum
compounded annually. At the end of the third year, Christo runs into financial
difficulty and withdraws R5 000 from the account. How much money will he have
saved at the end of the five-year period?
218
CHAPTER 10
STATISTICS
REPRESENTING UNGROUPED DATA
Ungrouped data is a set of individual values or observations. The data is discrete since the
values are distinct values. For example, the set of final examination marks of thirty university
students is referred to as ungrouped data. Another example could be the number of goals
scored by Kaizer Chiefs. Ungrouped data can be represented in different ways. The three most
common ways are: frequency tables, bar graphs and stem-and-leaf plots.
EXAMPLE 1
The American company Deloitte made a prediction that in 2014, the increase in
the number of smartphones being used world-wide would be the greatest for
people over 55 years.
In a survey conducted in a shopping mall during 2014, different people were
approached and asked what type of phone they were using. The following table shows the
ages of thirty people between the ages of 15 and 60 using smartphones.
[www2 deloitte com/content/dam/Deloitte/global/Documents/Technology-Media-Telecommunications/gx-tmt-2014prediction-smartphone pdf]
16
28
17
32
17
34
17
34
17
34
18
46
18
46
25
48
25
54
27
55
28
56
28
56
28
56
28
56
(a)
Draw a frequency table for this ungrouped, discrete data.
(b)
Represent the data in a stem-and-leaf plot.
(c)
Draw a frequency bar graph for this data.
(d)
Do the results of the survey prove that Deloitte was correct in their prediction?
28
56
Solutions
(a)
A frequency table shows the different observations and how many times they occur.
Age
16
17
18
25
27
28
32
34
46
48
54
55
56
Tally
|
||||
||
||
|
|||| |
|
|||
||
|
|
|
||||
Frequency
1
4
2
2
1
6
1
3
2
1
1
1
5
n = 30
219
(b)
In a stem-and-leaf plot, the tens digit is used as a “stem” and the units as a “leaf”.
Ensure that the “leaves” are equally spaced.
1
2
3
4
5
(c)
6
5
2
6
4
7
5
4
6
5
7
7
4
8
6
7
8
4
7
8
8
8
8
8
6
6
6
6
8
8
The individual ages are on the horizontal axis and the frequencies on the vertical axis.
It might be useful to place the ages into age groups rather than individual ages.
(d) Although the bar graphs indicate that there were a high number of over-55’s using
smartphones, there is not enough data to prove this prediction. The sample was too
small. It would be far more feasible to increase the number of people surveyed to well
in the millions to get a better idea. Also, one would have needed to compare sales in
previous years.
220
EXERCISE 1
(a)
The manager of a computer store assessed the quality of
service at his store based on the feedback from thirty
customers. The rating scale was as follows:
Extremely poor (0)
Poor (1)
Average (2)
Good (3)
Very good (4)
Outstanding (5)
The scores of the thirty customers are provided below.
0
3
(1)
(2)
(b)
4
4
3
5
4
3
1
4
4
2
2
1
4
1
5
3
4
4
2
3
4
5
4
2
2
4
Draw a frequency table for this ungrouped, discrete data.
Draw a frequency bar graph for this data.
A number of seeds of a particular variety of flower were sown. All germinated, but not
all at the same time. The following bar graph shows how many seeds germinated after
various number of days.
(1)
(2)
(3)
(c)
3
4
How many seeds were sown?
After how many days did the first seed germinate?
What percentage of seeds germinated within the first 8 days?
Donating blood can help to save the life of someone and even
yours should you be in an accident one day and require blood.
The ages of forty people who donated blood on a particular day
are provided below.
18
41
64
27
(1)
(2)
(3)
42
53
22
35
17
42
35
63
35
50
17
24
19
57
65
66
35
43
18
34
20
35
47
39
39
24
19
27
35
55
48
66
26
54
53
35
Draw a stem-and-leaf plot for this data.
Draw a frequency bar graph for this data. Group the data into appropriate age
groups (10-19 year olds, 20-29 year olds, etc).
Which age group donated the most blood?
221
(d)
(e)
The number of air conditioners sold by fifty sales representatives in the year 2015 are
recorded below:
25
13
20
18
3
22
13
3
22
14
19
9
7
27
4
(1)
(2)
(3)
(4)
Draw a stem-and-leaf plot for this data.
Draw a frequency bar graph for this data.
How many agents sold twenty or more air conditioners?
What percentage of the agents sold less than 20 air conditioners?
27
21
14
23
5
19
18
9
14
24
23
30
7
14
20
21
31
27
14
8
14
28
21
8
10
12
21
39
1
8
An app is a type of software that allows you to perform specific
tasks on your laptop or mobile phone. The word “app” is an
abbreviation for “application”. By the year 2016, there will be
almost 310 billion downloaded apps generating close to $74 billion
in revenue. With nearly two million apps already developed,
competition for someone wanting to develop a successful app is fierce. Having an app in
the top 25 will require at least 33 downloads per hour. Two companies recorded the
number of downloads per hour for one of their new apps over a period of 15 hours.
Company A
Company B
(1)
(2)
(3)
(4)
(5)
(f)
27
4
14
30
18
23
10
30
20
32
11
11
23
33
44
13
24
42
34
41
35
14
43
22
33
33
29
22
32
22
33
33
43
44
43
Using today’s exchange rates, convert $74 billion to rands, pounds and euros.
Draw a back-to-back stem-and-leaf diagram for the two companies.
What percentage of Company A’s downloads were more than 33 per hour?
What percentage of Company B’s downloads were less than 33 per hour?
In your opinion, which company has the better chance of success with their new
app? State a reason for your answer.
The following back-to-back stem-and-leaf diagram shows the average number of hours
spent per week on social networking websites by learners from two different classes.
8
(1)
(2)
(3)
(4)
(5)
7
7
CLASS A
7 5 5 5
1 1
1
0
1
0
0
0
9
0
1
2
3
1
0
1
0
2
0
3
0
CLASS B
3 4 6 6
1
7
8
9
How many learners are there in Class A?
How many learners are there in Class B?
How many learners in Class A spend exactly one hour per week on a social
networking website?
How many learners in Class B spend more than five hours per week on a social
networking website?
Which class spends more time, in total, on a social networking website?
222
(g)
The following bar graph shows the weekly social networking usage on smartphones by
different age groups in developed countries.
[Source: Deloitte Global Mobile Consumer Survey, Developed countries, May-July 2014]
(1)
(2)
(3)
(4)
In which age group does 44% use smartphones for social networking?
What percentage of the over 65-year olds do not use smartphones for social
networking?
If there are 200 000 people in the 25-34 year age group, how many will be using
smartphones for social networking?
If there are 375 000 people in the 18-24 age group using smartphones for social
networking, how many people in this age group are not using smartphones for
social networking?
MEASURES OF CENTRAL TENDENCY
Central tendency is the clustering of data around a central or middle value. A measure of
central tendency is a single value that attempts to describe a set of data by identifying the
central position within that set of data. As such, measures of central tendency are sometimes
called measures of central location. They are also classed as summary statistics. The mean
(often called the average) is most likely the measure of central tendency that you are most
familiar with, but there are others, such as the median and the mode.
THE MEAN
The mean or average of the data (referred to as x-bar or x ), is the sum of the data values
divided by the total number of data values (n). We refer to the data values as the x-values.
x=
sum of all x -values
x
=
total number of x -values
n
Whenever the number of data values is large and there are no extreme values (outliers), then
the mean is a good measure of central tendency.
EXAMPLE 2
Calculate the mean for the following data:
1
3
3
3
3
4
Solution
x=
 x = 1 + (3 × 4) + 4 + 5 + (9 × 3) = 49 = 4, 9
n
10
10
223
5
9
9
9
THE MEDIAN
The median is the middle-most number when the data values are written in ascending order.
In order to determine the correct position of the median in the data, use the following formula:
Position of median = 12 (n + 1) where n represents the number of data values.
If the data set contains an odd number of values, then the median will be part of the data set.
1 ( n + 1)
will be an integer. If the data set contains an even number of values, then the median
2
will not be part of the data set.
1 ( n + 1)
2
will not be an integer. The median will be the average
between the two middle numbers in the data set.
EXAMPLE 3
Determine the median for the following data set:
2
9
5
12
10
Solution
First arrange the values in ascending order:
2
5
9
10
12
There is an odd number of values and therefore the median will be part of the data set.
The position of the median = 12 (5 + 1) = 12 (6) = 3rd position.
The data value in the 3rd position is 9.
The median is therefore 9 and it divides the data into two equal halves.
25 9 10
12

lower half
upper half
EXAMPLE 4
Determine the median for the following data set:
3
9
10
12
15
18
Solution
The data is arranged in ascending order.
There is an even number of values and therefore the median will be not part of the data set.
The position of the median =
1
1
(6 + 1) = (7) = 3,5th position.
2
2
This means that the median is the average between the 3rd and 4th data values.
Median =
10 + 12 22
=
= 11
2
2
3
9 10

12
15 18


↑
11
The median is 11 and it divides the data into two equal halves.
It is not a value in the data set.
224
The median may be a better indicator of the most typical value if a data set has an outlier,
which is an extreme value that differs greatly from the other values.
EXAMPLE 5
Consider the following data set:
(a)
(b)
(c)
5
12
7
36
8
9
7
Determine the mean and the median.
Which value is an outlier?
Which measure of central tendency is more representative of the data set?
Solutions
(a)
Arrange the data in ascending order:
5
7
x=
7
8
9
12
36
5 + 7 + 7 + 8 + 9 + 12 + 36 84
=
= 12
7
7
There is an odd number of values and therefore the median will be part of the
data set.
1
1
The position of the median = (7 + 1) = (8) = 4th position.
2
2
The median is 8 and it divides the data into two equal halves.
It is a value in the data set.
8 9
12 36
5
7 7
(b)
36 is an extreme value compared to the other values and is an outlier.
(c)
The outlier has inflated the mean. It is therefore not a good value to use as a measure
of central tendency. The median is not affected by the outlier and is therefore a
much better measure than the mean.
THE MODE
The mode of a data set is the value that occurs most frequently. If two numbers tie for the
most frequent occurrence, the data set has two modes and is called bimodal. If three numbers
tie for the most frequent occurrence, the data set has three modes and is called trimodal. If a
data set has an outlier, the mode, like the median, may also be a better indicator of the most
typical value.
EXAMPLE 6
Calculate the mode for the following data:
1
2
2
2
2
3
5
6
6
7
8
8
8
8
9
10
12
Solution
There are two modes: 2 and 8
These values occur four times each in the data set. The data set is therefore bimodal.
225
EXAMPLE 7
Consider the data collected in Example 1. The ages of thirty people using smartphones
between the ages of 15 and 60 were recorded.
16
28
17
32
17
34
17
34
17
34
18
46
18
46
25
48
25
54
27
55
28
56
28
56
28
56
28
56
28
56
The stem-and-leaf diagram of this data is provided below:
1
2
3
4
5
6
5
2
6
4
(a)
(b)
(c)
7
5
4
6
5
7
7
4
8
6
7
8
4
7
8
8
8
8
8
6
6
6
6
8
8
Calculate the mean for this data.
Determine the median.
Determine the mode.
Solutions
x=
(a)
16+(17×4)+(18×2)+(25×2)+27+(28×6)+32+(34×3)+(46×2)+48+54+55+(56×5)
= 34, 27
30
(rounded off to two decimal places)
(b)
There is an even number of data values.
The position of the median = 12 (30 + 1) = 12 (31) = 15, 5th position.
The median is the average between the 15th and 16th data values.
It is not part of the original data set but it divides the data set into two equal halves.
1
2
3
4
5
6
5
2
6
4
7
5
4
6
5
7
7
4
8
6
7
8
4
7
8
8
8
8
8
6
6
6
6
8
8
16 17 17 17 17 18 18 25 25 27 28 28 28 28 28
28 32 34 34 34 46 46 48 54 55 56 56 56 56 56
The median =
(c)
28 + 28
= 28
2
The most frequently-occuring value is 28.
The mode is therefore 28.
226
EXERCISE 2
(a)
Find the mean, median and mode of the following sets of data values:
(1)
4 ; 13 ; 5 ; 7 ; 9 ; 6 ; 5
(2)
8 ; 22 ; 3 ; 18 ; 4 ; 14 ; 8 ; 5 ; 10 ; 8 ; 10
(3)
13 ; 2 ; 11 ; 2 ; 10 ; 4 ; 5 ; 10 ; 8 ; 10
(4)
9;1;4;4;2;8;5;2;5;5
(b)
The mean weight of ten people entering in a lift is 75 kg. The lift has a weight limit of
1 000 kg. Approximately how many more people can get into the lift assuming that the
mean weight remains at 75 kg?
(c)
The monthly salaries of nine employees in a small business are:
R15 400
R16 800
R86 300
R13 200
R16 900
R11 900
R17 100
R16 200
R16 900
(1)
Calculate the mean, median and mode for this data.
(2)
Which measure of central tendency is a sensible measure of the “typical”
monthly salary of an employee in this business? Explain.
(d)
A salesman at a shoe store sold eight pairs of men’s shoes one
morning. The sizes of the eight pairs of shoes were as follows:
8 12
6 10 8 12
8 7 12
9 8 12
Which measure of central tendency best describes the typical shoe
size for this data?
(e)
A teacher records the following results for an examination out of 100:
98 63 79 76 58 71 86 78 91 87
89 41 19 88 41 99 97 83 78 90
Which measure of central tendency best describes these results?
(f)
A dairy farmer has 32 cows for sale. The weights of these cows in kilograms are
recorded below. The total weight of the cows is 5 060 kg.
80
86
92
155
(1)
(2)
(g)
82
87
93
321
83
87
94
371
83
88
95
376
84
88
97
377
85
89
153
381
85
90
153
382
86
92
154
391
Calculate the mean and the median.
The farmer describes the cows to a buyer and states that the average weight is
over 158 kg. Which measure of central tendency did the farmer use to describe
the cows and does this measure describe the cows fairly?
The following stem-and-leaf diagram represents the ages of forty people who donated
blood. Refer to Exercise 1 no (c). The total of all the ages is 1 544.
1
2
3
4
5
6
(1)
(2)
7
0
4
1
0
3
7
2
5
2
3
4
8
4
5
2
3
5
8
4
5
3
4
6
9
6
5
7
5
6
9
7
5
8
7
7
5
5
9
9
Calculate the mean, median and mode for this data.
Comment on the usefulness of these measures of central tendency.
227
(h)
The following stem-and-leaf diagram represents the number of air conditioners sold by
fifty sales representatives. Refer to Exercise 1 no (d). The total number of air
conditioners is 843.
0
1
2
3
(1)
(2)
(i)
1
0
0
0
3
2
0
0
3
3
1
1
4
4
1
5
4
1
7
4
2
7
4
2
8
4
3
8
4
3
8
4
4
9
8
5
9
8
7
8
7
9
7
9
7
8
Calculate the mean, median and mode for this data.
Comment on the usefulness of these measures of central tendency.
Research was done on families having six children. The table
below shows the number of families in the study with the
indicated numbers of boys.
Number of boys
Frequency
(j)
4
3
1
9
0
1
1
24
2
45
3
54
4
50
5
19
6
7
(1)
(2)
Calculate the mean, median and mode for this data.
Comment on the usefulness of these measures of central tendency.
(1)
(2)
(3)
The mean of 3 ; 4 ; 8 ; 9 ; x is 7. Determine x.
The median of five consecutive natural numbers is 12. What is the mean?
The numbers 4 ; 6 ; 8 ; 9 ; x are arranged from smallest to biggest. If the mean
and the median are equal, determine x.
The mean of five numbers is 27. The numbers are in the ratio 1 : 2 : 3 : 4 : 5.
Determine the five numbers.
Write down three possible sets of five numbers such that the median is 4, the
mean is 5 and the mode is 3.
The mean of six numbers is 44 and the mean of five of these numbers is 46.
What is the sixth number?
(4)
(5)
(6)
MEASURES OF POSITION
Measures of position divide a set of data into equal groups. There are two measures of
position that we will discuss:
Quartiles: which divide the data set into four equal parts
Percentiles: which divide the data set into one hundred equal parts
Quartiles
Consider an ordered set of numbers whose median is m. The lower quartile is the median of
the numbers that occur before m and the upper quartile is the median of the numbers that
occur after m. The lower quartile is also called the first quartile ( Q1 ). The median is called the
second quartile ( Q 2 or M ) and the upper quartile is called the third quartile ( Q3 ). The three
quartiles divide the data into four quarters. One quarter of the data values are less than the
lower quartile and three quarters of the data values are less than the upper quartile.
The following steps may be used to determine the lower and upper quartiles:
(1)
(2)
(3)
Order the data (from smallest to biggest) and find the median ( Q 2 or M ).
Find the median of the lower half of the data (exclude the median of the entire set).
This is the lower quartile ( Q1 ).
Find the median of the upper half of the data (exclude the median of the entire set).
This is the upper quartile ( Q3 ).
228
EXAMPLE 8
Calculate the quartiles for the following sets of data:
(a)
1
3
4
5
6
7
8
8
9
9
10
(b)
2
3
4
5
5
5
6
7
7
8
9
(c)
95
120
125
140
105
12
142
60
9
10
10
10
135
130
Solutions
(a)
1
3
4
5
6
7
8
8
9
(11 values)
The position of the median = 12 (11 + 1) = 6
The median is 7 (the 6th value). It is a value in the data set.
1
3
4
5
6
7
8
8
9
9
10
The lower half of the data set is: 1
3
4
5 6
The lower quartile is the median of the lower half (consisting of 5 values).
The position of the lower quartile = 12 (5 + 1) = 3
The lower quartile is 4 (the 3rd value). It is a value in the lower half.
1
3
4
5
6
The upper half of the data set is: 8
8
9
9 10
The upper quartile is the median of the upper half (consisting of 5 values).
The position of the upper quartile = 12 (5 + 1) = 3
The upper quartile is 9 (the 3rd value). It is a value in the upper half.
8
8
9
9
10
Let’s consider the quartiles together:
Q2
Q1
(b)
2
3
4
5
5
5
6
Q3
7
7
8
9
10
10
(13 values)
Solution
The position of the median = 12 (13 + 1) = 7
The median is 6 (the 7th value). It is a value in the data set.
2
3
4
5
5
5
6
7
7
8
9
10
10
The lower half of the data set is: 2
3
4
5 5
5
The lower quartile is the median of the lower half (consisting of 6 values).
The position of the lower quartile = 12 (6 + 1) = 3,5
The lower quartile is the average between the 3rd and 4th values. It is not a value
in the lower half.
∴ Q1 =
4+5
= 4,5
2
229
The upper half of the data set is: 7
7
8
9 10
10
The upper quartile is the median of the upper half (consisting of 6 values).
The position of the upper quartile = 12 (6 + 1) = 3,5
The upper quartile is the average between the 3rd and 4th values. It is not a value
in the upper half.
8+9
= 8,5
2
Let’s consider the quartiles together:
∴ Q3 =
Q2
Q3
Q1
(c)
Arrange the data values in ascending order:
12
60
95
105
120
125
130
135
140
142
The position of the median = 12 (10 + 1) = 5,5
∴ Q2 =
120 + 125
= 122,5
2
(the average between the 5th and 6th values)
The lower half of the data set is: 12
60
95
105 120
The lower quartile is the median of the lower half (consisting of 5 values).
The position of the lower quartile = 12 (5 + 1) = 3
The lower quartile is the 3rd value. It is a value in the lower half.
∴Q1 = 95
The upper half of the data set is: 125
130
135 140
142
The upper quartile is the median of the upper half (consisting of 5 values).
The position of the upper quartile = 12 (5 + 1) = 3
The upper quartile is the 3rd value. It is a value in the upper half.
∴Q3 = 135
Q1
Q3
Q2
EXERCISE 3
(a)
Find the quartiles for the following sets of data:
(1)
2 3 5 8 10 12 13
(2)
1 4 6 7 11 13 15 15 20
(3)
3 6 7 9 14 18 20 21
(4)
5 6 8 9 16 19 23 25 28
(5)
6 7 9 9 10 12 16 21 23
230
30
26
27
28
30
32
(b)
(c)
(d)
(e)
(f)
(g)
(h)
The number of matches played by Argentina in past world cups are recorded below.
Determine the quartiles for the number of matches.
Year
Number of matches
1934
1
1958
3
1962
3
1966
4
1974
6
1978
7
1982
5
Year
Number of matches
1986
7
1990
7
1994
4
1998
5
2002
3
2006
5
2010
8
A salesman at a shoe store sold eight pairs of men’s shoes one morning. The sizes of the
eight pairs of shoes were as follows:
8 12
6 10 8 12
8 7 12
9 8 12
Determine the quartiles for this data.
A teacher records the following results for an examination out of 100:
98 63 79 76 58 71 86 78 91 87
89 41 19 88 41 99 97 83 78 90
Determine the quartiles for this data.
The estimated overall population percentage growth rates of South Africans over the
past 14 years are given below.
1,27
1,29
1,32
1,35
1,38
1,40
1,43
1,46
1,49
1,52
1,55
1,57
1,58
1,59
Determine the quartiles for this data.
Refer to Exercise 2 no (f). Determine the quartiles for this data.
Refer to Exercise 2 no (g). Determine the quartiles for this data.
Five data values are represented as follows: 2 x ; x + 1 ; x + 2 ; x − 3 ; 2 x − 2
(1) Determine the value of x if the mean of the data set is 15.
(2) Calculate the quartiles.
Percentiles
Percentiles are often used by academic institutions to compare student marks. Defining and
determining percentiles is not that straightforward as there are many different approaches.
There is no universally accepted definition of a percentile. Using the 65th percentile as an
example, the 65th percentile can be defined as the lowest score that is greater than 65% of the
scores. The 65th percentile can also be defined as the smallest score that is greater than or
equal to 65% of the scores. Percentiles are more appropriate with data sets containing a large
number of values. Note that the lower quartile is also the 25th percentile, the median is the
50th percentile and the upper quartile is the 75th percentile. We will use the following method
to determine percentiles.

Arrange the data in ascending order.

p
Compute index i , the position of the pth percentile using the formula: i = 100 ( n)

If i is not an integer, round up. The pth percentile is the value in the i th position.

If i is an integer, the pth percentile is the average of the values in positions i and
i +1 .
The following example involves a small number of data values. In the exercise which follows,
you will be given data sets with a large number of data values.
231
EXAMPLE 9
A Maths professor at a university posted a list of marks, without names, on the notice board
outside his office. The students were informed as to the percentile they were in. There are 45
students in his class and the marks are as follows:
66
58
24
(a)
(b)
(c)
86
78
42
65
65
54
78
50
55
32
36
54
52
67
68
69
55
65
85
72
88
87
57
80
28
64
84
90
70
68
98
92
61
73
95
75
64
33
76
56
32
82
Jaco scored in the 70th percentile. What is his mark?
Michael scored in the 20th percentile. What is his mark?
Dimpho scored in the 50th percentile. What is her mark?
Solutions
(a)
Arrange the marks in ascending order:
24 28 32 32 33 36 42 50 52
58 61 64 64 65 65 65 66 67
75 76 78 78 80 82 84 85 86
54
68
87
54
68
88
55
69
90
55
70
92
56
72
95
57
73
98
70
i = 100
× (45) = 31,5
All we now do is round this number up to 32 and the 70th percentile is the 32nd mark
which is 76%. Jaco therefore obtained 76% and scored better than 70% of all students.
(b)
20
i = 100
× (45) = 9
The 20th percentile is the average between the 9th and 10th mark: 52+254 = 53
There is no score of 53 in the data and therefore Michael will have obtained 54% and
will have scored better than 20% of all students.
(c)
50
i = 100
× (45) = 22,5
The 50th percentile is the 23rd mark which is 66.
Dimpho obtained 66% and scored better than 50% of all students.
EXERCISE 4
(a)
Tobacco use is a leading cause of death in the United States. Nicotine found in tobacco
is rapidly metabolised in the liver to a substance called cotinine. The levels of cotinine
in the body are measured in nanograms per millilitre (ng/ml). A nanogram is onebillionth of a gram. Consider the following cotinine levels of 50 smokers:
5
88
174
249
290
(1)
(2)
(3)
(4)
(b)
6
103
174
250
313
6
113
198
253
313
8
122
208
265
314
22
123
210
267
350
40
130
223
277
360
43
131
224
280
401
44
149
227
284
460
48
165
233
286
476
86
168
245
289
490
Calculate the 25th, 50th and 75th percentiles.
Calculate the 30th percentile.
Calculate the 65th percentile.
Calculate the 80th percentile.
A research survey was conducted to investigate how long people live, on average, in
different countries of the world. The table on the next page has the average life
expectancies of people in 216 different countries. [Source: CIA World Factbook]
232
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
Chad
Guinea-Bissau
South Africa
Swaziland
Afghanistan
Cent Afr Republic
Zimbabwe
Somalia
Lesotho
Mozambique
Nigeria
Namibia
Gabon
Malawi
Zambia
Mali
Tanzania
Uganda
Niger
Burkina Faso
Angola
Cameroon
Congo
Botswana
Sierra Leone
Ethiopia
Cote d’Ivoire
Liberia
Rwanda
Guinea
Burundi
Senegal
Benin
Ghana
Western Sahara
Mauritania
Djibouti
Haiti
Sudan
Comoros
Equatorial Guinea
Laos
Eritrea
Cambodia
Kenya
Togo
Sao Tome
Madagascar
Yemen
Kiribati
Vanuatu
Tuvalu
Burma
Nauru
Pakistan
Tajikistan
Russia
Papua New Guinea
Nepal
India
Guyana
Bhutan
Bolivia
East Timor
Belize
Mongolia
Ukraine
Turkmenistan
North Korea
Kyrgyzstan
Moldova
Kazakhstan
48,49
49,11
49,41
49,42
49,72
50,48
51,82
50,80
51,86
52,02
52,05
52,17
52,29
52,31
52,57
53,06
53,14
53,45
53,80
54,07
54,59
54,71
55,27
55,74
56,55
56,6
57,25
57,41
58,44
58,61
59,24
60,18
60,26
61,45
61,52
61,53
61,57
62,51
62,57
62,74
62,75
62,77
62,86
63,04
63,07
63,17
63,49
64,00
64,11
64,76
65,06
65,11
65,24
65,70
66,35
66,38
66,46
66,46
66,51
67,14
67,39
67,88
67,90
68,27
68,28
68,63
68,74
68,84
69,20
69,45
69,51
69,63
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
Bangladesh
Iran
Honduras
Iraq
Cape Verde
Suriname
Guatemala
Greenland
Azerbaijan
The Bahamas
Belarus
Fiji
Indonesia
Trinidad
Micronesia
Philippines
Marshall Islands
Palau
Nicaragua
Vietnam
Samoa
Peru
Turkey
Uzbekistan
Brazil
Egypt
Latvia
Grenada
Montserrat
Jamaica
Armenia
Estonia
El Salvador
Seychelles
Bulgaria
Malysia
Gaza Strip
Romania
Saudi Arabia
Solomon Islands
American Samoa
Oman
Barbados
Serbia
Maldives
Mauritius
Algeria
Columbia
China
Syria
Cook Islands
Hungary
Lebanon
West Bank
Tunisia
Macedonia
Tonga
Lithuania
Antigua
Aruba
Sri Lanka
Ecuador
Croatia
Slovakia
Morocco
Dominica
Poland
Brunei
French Polynesia
Paraguay
Uruguay
Mexico
233
70,06
70,35
70,71
70,85
71,00
71,12
71,17
71,25
71,32
71,44
71,48
71,59
71,62
71,67
71,80
71,94
72,03
72,06
72,18
72,41
72,66
72,73
72,77
72,77
72,79
72,93
72,93
73,30
73,41
73,43
73,49
73,58
73,69
73,77
73,84
74,04
74,16
74,22
74,35
74,42
74,44
74,47
74,52
74,56
74,69
74,71
74,73
74,79
74,84
74,92
74,92
75,02
75,23
75,24
75,24
75,36
75,38
75,55
75,69
75,93
75,94
75,94
75,99
76,03
76,11
76,18
76,25
76,37
76,39
76,40
76,41
76,66
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
United Arab Emirates
New Caledonia
Saint Lucia
Argentina
Kuwait
Georgia
Czech Republic
Dominican Republic
Slovenia
Albania
Libya
Cuba
Costa Rica
British Virgin Islands
Panama
Chile
Bahrain
Taiwan
United States
Guam
Portugal
Denmark
Gibraltar
Saint Helena
Bosnia
Puerto Rico
Wallis and Futuna
South Korea
Finland
Virgin Islands
Belgium
Luxembourg
Malta
Faroe Islands
Austria
Saint Pierre
Greece
United Kingdom
Jordan
Germany
Ireland
Norway
New Zealand
Isle of Man
Cayman Islands
Bermuda
Netherlands
Anguilla
Iceland
Israel
Switzerland
Sweden
Spain
France
Jersey
Canada
Liechtenstein
Italy
Australia
Hong Kong
Guernsey
Andorra
Northern Mariena
Saint Kitts
Gambia
Scotland
Wales
San Marino
Singapore
Japan
Macua
Monaco
76,71
76,94
77,04
77,14
77,28
77,32
77,38
77,44
77,48
77,59
77,83
77,87
77,89
77,95
77,96
78,10
78,29
78,48
78,49
78,50
78,70
78,78
78,83
78,91
78,96
79,07
79,12
79,30
79,14
79,47
79,65
79,75
79,85
79,85
79,91
80,00
80,05
80,17
80,18
80,19
80,32
80,32
80,71
80,76
80,80
80,82
80,91
80,98
81,00
81,07
81,17
81,18
81,27
81,46
81,47
81,48
81,50
81,86
81,90
82,12
82,24
82,50
82,51
82,53
82,54
82,55
83,01
83,07
83,75
83,91
84,43
89,68
(1)
(2)
(3)
(4)
(5)
Calculate the 25th, 50th and 75th percentiles.
Calculate the 90th percentile.
Calculate the 60th percentile.
Calculate the 45th percentile.
To what factors would you attribute high life expectancy averages in the top ranking
countries?
FIVE-NUMBER SUMMARIES AND BOX-AND-WHISKER PLOTS
A five-number summary is a method for summarising a distribution of data. The five
numbers are the minimum, the first quartile, the median, the third quartile, and the maximum.
These values have been selected to give a summary of a data set because each value describes
a specific part of a data set: the median identifies the centre of a data set; the upper and lower
quartiles span the middle half of a data set; and the highest and lowest observations provide
additional information about the actual dispersion of the data. This makes the five-number
summary a useful measure of spread.
In 1977, John Tukey published an efficient method for visually displaying a five-number
summary referred to as the box-and-whisker plot. A box-and-whisker plot is a visually
effective way of viewing a clear summary of one or more sets of data. It is particularly useful
for quickly comparing different data sets.
EXAMPLE 10
Determine the five-number summary and draw a box-and-whisker plot for the following data:
1
2
2
2
3
3
4
4
4
6
7
8
8
9
10
10
10
Solution
Five number summary:
Minimum:
1
Lower Quartile ( Q1 ):
Median ( M or Q2 ):
Upper Quartile ( Q3 ):
Box-and-whisker plot:
Maximum:
10
2+3
Q1 =
= 2, 5
2
Q2 = 4
8+9
Q3 =
= 8, 5
2
Q1
Q2
234
Q3
Note:
• Half of the values lie between the minimum value and the median.
• Half of the values lie between the median and the maximum value.
• One quarter of the values lies between the minimum value and the lower quartile.
• One quarter of the values lies between the lower quartile and the median.
• One quarter of the values lies between the median and the upper quartile.
• One quarter of the values lies between the upper quartile and the maximum value.
• Half of the values lie between the lower quartile and upper quartile.
Box-and-whisker plots provide useful information about how data is distributed. Take note of
the following three types of distributions that are frequently encountered in Statistics.
Type 1
Box-and-whisker plots where the data is positively skewed.
Data is positively skewed if there are some very high values which cause the
mean to be greater than the median.
Type 2
Box-and-whisker plots where the data is negatively skewed.
Data is negatively skewed if there are some very low values which cause the
mean to be less than the median.
Type 3
Box-and-whisker plots where the data is symmetrical.
Mean = Median
Data is symmetrical if the mean and median are equal.
EXERCISE 5
(a)
Consider the following data: 1 1 2 2 4
(1)
Determine the five-number summary.
(2)
Draw a box-and-whisker plot.
(3)
Calculate the mean.
(4)
How is the data distributed? Explain.
4 6
(b)
Consider the following data: 1 2 6 8 8 8
(1)
Determine the five-number summary.
(2)
Draw a box-and-whisker plot.
(3)
Calculate the mean.
(4)
How is the data distributed? Explain.
235
8
6
8
8
8
8
10
10
10
10
(c)
Consider the following data: 2 5 7 9 12 13
(1)
Determine the five-number summary.
(2)
Draw a box-and-whisker plot.
(3)
Calculate the mean.
(4)
How is the data distributed? Explain.
(d)
Consider the following data: 2 2 2 4 4 6
(1)
Determine the five-number summary.
(2)
Draw a box-and-whisker plot.
(3)
Calculate the mean.
(4)
How is the data distributed? Explain.
(e)
A salesman at a shoe store sold eight pairs of men’s shoes one morning. The sizes of the
eight pairs of shoes were as follows:
8 12
6 10 8 12
8 7 12
9 8 12
Draw a box-and-whisker plot for this data.
(f)
A teacher records the following results for an examination out of 100:
98 63 79 76 58 71 86 78 91 87
89 41 19 88 41 99 97 83 78 90
Draw a box-and-whisker plot for this data.
(g)
The estimated overall population percentage growth rates of South Africans over the
past 13 years are given below.
1,27
1,29
1,32
1,35
1,38
1,40
1,43
1,46
1,49
1,52
1,55
1,57
1,58
1,59
Draw a box-and-whisker plot for this data.
(h)
The prize money allocated to the first ten positions in the 2015 Comrades Marathon
were as follows:
6
15
8
8
10
10
10
Position 1
Position 2
Position 3
Position 4
Position 5
R350 000
R175 000
R130 000
R65 000
R50 000
Position 6
Position 7
Position 8
Position 9
Position 10
R30 000
R25 000
R22 000
R18 500
R16 500
[http://www.comrades.com/marathoncentre/faq/2-race-info/322-medals-and-prizes]
(1)
(2)
Draw a box-and-whisker plot.
How is the data distributed in the box-and-whisker plot? Explain.
(i)
(j)
Refer to Exercise 3 no (f). Draw a box-and-whisker plot for this data.
Refer to Exercise 3 no (g). Draw a box-and-whisker plot for this data.
(k)
The box and-whisker plots below summarise the final test scores for two Mathematics
classes from the same grade.
(1)
(2)
Describe the features in the scores that are the same for both classes.
The Head of Department considers the median of each class and reports that there
is no significant difference in the performance between them. Is this conclusion
valid? Support your answer with reasons.
236
(l)
Consider the following box-and-whisker plot:
The data set contains a total of nine numbers. The second and third number of the data
set are the same. The seventh and eighth numbers are different. The mean for the data
set is 40. Write down a possible list of nine numbers which will result in the above boxand- whisker plot.
MEASURES OF DISPERSION
Measures of dispersion help us to determine how data is spread around the mean or median.
This enables one to establish whether the data is grouped closely or scattered more widely.
There are three measures of dispersion that we will consider: range, interquartile range and
semi-interquartile range.
The range
The range is the difference between the highest and lowest scores in a data set and is the
simplest measure of spread. We calculate the range as follows:
Range = maximum value − minimum value
The disadvantage of the range is that a great deal of information is ignored when calculating
the range, since only the largest and smallest data values are considered. The range is greatly
influenced by the presence of just one unusually large or small value (outlier).
The interquartile range (IQR)
The interquartile range is the difference between the upper and lower quartiles. It spans 50%
of a data set and eliminates the influence of outliers because, in effect, the highest and lowest
quarters are removed. It is a good measure of the spread of the data either side of the median.
The semi-interquartile range
The semi-interquartile range is the half the difference between the upper and lower quartiles
and is also not affected by extreme scores. It is a good measure of spread for skewed
distributions.
EXAMPLE 11
Consider the following data:
1
2
2
2
3
3
4
4
4
6
7
8
8
9
10
10
30
(a)
Determine the range, interquartile range and semi-interquartile range.
(b)
Which measure of dispersion is more suitable for this data, the range or interquartile
range?
237
Solutions
(a)
Range = 30 − 1 = 29
2+3
= 2,5
2
8+9
= 8,5
Q3 =
Upper Quartile:
2
IQR = Q3 − Q1 = 8,5 − 2,5 = 6
Lower Quartile:
Q1 =
Semi-IQR = 12 (Q3 − Q1 ) = 12 (6) = 3
(b)
The range is too inflated due to the influence of the maximum value 30.
The interquartile range is a more realistic measure of dispersion.
EXERCISE 6
[Use your work done in Exercise 5 to save you time in this exercise]
Determine the range, interquartile range and semi-interquartile range for the following sets of
data. Refer to Exercise 5 (a)-(d).
(a) 1 1 2 2 4 4 6 6 8 8 10
(b) 1 2 6 8 8 8 8 8 8 10 10 10
(c) 2 5 7 9 12 13 15
(d) 2 2 2 4 4 6 6 8 8 10 10 10
(e) Determine the range, interquartile range and semi-interquartile range for the shoe sizes
in Exercise 5 (e).
(f) Determine the range, interquartile range and semi-interquartile range for the marks in
Exercise 5 (f).
(g) Determine the range, interquartile range and semi-interquartile range for the population
growth rates in Exercise 5 (g).
(h) Determine the range, interquartile range and semi-interquartile range for the prize
money in Exercise 5 (h).
(i) Determine the range, interquartile range and semi-interquartile range for the data in
Exercise 5 (i). Which measure of dispersion is more suitable for the data?
(j) Determine the range, interquartile range and semi-interquartile range for the data in
Exercise 5 (j). Which measure of dispersion is more suitable for the data?
(k) Consider the two classes A and B in Exercise 5 (k).
(1)
Compare the classes in terms of the range and interquartile range.
(2)
Compare the classes in terms of the semi-interquartile range.
(l) Six data values are represented as follows: 3x ; x + 4 ; 2 x + 2 ; 5x ; 4 x + 1; 6 x + 2
(1)
Calculate the value of x if the mean is 12.
(2)
Determine the interquartile range.
(m) In a data set made up of five numbers, the mean, median, mode and range are all equal.
Determine this data set.
(n) The table below contains the mean, median and range of the Mathematics final exam for
a large group of students.
Mean
Median
Range
56
51
86
The Mathematics teacher added 3 marks to each of the students’ marks. Write down the
mean, median and range for the new set of Mathematics.
238
REPRESENTING GROUPED DATA
In the section on representing ungrouped data, the values we dealt with were discrete.
Discrete data can only take on certain values. For example, the number of goals scored by a
football team is discrete since the team can score 1, 2, 3 or more goals, but not 2,5 goals. The
data is restricted to natural numbers. However, data can also be continuous and take on an
infinite number of real values within a range. For example, data containing the heights of
people is continuous since heights are not restricted to integer values. One person may be 1,63
m tall, while another may be 1,64 m tall. It is also possible for someone to have a height of
say 1,6399999 metres which lies between 1,63 m and 1,64 m.
Large sets of continuous data are grouped into class intervals. A class interval has a given
range and consists of class boundaries. The upper class boundary is the maximum possible
value which could be in the class interval and the lower class boundary is the minimum
value which could be in the class interval. Since the data we are dealing with is continuous,
the class intervals will overlap since there are no “gaps” in the data as in the case of discrete
data.
We will work with continuous grouped data and discuss the following concepts: the
estimated mean, the median class interval and the modal class interval. These concepts will be
explained in the following example.
EXAMPLE 12
Medical science has always recognised human growth and height as an important measure of
the health and wellness of individuals. Research into the average height of people in different
countries revealed that the tallest race of humans is the Nilotic peoples of Sudan having an
average height of 1,83 m.
The tallest man currently living is Sultan Kösen from Turkey who measures 2,51 m. The
average heights (ranging from 150-185 cm) of people in 135 countries have been grouped into
class intervals.
[Source: https://en.wikipedia.org/wiki/Template:Average_height_around_the_world]
Class intervals
Frequency
(average heights in cm)
(number of countries)
150 ≤ x < 155
12
155 ≤ x < 160
15
160 ≤ x < 165
19
165 ≤ x < 170
25
170 ≤ x < 175
33
175 ≤ x < 180
22
180 ≤ x < 185
9
(a) Calculate an estimated value for the mean.
(b) What is the modal class?
(c) In which class interval does the median lie?
Solutions
(a)
The data is continuous and the actual average heights per class interval are not known.
It is therefore impossible to calculate the actual mean for this data. We can, however,
calculate an estimated value for this mean using the following method.
Calculate the midpoint of each class interval, which represents the average of all
heights in that class interval. Simply calculate the average of the lower and upper class
boundaries. Then multiply the frequencies with the corresponding midpoint. Add up
the results, divide by the total frequencies and calculate the estimated mean for the data.
239
Class intervals
Frequency
150 ≤ x < 155
12
155 ≤ x < 160
15
160 ≤ x < 165
19
165 ≤ x < 170
25
170 ≤ x < 175
33
175 ≤ x < 180
22
180 ≤ x < 185
Totals
9
Midpoint
150+155
2
155+160
2
160+165
2
165+170
2
170+175
2
175+180
2
180+185
2
Freq × Midpt
= 152,5
12 ×152,5 = 1 830
= 157,5
15 ×157,5 = 2 362,5
= 162,5
19 ×162,5 = 3 087,5
= 167,5
25 ×167,5 = 4 187,5
= 172,5
33 ×172,5 = 5 692,5
= 177,5
22 ×177,5 = 3 905
= 182,5
9 ×182,5 = 1 642,5
135
22 707,5
22 707,5
= 168, 2037037 cm
135
The upper boundary of the class interval 150 ≤ x < 155 can have a value that
is extremely close to 155. There may be a height of 154,999999…
It therefore makes sense that the average of the class interval is the average
of the lower and upper boundaries.
Estimated mean =
Note:
(b)
Since 170 ≤ x < 175 contains the highest frequency of heights, this class interval will
be the modal class.
(c)
It is not possible to determine the actual median. There are 135 values and therefore
the position of the median is 12 (135 + 1) = 68 . The 68th value lies in the class interval
165 ≤ x < 170 (there are 46 values below 165 and 71 below 170).
This class interval is called the median class.
Please take note that with discrete data represented in class intervals, some difficulties arise,
especially when the actual data values are not known and just the frequencies are available.
Consider the class interval 150 ≤ x < 155 . Suppose that the actual unknown data values are
whole numbers. Then the possible values in the class interval will be 150, 151, 152, 153 and
154. This means that the upper boundary will be 154 rather than 155.
The midpoint would then be 150+2154 = 152 rather than 150+2155 = 152,5 since 155 is excluded. If
154 is not in the class interval, then the upper boundary will be one of the other values. This
complicates the calculation of the midpoint. Statisticians have developed advanced formulae
to calculate estimated values for the mean, mode and median in grouped data with discrete
values. This is not part of the school curriculum and therefore we will focus on estimating the
mean in continuous grouped data.
EXERCISE 7
(a)
A stopwatch was used to find the times that it took a group of athletes to run 100 m.
Class intervals
(Time in seconds)
10 ≤ x < 15
15 ≤ x < 20
20 ≤ x < 25
25 ≤ x < 30
(1)
(2)
(3)
Frequency
(number of athletes)
6
16
21
8
Calculate the estimated mean.
What is the modal class?
In which class interval does the median lie?
240
(b)
In a research survey, a gym measured the weights (in kg) of a number of members.
Class intervals
(weights in kg)
30 ≤ x < 35
35 ≤ x < 40
40 ≤ x < 45
45 ≤ x < 50
50 ≤ x < 55
55 ≤ x < 60
60 ≤ x ≤ 65
(1)
(2)
(3)
(c)
Calculate the estimated mean.
What is the modal class?
In which class interval does the median lie?
The raw data below shows an athlete’s different times in seconds in the 400 m.
43,0 43,1 45,3 44,8 44,9 46,3 44,8 46,3 46,1
45,4 44,7 43,1 44,9 45,3 45,2 45,5 45,6 45,0
45,1 46,2 45,9 43,2 43,3 43,8 43,9 43,7 45,3
45,7 44,7 46,2 45,7 44,9 45,0 45,5 46,0 46,9
(1)
(2)
Draw a stem-and-leaf diagram for this data.
Complete the following table:
Class interval
43, 0 ≤ x < 44, 0
44, 0 ≤ x < 45, 0
45, 0 ≤ x < 46, 0
46, 0 ≤ x < 47, 0
(3)
(4)
(5)
(d)
Frequency
(number of members)
11
13
15
17
19
26
36
Frequency
Calculate the actual mean, median and mode for this data.
Calculate the range and interquartile range.
Draw a box-and whisker plot for the data.
The number of litres of diesel purchased by 30 truck drivers at a petrol station is
presented below (litres are rounded off to the nearest whole number).
82 64
55
50
49
44
52
59
68
74
71 78
88
98
96
77
75
54
57
56
64 66
80
84
88
72
71
65
68
97
(1)
(2)
(3)
(4)
Draw a stem-and-leaf display for this data.
Organise the data into class intervals of your choice.
Calculate the actual mean and median for this data.
Calculate the interquartile range.
CONSOLIDATION AND EXTENSION EXERCISE
(a)
A small company pays their employees hourly rates. The hourly rates of eight
employees are as follows:
R36 R270 R90 R72 R54 R90 R54 R54
(1)
Calculate the mean, median, mode and range for this data.
(2)
Which measure would the employer use to claim that the staff were well-paid?
(3)
Which measure would the employee use to claim that the staff were badly-paid?
241
(b)
A gardener buys ten packets of seeds from two different companies. Each pack contains
25 seeds. The gardener records the number of plants which grow from each pack.
Company A:
25 25 10 25 11 25 25 25 13 25
Company B:
22 23 20 21 23 23 22 20 22 23
(1)
Which company does the mode suggest is best?
(2)
Which company does the mean suggest is best?
(c)
A fisherman records the number of fish caught over a number of fishing trips:
3 0 0 5 0 0 13
0
2
0
0
4
16
0
2
0
1
Why does the fisherman object to the mode and median of the number of fishes caught?
(d)
A school has to select one learner to take part in a Mathematics Quiz. Sandy and Paul
took part in six trial quizzes and the following are their scores:
Sandy: 29 25 22 28 25 27
Paul:
34 20 17 33 35 19
By using the mean and range, which learner qualifies to represent the school in your
opinion?
(e)
The following frequency bar graph shows the number of laptops sold at a computer
store per week for six weeks.
Calculate the mean number of laptops sold per week.
(f)
A traffic officer is trying to work out the mean number of parking tickets she has issued
per day. She produced the table below, but some of the data has been erased.
Tickets per day
0
1
2
3
4
5
6
Totals
Frequency
1
No of tickets × frequency
1
10
7
20
2
26
72
Complete the table for her and then calculate the mean, median and mode.
242
(g)
In a certain school, 60 learners wrote examinations in Maths and Science. Information
for each subject is provided below.
Maths
Science
Minimum
30
Minimum
30
Maximum
85
Range
55
Median
45
Upper quartile
70
Lower quartile
40
Interquartile range
30
Upper quartile
50
Median
55
(1)
Draw a box-and-whisker plot for both subjects.
(2)
The teacher argues that the number of learners who scored between 30 and 45 in
Maths is smaller than the number who scored between 30 and 55 in Science.
Does she have a valid argument? Explain.
(h)
The following table represents the percentage of monthly income spent on petrol and
car expenses by fifty people. Calculate the estimated mean, modal class and the interval
containing the median.
Percentage
12 ≤ p < 18
18 ≤ p < 24
24 ≤ p < 30
30 ≤ p < 36
36 ≤ p ≤ 42
Frequency
8
20
12
8
2
(i)
The mean height of a class of 30 learners is 164 cm. A new boy of height 148 cm joins
the class. Calculate the mean height of the class now.
(j)
After five matches, the mean number of goals scored by Orlando Pirates per match is
1,8. If Pirates scores three goals in their next match, what is the mean then?
(k)
The mean weight of 27 learners in a class is 62 kg. The mean weight of a second class
of 30 learners is 59 kg. Calculate the mean weight of all the learners.
(l)
The mean monthly salary of the eight people who work for a small company is R18 000.
When an extra employee is hired, this mean drops to R17 000. How much does the new
employee earn?
(m) Consider the following set of data values: x ; 2 x − 1 ; 2 x ; 2 x + 2 ; 3 x − 1
The inter-quartile range is 6. Determine the value of x.
243
CHAPTER 11
MEASUREMENT
In Grade 9 you studied the surface area and volume of cubes, cuboids (rectangular prisms),
triangular prisms as well as cylinders. You also discussed the effect on surface area and
volume when the dimensions are multiplied by a scale factor (k). In Grade 10, we will revise
this Grade 9 work and then we also study the surface area and volume of right pyramids,
cones and spheres.
SUMMARY OF FORMULAE FOR SURFACE AREA AND VOLUME OF PRISMS
Prism
Cuboid
(Rectangular prism)
Surface area
Sum of the areas of the six
rectangles:
Surface area
= 2ab + 2ac + 2bc
Volume
Area of a chosen base
multiplied by the distance
moved by the base (height).
Volume
= (ab) × c
= abc
Cube
Sum of the areas of the six
squares:
Surface area
= 2(a)(a) + 2(a)(a) + 2(a )(a )
= 6a 2
Triangular prism
Sum of the areas of two
triangles and three rectangles.
Surface area
1

= 2  (b × h)  + ad + bd + cd
2

= bh + ad + bd + cd
= bh + d (a + b + c)
Cylinder
Sum of the areas of two
circles and a curved surface.
If the cylinder is closed:
2
Surface area = 2πr + 2πrh
If open on top (or bottom):
Surface area = πr 2 + 2πrh
If open on top and bottom:
Surface area = 2πrh
244
Area of a chosen base
multiplied by the distance
moved by the base (height).
Volume
= a3
Area of a chosen base
multiplied by the distance
moved by the base (height).
Volume
1
= (b × h) × d
2
1
= bhd
2
Area of a chosen base
multiplied by the distance
moved by the base (height).
Volume = πr 2 h
Converting between units
Converting between lengths
×
10
km
10
10
÷
m
10
10
Converting between areas
×
100
10
cm mm
km
Converting between volumes
1000
km3
1000
÷
1000
m3
1000
cm
÷
1000
1000
1000
3
100
100
m
2
100
cm
100
2
mm2
Converting between capacities
×
×
1000
100
2
mm
3
m
kl
3
÷
l
cm
ml
3
1 ml = 1 cm3
1 kl = 1 m 3
EXAMPLE 1
A closed hollow prism made of sheet metal is shown.
Calculate:
(a)
the external surface area
(b)
the volume in m 3
(c)
the volume in cm3
Solutions
(a)
Let’s first calculate the external surface area of the top triangular prism ABCDEF
which is open at the bottom.
We need the length of AC:
AC 2 = (1,96) 2 + (2,12) 2
∴ AC 2 = 8,336
∴ AC = 8,336 = 2,887213189 m
Draw a net of the prism.
The triangular prism is made up of
two triangles and two rectangles
(bottom of prism is open)
1
Area of ΔABC = (2,12)(1,96) = 2, 0776 m 2
2
1
Area of ΔEFD = (2,12)(1,96) = 2, 0776 m 2
2
Area ABFE = (5,13)(1,96) = 10, 0548 m 2
ΔEFD
ΔABC
Area ACDE = (5,13)( 8,336) = 14,81140366 m 2
245
Surface area of top triangular prism
= 2, 0776 m2 + 2, 0776 m2 + 10, 0548 m2 + 14,81140366 m2
= 29, 02140366 m2
Now let’s calculate the surface area of the bottom rectangular prism (open on top)
Area BHGF = Area CIJD
= (5,13)(2, 79) = 14, 3127 m 2
Area FGJD = Area BHIC
= (2, 79)(2,12) = 5,9148 m 2
Area HIJC
= (5,13)(2,12) = 10,8756 m 2
Surface area of rectangular prism
= 2 × 14,3127 m 2 + 2 × 5,9148 m 2 + 10,8756 m 2
= 51,3306 m 2
Therefore, the surface area of the entire prism
= 29, 02140366 m 2 + 51,3306 m 2
= 80,35 m 2
(b)
Let’s first calculate the volume of the top triangular prism ABCDEF.
The base in this prism is ΔABC .
Volume
= area of base × distance moved by base
= area of base × height
1
= (2,12)(1,96) × 5,13
2
= 10, 658088 m3
246
Now let’s calculate the volume of the bottom rectangular prism.
Volume
= area of base × distance moved by base
= area of base × height
= (2, 79)(2,12) × 5,13
= 30,342924 m3
The volume of the entire prism
= 10, 658088 m3 + 30,342924 m3
= 41, 001012 m3
(c)
The volume in cm3 is 41, 001012 m 3 × 1000 × 1000 = 41 001 012 cm 3
EXAMPLE 2
A cylindrical drinking glass is made up of a solid glass base
and a top curved part made of glass and which is hollow and
open on top. Calculate:
(a)
the total volume of the drinking glass
(b)
the capacity of the drinking glass in l
(c)
the internal surface area of the glass
Solutions
Remember that the volume is the amount of space occupied by the prism whereas the
capacity is the amount of substance that the prism can hold.
(a)
Convert 16 mm to cm:
16 mm ÷10 = 1, 6 cm
Volume (space occupied by entire glass)
= area of base × distance moved by base
= π(4)2 ×10,6
= 532,81 cm3
(b)
Capacity (amount of liquid that the glass can contain)
= area of base × distance moved by base
= π(4)2 × 9
= 452,3893421 cm3
= 0, 4523893421 l
(c)
Internal surface area (open on top and therefore excludes the top circle)
= π(4)2 + 2π(4)(9)
= 276,46 cm 2
247
EXERCISE 1 (Revision)
In this exercise, answers must be rounded off to two decimal places where appropriate.
(a)
Consider the following three closed hollow prisms:
(1)
(2)
(3)
(b)
Calculate the volume ( cm3 ) of each of the three prisms.
Calculate the surface area ( cm 2 ) of each of the three prisms.
If the cylinder and cuboid (rectangular prism) are open on top, calculate the
surface area of these prisms.
Three solid wooden objects, a rectangular prism (A), a cube (B) and a cylinder (C), are
shown below.
20
cm
π
(1)
(2)
(3)
Show that A and B have the same volume.
Calculate the value of r for which C has the same volume as A and B.
Assuming that the radius of C is the same as the value calculated in (2), which
prism will have the largest surface area?
(c)
A company manufacturing solid chocolate bars has two new packaging containers that
will have same amount of chocolate inside. The one container is a triangular prism
with an equilateral triangle as a base. The other is a rectangular prism. The company
wants to cut down on the cost of the cardboard used for making a container.
Determine which container will be the least expensive to wrap.
(d)
(1)
(2)
(3)
(4)
The surface area of a rectangular prism is 136 cm 2 . If the length is 80 mm and
the width is 4 cm, calculate the height of the prism.
If the volume of a triangular prism is 1 400 cm3 and the height is 20 cm, then
calculate the area of the base.
The surface area of a cube is 384 cm 2 . Calculate the length of a side.
The surface area of a closed cylinder is (120π) cm 2 . If the height is 7 cm,
calculate the radius.
248
The holes on a minigolf course consist of a variety of
designs. An example is shown alongside. The diagram
below shows a hole consisting of a triangular prism
made of cement (A), a rectangular prism made of
wood (B) and a larger hollow rectangular prism made of
steel which is open on top (C) and is filled with water.
Area D is a flat rectangle with the small hole for the ball
to be putted into.
The external surface area of the structure (excluding the bottom of A, B and C and the
top of C) as well as the rectangle D (excluding the small hole) is covered with a green
material.
1m
(e)
(1)
(2)
(3)
Calculate the volume of the structure (A, B and C) in cm3 and l.
Calculate the capacity of the structure (the total amount of water that can be
contained in C) in ml and l.
Calculate the amount of green material used to cover the external surface area
of the structure and the area D. Express your answer in cm2 .
THE EFFECT OF MULTIPLYING DIMENSIONS BY A SCALE FACTOR
In Grade 9, you studied the effect on volume and surface area when multiplying any
dimension by a scale factor. You will have discovered that when the dimensions of a prism
are multiplied by a number k (called the scale factor), then the relationship between the
surface area and volume is as follows:
The surface area of the new prism formed after multiplying the dimensions of the original
prism by a scale factor k is equal to k 2 × surface area of original prism. (See Ex 2 no (a))
The volume of the new prism formed after multiplying the dimensions of the original prism
by a scale factor k is equal to k 3 × volume of original prism. (See Ex 2 no (a))
If k > 1 then the new prism formed is an enlargement of the original prism.
If 0 < k < 1 then the new prism formed is a reduction of the original prism.
The original prism and the enlarged (or reduced) prism are similar to each other.
EXAMPLE 3
The surface area of a cube is 2 400 cm2 and its volume is 8 000 cm3 . Determine the surface
area and volume of the cube formed if the dimensions of the original cube are multiplied by a
scale factor of 3.
Solution
Surface area of larger cube = 32 × 2 400 cm2 = 21 600 cm2
Volume of larger cube = 33 × 8 000 cm2 = 216 000 cm2
249
EXERCISE 2
(a)
A cylinder has a height of 6 cm and a diameter of 8 cm. The dimensions are doubled.
(1) What is the scale factor?
(2) Show that the surface area of the enlarged cylinder is k 2 × surface area of original.
(3) Show that the volume of the enlarged cylinder is k 3 × surface area of original.
(b)
The surface area of a cuboid is 0,0292 m 2 and its volume is 24 cm3 .
(1) Determine the surface area (in cm 2 ) and volume (in cm3 ) of the cuboid formed
if the dimensions of the original cuboid are multiplied by 5.
(2) Suppose that the volume of the original cuboid is increased by 8 times its value.
What is the surface area of the enlarged cuboid?
(c)
The surface area of a cube is 96 x 2 and its volume is 64 x3 . Determine, in terms of x:
(1) the surface area and volume of the cube formed if the dimensions of the original
cube are halved.
(2) the length of a side of the reduced cube.
(d)
A cylinder has a height of 8 cm and a radius of 7 cm. The height remains constant but
the radius is doubled.
(1) What is the volume of the enlarged cylinder?
(2) How does the volume of the larger cylinder relate to the volume of the original
cylinder?
(3) What is the surface area of the enlarged cylinder?
(4) How does the surface area of the larger cylinder relate to the surface area of the
original cylinder?
(e)
A cylinder has a height of b units and a diameter of 2a units.
(1) Determine the volume and surface area in terms of a and b.
(2) If you want to double the volume but keep the radius the same, by what
scale factor will the height increase?
(3) If the radius is doubled but the height stays the same, by what number will
the area of the base of the cylinder increase?
(4) If the radius is doubled but the height stays the same, by what number will
the area of the side surface of the cylinder increase?
(f)
Cylinder A and B are similar. The volume of A is 240 cm3 .
(1) Calculate the scale factor and hence the volume of B.
(2) Calculate the ratio of the surface areas of A and B.
(g)
Two soup tins are similar. Tin P can hold 500 grams of soup
while tin Q can hold 750 grams of soup. The height of tin P
is 11 cm.
(1) Calculate the height of tin Q.
(2) Calculate the ratio of the areas of the circular bases.
(h)
The heights of two similar cuboids are in the ratio 4:5.
(1) Calculate the ratio of the surface areas of the cuboids.
(2) Calculate the the ratio of the volumes of the cuboids.
250
240 cm3
500 g
750 g
PYRAMIDS, SPHERES AND CONES
A pyramid is a polyhedron in which three or more triangles
are based on the sides of the polygonal base and meet in one
point called the apex of the pyramid. Right pyramids are such
that the apex is perpendicularly above the centre of the regular base.
The right pyramid shown has a square base and four congruent
triangles meeting at the apex of the pyramid.
A sphere is a perfectly round object in three-dimensional space
that resembles the shape of a completely round ball. It is not a
polyhedron. The points on the surface of the sphere are the same
distance from the centre. The radius is the straight line from any
point on the sphere to its centre.
A right circular cone is similar to a pyramid in that it has an apex.
The difference is that it has a circular base.
Formulae for the surface area and volume of right pyramids, spheres and cones
In the CAPS curriculum, we focus on the surface area and volume of right pyramids with
bases that are either equilateral triangles or squares, cones and spheres.
Solid
Right triangular pyramid
Right square pyramid
Surface area
Volume
1
Sum of the area of the base V = (A × H)
3
(equilateral triangle) and
where:
three congruent triangles.
A = area of base
H = height
1
Sum of the area of the base V = (A × H)
3
(square) and four
where:
congruent triangles.
A = area of base
H = height
251
Sphere
S = 4πr 2 where r is the
radius of the sphere
Cone
S = πrl + πr 2 where:
l is the slant height
V=
4 3
πr
3
1
V = πr 2 h
3
πrl is the area of the
curved surface
πr 2 is the area of the
circular base
EXAMPLE 4
The Louvre Pyramid located in Paris is a large right square
pyramid made of metal and glass. It serves as the main
entrance to the Louvre Museum which houses famous
paintings such as the Mona Lisa. The length of one side of
the base is 35,4 m and the height is 21,6 m. The base of the
pyramid is open.
A
Calculate:
(a)
the exterior surface area of the pyramid (metal and glass).
(b)
the volume of the pyramid.
E
B
17,7 m
35
,4
G
m
C
Solutions
(a)
F
D
3 5 ,4
m
The exterior surface area consists of the sum of the areas of the four congruent triangles:
ΔABC , ΔACD , ΔAED and ΔABE
1
1
Area of ΔABC = (BC)(AG) = (35,4)(AG)
2
2
1
1
Now GF = CD = (35, 4) = 17, 7 m
2
2
2
2
AG = (17, 7) + (21, 6)2
∴ AG 2 = 779,85
∴ AG = 27,92579453 m
1
∴Area of ΔABC = (35,4)(27,9257953) = 494, 2865631 m 2
2
252
The total exterior surface area
= Area ΔABC + Area ΔACD + Area ΔAED + Area ΔABE
= 4 × 494, 2865631 m 2
= 1 977,15 m 2
(b)
1
1
The volume (V) of the pyramid = (A × H) = (A × 21,6)
3
3
A = Area of base BCDE = (35, 4)(35, 4) = 1 253,16 m 2
1
V = (1 253,16 × 21, 6) = 9 022,75 m3
3
EXAMPLE 5
Calculate the surface area and volume of the following closed solids:
(a)
(b)
(c)
Solutions
(a)
Radius (r) = 5 cm and slant height (l) = 13 cm
Use Pythagoras to get the height (h):
h 2 = (13) 2 − (5) 2
∴ h2 = 144
∴h = 12 cm
Surface area of cone (S) = π(5)(13) + π(5) 2 = (90π) cm 2 = 282, 74 cm 2
1
Volume of cone (V) = π(5) 2 (12) = (100π) cm3 = 314,16 cm3
3
(b)
Surface area of sphere = 4 π(5) 2 = (100 π) cm 2 = 314,16 cm 2
Volume of sphere =
(c)
4
 500π 
3
3
π(5)3 = 
 cm = 523, 60 cm
3
 3 
Surface area of hemisphere (half of a sphere)
= curved surface + circular base
1
= [4π(5) 2 ] + π(5) 2 = (75π) cm 2 = 235, 62 cm 2
2
1 4
250
Volume of hemisphere g = × π (5)3 =
π = 261,80 cm3
2 3
3
253
EXERCISE 3
(a)
(b)
(c)
Calculate the surface area and volume of the following closed solids. Round your
answers off to two decimal places.
(1)
(2)
(3)
(4)
(5)
(6)
The base of the given triangular prism is an equilateral
triangle. The height of the prism is 12 cm and the length
of a side of the triangular base is 10 cm. Calculate:
(1)
the area of the triangular base ( ΔBCD )
(2)
the area of ΔABC
(3)
the surface area of the prism
(4)
the volume of the prism
A frustum is formed by removing a small cone from a
similar larger cone. The height of the small cone is 20 cm
and the height of the larger cone is 40 cm. The diameter
of the larger cone is 30 cm.
Calculate:
(1)
the radius of the small cone
(2)
the volume of the frustum
254
The given solid made of clay consists of a cone
and a cylinder. The dimensions are shown in the
diagram. The solid is re-moulded into a sphere.
Calculate the radius of the sphere.
(e)
A solid right circular cone is placed centrally
within a container in the shape of a hemisphere.
The radius of the hemisphere is 20 cm and the
distance from the circular base of the cone to the
top of the hemisphere is 2 cm. Calculate the
volume of the right circular cone. Round off
your answer to the nearest whole number.
(f)
A barn is constructed as follows:
The room space is constructed as a right rectangular
prism with a square base. The length of one side of
the base of the prism is equal to 15 metres.
The height of the wall of the room is 20 metres.
The roof is constructed in the form of a right triangular
pyramid with a height of 10 metres. The base of the roof
is open.
70 cm
20 cm
(d)
Calculate:
(1) the total exterior surface area of the barn.
(2) the volume of the barn.
CONSOLIDATION AND EXTENSION EXERCISE
(a)
A closed hollow prism is made up of a rectangular
prism and a prism with a trapezium base.
The formula for the area of a trapezium is:
1 (sum
2
of the || sides) × ⊥ height
Calculate, rounded off to two decimal places:
(1)
the volume of the prism
(2)
the external surface area of the prism
255
(b)
(c)
A closed prism of height 8 cm has a regular hexagonal
base with each side equal to 5 cm. The hexagonal base
is made up of six equilateral triangles. Calculate:
(1)
the area of the hexagonal base ABCDEF.
(2)
the volume of the hexagonal prism.
A cardboard sweet box has the form of a cylindrical
wedge. The dimensions are shown in the diagram.
The depth of the box is 7 cm.
(1)
Calculate the area of top face ABC.
(2)
Calculate the volume of the box.
(3)
If a new similar box is created to be
twice the volume of the original box,
calculate the depth of the new box.
60°
(d)
A rectangular water tank is 12 m long, 11 m wide
and 10 m high. The water depth is 8 m. A large metal
ball of diameter 7 m is dropped into the water and it
sinks to the bottom. Calculate the rise in the water
level (rounded off to three decimal places).
(e)
A right cone is fits exactly into a right square
pyramid. The radius of the cone’s base is r units.
The height of both prisms is h units.
(1)
Determine the ratio of the area of the
circle base to the area of the square base.
(2)
Hence write down the ratio of the cone’s
volume to the square pyramid’s volume?
(3)
(4)
What is the square pyramid’s volume
in terms of r and h?
h
A
1
Hence show that the cone’s volume is πr 2 h
3
256
F
r
E
B
D
C
CHAPTER 12
PROBABILITY
THE PROBABILITY SCALE
In Grade 8 and 9 you were introduced to the concept of probability and the probability scale.
Remember that the probability of an event happening is a number in the interval [0 ; 1] .
We may regard an event with a probability of 0 as impossible and events with probabilities
close to 0 as unlikely. Events with a probability of 1 are certain to happen and events with
probabilities close to 1 are likely to happen.
0
1
6
3
4
1
2
1
−20°C
We can express probabilities as fractions, decimals or percentages.
EXPERIMENTS, SAMPLE SPACES AND EVENTS
An experiment is a natural or man-made occurrence that can have different possible
outcomes. Examples include: tossing a coin, rolling a die or tomorrow’s weather. An event is
a collection of outcomes that satisfy a certain condition. If you roll a die, the die can land on
1, 2, 3, 4, 5 or 6. These are the six outcomes of the experiment. The outcomes can be
described as ‘the die lands on an even number’.
The probability of an event occuring
The probability of an event can be calculated as follows:
number of outcomes in the event
Probability of event =
total number of outcomes in the experiment (sample space)
EXAMPLE 1
A six-sided die is rolled. What is the probability that it lands on an even number?
Solution
Outcomes: { 1 ; 2 ; 3 ; 4 ; 5 ; 6}
3 1
Probability = =
6 2
Outcomes in event:
{2;4;6}
EXAMPLE 2
A coin is tossed twice. What is the probability of getting heads twice?
Solution
Outcomes:
{ HH ; HT ; TH ; TT}
Outcomes in event:
1
Probability =
4
257
{ HH }
The relative frequency of an event
When an experiment is repeated many times, the number of times a certain event takes place
is called its frequency. When this is divided by the number of times the experiment was
repeated, the result is called the relative frequency of the event:
frequency of event
Relative frequency of event =
Total number of times experiment is repeated
When the experiment is repeated many times, the relative frequency should eventually
approach the calculated or theoretical probability of the event.
EXAMPLE 3
A coin is tossed 100 000 times. The coin lands on heads 49 996 times. Calculate the relative
frequency of this event and establish whether it approximates the theoretical probability.
Solution
1
= 0,5
2
frequency of event
49 996
=
= 0, 49996
Relative frequency =
Total number of times experiment is repeated 100 000
Probability that the coin lands on heads =
Since 0, 49996 ≈ 0, 5 , the relative frequency approximates the theoretical probability.
EXAMPLE 4
A die is rolled 12 000 times. Approximately how many times do you expect the die to land on
a factor of 6.
Solution
Outcomes: { 1 ; 2 ; 3 ; 4 ; 5 ; 6 }
Factors of 6: { 1 ; 2 ; 3 ; 6 }
4 2
=
6 3
frequency
f
2
=
≈
Relative frequency =
number of repetitions 12 000 3
2
∴ f ≈ ×12 000 = 8 000 times
3
EXAMPLE 5
Theoretical probability =
The letters of the word EXCELLENCE are written on different cards and placed in a box.
(a) Determine the probability of taking out, at random, the letter E.
(b) Determine the probability of taking out, at random, the letter X or C.
(c) Suppose that the letter N is taken out of the box and not placed back into the box.
What is the probability of now taking out one letter E?
Solutions
(a)
Probability = 104 =
(b)
Probability of X or C = 101 + 102 = 103
If the letter N is not replaced, then there will be 9 letters to choose from.
Probability = 94
(c)
2
5
[there are 4 E’s out of a total of 10 letters in the given word]
258
EXERCISE 1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
70 tickets were sold in a competition. The prize is a smartphone. Mpho decided to buy
12 tickets. What is the probability that he will:
(1) win the prize?
(2)
not win the prize?
A six-sided die is thrown. Determine the probability of:
(1) throwing a 6
(2)
throwing a 3 or a 4
(3) throwing an even number
(4)
not throwing a 2
A container is filled with 5 blue blocks, 8 red blocks, 6 green blocks and 9 white
blocks. A block is taken out of the container at random. Find the probability of taking
out:
(1) a blue block
(2)
a white block
(3) a red or green block
(4)
a brown block
(5) any block that is not white
A cupboard contains 5 shirts, 3 pairs of jeans and 8 pairs of socks.
(1) What is the probability of taking out, at random, one pair of socks?
(2) What is the probability of taking out, at random, a shirt or pair of jeans?
(3) What is the probability of not taking out a pair of jeans?
(4) Assuming that you have already taken out a pair of socks from the cupboard and
put them on your feet. What will the probability now be of taking out a shirt?
There are 52 cards in a standard deck of cards of which 13 are hearts. A card is drawn
at random, returned, and then the deck is reshuffled. This is repeated 50 000 times.
Which of the following do you consider to be the most reasonable answer for the
number of times that a card of hearts was chosen? Motivate your answer.
A. 31 210
B. 2003
C. 12 685
D. 25 443
The letters of the word IMAGINATION are written on different cards and placed in a
box.
(1) Determine the probability of taking out, at random, the letter I.
(2) Determine the probability of taking out, at random, the letter G or N.
(3) Suppose that the letter T is taken out of the box and not placed back into
the box. What is the probability of now taking out one letter A?
A card is drawn from a pack of 52 cards. Determine the probability of drawing:
(1) a heart
(2)
a jack of clubs
(3)
an ace
(4) a king or queen
(5)
neither a heart or a spade
Complete the following table. Determine the probabilities in common fraction form.
Scenario
A standard six-sided die
is rolled.
Additional
information
No additional
information.
Two standard six-sided
dice are rolled (one green
and one red)
A couple has two
children
The number on the
green is less than the
number on the red.
No additional
information.
A couple has two
children
A couple has two
children
The oldest child is
male.
At least one of the
children is male.
259
Event
The die lands
on a number
less than 3.
The sum of the
two numbers is
seven.
At least one of
the children is
female.
Both children
are male.
Both children
are male.
Probability
of event
(1)
(2)
(3)
(4)
(5)
(i)
A dart is thrown at random onto a board that has the
shape of a circle as shown in the figure. Calculate the
probability that the dart will hit the shaded region.
The two circles are concentric (have the same centre).
(j)
An arrow is shot at random onto rectangle ABCD.
E, F, G and H are the midpoints of the sides of rectangle
ABCD. Let ED = x and DH = y .
Calculate the probability of the arrow:
(1) landing in ΔAEF .
(2) landing in the shaded region.
(3) landing in either in ΔCGH or the unshaded area.
PROBABILITY IN TERMS OF SETS – VENN DIAGRAMS
The set of all possible outcomes of an experiment is called the sample space and is denoted
by the symbol S. When rolling a die, the sample space is given by the set
S = {1 ; 2 ; 3 ; 4 ; 5 ; 6} . An event is a subset of the sample space and is denoted by a given
capital letter A, B, C and so forth. When rolling a die, if A is the event in which the die lands
on an even number, then A = {2 ; 4 ; 6} . We use a special diagram, called a Venn diagram to
represent events in a sample space. The sample space is represented by a rectangle and the
events by circles inside the rectangle. There are three types of Venn diagrams:
(1) Venn diagrams showing the actual outcomes
(2) Venn diagrams showing the number of outcomes
(3) Venn diagrams showing probabilities
EXAMPLE 6
A die is rolled. Let A be the event in which the die lands on an even number. Draw a Venn
diagram showing the:
(a)
outcomes
(b) number of outcomes
(c) probabilities
Solutions
(a)
S
(b)
A
3
3
The actual outcomes
are shown. The three
even numbers 2, 4 and
6 are written in the
circle with the odd
numbers 1, 3, and 5 on
the outside of the circle.
The number of outcomes
are shown. The number
3 for there being 3 even
numbers is written inside
the circle and the
number 3 for there being
3 odd numbers is written
outside of the circle.
260
S
(c)
1
2
A
1
2
The probability of
getting an even
number is 12 . This
is written inside
the circle. The 12
outside the circle
is for the odd
numbers.
The formula for the probability of an event
n(E)
n(S)
where P(E) is the probability of event E occurring, n(E) is the number of outcomes in E and
n(S) is the number of outcomes in the sample space.
The probability of event E occurring is given by the formula:
P(E) =
EXAMPLE 7
The letters of the word ENGLISH are written on cards and placed in a hat. One card is drawn
randomly. Let A be the event in which a consonant is drawn.
(a)
Draw a Venn diagram showing all outcomes.
(b)
Write down the outcomes of the sample space (S) and event A in set form.
(c)
Write down n(A) and n(S)
(d)
Calculate P(A)
Solutions
(a)
(b)
(c)
(d)
Sample space S = {E ; N ; G ; L ; I ; S ; H}
Event A = { N ; G ; L ; S ; H}
n(A) = 5
n(S) = 7
n(A) 5
P(A) =
=
n(S) 7
Sometimes we can consider more than one event simultaneously. When doing this we must
establish whether the events overlap (share outcomes or not).
EXAMPLE 8
A 12-sided dodecahedral die is rolled.
The following events are defined:
A = {multiples of 3}
B = {factors of 9}
C = {multiples of 5}
(a)
(b)
(c)
List the outcomes in set form.
Draw a Venn diagram to represent these events.
Determine P(A) , P(B) and P(C) .
Solutions
(a)
A = {3 ; 6 ; 9 ; 12}
B = {1 ; 3 ; 9}
C = {5 ; 10}
Notice that A and B have outcomes 3 and 9 in common whereas C has no outcomes
in common with A or B.
(b)
(c)
261
n(A) 4 1
= =
n(S) 12 3
n(B) 3 1
P(B) =
= =
n(S) 12 4
n(C) 2 1
P(C) =
= =
n(S) 12 6
P(A) =
EXERCISE 2
(a)
(b)
(c)
A six-sided die is rolled. Let A be the event “getting an odd prime number”.
(1) Write down the set A, n (A) and P(A) .
(2) Draw a Venn diagram showing the actual outcomes.
(3) Draw a Venn diagram showing the number of outcomes.
(4) Draw a Venn diagram showing the probabilities.
The letters of the word RANDOMLY are written on cards and placed in a hat. One card
is drawn randomly. Let A be the event in which a vowel is drawn.
(1) Draw a Venn diagram showing all outcomes.
(2) Write down the outcomes of the sample space (S) and event A in set form.
(3) Write down n (A) and n(S)
(4) Calculate P(A)
A twelve-sided die is rolled. The following events are defined:
A = {the first five natural numbers} B = {the first two multiples of 5}
(1) Write down the set A, n (A) and P(A) .
(2) Write down the set B, n (B) and P(B) .
(3) Draw a Venn diagram to represent the actual outcomes.
(4) Draw a Venn diagram to represent the number of outcomes.
(5) Draw a Venn diagram to represent the probabilities.
(d) A twelve-sided die is rolled. The following events are defined:
A = {the first five prime numbers}
B = {the first two multiples of 6}
(e)
(f)
(1) Write down the set A, n (A) and P(A) .
(2) Write down the set B, n (B) and P(B) .
(3) Draw a Venn diagram to represent the actual outcomes.
(4) Draw a Venn diagram to represent the number of outcomes.
(5) Draw a Venn diagram to represent the probabilities.
Consider the given Venn diagram.
(1) Write down the sets S, A, B and C.
(2) Write down n (A) , n (B) and n (C) .
(3) Calculate P(A) , P(B) and P(C) .
(4) Redraw the Venn diagram showing the
number of outcomes.
(5) Redraw the Venn diagram showing the
probabilities.
In a survey conducted at a music store in Johannesburg, it was found that 120 people
bought only Trance music, 150 bought only Deep House music and 100 people bought
both. Twenty people did not buy either Trance or Deep House music.
Let T = {Trance music} and D = {Deep House music}
(1)
(2)
(3)
(4)
(5)
(6)
Draw a Venn diagram to represent events T and D.
Determine n (T) , n (D) and n(S) .
Calculate the probability of selecting, at random, a person who likes only Deep
House music.
Calculate the probability of selecting, at random, a person who likes only Trance
music.
Calculate the probability of selecting, at random, a person who likes neither types
of music.
Calculate the probability of selecting, at random, a person who likes both types of
music.
262
DERIVED EVENTS (COMPLEMENT, INTERSECTION AND UNION)
The complement of an event
The complement of an event A is the event consisting of
all outcomes that are in the sample space, but not in A.
We write the complement of A as not A.
For example, if you roll a die and A is the event in which
the die lands on an even number, then:
S = {1 ; 2 ; 3 ; 4 ; 5 ; 6}
A = {2 ; 4 ; 6}
not A = {1 ; 3 ; 5}
The intersection of events
The intersection of two events, event A and event B, is
the event consisting of all outcomes that are in both A
and B simultaneously. We write this intersection of
event A and event B as A and B.
For example, if you roll a die and A is the event in which
the die lands on an even number and B is the event that the
die lands on a prime number, then:
A = {2 ; 4 ; 6}
B = {2 ; 3 ; 5}
A and B = {2}
The union of events





The union of two events, event A and event B, is the event
consisting of all outcomes that are in at least one of these
events. The union consists of outcomes that are either in A,
or in B, or in both. This basically means that we put all
of the outcomes of A and B together by uniting them into
one big set. We write the union of event A and event B as
A or B. For example, if you roll a die and A is the event in
which the die lands on an even number and B is the event that the
die lands on a prime number, then:
A = {2 ; 4 ; 6}
B = {2 ; 3 ; 5}
A or B = {2 ; 3 ; 4 ; 5 ; 6}
EXAMPLE 9
In a certain experiment, the sample space is S = {a ; b ; c ; d ; e ; f ; g ; h} .
Events A and B are defined as follows:
A = {a ; b ; c ; d ; e}
B = {d ; e ; f ; g ; h}
(a)
Represent all of the outcomes in a Venn diagram.
(b)
List the outcomes in:
(1) not A
(2)
A and B
(3)
A or B
What is the value of:
(1) n(not A)
(2)
n(A and B)
(3)
n (A or B)
Determine:
(1) n(not A and B)
(3) n(not A or B)
(5) P(not A or not B)
(2)
(4)
(6)
P(not A and B)
P(not A or B)
P(not A and not B)
(c)
(d)
263
Solutions
(a)
(b)
(1) not A = { f ; g ; h}
(2)
A and B = {d ; e}
(1) n(not A) = 3
(2)
n(A and B) = 2
(2)
P(not A and B) =
(4)
P(not A or B) =
(6)
not A and not B = {
}
P(not A or not B) =
0
8
(3) A or B = {a ; b ; c ; d ; e ; f ; g ; h}
(c)
(3) n(A or B) = 8
(1) not A = { f ; g ; h}
(d)
B = {d ; e ; f ; g ; h}
3
8
∴ not A and B = { f ; g ; h}
(3) not A = { f ; g ; h}
B = {d ; e ; f ; g ; h}
5
8
∴ not A or B = {d ; e ; f ; g ; h}
(5) not A = { f ; g ; h}
not B = {a ; b ; c}
not A or not B = {a ; b ; c ; f ; g ; h}
P(not A or not B) =
6
8
=
3
4
EXAMPLE 10
The diagram shows the subjects taken by the matric
group in Kwazamakuhle FET school with respect to
Geography (G) and Accounting (A).
(a) How many learners take:
(1)
both Geography and Accounting?
(2)
neither Geography nor Accounting?
(3)
at least one of the subjects?
(4)
Geography but not Accounting?
(b) If a learner is chosen at random from this group, determine:
(1)
P(not A)
(2)
P(G or (not A))
(3)
P((not G) and A)
(4)
P(not(G and A)).
Solutions
(a) (1)
(3)
30 learners
(2)
90 learners
10 + 30 + 20 = 60 learners
(4)
10 learners
264
=0
(b) (1)
n(not A) = 10 + 90 = 100
100 2
∴ P(not A) =
=
150 3
n(S) = 150
(b) (2)
Mark G with  and (not A) with ×
Take all numbers marked with either 
or × or both. This is a total of:
10 + 30 + 90 = 130 learners
130 13
∴ P(G or (not A)) =
=
150 15
×
×
Mark (not G) with  and A with ×
Take the number with both markings.
∴ n((not G) and A) = 20 learners
20
2
∴ P((not G) and A) =
=
150 15
(3)
×
×
n(G and A) = 30
∴ n(not(G and A)) = 10 + 20 + 90 = 120
120 4
∴ P(not(G and A)) =
=
150 5
EXAMPLE 11
(4)
In a group of 50 people, 30 can speak Afrikaans (A) and 40 can speak IsiZulu (I). There are
25 of these people that can speak both languages.
(a)
Draw a Venn diagram showing this information.
(b)
How many of these 50 people cannot speak either of the two languages?
(c)
If a person is chosen from this group at random, find the probability that this person
can speak:
(1) at least one of the two languages
(2) IsiZulu but not Afrikaans
Solutions
(a)
Start in the middle with the intersection and then work your way outwards.
40 − 25
30 − 25
50 − 5 − 25 − 15
(b)
(c)
5 people
5 + 25 + 15 45 9
=
=
50
50 10
15 3
(2) Probability =
=
50 10
(1) Probability =
265
EXERCISE 3
(a)
(b)
(c)
(d)
(e)
(f)
A twelve-sided die is rolled.
(1) Write down the sample space in set form.
(2) Determine the probability that the die will land on an even number.
(3) Determine the probability that the die will land on a prime number.
(4) Determine the probability that the die will land on an even and prime number.
(5) Determine the probability that the die will land on an even or prime number.
(6) Determine the probability that the die will land on a prime number, given that the
number is greater than 8.
You are given the sample space S = {a ; b ; c ; d ; e ; f ; g} with the following events:
A = {a ; b ; c}
B = {e ; f }
C = {c ; d ; e}
Determine:
(2)
P(A and C)
(3)
P(A or C)
(1) P(A)
(4) P(B or C)
(5)
P(not A)
(6)
P(not B)
A = {a ; b ; c ; d ; e} , not A = { f ; g ; h} and B = {c ; d ; e ; f } are given. Determine:
(3)
(1) the sample space S (2)
P(A)
P(A and B)
(4) P(A or B)
(5)
(6)
P(not A)
P(not B)
A twelve-sided die is rolled. Suppose that the following events are given:
A = {multiples of 3}
B = {factors of 12}
Determine:
(2)
(3)
(1) n(A and B)
n (A or B)
n(not A)
(4) n(not B)
(5)
(6)
P(A and B)
P(A or B)
(7) P(not A)
(8)
(9)
P(not B)
n((not A) and B)
(10) P((not A) and B)
(11) n(A and (not B))
(12) P(A and (not B))
(13) n((not A) or B)
(14) P((not A) or B)
(15) n(A or (not B))
(16) P(A or (not B))
(17) n(not(A and B))
(18) P(not(A and B))
P((not A) and (not B))
(19) n(not(A or B))
(20) P(not(A or B))
(21)
In a recent sports survey, it was found that 120 people enjoy watching cricket only, 95
people enjoy watching rugby only and 45 people enjoy watching both sports. There
were 40 people who don’t watch either sport.
(1) Draw a venn diagram to illustrate this information.
(2) How many people watch cricket in total?
(3) How many people watch rugby in total?
(4) How many people were there in the survey?
(5) Determine the probability that a person selected watches both sports.
(6) Determine the probability that a person selected watches none of the sports.
(7) Determine:
(ii)
(i)
P(C or (not R))
P(R and (not C))
In a survey on brain diseases conducted by medical researchers, it was found that of a
total of 1 520 genes, 454 are associated with Alzheimer’s disease, 1 091 are associated
with Multiple Sclerosis and 40 genes are associated with both diseases.
(1) Draw a venn diagram to illustrate this information.
(2) How many genes are not associated with either of the diseases?
(3) Determine the probability that a gene, selected at random, will be associated with
Alzheimer’s disease and Multiple Sclerosis.
(4) Determine the probability that a gene, selected at random, will be associated with
at least one of the diseases.
(5) Determine the probability that a gene, selected at random, will be associated with
only one of the diseases.
266
(6)
(g)
(h)
(i)
Determine:
(ii)
(i)
P((not A) or (not M))
P((not A) and (not M))
A travel agency recorded the travel destinations of 60 South African tourists last month.
Of the 60 tourists, 30 visited Europe, 28 visited the UK, 9 visited America and 12
visited both Europe and the UK. Five tourists did not visit America, Europe or the UK.
(1) Draw a Venn diagram to represent this information.
(2) Determine the probability that a tourist, selected at random, visited Europe only.
(3) Determine the probability that a tourist, selected at random, visited America.
(4) Determine the probability that a tourist, selected at random, visited both Europe
and the UK.
(5) Determine:
(ii)
(i)
P((not E) or A)
P((not A) and UK)
The probability that Tumi will not see a movie today is 0,3. The probability that he will
go to a restaurant today is 0,6. The probability of him seeing a movie and going to a
restaurant today is 0,4. Determine the probability that he:
(1) doesn’t go to a movie or a restaurant.
(2)
only goes to a movie.
(3) only goes to a restaurant.
(4)
doesn’t go to a movie.
(5) doesn’t go to a restaurant.
(6)
goes to either one or the other.
100 boys – 60 from school P and 40 from school K – were included in a survey in
which they were asked whether or not they liked Mathematics. The results were as
follows:
Like Mathematics
16
40
School K
School P
Don’t like Mathematics
24
20
A learner is chosen at random from this group. Suppose that the following events are
A = {learners from school P} and B = {learners who don't like Maths}
defined:
Describe the event (not A) and B in words.
Calculate:
(i)
(ii)
(iii)
P(A)
P(A or B)
P(A and (not B))
The following diagram shows the gender and qualification of 10 applicants for a job.
A stands for Accounting, E stands for Economics and M stands for Management.
(1)
(2)
(j)
A
(1)
(2)
(3)
M
E
A
M
A
E
M
E
M
Determine the probability that a randomly selected applicant is female and has an
Economics qualification.
In order to appoint a candidate for the job, the company demands that the person
is female or has a qualification in Management. How many applicants qualify for
appointment?
A person is selected randomly from the applicants. Let event A be the event in
which a male applicant is chosen and event B the event in which an applicant
with an Accounting qualification is chosen. Determine:
n (A or B)
(ii)
n(not (A or B))
(iii)
P(A and B)
(i)
(iv)
P(A and (not B))
(v)
P(not A or B)
[Hint: A Venn diagram is useful for the questions in (3)]
267
FURTHER PROBABILITY QUESTIONS USING VENN DIAGRAMS
The following events represented by shaded regions in a Venn diagram are useful:
A and (not B)
A or (not B)
not (A and B)
(not A) and B
(not A) or B
not (A or B)
EXAMPLE 12
Given two events, A and B, with P(A) = 0, 4 , P(B) = 0, 5 and P(A and B) = 0,1
Determine P((not A) and B) .
Solution
0,4 − 0,1
Remember
that all
probabilities
add up to 1
0,5 − 0,1
1 − 0,3 − 0,1 − 0,4
Now mark (not A) with  and B with ×
Take the number marked with both  and
∴ P((not A) and B) = 0, 4
×
×
×
EXAMPLE 13
If P(A) = 0,3 ; P(B) = 0, 6 and P(A or B) = 0, 7 , find P(A or (not B)) .
Solution
We are not given the intersection (A and B)
so let P(A and B) = x
P(A or B) = 0, 7
∴ (0,3 − x) + x + (0, 6 − x) = 0, 7
0,3 − x x 0,6 − x
∴ 0,9 − x = 0, 7
∴ x = 0, 2
Now update the Venn diagram.
To locate A or (not B):
Mark A with  and (not B) with ×
Take all numbers marked with either  or × or both.
∴ P(A or (not B)) = 0,1 + 0, 2 + 0, 3 = 0, 6
268
×
0,1
0,2 0,4
×
0,3
EXERCISE 4
(a)
Consider the following Venn diagrams:
A:
C:
B:
(b)
(c)
(d)
(e)
(f)
D:
In each case, choose the diagram in which the shaded region represents the event:
(1) (not X) and Y
(2)
not (X and Y)
(3)
(not X) or Y
(4) (not X) and (not Y)
(5)
(not X) or (not Y)
(6)
not (X or Y)
What can you conclude about [not (X or Y)] and [(not X) and (not Y)]?
What can you conclude about [not (X and Y)] and [(not X) or (not Y)]?
X and Y are events such that P(X) = 0,6 ; P(Y) = 0,3 and P(X and Y) = 0, 2 .
Draw a Venn diagram and hence find the value of P((not X) or (not Y)) .
X and Y are events such that P(X) = 0,35 ; P(Y) = 0,7 and P(X and Y) = 0, 2 .
Draw a Venn diagram and hence find the value of P((not X) and (not Y)) .
Two events, A and B, are such that P(A) = 0,5 ; P(not B) = 0, 7 and P(A and B) = 0, 2
Draw a Venn diagram and hence determine P((not A) or B) .
If P(A) = 0,6 ; P(B) = 0,5 and P(A or B) = 0,9 , determine:
P((not A) and B)
P(A or (not B))
(1) P(A and B)
(2)
(3)
In a class of 28 learners, 20 take Science and 15 take Biology. There are 3 learners
who don’t take either of these two subjects. Draw a learner at random. Let C be the
event in which the learner chosen takes Science and B the event in which the learner
takes Biology. Determine:
P(C and B)
P((not C) and B)
(2)
(3)
(1) P(C or B)
(4) P((not B) or (not C))
THE FUNDAMENTAL LAWS OF PROBABILITY
The following two laws are always valid and form the basis of probability theory. When used
correctly, they can make many probability problems much easier to solve.
Consider the following Venn diagram:
6
5
2
P(A) =
P(B) =
P(A and B) =
10
10
10
9
P(A or B) =
10
6 5 2
9
Now P(A) + P(B) − P(A and B) = + − =
10 10 10 10
Therefore we can conclude that P(A or B) = P(A) + P(B) − P(A and B)
4
6
4
and 1 − P(A) = 1 − =
Also P(not A) =
10
10 10
Therefore we can conclude that P(not A) = 1 − P(A)
269
Summary of the two rules of probability
P(A or B) = P(A) + P(B) − P(A and B)
P(not A) = 1 − P(A)
or
or
P(A and B) = P(A) + P(B) − P(A or B)
P(A) = 1 − P(not A)
EXAMPLE 14
If P(A) = 0, 7 ; P(B) = 0,5 and P(A and B) = 0, 4 , determine:
P(A or B)
P(not A)
P(not (A or B))
(a)
(b)
(c)
Solutions
(a)
P(A or B) = P(A) + P(B) − P(A and B)
∴ P(A or B) = 0,7 + 0,5 − 0,4 = 0,8
(c)
P(not (A or B)) = 1 − P(A or B)
∴ P(not (A or B)) = 1 − 0,8 = 0, 2
(b)
P(not A) = 1 − P(A)
∴ P(not A) = 1 − 0,7 = 0,3
EXAMPLE 15
If P(not A) = 0, 25 ; P(A or B) = 0,8 and P(A and B) = 0,15 , determine:
P(A)
P(B)
P(A and (not B))
(a)
(b)
(c)
P((not A) or B)
(d)
Solutions
(a)
P(A) = 1 − P(not A)
∴ P(A) = 1 − 0,25 = 0, 75
(b)
P(A or B) = P(A) + P(B) − P(A and B)
∴ 0,8 = 0,75 + P(B) − 0,15
∴ P(B) = 0, 2
(c)
Use a Venn diagram:
(d)
Use a Venn diagram:
×
×
×
×
∴ P((not A) or B) = 0,15 + 0, 05 + 0, 2 = 0, 4
∴ P(A and (not B)) = 0, 6
Mutually exclusive events
Two events are mutually exclusive if they do not share any common outcomes and can
therefore never both take place at the same time. If events A and B are mutually exclusive,
then A and B = ∅ (the empty set). Also P(A and B) = 0 which means that:
P(A or B) = P(A) + P(B) − P(A and B)
∴ P(A or B) = P(A) + P(B) − 0
∴ P(A or B) = P(A) + P(B)
For mutually exclusive events:
P(A and B) = 0 and P(A or B) = P(A) + P(B)
n(A and B) = 0
P(A and B) = 0
P(A or B) = P(A) + P(B)
270
Exhaustive and complementary events
Two events A and B are said to
be exhaustive if, together,
they cover all elements of the
sample space, i.e. P(A or B) = 1
P(M or N) =
6
=1
6
P(X or Y) =
8
=1
8
Two events, A and B, are said to be
complementary if they are both
exhaustive and mutually exclusive.
∴ P(A or B) = 1 and P(A and B) = 0
P(A or B) = P(A) + P(B) − P(A and B)
∴1 = P(A) + P(B) − 0
P(A or B) =
∴ P(A) + P(B) = 1
6
=1
6
P(A and B) =
0
=0
6
For complementary events:
P(A and B) = 0 (mutually exclusive) and P(A or B) = 1 (exhaustive) and P(A) + P(B) = 1
EXAMPLE 16
If P(A) = 0,3 and P(B) = 0, 4 where A and B are mutually exclusive events, determine:
P(A and B)
P(A or B)
P(not (A or B))
(a)
(b)
(c)
Solutions
(a)
P(A and B) = 0
P(not (A or B)) = 1 − P(A or B)
∴ P(not (A or B)) = 1 − 0,7 = 0,3
(b)
P(A or B) = P(A) + P(B) = 0,3 + 0,4 = 0, 7
(c)
EXAMPLE 17
A and B are mutually exclusive events with P(not A) = 0,3 and P(A or B) = 0,8
(a)
Determine P(B) .
(b)
Explain why A and B are not complementary.
Solutions
(a)
(b)
P(A) = 1 − P(not A) = 1 − 0,3 = 0, 7
P(A or B) = P(A) + P(B)
∴ 0,8 = 0,7 + P(B)
∴ P(B) = 0,1
P(A) + P(B) = 0, 7 + 0,1 = 0,8 ≠ 1
Also P(A or B) = 0,8 ≠ 1 which means that
A and B are not exhaustive.
Therefore A and B are not complementary
(mutually exclusive but not exhaustive)
271
EXAMPLE 18
If P(A) = 0, 6 and P(B) = 0, 5 , prove that A and B cannot be mutually exclusive.
Solution
If A and B are mutually exclusive then: P(A or B) = P(A) + P(B) = 0, 6 + 0,5 = 1,1
This is impossible since probabilities can never exceed 1. Therefore the events can never be
mutually exclusive.
If two events are not mutually exclusive then
there will be an intersection and P(A and B) ≠ 0 .
The two events are then said to be inclusive.
In Example 18, the two events are inclusive.
There is an intersection and P(A and B) = 0,1 ≠ 0
Note:
EXERCISE 5
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
If P(A) = 0, 5 ; P(B) = 0, 7 and P(A and B) = 0, 3 , determine:
(1) P(A or B)
(2)
(3)
P(not A)
(4) P(not (A or B))
(5)
(6)
P(not (A and B))
If P(A) = 74 ; P(A or B) =
(l)
and P(A and B) = 17 , determine:
(2)
(3)
(1) P(B)
P(not A)
P(not (A or B))
(4) P((not A) or B)
(5)
(6)
P(A or (not B))
P((not A) and B)
If P(not A) = 0, 4 ; P(A or B) = 0, 9 and P(A and B) = 0, 2 , determine:
(1) P(A)
(2)
(3)
P(B)
P(A and (not B))
(4) P((not A) or B)
(5)
(6)
P(A or (not B))
P((not A) and B)
If P(A) = 0,38 ; P(B) = 0, 45 and P(not(A or B)) = 0, 4 , determine:
(1) P(A or B)
(2)
(3)
P(A and B)
P(A and (not B))
If P((not B) and A) = 0,3 ; P(A and B) = 0,1 and P(not (A or B)) = 0, 2 , determine
(1) P(A or B)
(2)
(3)
P(B or (not A))
P(B and (not A))
A and B are mutually exclusive events such that P(A) = 0, 6 and P(B) = 0, 3 .
Determine P(A and B) and P(A or B) .
A and B are mutually exclusive events such that P(A) = 0, 25 and P(not B) = 0,58 .
Determine P(A or B) .
The events C and D are mutually exclusive with P(not C) = 0,3 and P(C or D) = 0,8 .
Determine P(D) .
If P(A) = 0, 4 and P(A or B) = 0,5 , determine P(B) if:
(1) A and B are mutually exclusive
(2)
P(A and B) = 0,3
Determine whether the following events are complementary if S = {a ; b ; c ; d ; e ; f ; g}:
A = {a ; b ; c ; d } and B = {e ; f }
(1) A = {a ; b ; c ; d } and B = {d ; e ; f ; g}
(2)
(3)
(k)
6
7
P(not B)
P((not A) and B)
A = {a ; b ; c ; d } and B = {e ; f ; g}
If P(A) = 0, 25 ; P(B) = 0, 5 and P(A or B) = 0, 625
(1) Calculate P(A and B)
(2) Are events A and B are complementary? Give reasons.
A smoke detector system in a large warehouse uses two devices, A and B. If smoke is
present, the probability that it will be detected by device A is 0,95. The probability that
it will be detected by device B is 0,98 and the probability that it will be detected by both
devices simultaneously is 0,94. What is the probability that the smoke will not be
detected?
272
CONSOLIDATION AND EXTENSION EXERCISE
(a) In a survey on internet usage, it was found that out of a total of 27 people, 16 use ADSL
lines, 15 use wireless internet and 3 use neither of the two. What is the probability that a
person chosen at random uses both internet connections?
(b) Simon and his girlfriend decide to go out one evening. They can see a movie, go to a
restaurant or do both. The probability of them seeing a movie is 0,6. The probability
of them going to a restaurant is 0,7. The probability of them seeing a movie without
going to a restaurant is 0,2. What is the probability of them not seeing a movie and not
going to a restaurant?
(c) If n(S) = 46 , n(A) = n(B) = 19 and n(not (A or B)) = 11 , determine:
(1) P(A and B)
(2)
(3)
P(A or B)
P((not A) and B)
(d) Determine whether the following events are complementary:
(1) A and B, if P(A) = P(not B) = 52
(2) C and D, if P(C) = 12 , P(D) = 23 and P(not (C or D)) = 0
9
8
(3) F and G, if P(not F) = 11
, P(G) = 11
and F and G are mutually exclusive.
(e) G and H are inclusive events in a sample space S. If it is given that P(G or H) = 43 ,
P(G) = 25 and P(H) = 12 , determine P(G and H) and the value of P(not G) .
(f) Two events, K and L, are such that P(K) = 0, 7 ; P(L) = 0, 4 and P(K or L) = 0,8 .
Determine:
(1) P(K and L)
(2) P(K and (not L))
(3) n(L) if K = {a ; b ; c ; d ; e ; f ; g }
(g)
(h)
(i)
Given event A = {1; 2 ; 3 ; 4} , event B = {5 ; 6} and P(A or B) = 23 .
(1)
Determine the values of P(A) and P(B) .
(2)
A third event C is mutually exclusive with A as well as B and P(B or C) = 13 .
Determine P(A or C) .
The diagrams below represent a class of learners. A is the set of girls and B is the set of
learners that like rugby.
Indicate the diagram representing:
(1) the girls who like rugby
(2)
the boys who like rugby
(3) the girls who dislike rugby
(4)
the boys who dislike rugby
(6)
P((not A) and B)
(5) P(A and B)
(7) P(A and (not B))
(8)
P((not A) and (not B))
You are given a group of eight students studying different degrees in Management.
Consider the gender and degree in Logistics (L), Marketing (M) or Human Resources
(HR)) that a student from this group is registered for:
L
M
HR
L
M
L
HR
M
Calculate the probability that a student selected at random from this group:
(1) is male
(2)
is female and studies Marketing
(3) studies Logistics or is male
(4)
must be female and studies Logistics
273
CHAPTER 13
SOLUTIONS TO EXERCISES
CHAPTER 1 (ALGEBRAIC EXPRESSIONS)
EXERCISE 1
(a)
(b)
(1)
2; 9
(2)
3
(4)
−3; ; 0; 2; 9
4
(1)
Irrational
(2)
(5)
Neither
(6)
(c)
(1)
(5)
(d)
(1)
(5)
Mixed
Terminating
4
9
28
225
0; 2; 9
(5)
(2)
(6)
(2)
(6)
−3;0; 2; 9
(3)
2
3
−3; ; 2 ; 9 ; 0; 2
4
(4)
Rational
(8)
Irrational
(6)
Irrational
Irrational
(3)
(7)
Rational
Rational
Mixed
Terminating
7
33
124
999
(3)
Natural
(4)
13
90
371
−
330
(3)
(7)
Integer
1751
90
106
−
45
(4)
(8)
EXERCISE 2
(a)
(1)
(5)
(b)
(1)
(c)
(1)
(5)
7 and 8
(2)
5 and 6
1 and 2
(6)
3 and 4
19
(There are other possibilities)
30
9,236
(2)
67,24
80,000
(6)
34,2785
(3)
3 and 4
(2)
−8 and −7
(4)
3,1409 (There are other possibilities)
(3)
(7)
4,3769
5,55556
(4)
(8)
17,24740
9, 425
EXERCISE 3
(a)
(1)
(2)
0
1
3
2
(8)
(b)
(1)
(1)
6
7
8
−1
0
1
0
1
2
3
2
3
4
4
{ x : 0 ≤ x ≤ 4 ; x ∈}
(2)
5
5
6
6
(7)
7
(3)
(4)
4
(5)
−3
10
−5
5
7
−6
(8)
6
(d)
(1)
[ −7 ; 9 )
(2)
( −2 ;11]
(3)
2
[−6; ∞ )
274
{x : x ≥
(4)
(6)
(7)
1
3
4
7
(2)
3
4
0
{ x : −6 ≤ x < 4 ; x ∈}
7
−1
1
4 12
{ x : x < −7 ; x ∈ }
−2
12
−3
(5)
2
11
10
9
(3)
(3)
(c)
5
−4
(4)
(6)
4
3
8
(4)
1

 −∞ ;1 
2

5 ; x ∈}
5
6
7
EXERCISE 4
(a)
(b)
(c)
(1)
3x 2 + 9 x
(2)
−9a4 + 18a3 − 3a2
(3)
x2 + 7 x + 10
(4)
x2 − 7 x + 10
(5)
x2 + 3x − 10
(6)
x2 − 3x − 10
(7)
6 x2 + 7 x − 3
(8)
21m2 + 22m − 8n2
(9)
6 x8 − 5 x 4 y 2 − 6 y 4
(10)
8 x 8 − 16 x 4 y 3 + 6 x 4 y 5 − 12 y 8
(1)
(4)
x3 + 3x 2 + 5 x + 3
x3 − 3 x 2 + 5 x − 3
(3)
(6)
2 x 3 − 2 x 2 − 10 x + 4
4a3 + 5a 2b − 5ab2 + 2b3
(3)
− 4 x 2 + 16 xy − 19 y 2
2 x3 − 10 x 2 + 14 x − 4
(2)
(5)
6 x 3 + 10 x 2 y − 7 xy 2 + y 3
(7)
27 x 3 − 8 y 3
(8)
27 x 3 + 8 y 3
(1)
− x 2 − 6 xy
(2)
34 y 2 + 16 y − 7
(4)
2 x6 − 9 y 2
(5)
18a 3 + 45a 2 b − 2ab 2 − 5b3
EXERCISE 5
(a)
(b)
(1)
x2 − 49
(2)
x2 − 9
(3)
4 x2 − 1
(4)
81x2 − 16
(5)
9x2 − 4 y2
(6)
16a6b2 − 9
(7)
36 − 9 x 8 y 2
(8)
9 x 2 − 12 x + 4 − y 2
(9)
1 − 2a 4 + a8
(1)
x2 + 10 x + 25
(2)
x2 − 10 x + 25
(3)
4a2 + 12a + 9
(4)
4a2 − 12a + 9
(5)
a 2 − 8ab + 16b2
(6)
a2 + 6ab + 9b2
(7)
9a 2 − 30ab + 25b2
(8)
3 x 2 − 18 xy + 27 y 2
(9)
4m 2 − 32mn + 64n 2
(10)
x 6 − 6 x 3 y 6 + 9 y12
(11)
8a 3 + 36a 2 b + 54ab 2 + 27b3
(12)
8a 3 − 36a 2 b + 54ab 2 − 27b3
EXERCISE 6
(a)
(b)
(1)
6 x ( x 2 + 2)
(2)
2 x 2 (3 x + 2)
(3)
5 x ( x 2 + 1)
(4)
6 x 2 (2 x − 3)
(5)
3( x 2 − 3 y + 4 xy )
(6)
8 ab ( ab − 8)
2 3
2 5
4
(7)
(1)
( a + b )( x + y )
(2)
( x + y )( k + p )
(3)
( q + r )(3 p − 4 m )
(4)
( m − 3n )(7 k − 3 p )
(5)
( x − y )( x − y − 3)
(6)
( a + c ) 4 [1 + ( a + c ) ]
(7)
(m − n)2 (m − n) 4 + 1
(8)
( m − 3n )(7 x + 4 y )
(9)
( m − 3n )(7 x − 4 y )
(10)
(m + 3n)(7 x − 4 y )
(11)
2(3 p + q)( x − 2 y )
(12)
( a − b)(1 + p )
(14)
2 x(3a − b)( x + 6)
(15)
(a − 3b)(1 + c − d )
(2)
(5)
( x + 6)( x − 6)
(3)
(6)
(3 x + 2)(3 x − 2)
(9)
( a 2 + 4)( a + 2)( a − 2)
(3)
a (9a + 7b)(9a − 7b)
(6)
(9)
25 p 4 q − 100 p 4 q 3
(13)
4 m n (4 m n − 2 mn + 9)
3
( a − 2)(4 x − 1)
EXERCISE 7
(a)
(b)
(1)
(4)
( x + 4)( x − 4)
( x − 1)( x + 1)
4
3
4
3
(13 x + 10)(13 x − 10)
2
(7)
(10 x + y )(10 x − y )
(8)
x ( x − 1) x + 1)
(1)
( n 4 + 9)( n 2 + 3)( n 2 − 3)
(2)
3(2 x + 5 y )(2 x − 5 y )
(4)
(7)
3a (3a + b )(3a − b )
(a + b + c)(a + b − c)
2
4
4
(5)
(8)
4 y ( x + 2)( x − 2)
(2)
(5)
(8)
(11)
(14)
(2)
(5)
( x − 12)( x + 7)
4 x( x + y )
(4 a + 11b )(4 a − 11b )
(9a − 4b)(a + 4b)
EXERCISE 8
(a)
(b)
(1)
(4)
(7)
(10)
(13)
(1)
(4)
( x − 7)( x − 4)
( x − 6)( x + 2)
( a − 7)( a − 5)
( m − 6)( m + 1)
( x + 4)( x + 3)
2( x − 3)( x + 2)
6(a + 5)(a − 1)
( a − 5)( a − 4)
( a − 8)( a + 6)
( m + 6)( m − 1)
( x − 4)( x − 3)
3( x − 9)( x + 2)
x ( x + 4)( x − 2)
275
(3)
(6)
(9)
(12)
(15)
(3)
(6)
( x + 3)( x + 3)
( a − 12)( a + 1)
( m + 3)( m + 2)
( m − 3)( m − 2)
( x + 6)( x − 2)
4( a + 5)( a − 2)
5( x − 6)( x − 3)
(c)
(d)
The only product option for the last term is 5 × 1 . The signs in the brackets have to be the same. There is no
way to obtain the middle term 4 x using the option of 5 × 1 .
(1)
The product of the lasts is not + 6 . The signs in the brackets are supposed
to be the same.
(2)
The product option 6 × 4 is incorrect because + 6 − 4 ≠ 10 .
EXERCISE 9
(a)
(b)
(c)
(1)
(4)
(7)
(10)
(1)
(4)
(4 p − q )( p + 2 q )
(2)
(5)
(8)
(11)
(2)
(5)
(1)
10( x − 4 y )( x + 3 y )
(4)
( x 2 + 3 y 2 )( x 2 + 2 y 2 )
(3 x + 1)( x + 1)
(6 x − 1)(3 x + 1)
(2 x − 3)( x − 2)
(10 x − 3)(2 x + 3)
2(2 x − 1)( x + 3)
(5 m + n )(2 m − 3n )
(3)
(6)
(9)
(12)
(3)
(6)
(2)
a ( a − 3b )( a − 3b )
(3)
( x 2 − 8)( x + 2)( x − 2)
(5)
(a + b + 8)( a + b − 4)
(2)
( x − y )( r + t )
(3)
( p − q )(2 + n )
(6)
(9)
( x − y )(b − a )
(12)
( x + 2)( x 2 − 2)
(2 x − 1)( x − 1)
(2 x + 1)( x − 3)
(3 x + 2)(2 x − 5)
(6 x − 5)(3 x + 2)
3(5 x − 1)( x − 1)
(4 x − 1)(3 x − 1)
(5 x + 4)( x + 2)
(3 x − 7)(2 x + 3)
(5 + 3 x )(3 − 2 x )
( x − 3 y )( x + 2 y )
(4 a + 5b )(3a + 2b )
EXERCISE 10
(a)
(1)
( x + y )( p + k )
(4)
(7)
(b)
(5)
(8)
( a − 2)( a + b )
( x + y )(b − a )
2
( x + 1)(4 x + 5)
( a − b )(1 − c )
2
( x − y )(3a − 1)
(10)
( x + y )(3a − 1)
(11)
( x − 2)( x − 2)
(13)
(16)
(1)
(4)
(2 x − 3)( x 2 − 3)
(14)
3( x + 1)( x − 1)( x + 3)( x − 3) (15)
(b + 1)(a − 1)
(2)
(5)
( x − b)( x − a )
( x − 3)(3 x + 2)
2ap(3a + 2 p)( x − y )
2(3x − 1)( x + 3)( x − 3)
(a + b)(a − b − 1)
(3)
( a − 2b + 4 x )( a − 2b − 4 x )
EXERCISE 11
(a)
(1)
(3 x − 1)(9 x 2 + 3 x + 1)
(2)
(2 x + 1)(4 x 2 − 2 x + 1)
(3)
(4 x − y )(16 x 2 + 4 xy + y 2 )
(4)
(5 − 9 x)(25 + 45 x + 81x 2 )
(5)
1  2 2 1
1

 ab −   a b + ab + 
2
2
4


(6)
5( x + 2)( x 2 − 2 x + 4)
(7)
8 a ( a − 2)( a 2 + 2 a + 4)
(8)
−( x + 3)( x 2 − 3x + 9)
1
 1 2

 x + 6   x − 2 x + 36 
3
 9

1  2
1 

 x + x  x − 1 + 2 
x



(9)
(11)
(b)
( x − 2)( x + 2)( x 4 + 4 x 2 + 16)
(1)
(10)
(3 − a )(3 + a 2 )
(12)
1  2
1 

 x − x  x + 1 + 2 
x



(2)
( x − 2)( x 2 + 2 x + 4)( x + 2)( x 2 − 2 x + 4)
EXERCISE 12
(a)
(2a + 5)( a − 3)
2
(b)
5(a − 3)(a − 1)
(c)
x 2 (1 − x )(1 + x )
(d)
(2 x − 3)(4 x + 6 x + 9)
(e)
( x − 3)( x − 2)
(f)
2(3x + 5)( x − 2)
(g)
(2 x − 1)(2 x + 1)( x − 1)
(h)
( x − 2)( x + 1)
(i)
2(2 a − b )(4 a 2 + 2 ab + b 2 )
(j)
7 x7 y7 ( y7 − 2 x7 )
(k)
4(2 x − 5)(2 x + 5)
(l)
(1 − x )(1 + x + x 2 )
(m)
6(2 x 2 − 3 xy + 1)
(n)
6(2 x − y )( x − y )
(o)
3(2 y + 3)( y − 2)
2
4
2
4
(p)
4 x y (2 x y − 1)
(q)
(2 x − y )(2 x + y − 3)
(r)
( y − 1)(2 x − 6)(2 x + 6)
(s)
(5 − x + y )(5 + x − y )
(t)
(a − b)(a − b + 3)
(u)
( x + a )( x − a + 4 x 2 )
EXERCISE 13
(a)
(1)
x −1
(2)
a−2
(3)
276
p−2
2
(4)
m +1
m
(b)
(5)
2(a − 1)
(6)
x+2
x
(7)
x−2
2
(8)
x+3
2x + 1
(9)
3( p − 9)
2p
(10)
3p − 2
3
(11)
−3 − k
k
(12)
3
3k − 2
(13)
−2
(14)
x−2
(15)
x−3
x+3
(2)
x2
x +1
(3)
(5)
k2 + 3
2
(6)
(1)
(4)
( x + 3)( x + 4)
2x
−k −1
6
EXERCISE 14
4x + 1
(a)
(1)
6
(6)
(9)
−12 x
( x + 3)( x − 3)
(10)
(1)
(4)
(c)
18 x y
2
5 − 3xy − 6 y
6 xy
(1)
1
6
9 x2 − 1
9
x4 − 1
x2
2x 2 + x
−2 x + 3 x − 10
2 x ( x − 2)
(2)
36 x 2 − y 2
6
(2)
−5 x 2 + 4 x − 8
( x + 2)( x 2 + 2)
4x2 y
16 x 2 − 8 x + 1
16
x 4 + 8 x 2 + 16
(6)
x3
2x − 3
x
(8)
3
2
11x + 1 − 5 x 3 y 2 + 4 x 2 y
(3)
x6 + 1
(5)
(4)
2
7x + 2
x ( x + 1)
(11)
(2 x + 1) 2
3x2 − 4 x
3x − 2
x
(7)
−2 x 2 − 3 x + 3
(13)
3x + 2
(3)
3
(5)
(12)
(b)
21x 2 y − 4 x 2 + 6 y
(2)
( x − 2 y )( x − 2)( x + 2)
8
−7
3( x − 1)
4 x2
EXERCISE 15
(a)
(1)
x−2
x−3
(2)
x−2
x+3
(3)
2 x2 + 5x − 1
( x − 3)( x + 3)
(4)
(5)
−9 x − 5
(2 x + 1)( x + 5)
(6)
−x − 6
x
(7)
x2 + 2x − 2
6 x ( x + 1)
(8)
(9)
(b)
(1)
x2 + 2x + 4
( x − 4)( x + 1)
x3 − 2 x − 8
x ( x − 2) 2
2 x2 − 3x + 1
(2 x + 1)(4 x 2 − 2 x + 1)
Line 2: 3x× (+1) is not equal to 3x + 1 (2)
Line 1: Identified the LCD incorrectly
2
3x − 6 x − 2
3 x (3 x + 1)
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(b)
(c)
(d)
(1)
Rational
(5)
Rational
(9)
Rational
(1)
7 and 8
1
1 ;2;3
2
(1)
− 2 −1 0 1 2
(2)
(6)
(10)
(2)
3
Rational
Irrational
Rational
−8 and −7
(2)
4
(3)
(e)
(1)
(3)
(7)
(11)
(3)
−2 −1 0
(4)
4 5 6 7
6 x 2 − 5 xy − 6 y 2
(2)
Rational
Rational
Rational
2 and 3
− 5
x8 − 1
277
1
(4)
(8)
(12)
Irrational
Rational
Rational
2 3
(3)
5
−12 x 2 + 60 x − 75
(f)
(g)
(4)
27 x 3 − 64 y 3
(5)
−2x 2
(7)
10 x 4 − 24 x 2 y 3 + 12 y 6
(8)
125a3 − 150a 2b + 60ab 2 − 8b3
(9)
(1)
(3x + 4)( x − 3)
(2)
3( x − 4)( x + 1)
(3)
−(3 x − 2)( x + 6)
(i)
x4
(5)
( x − 4)( x + 4 x + 16)
(6)
( x − 8)( x + 8)
(7)
4( x − 5)( x + 5)
(8)
4 x( x − 25)
(9)
1( x 2 + 64)
(10)
( x − 3)( x − 2)( x + 2)
(11)
( x − 3)( x 2 + 4)
(12)
−( x − 8)( x + 8)
(13)
2
(3 x − 1)( x + 2)
(14)
(3 x − 1)( x − 2)
(15)
2(1 − 2 x )(1 + 2 x + 4 x 2 )
(1)
x+2
2
(2)
−27 y − 24 xy − 68 x 2
12 xy
(3)
(4)
x2 − 2
2x
(5)
−(2 − x)2
(8)
2
x + 2x + 4
2
9
4x 2
2 x2 − 8x + 3
2
( x − 3) ( x − 2)
33 x 3 + x + 6
12 x 2
(6)
5 x 2 − 3x + 4
x ( x − 2)
(9)
x( x − 2)( x + 2)
16
22
7
(1)
π = 3,141592654..... Irrational
(3)
π is an irrational number but
(1)
−9 x 4 + 21x 2 − 4
(2)
(3 x − 3 x 2 + 2)(3 x + 3 x 2 − 2)
(j)
( x + 1)( x 2 + x + 1)
(k)
x−2
x
(l)
(1)
( p − 8)( p − 4)
(2)
(a + b − 4)(a + b − 8)
(2)
364
(2)
(q)
a − b = ± 18
2009
x2 +
1
(n)
(p)
(r)
(1)
(s)
125
x( x − 8)( x + 8)
= 51
x2
a 2 + b 2 = 14
(1)
a = 2, b = 3, p = 12
(m)
2
( a + 3) 2 − 4b 2
(4)
(7)
(h)
(6)
(1)
1
( x − 4)( x + 2)
2
22
22
is rational. In fact,
is a rational approximation of π .
7
7
(2)
y−x
xy
(2)
(t)
1
(3 x − 2)(2 x − 1)
2
1
2
(o)
2
(3)
1
(2 x − 3)( x − 2)
14
CHAPTER 2 (EXPONENTS)
EXERCISE 1
(a)
(1)
16 p 5
(2)
36x 3 y 7
(3)
512
(4)
326
(5)
(9)
16 807
(6)
(10)
105
515
8
(7)
(11)
(8)
(12)
1213
32
(13)
9x 6 y10
(14)
16 a 6
(15)
(16)
72 x13
(17)
215. 6 21
(21)
64
(25)
(b)
3
25
2a
(18)
(22)
−b 6
5
16
(26)
310. 2 9
36
 6a 2 b3 


(19)
1
256
(23)
x2
(27)
y2
2
x2
2a8
3
(24)
1
x4
x4
y4
1
4a 4 b 2
(1)
2
(2)
1
(3)
18
(4)
(6)
1
64
(7)
1
4
(8)
121
(9)
278
(20)
1
x5
4
x2
(5)
(10)
1
a4
1
16 x2
3
(11)
(d)
(12)
1
27a
(13)
3
(16)
49
(17)
18
(18)
(21)
16x 4
(22)
27
125
(23)
(26)
ab7 c
(27)
(31)
(c)
a
3
(1)
(6)
(11)
(16)
3a 5 c
2b16
1
1
(32)
1
3x 2 y 2
1
2
(28)
(33)
1
9
2
8
125
3
5
(14)
1
64
(15)
x4
(19)
36
(20)
x4
2
(24)
2 187
(25)
(29)
y11
x6
(3)
(8)
(13)
(18)
y 22
1
16
128
a 90 b 30
(4)
(9)
(14)
(19)
1
−32
2 187
1
a12
16a12
(21)
243x 5 y 5
(22)
− 16 x 8 y12
(23)
128 x14 y 21
(24)
− 243 x15
(1)
6x3
(2)
5 x5
(3)
− x4
(4)
(6)
−8a5b5
(7)
5 a10 + 10b10 (8)
−4x 7
(11)
12 x 5 − 4 x 6
1
125
8
27
−64
−32a
15
x4
3x 4
2
x12
(2)
(7)
(12)
(17)
−1
(30)
y5
(5)
(10)
(15)
(20)
−1
−a9
6 x3 y 4
(5)
16 x10 + 3 x 9
(9)
2x4 y2
(10)
−28a 6
(4)
8
(5)
3
(9)
63
(10)
4
9
99 33
−a b
−2 187
− 81x 4 y 4
EXERCISE 2
(a)
(b)
(c)
(1)
37
(2)
27
(3)
(6)
1
125
(7)
4
25
(8)
(11)
3
2
(12)
49
(13)
27
4
(14)
512
3
(15)
2
(1)
22 x
(2)
25 x
(3)
33 x
(4)
2 x+ 3
(5)
53 x. 35 x
(6)
34 x
(7)
310 x
(8)
1
1
2
1
5
(9)
1
(10)
1
(11)
5
(12)
1
(13)
(1)
3
(2)
8
(3)
(6)
8
(7)
7
(8)
(11)
3
4
(12)
1
6
(13)
5
(3)
(3)
(14)
2
(15)
22 x
(4)
4
(5)
36
(9)
2p
(10)
(14)
1
5
(15)
2
(4)
3
(5)
2a
m
5
(4)
(5)
1
3
(5)
1
2
(5)
6x
9
1
25
3x
1
125
1
18
EXERCISE 3
(a)
(1)
(6)
3
3y
(2)
(b)
(1)
2x2
(2)
(6)
1
2
x
5
x2
7
EXERCISE 4
(a)
(b)
(1)
1
(2)
(6)
8
(7)
(1)
(6)
3x − 1
(2)
(7)
4x − 7
1
2
3
2
2x + 3
3x + 2
(3)
(8)
(3)
(8)
2
5
1
4
5x − 6
2x − 3
279
(4)
1
(9)
1
6
(4)
(9)
6x − 6
5x + 1
EXERCISE 5
(a)
(b)
(c)
1
2
(1)
x=0
(2)
x =1
(3)
x=3
(4)
x=2
(5)
x=
(6)
x=4
(7)
x=
7
3
(8)
x=
1
8
(9)
x=3
(10)
x=5
(11)
x=4
(12)
x=2
(13)
x = −2
(14)
x = −2
(15)
x = −3
(16)
x = −2
(17)
x = −2
(18)
x=3
(19)
x=3
(20)
x=
(1)
x =1
(2)
x = − 32
(3)
x=−
(4)
x = −4
(5)
x=5
(6)
x=
3
2
(7)
x=4
(8)
x=3
(9)
x = −9
(10)
x=2
(11)
x=2
(12)
x = −1
(13)
x = −2
(14)
x=
(15)
x=0
(1)
x=2
(2)
x=2
(3)
x=2
(4)
x =1
(5)
x=
x = 25
x=9
x = 256
x = 81
x = 36
x=9
(2)
(7)
(2)
x=8
x=4
x = 256
(3)
(8)
(3)
x =1
x = 64
x = 2 401
x =1
(4)
(9)
(4)
x = 32
x=3
x = 256
x = 16
(5)
x=8
(5)
x = 214
x =1
(5)
−1
3
2
1
5
3
2
3
2
EXERCISE 6
(a)
(b)
(1)
(6)
(1)
(6)
CONSOLIDATION AND EXTENSION EXERCISE
(a)
9
(1)
4x
2
(2)
3
(3)
b+a
ab
(4)
1
a+b
(6)
18x16
(7)
11x8
(8)
216x9
(9)
− 1024x10 y 5 (10)
(11)
625
8
(12)
1
2
(13)
8
3
(14)
2
(16)
11x − 2
(17)
2x − 9
(18)
5x
(1)
x = −1 (2)
x=
(6)
x=
(c)
(1)
43 x
(2)
3 . 4x
(d)
(1)
223
(2)
635
(e)
(1)
226
(2)
330 > 4 20
(b)
5
3
(3)
x = 16
(4)
x = 25 (5)
(15)
3 . 2 21
CHAPTER 3 (NUMBER PATTERNS)
EXERCISE 1
(a)
(1)
Tn = 3n + 3 ; T100 = 303
(2)
(3)
; T100 = 498
Tn = 6n + 4 ; T100 = 604
Tn = −5n + 10 ; T100 = −490
Tn = −5n − 1 ; T100 = −501
Tn = −6n + 1 ; T100 = −599
Tn = 2 n + 12 ; T100 = 200 12
Tn = 0, 2n + 0,3 ; T100 = 20,3
Tn = −10n + 11 ; T100 = −989
(4)
(5)
(7)
(9)
(11)
(13)
(15)
(17)
Tn = 5n − 2
(6)
(8)
(10)
(12)
(14)
(16)
(18)
280
; T100 = 405
Tn = 4n − 1 ; T100 = 399
Tn = 7n − 3 ; T100 = 697
Tn = −3n + 3 ; T100 = −297
Tn = −4n + 9 ; T100 = −391
Tn = 12 n + 3 ; T100 = 53
Tn = 34 n − 12 ; T100 = 74 12
Tn = 6n − 19 ; T100 = 581
Tn = −n + 14 ; T100 = −86
Tn = 4n + 5
4 x8
1
x = 625 x = 81
2
3
(3)
9 y6
(1)
; T45 = 311
Tn = −3n + 22 ; T65 = −173
(2)
(1)
(1)
T1 = 5 ; T2 = 14 ; T3 = 23 ; T4 = 32
Tn = (3n + 1)(8n − 1)
(2)
(2)
T50 = 60 249
(b)
(1)
(c)
(d)
(e)
Tn = 7n − 3
(2)
; T90 = 627
n = 45 ; T45 = −113
n = 110 ; T110 = 996
n = 90
EXERCISE 2
(a)
(b)
(c)
(1)
Tn = 2 × 2 n −1 ; T10 = 1 024
(2)
Tn = 1 × 3n −1 ; T10 = 19 683
(3)
Tn = 4 × 3n −1 ; T10 = 78 732
(4)
Tn = 32 × ( 12 ) n −1 ; T10 =
(5)
Tn = ( − 2) × (3) n −1 ; T10 = −39 366
(6)
Tn = ( 12 ) × (2) n −1 ; T10 = 256
(7)
Tn = 16 × ( 14 ) n −1 ; T10 =
1
16 384
(8)
Tn = ( 12 ) × ( 12 ) n −1 ; T10 =
(9)
Tn = 28 × ( 14 ) n −1 ; T10 =
7
65 536
(1)
Tn = n 2
(2)
Tn = n 2 + 1
(3)
Tn
(5)
Tn
Tn =
; T100 = 10 000
2
= n + 3 ; T100 = 10 003
= n 2 − 1 ; T100 = 9 999
(4)
Tn
(6)
Tn
1
16
1
1 024
; T100 = 10 001
= n + 4 ; T100 = 10 004
= n 2 − 2 ; T100 = 9 998
2
n −1
1× 3
n2 + 4
CONSOLIDATION AND REVISION EXERCISE
(a)
(1)
Tn = 9n − 2
; T300 = 2 698
(2)
n = 50 ; T50 = 448
(b)
(c)
(1)
(1)
(3)
(5)
(7)
(9)
Tn = −3n + 1 ; T145 = −434
Figure 4:
14
Figure 8:
26
Figure n:
Tn = 3n + 2
n = 90 Figure 90 will contain 272 dots
T900 = 4 500
(2)
(2)
(4)
(6)
(8)
(10)
n = 130 ; T130 = −389
Figure 4:
20
Figure 8:
40
T186 = 560
Figure n:
Tn = 5n
n = 130 Figure 130 will contain 650 lines.
(d)
(1)
Design 1:
2 = T1 = (1) 2 + (1)
Design 2:
6 = T2 = (2) 2 + (2)
Design 3:
12 = T3 = (3) 2 + (3)
Design 4:
20 = T4 = (4) 2 + (4)
Design 5:
30 = T5 = (5) 2 + (5)
(2)
Design n:
Tn = n 2 + n
(e)
(1)
Tn =
(f)
(g)
T999 = 1
(1)
E
4n + 2
5n − 2
T1000 = 1500
(2)
W
(3)
T20 = (20) 2 + (20) = 420
(2)
T20 =
(3)
497
41
49
CHAPTER 4 (EQUATIONS AND INEQUALITIES)
EXERCISE 1
(a)
(1)
x = − 32
(2)
x = −8
(3)
p = −8
(4)
x=4
(b)
(5)
(1)
m=2
x = −24
(6)
(2)
x=3
x = −6
(3)
x = − 12
(4)
y = 23
− 75
(8)
k =1
(5)
a =8
(6)
m = 24
(7)
x=
(2)
x = −4 ; x = 8
(3)
x = 0 ; x = −3
(4)
x=
(2)
(6)
x = 7 ; x = −7
(3)
(7)
x = 10 ; x = −10
x = −2 ; x = 2
(4)
(8)
x =0 ; x=9
x=0;x=4
EXERCISE 2
(a)
(b)
(1)
x = 3 ; x = −6
(5)
x=
(1)
(5)
x = 3 ; x = −3
x=0;x=
(9)
x = 4 ; x = −2
(10)
x =2 ; x =3
(11)
x = − 13 ; x = 2
(12)
x=
5
2
; x = −2
(13)
x = −8 ; x = 1
(14)
m = 6 ; m = −1
(15)
x = −6 ; x = 2
(16)
x=
2
3
; x = − 23
5
2
; x = − 13
32
5
1
9
x=0;x=
5
2
281
p = 9 ; p = −5
(17)
(21)
(25)
non-real solution
x = 12
(18)
non-real solution
(22)
(26)
x = 4 ; x = −2
(19)
x = − 32 ; x = 12
(20)
non-real solution
(23)
x = 2 ; x = −1
(24)
x = 3 ; x = −3
(3)
(7)
no solution
x = 54
(4)
(8)
x=0
x = −1
(3)
(7)
real numbers
x=0
(4)
no solution
(c)
x =1 ; y = 2
(d)
x =1 ; y =1
(3)
(3)
x =1 ; y =1
x=2;y=3
(4)
(4)
x = 5 ; y = −2
x = 7 ; y = 14
(3)
x=0 ; x=−
(4)
x=−
(3)
g=
(3)
v=
x=4
EXERCISE 3
(a)
(b)
(1)
(5)
x = −18
x = −5
(2)
(6)
no solution
(1)
(5)
x = 6 ; x = −3
x = 13 ; x = −1
x=4
(2)
(6)
x = 4 ; y =1
(b)
x = −2 ; y = −4
x=
15
4
x=0;x=
24
9
EXERCISE 4
(a)
(e)
x =8 ;y =2
1
2
(f)
x=
;y=3
(2)
(2)
(6)
x = 4 ; y = 10
x = −2 ; y = 4
x = 7 ; y = 14
EXERCISE 5
(a)
(b)
(c)
x = 6 ; y = −2
x =1 ; y = 0
x = 4 ; y = −3
(1)
(1)
(5)
R24
EXERCISE 6
(a)
(1)
(b)
(1)
v −u
t
c −b
x=
a
a=
(1)
(d)
(1)
(e)
(1)
(2)
x = 3y ; x = 2 y
(5)
(c)
(2)
(6)
2S − nL
n
uf
v=
u− f
a=
(2)
E
m
c=
v2 − u 2
2a
c
x=
3a − 2b
3
3
x=− y;x= y
2
2
5F − 160
C=
9
s=
v 2 − 2 as
(2)
u=
(2)
x > −8
b
a
1
1
y;x= y
2
2
mv 2
Fr
Fgr
m
EXERCISE 7
(a)
(1)
x ≤ −3
(4)
x≥5
−3
(3)
x <1
−8
1
5
(b)
(1)
x < −6
(2)
y > −6
(5)
−6
(c)
(1)
−1 ≤ x < 6
(4)
−2 < x < 2
(3)
y ≤ − 12
− 12
−6
−6
(4)
x ≥ −6
x≥0
0
−1
6
−2
2
(2)
−5 < x < 2
(5)
−4 ≤ x < 2
−5
2
−4
2
282
(3)
−3 ≤ x ≤ 2
(6)
0< x≤4
−3
0
2
4
CONSOLIDATION AND EXTENSION EXERCISE
x=0 ;x=4
(4)
x = 2 ; x = −2
(7)
x = −2
(8)
x=0;x=
x = 5 ; x = −5
(11)
x = −2
(12)
x=
(14)
−2 < x ≤ 5
(15)
x ≥ 17
33
(2)
x=3
(3)
x = ±3
(4)
x=0 ;x=9
(2)
(2)
(2)
x = 4 ; x = −2 ; x = 3 ; x = −1
x = −1
(2)
x = − 12
(5)
x = 5 ; x = −2
(6)
x=2;x=−
(9)
x=
(10)
(13)
x > −6
(1)
−12 <
(1)
(1)
(1)
x=2
x<6
p =8 ;p =3
x = 1 ; y = −2
x = 56 ; y = 90
4
17
3
2
x = 1 ; y = −2
x = 56 ; y = 90
(3)
3
x≤
(h)
(1)
real numbers
(2)
1
(i)
23
120
(2)
(j)
(1)
(3)
(1)
(k)
(1)
x = −1
(l)
(n)
152
10 metres
–2
x=
; x = −3
2
− 103
(g)
y
5
2
1
2
>
(b)
(c)
(d)
(e)
(f)
(3)
(1)
1
2
x
2
1
6>
4
–2
–3
>
(a)
21
–4
–
>
6
x = 0 ; x = −1
x=2
2y − 5
x=
; y≠4
y−4
(m) 200 of the R200 tickets were sold and 50 of the R300 tickets were sold.
x=3
(o)
(2)
(2)
CHAPTER 5 (TRIGONOMETRY)
EXERCISE 1
hypotenuse
adjacent
θ
(a)
opposite
hypotenuse
(b)
θ
adjacent
opposite
k
m
k
tan K =
g
sin K =
(1)
(3)
(2)
cos θ =
p
q
r
q
(6)
tan α =
p
r
(1)
EF = 676 = 26
(2)
(e)
(1)
sin N =
(f)
(1)
(1)
sin θ =
(5)
cos α =
(d)
r
q
k
m
r
tan θ =
p
8
17
24 4
cos L =
=
30 5
(3)
10 5
=
26 13
3k 3
tan C =
=
(2)
4k 4
16 4
cos B =
=
20 5
sin θ =
(4)
cos G =
(5)
(c)
(2)
cos θ =
(6)
(4)
24 12
=
26 13
(2)
Both equal
(3)
(7)
(11)
(15)
(3)
(3)
2,05
11,2
0,04
0,57
1
2,91
tan θ =
4
5
g
m
g
sin G =
m
cos K =
g
k
p
sin α =
q
tan G =
10 5
=
24 12
(3) Triangles are similar
EXERCISE 2
(a)
(b)
(c)
(1)
(5)
(9)
(13)
(1)
(1)
(5)
0,92
1,41
0,31
1,84
1
0,05
0,19
(2)
(6)
(10)
(14)
(2)
(2)
(6)
0,93
4,53
0,04
−2, 26
1
1,43
0,73
283
(4)
(8)
(12)
0,69
−2
5,14
(4)
(4)
1
0,62
EXERCISE 3
(a)
(b)
(c)
(d)
θ = 60°
θ = 0°
(1)
(5)
(1)
(5)
(1)
(2)
(6)
(2)
(6)
(2)
θ = 23,6°
θ = 19,0°
x = 62,7°
θ = 23,2°
θ = 30°
θ = 45°
θ = 83,7°
θ = 11,3°
x = 15,0°
(3)
θ = 59,58°
(4)
θ = 90°
(3)
θ = 15,7°
(4)
θ = 12,9°
(3)
x = 29,6°
(4)
x = 38,7°
(3)
1
3
(4)
−
(7)
−
(8)
1
2
(3)
(7)
(11)
x = 60°
x = 45°
x = 30°
EXERCISE 4
(a)
(c)
3 3
2
(1)
1
(2)
(5)
3 3
2
(6)
(9)
0
(10)
(1)
(5)
(9)
x = 30°
x = 60°
x = 25°
(2)
(6)
(10)
x = 60°
x = 30°
(b)
(2)
(2)
(2)
(g)
(1)
(2)
B̂ = 33, 7°
h
sin R =
n
(3)
BC = 28,7mm
AB = 7, 2
(4)
P̂ = 47, 6°
DE = 5,8 m
(2)
θ = 44, 4°
3
2
7
2
x = 22,5°
1
2
1
4
(4)
(8)
(12)
x = 45°
x = 15°
(3)
AC = 4,9 m
x = 11,25°
EXERCISE 5
(a)
(c)
(d)
(e)
(f)
PQ = 2,2
θ = 37,9°
(1)
α = 53,1°
(1)
AC = 20,5
(1)
AC = 70,4 m
(i)
(1)
BC = 6
(j)
(1)
sin P =
h
m
(2)
AB = 5, 2
AC = 11,4
θ = 36,9°
AB = 18,8
θ = 41,8°
(2)
(h)
BC = 2, 2
EXERCISE 6
(a)
(c)
(d)
h = 0,8 km
θ = 20°
(1)
(b)
(2)
(1)
No
BC = 18,5 m
EXERCISE 7
(a)
(d)
Quad 1
Quad 2
(b)
(e)
Quad 3
Quad 3
(c)
(f)
Quad 3
Quad 2
(b)
(1)
(d)
(1)
EXERCISE 8
16
25
16
9
(2)
3
2
4
3
12
(a)
(1)
(c)
(1)
(e)
(1)
1
(2)
144
(f)
(1)
(g)
(1)
3
(2)
27
(h)
b = 7,5
(2)
7
17
49
25
(2)
1
(2)
38
(2)
1
(i)
EXERCISE 9
(a)
(1)
(b)
(1)
5
3
k
g
(2)
(2)
3
4
j
k
(3)
(3)
5
4
g
j
(4)
(4)
4
3
g
j
284
(5)
(5)
5
3
j
k
(6)
(6)
a2
b2 + a2
5
4
k
g
(c)
(d)
(e)
(f)
2
(1)
3
(2)
2
(3)
(6)
4
(7)
1
(8)
(1)
(6)
(1)
(1)
1,47
3,64
x = 60°
(2)
4,81
(3)
1
(4)
4
3
0,49
x = 60°
(2)
(2)
θ = 56, 44°
1
(9)
2
(4)
2,22
(5)
2
(5)
1,27
x = 30°
(3)
(3)
x = 77,16°
3
x = 13,28°
Trigonometric functions
EXERCISE 10
(a)
y
(1)
(2)
4
3
y = −2sin θ
2
(3)
1
90°
180°
360°
270 °
θ
-1
-2
y = 3sin θ
-3
-4
(b)
(1)
y
(2)
4
y = −3cos θ
3
2
(3)
1
90 °
180 °
-1
-2
360°
270 °
θ
y = 2cos θ
-3
-4
(c)
(1)
y
(2)
2
1
1
y = cos θ
2
(3)
180°
90 °
-1
270°
θ
360°
y = 3sin θ : max = 3
min = −3
y = −2 sin θ : max = 2
min = −2
y = 3sin θ : range: y ∈ [ −3 ; 3]
amplitude = 3
period = 360°
y = −2 sin θ : range: y ∈ [−2 ; 2]
amplitude = 2
period = 360°
y = 2 cos θ : max = 2
min = −2
y = −3cos θ : max = 3
min = −3
y = 2 cos θ : range: y ∈ [−2 ; 2]
amplitude = 2
period = 360°
y = −3cos θ : range: y ∈ [ −3 ; 3]
amplitude = 3
period = 360°
y = 12 cos θ : max =
1
2
min = − 12
y = − sin θ : max = 1
min = −1
y=
1
cos θ
2
: range: y ∈ [ − 12 ; 12 ]
amplitude =
1
2
period = 360°
y = − sin θ : range: y ∈ [−1 ; 1]
amplitude = 1
period = 360°
y = − sin θ
-2
EXERCISE 11
(a)
(1)
y
(2)
4
3
y = sin θ + 2
2
(3)
1
90°
180 °
θ
360 °
270 °
-1
-2
y = cos θ − 1
-3
-4
(b)
(1)
(2)
y
5
4
y = − cos θ + 3
(3)
3
2
1
-1
90°
180 °
360 °
270 °
-2
-3
-4
y = − sin θ − 2
285
θ
y = sin θ + 2 : max = 3
y = cos θ − 1 : max = 0
min = 1
min = −2
y = sin θ + 2 : range: y ∈ [1 ; 3]
amplitude = 1
period = 360°
y = cos θ − 1 : range: y ∈ [−2 ; 0]
amplitude = 1
period = 360°
y = sin θ + 2 : max = 3
y = cos θ − 1 : max = 0
min = 1
min = −2
y = sin θ + 2 : range: y ∈ [1 ; 3]
amplitude = 1
period = 360°
y = cos θ − 1: range: y ∈[−2 ; 0]
amplitude = 1
period = 360°
y
(c)
(1)
y = 2sin θ + 4 : max = 6 min = 2
y = −3cos θ − 1 : max = 2 min = −4
y = 2sin θ + 4 : range: y ∈ [2 ; 6]
amplitude = 1
period = 360°
y = −3cos θ − 1 : range: y ∈ [−4 ; 2]
amplitude = 3
period = 360°
(2)
y = 2sin θ + 4
7
6
5
4
3
2
1
(3)
90°
-1
-2
-3
-4
-5
270°
180°
θ
y = −3cos θ − 1
EXERCISE 12
y
(a)
θ = 90°
3
y = − tan θ
(135° ;1)
1
.
45 °
135 °
90 °
.
225 °
3 5°
2 0°
360°
-3
y
θ = 90°
2
.
1
(d)
θ = 270°
1
y = − tan θ
2
.
1
(45° − )
2
-1
135°
90°
.
180 °
1
(225° − )
2
y
θ = 90°
(225° ; 3)
90 °
135 °
180 °
225°
3 5°
270 °
.
.
(135° ; − 3)
y
θ = 90°
(135° ; 0)
45 °
90 °
360°
θ
(315° ; 3)
θ = 270°
(45° ; 2)
-1
-2
-3
-4
-5
-2
(e)
45 °
θ
270°
225°
θ = 270°
.
(45° ; 3)
5
4
3
2
1
1
(135° ; )
2
45°
.
-1
-2
-3
-4
-5
(225° ; − 1)
θ = 90°
y = 3tan θ
θ
-2
(c)
5
4
3
2
1
(315° ;1)
180
(45° ; − 1)
y
(b)
.
.
2
-1
θ = 270°
.
.
y = tan θ + 1
135 °
(225° ; 2
.
(315° ; 0)
180°
225 °
3 5°
270°
360 °
θ
θ = 270°
3
2
1
y = tan θ − 2
45°
−1
90°
−2
−3
−4
180°
135°
(135° ; − 3)
.
.
.
225°
270°
315°
(135° ; − 1)
(315° ; − 3)
(180° ; − 2)
θ
360°
.
−5
(f)
y
θ = 270°
.
y = −2 tan θ − 1
45°
90 °
.
135 °
180 °
(g)
.
(135° ;1)
-1
-2
-3
-4
-5
-6
(g)
θ = 90°
4
3
2
1
(315° ;1)
225 °
270 °
3 5°
360 °
.
-1
-2
-3
-4
-5
-6
-7
(225° ; − 3)
(45° ; − 3)
y = 2 tan θ :
180°
y = −3sin θ − 3 :
360°
θ
y
6
5
4
3
2
1
y = 2 tan θ
θ = 90°
θ = 270°
.
.
(45° ; 2)
45 °
(225° ; 2)
90 °
.
135°
180 °
225°
(135° ; − 2)
y = −3sin θ − 3
EXERCISE 13
(a)
(b)
(1)
(3)
(1)
(2)
(3)
Amplitude = 1
Range is y ∈ [0 ; 2]
a =1 q =1
Range is y ∈ [−3 ; − 1]
Amplitude = 1
Amplitude = 2
Range is y ∈ [−1 ; 3]
a = −1 q = −2
(2)
Amplitude = 2
m = −2 n = 0
(4)
Period = 360°
Period = 360°
m = −2 n = 1
(4)
286
Range is y ∈ [−2 ; 2]
2 0°
θ
(c)
y = 2 tan x for x ∈ [0° ; 270°)
y = 2 tan x :
y = −3cos x + 1 for x ∈ [0° ; 270°]
Range is y ∈ (−∞ ; ∞)
y = −3cos x + 1
Range is y ∈ [−2 ; 4]
No Amplitude
Period = 180°
Amplitude = 3
Period = 360°
(1)
(2)
EXERCISE 14
(a)
(b)
(c)
(1)
(4)
(5)
(8)
(1)
(4)
(1)
(4)
OA = 2 OB = 1
CD = 3
(2)
(3)
EF = 2
90° ; 270°
90° ; 270°
0° ; 360°
(i)
(ii)
(iii) 0° ; 360°
(iv)
90° < x < 270°
(vi)
0° ≤ x ≤ 90° ; 270° ≤ x ≤ 360°
(v)
(vii) 0° ≤ x ≤ 90° ; 270° ≤ x ≤ 360° (viii) 90° < x < 270°
x = 60°
180° < x < 360°
180° < x < 360°
(6)
(7)
t=2
k < −1 or k > 1
(9)
−4
x = 135°
(2)
(3)
0° ≤ x < 90° ; 135° ≤ x < 270°
x = 0° ; x = 180°
180° ≤ x ≤ 360°
x = 90° ; x = 180° (2)
0° ≤ x ≤ 90° ; 180° ≤ x ≤ 360°
(3)
0° < x < 360° ; x ≠ 180°
(5)
x = 0° ; x = 360°
(6)
0° ≤ x ≤ 180° ; x = 360°
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(b)
(d)
(1)
(1)
(5)
(1)
(5)
(1)
(e)
(c)
(f)
(g)
(h)
(i)
(j)
(l)
(n)
−0,72
θ = 11,79°
θ = 23,58°
θ = 21,14°
(2)
(2)
1,27
θ = 60°
0,39
(2)
(6)
(2)
(1)
0
(2)
(5)
2
(6)
θ = 30°
θ = 45°
PQ = 25, 41
PR = 18,88
x=5
(1)
CD = 3,91 m
(1)
QR = 30
YA = 21,125cm
θ = 30°
(1)
h = 3,13 m
(1)
(5)
(1)
(3)
(3)
1,28
(4)
(4)
3,69
θ = 53,13°
θ = 33,69°
θ = 71,52°
(3)
θ = 75,52°
(4)
θ = 98,60° s
(3)
1,96
1
2
(4)
3
(2)
(6)
(2)
0,88
1
4
1
2
θ = 45°
θ = 60°
K̂ = 49,18°
(2)
θ = 11,54°
(3)
θ = 47,16°
(3)
(7)
θ = 30°
θ = 30°
(4)
(8)
θ = 30°
θ = 60°
x = 14,59
(3)
x = 68, 68°
(4)
x = 41, 27°
(2)
QR = 30
(k)
(1)
EA = 19,5cm
(2)
YS = 3,125
(2)
8,66 m
(m)
(1)
θ = 60°
(2)
OQ = 4
(b)
(1)
(c)
(1)
The graph of y = x is reflected in
(2)
the x-axis and then translated 4
units up.
The graph of y = x is stretched
vertically by a factor of 3 and
then translated down by 6 units.
(3)
y = −x + 4
CHAPTER 6 (FUNCTIONS)
EXERCISE 1
(a)
(1)
y = 32 x
(2)
(2)
y = 32 x
y= x
(1 ; 32 )
(1 ; 1)
(1 ; 14 )
y = 14 x
y = − 32 x
y = −x
y = − 32 x
y = − 14 x
(1 ; − 14 )
(1 ; − 1)
(1 ; − 32 )
(d)
(1)
(5)
F
E
(2)
(6)
D
H
(3)
(7)
G
B
(4)
(8)
A
C
y = 3x − 6
4
2
−6
287
4
EXERCISE 2
(a)
(1)
y = 4 x2
(2)
y
(b)
(1)
(2)
y = − x2
(c)
(1)
(2)
y = −4 x 2
y = 4 x2
y=
3 2
x
2
y = 3x 2
y=
y = x2
y = − x2
x
(3)
(d)
concave up ( a > 0 )
(1)
y = x2 + 3
concave down ( a < 0 )
(2)
y = x2 + 3 :
y = −4 x 2
(3)
y-int: (0 ; 3)
x-int: none
y = x2 − 4 :
y-int: (0 ; − 4)
x-int: (−2 ; 0) (2 ; 0)
(1 ; 4)
−2
1
y = − x2 y = − 1 x2
2
4
(3)
y = x2 − 4
3
1 2
x
4
2
First two concave up
Third one convave down
−4
(e)
(1)
(2)
4
y = −4 x 2 + 4
−1
y = −4 x 2 + 4
y-int: (0 ; 4)
x-int: (−1 ; 0)
(f)
y
(1)
8
(1 ; 0)
1
y = − x2 + 8
2
2
y = −x − 2
1
−2
y-int: (0 ; − 2)
x-int: none
−4
x
4
y = − x2 − 2
(g)
(1)
(2)
(3)
(4)
(5)
(6)
(2)
y-int: (0 ; 8)
x-int:
(−4 ; 0)
(4 ; 0)
(c)
(1)
x = 0 ; y = −1
(2)
(3)
(−4 ; 0)
C
E
D
F
B
A
EXERCISE 3
(a)
(1)
y=
5
x
( 5 >1)
(b)
(2)
5
y=−
x
y=
y=
5
x
(1)
x=0 ; y=2
(2)
(3)
(−1 ; 0)
(1 ; 4)
1
x
y=2
4
y = − −1
x
(−4 ; 0)
y = −1
(−1 ; 0)
y=
(3)
see (2)
2
+2
x
288
(1 ; − 5)
(d)
(e)
(1)
B
(2)
A
(3)
C
(4)
D
(1)
(2)
(3)
y = 6 x (smallest base)
y=0
(1 ; 5)
y=2
(−1 12 ; 0)
y=
3
+2
x
EXERCISE 4
(a)
(1)
(2)
(3)
y = 6 x (largest base)
y=0
(b)
y = 6x
x
1
y = 
6
( −1 ; 6)
(1 ; 6)
1
y = 
4
y = 4x
(1 ; 4)
y = 2x
1
y = 
2
(1 ; 2)
(c)
(1)
(2)
y = 2.2 x
y-int: (0 ; 2)
y = 4.2 x
y-int: (0 ; 4)
1
y = 3 
2
y=0
(3)
x
x
( −1 ; 4)
( −1 ; 2)
y = 4 2x
1
y = 3 
2
x
(1 ; 8)
y = 2 2x
x
y-int: (0 ; 3)
4
(1 ; 4)
3
(1 ; 1 12 )
2
x
(d)
(1)
1
y = −   : y-int: (0 ; − 1)
2
(3)
x
1
y = −3   : y-int: (0 ; − 3)
2
(2)
y=0
(4)
1
y =   is vertically stretched to form
2
1
y = − 
2
(−1 ; − 2)
x
−1
−3
1
y = −3  
2
x
x
x
1
the graph of y = 3   and then
2
( −1 ; − 6)
x
1
1
y = 3   is reflected in the y-axis to form the graph of y = −3  
2
2
(e)
(1)
(2)
(4)
(5)
(6)
x-int: (1 ; 0)
y-int: (0 ; − 1)
y = −2
x
y = 2x + 1
(3)
y = 2x − 2
(0 ; 2)
y =1
y = 2 x − 2 is the graph of y = 2 x
shifted 2 units down.
see diagram
The equation 0 = 2 x + 1 has no solution.
(0 ; − 1)
(1 ; 0)
y = −2
289
(f)
(1)
(2)
(0 ; − 3)
(−1 ; 0)
y-int:
x-int:
y = −4
(3)
( −1 ; 0)
x
(4)
x
1
1
y =   − 4 is the graph of y =  
4
 
4
shifted 4 units down.
x
(0 ; − 3)
1
y =  −4
4
y = −4
(g)
(1)
y = 2 4x − 2
(2)
(3)
(1 ; 10)
(4)
( −1 ; 10)
( −1 ; 6)
(1 ; 6)
y = 2 4x + 2
x
1
y = 2  − 2
4
(0 ; 4)
x
(0 ; 4)
(0 ; 0)
(0 ; 0)
y=2
y=2
y = −2
y = −2
(h)
(1)
1
y = 2  + 2
4
B
(2)
A
(3)
E
(4)
F
(5)
D
(6)
G
(7)
C
EXERCISE 5
(a)
(1)
(2)
(3)
y=
(1 ; 2)
y = 2x
(1 ; 2)
( −1 ; 2)
y = 2x
(6)
( −1 ; 3)
(1 ; 2)
(7)
y = x2 + 2
(8)
2
y = −2 x + 2
(1 ; 3)
(9)
(10)
y =1
( −1 ; 2)
(1 ;
1
)
2
(2 ; 0)
(11)
(12)
1
y = −  +1
2
(1)
(1 ; 2)
(0 ; 0)
(0 ; − 2)
x
y = −2 x − 2
y=
1
x
3
1
y = − x +1
3
(0 ; 1)
(3 ; 0)
(3)
3
y=
x
(4)
y=
(1 ; 3)
(1 ;
1
)
3
y=
( −1 ; − 3)
(6)
y=
y = −2
x=2
(2)
(1 ; 13 )
2
y = − +1
x
y =1
(1 ; 0)
( −1 ; 0)
(0 ; 0)
(5)
(0 ; 2)
( −1 ; 0)
(0 ; 2)
(b)
y=2
(1 ; 2)
( −1 ; − 2)
(5)
(0 ; 1)
2
x
(4)
y = 2 x2
(3 ; 0)
y = −1
( −1 ; − 4)
290
( −1 ; 13 ) (1 ; 13 )
1
x
3
(7)
3
−1
x
1 2
x
3
(8)
(0 ; 1 13 )
(1 ; 2)
y =1
y = − x2 + 9
( −3 ; 0)
(0 ; 9)
(3 ; 0)
(9)
(10)
(11)
(12)
x
1
y =   +3
3
(0 ; 4)
( −1 ; 6)
y = 3.3x − 3
(1 ; 6)
y = 3.2 − 3
(1 ; 3)
(0 ; − 1)
(0 ; 0)
(0 ; 0)
y=3
y = 2x − 2
x
(1 ; 0)
y = −2
y = −3
y = −3
EXERCISE 6
(a)
(b)
(c)
(d)
(1)
(5)
(9)
(13)
(1)
(1)
2
1
4 x2 − 2 x + 2
4 x 2 − 10 x + 5
x = −3 or x = 3
−2 p + 6
(2)
(6)
(10)
(14)
(2)
(2)
4
(3)
2
(7)
2
2x − x + 3
(11)
2 x 2 + 4 xh + 2h 2 − x − h + 1
x = −5 or x = 5
(3)
(1)
(3)
(5)
f is vertically stretched by a factor of 3 to form g
f is shifted 2 units up to form g
f is vertically stretched by a factor of 4 to form g
7
2a 2 − a + 1
2x2 + x + 1
x = −6 or x = 1
(4)
(8)
(12)
(15)
(4)
11
8x2 − 2 x + 1
2 x2 − 5x + 4
4 xh + 2h 2 − h
x=3
− 72
(2)
(4)
(6)
f is reflected in the x-axis to form g
f is shifted 4 units down to form g
f is reflected in the x-axis to form g
EXERCISE 7
(a)
(b)
(c)
(d)
(e)
(f)
x ∈ (−3 ; 1]
y ∈ [−4 ; 2)
x ∈ (−∞ ; ∞)
y ∈ ( −∞ ; ∞ )
f:
x ∈ (−∞ ; ∞)
y ∈ [0 ; ∞)
x ∈ (−∞ ; ∞) x ≠ 0
y ∈ (−∞ ; ∞)
x ∈ (−∞ ; ∞) x ≠ 0
y ∈ (−∞ ; ∞)
x ∈ (−∞ ; ∞)
f:
y ∈ (−∞ ; 1)
g:
y≠0
y≠2
x ∈ (−∞ ; ∞)
y ∈ (−∞ ; − 3]
g:
x ∈ (−∞ ; ∞)
y ∈ (2 ; ∞)
EXERCISE 8
(a)
(1)
(5)
(b)
(1)
A(2 ; 0) B(0 ; − 4)
E(1 ; − 2)
(2)
(6)
(c)
3x − y = 4
2x − y = 5
(1 13 ; 0)
(2 12 ; 0)
(0 ; − 4)
(0 ; − 5)
(d)
(−1 ; − 7)
(2)
C(0 ; − 1) D(−1 ; 0)
(3)
2
4 x − 2 y = 8 increases for all real values of x
x + y = −1 decreases for all real values of x
(1)
y = x+3
y = −x − 5
(2)
y = 2x + 6
(3)
(4)
y = − 32 x − 6
(5)
y = x+2
(6)
y = − 12 x + 6
(1)
(2)
y = −1
x = −1
(4)
−1
Gradient is 0
Gradient is undefined
(−1 ; − 7)
EXERCISE 9
(a)
(1)
A(0 ; − 2)
B( − 1 ; 0)
C(1 ; 0)
(2)
(5)
(8)
x>0
A(0 ; − 2)
y = −2 Grad = 0
291
(3)
(6)
(9)
x<0
(4)
−2
x=0
(7)
x ∈ (−∞ ; ∞) y ∈ [−2 ; ∞)
x ∈ (−∞ ; ∞) y ∈ {−2}
(b)
(1)
(2)
(5)
(7)
A(0 ; 9)
B( − 3 ; 0)
C(3 ; 0)
(c)
(d)
(8)
(9)
(1)
x<0
(3)
9; 0
(6)
For f : x ∈ (−∞ ; ∞)
y ∈ (−∞ ; 9]
x>0
(4)
−2
(0 ; 9) ; (0 ; 0)
For g: x ∈ (−∞ ; ∞)
y ∈ ( −∞ ; 0]
h( x) = x 2 + 12
h is the graph of g reflected in the x-axis and then shifted 12 units up.
x = −3 Gradient is undefined
(10) x ∈ {−3} y ∈ ( −∞ ; ∞ )
A(0 ; − 1) B(0 ; − 9) C( − 3 ; 0) D(3 ; 0)
(2)
E( − 2 ; − 5) F(2 ; − 5)
(1)
f ( x) = 4 x + 2
(2)
f ( x) = − x − 1
(3)
f ( x) = x 2 − 9
(4)
f ( x) = − x 2 + 16
(5)
f ( x) = 2 x 2 − 8
(6)
f ( x) = − x 2 + 25
(3)
x ∈ (−∞ ; ∞) x ≠ 0
(6)
y = −x + 4
2
2
EXERCISE 10
(a)
(b)
(1)
q=4
(2)
(4)
( −∞ ; 0) ; (0 ; ∞)
(5)
(1)
(5)
(9)
(c)
(1)
12
x
32
f ( x) =
−4
x
−6
f ( x) =
−2
x
x ∈ ( −∞ ; 0)
y ∈ (0 ; ∞)
f ( x) =
f ( x) =
(2)
(6)
(10)
(2)
−8
x<0
x
−2
+4
x
x=0 y=4
f ( x) =
4
+2
x
−8
f ( x) =
−4
x
1
f ( x) = − + 1
2x
x ∈ ( −∞ ; 0)
y ∈ (−∞ ; − 3)
f ( x) =
f ( x) =
(3)
(7)
(11)
(3)
8
−3 x < 0
x
−15
x
5
f ( x) = − 3
x
4
f ( x) = + 2
x
x ∈ (0 ; ∞)
y ∈ ( −∞ ; 5)
f ( x) =
y ∈ (−∞ ; ∞) y ≠ 4
(4)
(8)
−5
+3
x
4
f ( x) = + 1
x
f ( x) =
−18
+5 x > 0
x
f ( x) =
EXERCISE 11
x
(a)
f ( x) = 2 x − 4
(b)
1
g ( x) =   − 1
3
(c)
h ( x ) = −3 x + 3
(d)
f ( x ) = 3.2 x
(e)
f ( x) = −2.3x + 18
(f)
1
g ( x ) = 4.   − 4
4
x
EXERCISE 12
(a)
(b)
(c)
(d)
(e)
(f)
(1)
(3)
(5)
(1)
(5)
(1)
(4)
(1)
(3)
(7)
OA = 2
OD = 2
RS = 1
OP = 5
OP = 4
EF = 2
OB = 2 OC = 4
O F = 4 DE = 4
OS = 3
OQ = 5
OH = 8
AB = 6
CD = 2
CT = 6
x≤4
AB = 7
(5)
x<4
x < −5
DF = 2
GH = 3
AB = 12
(6)
(2)
(6)
CD = 6
(2)
(2)
(4)
(7)
(3)
AP = 6 BC = 2
O K = 1 G H = 1 OG = 5 KH = 5
(8)
x<2
x>3
(4)
DC = 2
OF = 2
(2)
(6)
PQ = 4
x ≤ −2
O F = 4 EF = 7
(5)
JL = 10
OV = 11
x < − 3 or x > 3
(4)
(8)
(1)
OA = 1
(2)
BC = 2 12
(3)
OD = 2
(4)
1
2
F( ; 2)
(5)
0 < x ≤1
(6)
x<0
(8)
0< x<
(1)
(4)
All real values of x
(2)
(5)
x < −1
x<0
(3)
(6)
x ≥ −1
x≥0
GH = 4
OA = 1
OU = 5
(6)
OQ = 4
x <1
x ≤ − 4 or x ≥ 3
1
2
CD = 1
(3)
(7)
292
OE = 2
or x > 1
(7)
x>0
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(1)
(2)
x ∈ (−∞ ; ∞)
(0 ; 2)
(4)
−2
(6)
(0 ; 1)
(7)
(−1 ; 0)
y ∈ [−1 ; ∞)
f ( x) = x 2 − 1
(5)
g ( x) = x − 1
−1
(3)
y = x2 + 2
(−1 ; 3)
(1 ; 3)
(0 ; 2)
1
(0 ; − 2)
−1
(−1 ; − 3)
(1 ; − 3)
y = − x2 − 2
(b)
(1)
(2)
y = 3.3x − 1
( −1 ; 4)
y ∈ (−1 ; ∞)
x ∈ (−∞ ; ∞) x ≠ 0
y =1
(0 ; 2)
( −1 ; 0)
(3)
(4)
3
y = − +1
x
(3 ; 0)
y = −1
h is the graph of f stretched
vertically by a factor of 3 and
then shifted 1 unit down.
(c)
(d)
(1)
f ( x) = x 2 + 2
(5)
g ( x) =
(6)
(7)
f:
x<0
(1)
q = −1
(2)
p is the graph of g reflected in
the x-axis and then shifted 1
unit up.
(3)
(0 ; 2)
1
+ 2 ; A( − 12 ; 0)
x
x ∈ (−∞ ; ∞)
y ∈ [2 ; ∞)
x=0
(4)
x ∈ (−∞ ; ∞) x ≠ 0
g:
3
4
(2)
AB =
(2)
A(3 ; 0) ; C(−3 ; 0)
y=2
y ∈ (−∞ ; ∞) y ≠ 2
(3)
OC = 3
(4)
OT = 1
(3)
g ( x) = x 2 − 9
(4)
AC = 6
(8)
x ≤ −3 or x ≥ 3
x
(1)
1
g ( x) =   − 2
2
f ( x) = − x + 3
(5)
DE = 7 ; DF = 4
(9)
−4 < x < 3
(1)
y ∈ (−∞ ; 6]
(5)
(e)
(6)
ST = 8
(7)
OM =
3
2
(3)
g ( x) =
(3)
(7)
−1 ≤ x ≤ 1
t < 12
x
(f)
(4)
(g)
(1)
(5)
(5)
1
f ( x ) = −2   + 6
3
All real values of x
(2)
(6)
−1 < x < 2
t > 12
(2)
x
1
h( x ) = 2   − 6
3
−2 ≤ x ≤ 2
x ≥ −2
6
+6
x
(4)
x < −2
CHAPTER 7 (EUCLIDEAN GEOMETRY)
EXERCISE 1
(a)
(1)
(b)
(d)
(1)
(2)
a = 70°
h = 100°
a = 40°
y = 56°
c = 70°
e = 70°
f = 20°
b = 70°
d = 110°
i = 60°
(2)
b = 20°
c = 80°
d = 100°
e = 80°
(f)
BC = 674 cm
g = 40°
EXERCISE 2
(a)
 = 70°
B̂ = 110°
Ĉ = 70°
D̂ = 110°
(c)
K̂ = 63°
L̂ = 117°
M̂ = 63°
N̂ = 117°
293
(b)
ˆ =Q
ˆ = Sˆ = 60°
R̂ = Sˆ 1 = Q
1
2
2
(d)
(e)
(f)
(g)
(h)
(i)
B̂ = 65°
Ê 2 = 65°
M̂ 2 = 115° N̂1 = 115°
ˆ =D
ˆ = 35°
ˆ =D
ˆ = 35°
 = 110°
B
B
1
1
2
2
ˆ
ˆ
ˆ
ˆ
C = L 2 = 60°
M 2 = N 2 = 120°
ˆC = M
ˆ =A
ˆ = 60°
ˆ = 120°
ˆ =B
ˆ =M
ˆ = 60°
A
D
2
1
2
1
D̂1 = 68°
(2)
D E = 50 cm
300 cm
EXERCISE 4
(a)
ˆ = Sˆ = Q
ˆ = 35°
Q
2
1
1
Rˆ = Pˆ = 110°
(b)
Â1 = 70°
 2 = 20°
B̂1 = 20°
AD = 8 cm
A E = 5 cm
(c)
(j)
(m)
F̂1 = 80°
b = 20°
a = 80 °
(1)
(d)
Ĉ 2 = 20°
D̂1 = 70°
AB = 6 cm
(e)
PR = 20 cm
AC = B D = 5 x
d = 20 °
c = 20 °
A E = 8 cm
Ĉ1 = 70°
B̂2 = 70°
(2)
AC = 28 cm
(b)
θ = 108 °
e = 140 °
(3)
B̂1 = 70°
EXERCISE 5
(a)
(d)
(1)
x = 20 °
β = 115°
β = 36°
α = 72 °
(c)
x = 45 °
(2)
19, 3
EXERCISE 6
(a)
C E = 60 cm
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(h)
R̂ = 45°
Ŝ2 = 45°
ˆ = Eˆ = 72°
QER
2
Ê 2 = 135°
(m)
T̂2 = 135°
(1)
(n)
(2)
CHAPTER 8 (ANALYTICAL GEOMETRY)
EXERCISE 1
(a)
(c)
AB = 13 ; CD = 5 ; EF = 58 ; GH = 5
AC = BC
(d)
x = −1
(b)
(e)
AB = BC
(1)
y = 3 or y = 1
EXERCISE 2
(a)
Midp AB = (4 ; 2)
(b)
(c)
(d)
(1)
(2 ; 1)
x = −5 ; y = 3
(1)
F( − 4 ; 3)
EXERCISE 3
(a)
(1)
grad AB = 1
(b)
grad AB = −3
(c)
(d)
(e)
(1)
Parallel
a=3
(1)
mFR = mRN
(g)
(1)
 −3 −1 
Midp EF = 
; 
 2 2 
Midp CD = (−3 ; 2)
(2)
(2)
(e)
4,2
x = −4 ; y = 8
(2)
gradAB = 53
(3)
x = 4 ; y = −7
(3)
grad AB = 1
B(0 ; 5)
gradCD = 75
grad EF is undefined
(2)
(2)
(f)
Neither
x = 13
(2)
x = − 23
4
(h)
x = 4; y =5
(i)
mBD = −
(j)
(1)
(2)
P(1 ; 5)
3,6
 −5 
Midp GH =  3 ;

2 

(4)
grad AB = − 53
grad GH = 0
(3)
Perpendicular
(3)
x = −17
(3)
mAB × mBC = 15 × −5 = −1
b =1
x = ±3
2
3
294
(4)
13
EXERCISE 4
(a)
(c)
(d)
(e)
(f)
Diagonals bisect
(b)
Diagonals bisect at right angles
Diagonals bisect and one interior angle is a right angle
(1)
Diagonals bisect
(2)
Interior angle is not 90°
B( − 4 ; 1)
C( − 6 ; 4) ; D( − 2 ; 0)
(2)
AP is not perpendicular to DP
(1)
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(b)
(c)
(1)
(1)
(d)
(1)
14,5
(2)
(2)
M(7 ; 4)
Isosceles
(3)
(3)
P(16 ; 7)
No pair of sides form a right angle
grad AC = 2
(2)
BC = 5, 6
(3)
M
(3)
grad UO = grad UR = − 34
(3)
grad SE × grad AS = 23 × − 32 = −1
AB = 6, 3
b=2
( 12 ; −2 )
(4)
(5)
grad BD ×grad AC = − × 2 = −1
(e)
(1)
U(−4 ; 3)
(2)
grad RU = − 34
(f)
(1)
a = 32 ; k = −2
(2)
D( − 3 ; 0)
(g)
(1)
12
(2)
V(−2 ; 3)
(h)
(i)
Diagonals bisect
N(3 ; 5)
(1)
(2)
grad MP × grad LN = − 23 × 23 = −1 ; LMNP is a rhombus
(3)
D(0 ; 2)
1
2
(4)
ˆ =N
ˆ = Lˆ = Pˆ = 90°
grad LM × grad MN = −5 × 15 = −1 ; M
(j)
b = 3 ; S(0 ; 6)
(k)
(l)
(1)
(1)
B(6 ; 2)
y = x+3
(2)
(2)
C(6 ; 2 + r )
(3)
r =3
y = −x −1
CHAPTER 9 (FINANCIAL MATHEMATICS)
EXERCISE 1
(a)
(b)
(c)
(d)
(e)
(h)
(i)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
R 5 400
R17 760
R649 800
R285 132,38
R485 717,81
11,4%
25%
(2)
(2)
(2)
(2)
(f)
(2)
(2)
R6 045,20
R740
R912 922,21
R279 793,50
5 years, 5 months
9,5%
6,1%
(2)
R22 800
(g)
9 years, 1 month
EXERCISE 2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
R603,50 ; R23 426
(1)
R886,67
15%
12,9%
R2 250 ; 25%
R6 600 ; 22%
R58 064,52
R216 397,55
(3)
R60 800
EXERCISE 3
(a)
(e)
(f)
R8 212,89
(b)
(1)
R16 310,19 (2)
10,5%
(g)
R20,94
R51 428,59
R5 340
(c)
No
(h)
9,1%
(d)
R90 212,12
EXERCISE 4
(a)
$645 = R7 901,25
(b)
30£ = R 510
(c)
(d)
(g)
$265,96 = R3 000
(1)
1£ = R18,24
(e)
(2)
£4 000 = R 72 960
£1 619,50
(f)
295
R2,82 = ¥50
¥20 454,55 = R4 500
150
(h)
(i)
(j)
Germany
R177 984
R145 640
EXERCISE 5
(a)
(e)
(f)
R50 229 570
(1)
115,4%
2%
(b)
(2)
R332 634 162
R60 000
(c)
R4 319
(d)
R6 175 634 873
(b)
(f)
R1 279 685,86
R6 564,25
(c)
R74 032,86
(d)
R698 273,39
EXERCISE 6
(a)
(e)
R34 474,39
R188 593,73
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(b)
(c)
(d)
(g)
(k)
(1)
R30 693,85
(1)
R434 782,61
(1)
R=33,3%
R167 568 672 700
R261 920,83
R34 975,41
(2)
(2)
(2)
(e)
(h)
(l)
R34 377,11
R294 588,35
26%
(f)
(i)
(m)
R 16 362
10,4%
R19 734,32
$578, 03 = R10 000
39 900 000%
(j)
R14 060,20
R13 008,69
CHAPTER 10 (STATISTICS)
EXERCISE 1
(a)
(1)
Score
0
1
2
3
4
5
(2)
Frequency
1
3
5
6
12
3
n = 30
(b)
(1)
16 seeds
(c)
(1)
1
2
3
4
5
6
(2)
5 days
7
0
4
1
0
3
7
2
5
2
3
4
8
4
5
2
3
5
(3)
75%
8
4
5
3
4
6
9
6
5
7
5
6
(2)
(d) (1)
9
7
5
8
7
(3)
0
1
2
3
1
0
0
0
3
2
0
0
3
3
1
1
4
3
1
9
4
4
1
5
4
1
7
4
2
7
5
5
9
9
30-39 year olds
7
4
2
296
8
4
3
8
4
3
8
4
4
9
8
5
9
8
7
8
7
9
7
9
7
8
(2)
(e)
(1)
(3)
(4)
58%
Check the current exchange rates using GOOGLE.
COMPANY A
4 3
3 2 2
3 3 3 2
4 2
(2)
(f)
(g)
21 agents
1
2
0
1
1
2
3
4
0
0
2
3
COMPANY B
1
3 4 9
3 3 4 5
3 3 4
(3)
20%
(4)
46,7%
(5)
Company B.
(1)
(1)
16
45-54
(2)
(2)
16
72%
(3)
(3)
2
132 000
(4)
(4)
11
125 000
(5)
Class B
EXERCISE 2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(1)
Median = 6
Mode = 5
x =7
(2)
Median = 8
Mode = 8
x = 10
(3)
Median = 9
Mode = 10
x = 7, 5
(4)
Median = 4,5
Modes = 2 ; 4
x = 4, 5
3 people
(1)
Median = R16 800
Mode = R16 900
x = R23 411,11
(2)
The mean has been inflated by the outlier (R86 300). The median and mode are better measures.
Median and mode
Mean
x = 158
Median = 92
(2)
Mean
(1)
(1)
Median = 35
Mode = 35
x = 38, 6
(2)
All three measures are useful since most of the ages are in the 30-year old group.
(1)
Median = 18
Mode = 14
x = 16,86
(2)
The mean and median are probably the best measures. The mode is too low.
x =3
Median = 3
Mode = 3
(1)
(2)
All three measures are the same indicating that 3 boys are typical of a family with six children.
x = 11
x = 13
x = 12
(2)
(3)
(4)
9 ; 18 ; 27 ; 36 ; 45
(1)
(5)
3 ; 3 ; 4 ; 5 ; 10
3;3;4;6;9
3;3;4;7;8
(6)
34
EXERCISE 3
(a)
(1)
M=8
Q1 = 3
Q3 = 12
(2)
M = 11
Q1 = 5
Q 3 = 15
(3)
M = 11, 5
Q1 = 6, 5
Q 3 = 19
(4)
M = 17, 5
Q1 = 8
Q3 = 25
(5)
M = 18,5
Q1 = 9
Q3 = 27
Q3 = 7
(c)
M = 8 12
Q1 = 7 34
Q3 = 8 34
Q 3 = 89, 5
Q3 = 154,5
(2)
M = 13
(e)
M = 1, 445
M = 35
Q1 = 1,35
Q3 = 1,55
Q1 = 25
Q 3 = 51, 5
(b)
M=5
(d)
(f)
(h)
M = 81
Q1 = 67
M = 92
Q1 = 86
x = 11
(1)
Q1 = 3
(g)
Q1 = 10
Q 3 = 21
EXERCISE 4
(a)
(1)
(2)
The 25th percentile is the 13th value which is 113.
The 50th percentile is the average of the 25th and 26th values: 216,5
The 75th percentile is the 38th value which is 284.
The 30th percentile is the average of the 15th and 16th values: 126,5
297
(b)
(3)
(4)
(1)
(2)
(3)
(4)
(5)
The 65th percentile is the 33rd value which is 253.
The 80th percentile is the average of the 40th and 41st values: 289,5
The 25th percentile is the average of the 54th and 55th values: 66, 03
The 50th percentile is the average of the 108th and 109th values: 74,1
The 75th percentile is the average of the 162nd and 163rd values: 78, 485
The 90th percentile is the 195th value which is 81,17.
The 60th percentile is the 130th value which is 75,55.
The 60th percentile is the 98th value which is 72,93.
The top ranking countries tend to have better medical facilities and the employment figures are higher,
meaning less poverty.
EXERCISE 5
(a)
(b)
(c)
(d)
(1)
(2)
Maximum = 10
Minimum = 1
Q2
Q1
Maximum = 10
(1) Minimum = 1
(2)
Q1
(1) Minimum = 2 Maximum = 15
Q2
Q1
(2)
(1) Minimum = 2 Maximum = 10
(2)
Q2
Q1
(e)
M = Q2 = 4
Q1 = 2
(3)
x = 4, 7
(4)
Mean > Median
Q3
Q1
Q2
Q3
M = Q2 = 8
Q1 = 7
(3)
x = 7, 25
(4)
Mean < Median
M = Q2 = 9
Q1 = 5
(symmetrical)
Q 2 Q3
(f)
Q2
Q1
Q1
Q2
Q3
(g)
298
(negatively skewed)
Q1 = 3 Q3 = 9
x =6
(3)
(4)
Mean = Median
8 12 8 34
7 12
Q3 = 9
(symmetrical)
(3)
(4)
Q3
(positively skewed)
Q3 = 13
x =9
Mean = Median
Q3
M = Q2 = 6
Q3 = 8
Q3
(h) (1)
Q1 Q 2
(i)
Q1 Q 2
(j)
(l)
Q3
Q2
Q1
(k) (1)
(2)
Mean > Median
(positively skewed)
(2)
Q3
Q3
Maximum values and medians are the same.
Not valid. Although both classes have the same median which means that half the scores are above 80
and half are below 80, none of the scores for Class B are below 66. For Class A, one quarter of the
scores are below 66, Therefore Class B performed better than Class A.
A possible set of nine numbers are:
10; 20; 20; 34; 45; 47; 51; 53; 80
EXERCISE 6
(a)
Range = 10 − 1 = 9
IQR = 8 − 2 = 6
Semi-IQR = 12 (8 − 2) = 3
(b)
Range = 10 − 1 = 9
IQR = 9 − 7 = 2
Semi-IQR = 12 (9 − 7) = 1
(c)
Range = 15 − 2 = 13 IQR = 13 − 5 = 8
(d)
Range = 10 − 2 = 8
IQR = 9 − 3 = 6
Semi-IQR = 12 (9 − 3) = 3
(e)
Range = 10 − 6 = 4
IQR = 8 34 − 7 34 = 1
Semi-IQR = 12 (8 43 − 7 43 ) =
(f)
Range = 99 − 19 = 80
IQR = 22,5
Semi-IQR = 12 (22, 5) = 11, 25
(g)
Range = 0,32
IQR = 0, 2
Semi-IQR = 12 (0, 2) = 0,1
(h)
Range = 333 500
IQR = 108 000
Semi-IQR = 12 (108 000) = 54 000
(i)
Range = 311
IQR = 68, 5
Semi-IQR = 12 (68, 5) = 34, 25
(j)
Range = 49
IQR = 26,5
(k)
(l)
(m)
(n)
Semi-IQR = 12 (13 − 5) = 4
1
2
Semi-IQR = 12 (26, 5) = 13, 25
IQR = 84 − 66 = 18
(1) Class A
Range = 96 − 30 = 66
Class B
Range = 96 − 66 = 30
IQR = 90 − 72 = 18
The range for Class A is greater than the range for Class B. The minimum value for Class A is an
outlier and has affected its range. Both classes have the same interquartile range.
(2) Semi-IQR for both classes is 9.
(1) x = 3
(2)
IQR = 15 − 8 = 7
Possible numbers are:
2 5 5 6 7
2 4 4 4 6
1 6 6 8 9
Mean = 59
Median = 54
Range = 86
EXERCISE 7
(a)
(b)
(c)
(1) 20,54
(1) 51,33
(1)
43
44
45
46
(2)
(2)
0
7
0
0
1
7
0
1
1
8
1
2
(3)
(3)
20 ≤ x < 25
60 ≤ x ≤ 65
2
8
2
2
3
9
3
3
7
9
3
3
8
9
3
9
299
20 ≤ x < 25
50 ≤ x < 55
9
4
5
5
6
7
7
9
(2)
Class interval
Frequency
8
7
14
7
43, 0 ≤ x < 44, 0
44, 0 ≤ x < 45, 0
45, 0 ≤ x < 46, 0
46, 0 ≤ x < 47, 0
(3) Actual mean = 45,01
(4) Range = 3,9
Q1
(5)
(d)
(1)
4
5
6
7
8
9
4
0
4
1
0
6
9
2
4
1
2
7
4
5
2
4
8
(3) Actual mean = 70, 07
Median = 45,15
IQR = 1
Q2
5
6
4
8
The modes are 44,9 and 45,3
Q3
6
8
5
8
7
8
7
9
8
Median = 69,5
IQR = 80 − 57 = 23
(4)
CONSOLIDATION AND EXTENSION EXERCISE
x = R90
(1)
Median = R63
Mode = R54
Range = R234
(2)
The mean since it is the highest measure of central tendency.
(3)
The mode since it is the lowest measure of central tendency.
(b) (1)
Company A since its mode of 25 is higher and more frequent than the mode for Company B.
(2)
Mean for Company A = 20,9
Mean for Company B = 21,9
The mean suggests that Company B is the best since the mean is the higher of the two.
Median = 0
(c) Mode = 0
The mode and median suggest that he caught no fish.
x = 26
(d) Sandy:
Range = 7
Paul:
x = 26,3
Range = 18
For Paul, the mean is slightly higher than the mean for Sandy. The range is significantly higher for Paul
indicating that Sandy tends to be more consistent than Paul. This suggests that Sandy is the best to consider.
(e) x = 6, 5
(f)
Tickets per day
Frequency
No of tickets × frequency
(a)
0
1
2
3
4
5
6
Totals
1
1
10
7
5
2
0
26
Median = 3
x = 2, 77
(g) (1)
Maths:
0
1
20
21
20
10
0
72
Mode = 2
Science:
Q1
Q2
Q1
Q3
(2) No, the argument is invalid.
50% of the learners scored between 30% and 55% in Science.
50% of the learners scored between 30% and 45% in Maths.
Therefore, the numbers will be equal.
300
Q2
Q3
(h)
(i)
(l)
Estmated mean = 24,12
x = 163, 48
R9 000
(j)
(m)
Median class: 18 ≤ p < 24
x =2
(k)
x =5
Modal class: 18 ≤ p < 24
x = 60, 42
CHAPTER 11 (MEASUREMENT)
EXERCISE 1
(a)
(b)
(c)
(d)
(e)
(1)
Cuboid: V = 126 cm3
Cylinder: V = 318, 09 cm3
Triangular prism: V = 18 cm3
(2)
Cuboid: S = 162 cm2
Cylinder: S = 268, 61 cm 2
Triangular prism: S = 63 cm 2
(3)
Cylinder: S = 204,99 cm 2 Cuboid: S = 120 cm2
(1)
Volume of A and B is 8 000 cm3
The triangular prism
3 cm
(1)
(2)
70 cm2
(3)
3
(1)
7 875 000 cm = 7 875 l
(2)
(2)
r = 20 cm
8 cm
(4)
(3)
C
(3)
189 105,63 cm2
5 cm
3000 000 ml = 3 000 l
EXERCISE 2
(a)
k =2
(1)
(2)
Original: S = (64π) cm 2
Enlarged: S = (2) 2 × (64π) cm 2
(3)
Original: V = (96 π) cm 3
Enlarged: V = (2)3 × (96π) cm 3
(1)
(1)
(1)
7 300 cm 2 ; 3 000 cm3
(2)
1 168 cm 2
24x 2 ; 8x3
(1 568π) cm 3
(2)
(2)
2x
Vnew = 4 × Voriginal
(3)
(616π) cm 2
(4)
Snew = 4, 2 × Soriginal
(e)
(1)
(4)
(2)
scale factor of 2
(3)
4 times
(f)
(1)
(2)
SA : SB = 4 : 9
(g)
(1)
V = πa 2b ; S = 2πa 2 + 2πab
2 times
3
k = ; 810 cm3
2
hQ = 12,59 cm
(2)
Area of base P : Area of base Q = 1 : 1,31
(h)
(1)
Senlarged : Soriginal = 25 :16
(2)
Venlarged : Voriginal = 125 : 64
(b)
(c)
(d)
EXERCISE 3
(1)
684, 00 cm2 ; 1 045,33 cm3
(2)
7 539,82 cm 2 ; 37 699,11 cm3
(3)
26, 62 m 2 ; 12,57 m3
(4)
1 809,56 cm2 ; 7 238,23 cm3
(5)
84,82 m 2 ; 56,55 m3
(6)
1 403,07 cm 2 ; 19 893,61 cm3
(1)
43,30127019 cm 2
(2)
63, 78675411 cm 2
(3)
234, 66 cm2
(4)
173, 21 cm3
(c)
(1)
7,5 cm
(2)
8 246,68 cm3
(d)
30,81 cm
(e)
1 432,57 cm3
(f)
(1)
(2)
5250 m3
(a)
(b)
1800 m 2
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(1)
64 000 cm3
(2)
9 740, 62 cm2
(b)
(1)
64,95190528 cm 2
(2)
519, 62 cm3
(c)
(d)
209, 4395102 cm 2
(1)
1,361 cm
π
(1)
4
(2)
1 466,08 cm3
(3)
8,82 cm
(2)
π
4
(3)
V=
(e)
301
4 2
r h
3
CHAPTER 12 (PROBABILITY)
EXERCISE 1
6
58 29
(a)
(1)
(2)
=
35
70 35
1
2 1
(2)
(b)
(1)
=
6
6 3
5
9
(2)
(c)
(1)
28
28
8 1
8 1
(2)
(d)
(1)
=
=
16 2
16 2
(e)
12 500
C is the closest answer
3
3
(f)
(1)
(2)
11
11
13 1
1
(2)
(g)
(1)
=
52 4
52
2 1
3 1
(2)
(h)
(1)
=
=
6 3
15 5
1
192π 3
(i)
(j)
(1)
=
8
256π 4
EXERCISE 2
(a)
(b)
(c)
(1)
A = {3 ; 5}
(2)
A
3
1
5
n(A) = 2
2
4
(3)
(3)
(4)
(4)
2 1
=
10 5
4
1
=
52 13
3
4
1
(2)
2
(3)
(3)
(3)
P(A) =
(4)
(4)
(4)
5
6
0
=0
28
5 1
=
15 3
8
2
=
52 13
1
2
5
(3)
8
(4)
6
A = {1 ; 2 ; 3 ; 4 ; 5}
(5)
2
6
(1)
(3)
n(A) = 2 ; n(S) = 8
P(A) =
5
12
(4)
B = {5 ; 10}
(2)
(4)
A = {2 ; 3 ; 5 ; 7 ; 11}
n(A) = 5
(3)
P(A) =
(1)
P(A) =
5
12
B = {6 ; 12}
(2)
n(A) = 5
(3)
P(A) =
(4)
5
14
B = {d ; e ; f ; g ; h ; i}
n(B) = 6
P(B) =
C = { j ; k}
n(C) = 2
6 3
=
14 7
(5)
1
12
1
12
2 1
=
12 6
6
12
P(B) =
n(A) = 2
(5)
S = {a ; b ; c ; d ; e ; f ; g ; h ; i ; j ; k ; l ; m ; n}
(2)
P(B) =
(5)
(4)
A = {a ; b ; c ; d ; e}
2 1
=
8 4
n(A) = 2
5
12
(e)
4
6
S = {R ; A ; N ; D ; O ; M ; L ; Y} Event A = {A ; O}
4
12
(d)
26 1
=
52 2
1
3
(5)
(2)
n(A) = 5
(3)
19
28
(5)
2 1
=
6 3
(3)
S
(1)
(1)
3 1
=
6 2
14 1
=
28 2
13
16
(3)
P(C) =
2 1
=
14 7
S
B
A
3
14
2
14
4
14
3
14
302
C
2
14
2
12
5
12
2 1
=
12 6
(f)
(1)
(2)
(3)
(6)
n(T) = 220
150 5
=
390 13
100 10
=
390 39
n(D) = 250
n(S) = 390
(4)
120 4
=
390 13
(2)
6 1
=
12 2
(4)
4
7
(5)
4
7
6 3
=
8 4
(5)
20
2
=
390 39
EXERCISE 3
(a)
(1)
(4)
(b)
(1)
(6)
(c)
(d)
1
12
3
7
5
7
(5)
(2)
5
12
1
7
(6)
(3)
(1)
S = {a ; b ; c ; d ; e ; f ; g ; h}
(5)
3
8
(6)
4 1
=
8 2
(1)
3
(2)
7
(6)
7
12
(7)
(11)
1
(12)
(16)
(21)
(e)
S = {1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 ; 10 ; 11 ; 12}
(1)
9 3
=
12 4
5
12
(17)
8 2
=
12 3
1
12
9
(2)
(5)
(7)
(f)
205 41
=
300 60
(1)
(6)
(g)
(i)
(i)
1 480 37
=
1 520 38
(1)
(2)
5
8
(3)
3
8
(4)
(3)
8
(4)
6
(5)
(8)
6 1
=
12 2
(9)
3
(10)
(13)
11
(14)
11
12
(15)
9
(18)
9 3
=
12 4
(19)
5
(20)
5
12
(3)
140
40 2
=
300 5
165
45
3
=
300 20
(6)
(ii)
95 19
=
300 60
(2)
15
(3)
(4)
1 505 301
=
1 520 304
(5)
15
3
=
1 520 1 024
18 3
=
60 10
12 1
=
60 5
(3)
9
3
=
60 20
(3)
(6)
0,2
0, 3 + 0, 4 + 0, 2 = 0,9
(ii)
(2)
(4)
(h)
1
12
5
7
(3)
(5)
(i)
30 1
=
60 2
(ii)
28 7
=
60 15
(1)
(4)
0,1
0, 2 + 0,1 = 0,3
(2)
(5)
0,3
0,3 + 0,1 = 0, 4
303
40
1
=
1 520 38
1 465 293
=
1 520 304
5
12
3 1
=
12 4
3 1
=
12 4
(4)
300
(i)
(1)
(2)
(j)
(1)
(3)
Learners that are not in School P that dislike Maths. In other words, learners in School K that
dislike Maths.
60 3
84 21
40 2
(ii)
(iii)
=
=
=
(i)
100 5
100 25
100 5
2 1
6 1
7
(2)
=
+ =
10 5
10 10 10
2 1
(i)
5
(ii)
5
(iii)
=
10 5
2 1
8 4
(iv)
(v)
=
=
10 5
10 5
EXERCISE 4
(a)
(1)
B
(2)
(b)
C
(3)
D
(c)
(4)
A
(5)
(e)
(1)
(f)
(1)
0,15
0,2
25
28
(2)
(2)
(6)
A
(d)
×
0,8
C
×
×
0,7
0,3
10 5
=
28 14
(3)
0,5
3
7
0,5
0,23
0,7
(3)
(3)
0,7
5
28
(4)
18 9
=
28 14
EXERCISE 5
(a)
(1)
(b)
(1)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(1)
(1)
(1)
0,9
0,67
0,1
(1)
(1)
(1)
0,01
0,9
3
7
0,6
0,6
0,8
(2)
(2)
(2)
(2)
(2)
(3)
(3)
(3)
0,3
1
7
0,4
0,15
0,4
(2)
(2)
not complementary
not complementary
(3)
0,1
(2)
0,4
not complementary
0,125
(4)
(4)
(4)
0,1
4
7
0,6
(5)
(5)
(5)
0,7
5
7
0,7
(6) 0,4
2
(6)
7
(6)
0,2
(3)
complementary
(3)
16 8
=
46 23
not complementary
CONSOLIDATION AND EXTENSION EXERCISE
(a)
(d)
(e)
(f)
7
27
(1)
3
5
(1)
(g)
(1)
(h)
(1)
(5)
(i)
(1)
(b)
0,1
(c)
(1)
complementary
(2)
0,3
(3)
4
P(A) = ;
9
Diagram 2
Diagram 2
3
8
(2)
0,4
2
P(B) =
9
(2)
Diagram 1
(6)
Diagram 1
2 1
(2)
=
8 4
(2)
(3)
(7)
(3)
3
(2)
46
not complementary
4
5
9
Diagram 4
Diagram 4
4 1
=
8 2
304
(4)
(8)
(4)
35
46
(3)
Diagram 3
Diagram 3
1
8