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Assemblies of God High
School
For the year 2011
Photocopy and electronic transmission is only
permitted if it is done within Assemblies of God
High School network.
Any one else would be an infringement
of the copyright laws of Fiji.
We trust you that you would not circulate this ebook to anyone else outside the network of the
Assemblies of God High School.
The Assemblies of God High school network
means:
•
•
•
•
Assemblies of God High School Teachers.
Assemblies of God High School Students.
Computers at Assemblies of God High School.
Computers at home of students and teachers
of Assemblies of God High School.
Vinaka
Kishore Kumar
[Author/Publisher/Full Registered Teacher/Education Consultant]
1
Fiji School Leaving Certificate
Mathematics
FORM 6
E-BOOK
The perfect aid for better grades!
Covers all course fundamentals-supplements any class text
Teaches effective problem solving
154 examples showing step-by-step working
Ideal for independent study.
• Provides ongoing support.
Kishore Kumar
2
With ongoing educational support from
Presents
Fiji’s First Fiji School Leaving Certificate
Physics Text Book
National Youth Party together with Kishore Kumar Publication are educational sponsors of the
Yellow Ribbon Project in Fiji. The sponsorship deal includes donation of text books published by
Kishore Kumar Publication. We write and publish books in Fiji for the people of Fiji and help build a
better Fiji by giving this new generation a new hope of accessing text books in the field of
mathematics, physics, computer, education, chemistry and Christian religion at a very low cost. Our
books used in secondary schools are approved by the Ministry of Education.
Other Books written by the Author and Publisher:
•
•
•
•
•
•
•
•
•
•
Fiji School Leaving Certificate Form 6 Physics
Fiji School Leaving Certificate Form 5 Physics
Fiji School Leaving Certificate Form 6 Mathematics
Fiji School Leaving Certificate Form 5 Physics
Fiji School Leaving Certificate Computer Studies
Fiji Seventh Form Mathematics
Fiji Seventh Form Physics
Fiji Seventh Form Computer Studies
Fijian Education: Problems and Solutions
Introduction to Christianity
© 2011 Kishore Kumar Publication
All rights reserved. No part of this book may be reproduced in any form or by any
means without permission in writing from the publisher, Kishore Kumar Publication,
P. O. Box 17773, Suva, Fiji.
E-mail: [email protected]
[email protected]
Kishore Kumar: 679 – 9961207
Nayagodamu R. Korovou: 679 – 9292394
3
Dedication
Christianity was introduced in Fiji in the 1830s in Levuka and Lau Islands. The
greater area of Fiji changed from cannibalism to the People of God and by the 1870s,
100% of the Fijians accepted Christianity. Fiji is the only nation in the world where
cannibalism was done away within a twinkling of an eye and Christianity was
accepted within the twinkling of an eye as well. No nation has ever become divinely
civilized overnight except, Fiji.
This book is dedicated to those early missionaries who entered our shores and in spite
of the threat of being eaten alive, they came to do the Will of God and that will was to
educate the ancient Fijians about the God who has created us all and our reason of
being here on earth. Our life on earth is an opportunity for us to prepare to meet God
one day. It was then and there when the ancient Fijians came to know the only true
and living God who looks like us because we are His children. It was the
missionaries who helped ancient Fijians to gain further light and knowledge that after
we die, we will be judged by God and rewarded according to our works on earth.
After the ancient native Fijians accepted the only true and living God, Fiji was blessed
richness of the earth that included minerals such as gold.
Some riches beneath the earth
have not surfaced. It will be
done in day when the children
of the ancient natives of Fiji
would once again acknowledge
God and be obedient to His
commandments.
The Colonial Government
refused to provide education to
the children of indentured
labourers from India. It was the
missionaries and those ancient
converted Fijians who pleaded
and soften the hearts of the
Colonial Government to allow
children of indentured labourers
to receive education.
I am one of descendants of the
indentured labourers and as a
result, I have written this book. Thanks to those early missionaries and native Fijians
because of their love that they received after accepting Lord Jesus Christ. Today the
descendants of those indentured labourers are being fruitful in Fiji and have and are
still given free primary and secondary education by the Government of Fiji. Now the
children of those indentured labourers are receiving free books from the Government
of Fiji. No Government in any part of the world has been so kind like Fiji to provide
free education and free books to primary and secondary school students including the
children of the indentured labourers and allow them full citizenship. This book is
dedicated to the Government of Fiji as well. Fiji the way the world should be.
4
Preface
Dear Teachers, Students and Parents/Guardians:
This Physics book contains detailed explanation and shows students various ways of
solving a single type of problem and may be substituted in place of a note book.
Experienced teachers do acknowledge that at least 45% of school time is used for writing
notes, which teachers may dictate to their students or students may copy the notes from the
board, which has been written by their respective teachers. Such usage of time is
undeniably a concern because students taking physics do not come to school to copy notes
but to solve physics problems. The only way to learn physics is to do Physics and the best
of doing Physics is by solving problems.
This book ensures that the 45 % of the school time that has been traditionally used for
copying or writing notes will now be used constructively for solving problems and engage
students in classroom discussions. With the availability of this Physics book, there
would be no need for students to rewrite notes and hence, such time now can be used
to enhance their knowledge and performance of understanding by solving problems.
From 2002 and 2004 I conducted a thorough research in Fijian Education and the work was
published in 2005 titled, “Fijian Education: Problems and Solutions.” One of the findings
or problems in Fijian Education is that some slow writers get frustrated because they
cannot cope with their peers (Kumar, K. 2005, 19-40).
It is also a challenge for teachers to cope with or tolerate slow writers because they have to
move on with time. Fortunately, my thesis dealt with identifying problems and
recommending significant solutions that would enhance Fijian Education.
You would agree that previously, schools had to buy at lest 3 different Physics books from
overseas and then use only those contents that are part of our curriculum. This is a concern
in terms of cost and management. This book solves both concerns. This book maps
perfectly onto the Fiji School Leaving Certificate Form Six Curriculum and the same
curriculum maps perfectly onto this book.
Since the implementation of our FSLC Form Six Curriculum in 1988, we did not have a
single physics book that maps directly onto our curriculum or our curriculum mapping
directly onto such books. This physics book is the first of its kind and it maps directly onto
our curriculum for Fiji School Leaving Certificate Examination Physics.
This book is a vital instrument to help learner’s gain better understanding of Physics. All
students and teachers would find this book as an important tool to both sustain our
curriculum and the teaching and learning process.
^|á{ÉÜx ^âÅtÜ
Full Registered Teacher
Author & Publisher
Education Consultant
About the Author: Kishore Kumar is a member of The Church of Jesus Christ of Latter Day Saints
and he believes that failure is a constructive method that shows our weaknesses so that we can
improve from thereon. Failure does not necessary mean end of success. Failure means things that
we need to get it done right to gain success. Failure should be seen as a constructive and
worthwhile experience because it teaches us lessons in humility and areas we need to improve on.
5
Acknowledgements
The author of this book acknowledges the following organisations and people for their
unconditional support and inputs.
Fiji National University for technical knowledge support
Box Hill Institute-Melbourne Australia for general technical support
Kahuku High School-Hawaii for structural recommendation
National Youth Party for ongoing educational support
Navua High School now known as Vashist Muni Memorial College for providing
humble but sure leadership training and education and for helping to build a better
Fiji.
2006-2009
physics students and teachers of Bhawani Dayal Arya College for
constructive inputs
Government Printer for professional assistance and work
Fiji Teachers Union general recommendations
Fijian Teachers Union for general recommendations
Ministry of Education of the following countries:
Fiji
Papua New Guinea
Vanuatu
Solomon Islands
Tonga
Samoa
Kiribati
Tuvalu
My God, for His continuous divine support during my most difficult times
for helping me to complete this book. “Behold, I am the LORD, the God of all
flesh: is there any thing too hard for me?” (Bible | Old Testament | Jeremiah 32:27)
6
Mathematical System
Introduction
Modular system is an important part of algebra and it is used by various types of
businesses for the purpose of coding so that they can achieve their business
objectives. The pin number found in both the Vodafone Fiji and Digicel Fiji recharge
cards use modular system to formulate 12 digit numbers. The encoding function is
known only by the company that manufactures recharge cards.
It is very difficult to break such codes but codes can be broken. For security reasons,
greater management control are taken to make sure that a code made up of 12 digits
becomes almost impossible to break.
One of the famous equations used for making codes is linear equation. Linear
equations are derived from long division that you will learn later in this course. Since
this topic is discussed under basic mathematics, at this point, only basic concept of
long division will be discussed.
Division Algorithm
Let a and b be integers with b > 0. Then there exist unique integers q and r with the
property that a = bq + r where 0 ≤ r ≤ b.
The integer q in the division algorithm is called the quotient upon dividing a by b; the
integer r is called the remainder upon dividing a by b.
If a = 21 and b = 4, then a ÷ b → 21 ÷ 4 = 5R1
5
4 21
− 20
1
q = 5 r =1
a = bq + r → 21 = 4 × 5 + 1
In USA, some states use linear functions to encode the month and date of birth into
three-digit number that is incorporated into drivers license numbers.
All New York licenses issued prior to September of 1992, the last two digits indicate
the year of birth and the three preceding digits code the month and date of birth. For
male drivers these three digits are a = 63m + 2b, where m denotes the number of the
month of birth and b is the date of birth. So, since 786 = 63 × 12 + 2 × 15 , is a license
that ends with 78674 indicates that the holder of the license is a male born on 1974.
Month and day are coded in 786 [the first three digits] but we know from the working
that 786 = 63 × 12 + 2 × 15 gives a December as the month and 15 as the day born. In
case of more than one person having the same date of birth, a tie breaking digit is
placed before the two digits for the year of the birth. If two license holders were born
on 15th of December, 1974, a check digit before the year of birth is placed. 78674
would become 786074 for the first person and 786174 for the second person.
7
Modular Arithmetic
Modular Arithmetic is another application of division algorithm.
Definition:
a modulo n [sometimes read as a mod n]
Let n be a fixed positive integer. For any integer a, a modulo n [a mod n] is the
remainder upon dividing a by n.
8 mod 3 = 2: means that when 8 is divided by 3, the remainder is 2.
38
2
3 8
−6
2 → remainder
Modular arithmetic is used by the recharge card makers of both Vodafone Fiji and
Digicel Fiji to assign an extra digit to make sure that recharge pin codes are not
decoded by their customers. However, since these types of company’s deals with
millions of dollars each year, their modular arithmetic system is very complex and is a
combination of at least one linear system and one non linear system. To maintain
integrity of their recharge cards, such companies continually changes their modular
arithmetic system and hence, preventing it from being compromised.
The above information is to let you know that though you are studying basic modulo
system, its application is beyond measure and it is used in Fiji for writing codes in the
form of numbers, alphabets and combination of both.
If you have some knowledge of computer software then you would know that
software comes with a key [pin number usually consist of numbers and alphabets] that
must be entered for installation and updating software online and to enjoy the
complete features found in the software. Software has a check digit. Microsoft
operating systems such as Windows XP, Windows Vista has at least two check digits.
If you make an error when entering the keys, the error will be detected and the
software will not be loaded. Postal services use check digits for money orders and
many other companies use modulo of a certain numerical value to identify numbers
when assigning check digits.
Banks in Fiji do not use any check digit when assigning pin numbers for their
customer’s access card. ANZ bank in Fiji allows their customers to change their pin
number and select any four digit number that they can always remember. This is very
dangerous because any 4 digit numbers can be easily guessed. However, ANZ bank
has programmed their ATM and it acts as a security check digit. This means if you
enter your pin wrongly for 3 times in row, or someone else takes your card and enters
incorrect pin numbers 3 times in a row, the ATM will swallow the access card and
you will have to contact ANZ bank for both explanation and retrieval of your card.
8
Modular System
It was this century when the word group was clearly defined and accepted by
mathematicians. Group: one-to-one functions on finite sets that could be grouped
together to form a closed set.
To combine two numbers, binary operation is needed.
Definition
Binary Operation
Let S be a set. A binary operation on S is a function that assigns each ordered pair of
elements of S an element of S.
Definition
Group
Let S be a nonempty set together with a binary operation ∗ that assigns to each
ordered pair (x, y) of elements of S an element in S denoted by xy. The binary
operation is usually multiplication and hence, for the above definition, the binary
operation ∗ stands for multiplication.
We can say that S is a group under this operation if the following first four conditions
are satisfied.
1. Closure: When any pair of elements is combined and the result is not
outside the set.
2. Identity: There is an element I that is called the identity element in S,
such that AI = IA = A for all A in S .
3. Inverse: For each element A in S, there is an element B in S that is
called the inverse of A such that AB = BA = I .
4. Associative: If S is a set and A, B and C are elements of set S then it
follows that ( A ∗ B ) ∗ C = A ∗ ( B ∗ C ) .
5. Abelian: If set S has the property that a ∗ b = b ∗ a for every pair of
elements a and b in S, then the group is called Abelian.
The above can be stated that a group is a set that obeys the associative property and
every element has an inverse and any pair of element can be combined without
moving outside the set.
9
Elementary Properties of Groups
Uniqueness of the Identity
In a group S, there is only one identity element.
Uniqueness of Inverse
For each element a in a group S, there is a unique element b in S such that
ab = ba = I
CONDITIONS OF A GROUP
For any mathematical system to be declared a group, the following conditions must
be satisfied.
•
After any operation, the product must belong to the set. You know that natural
numbers are counting numbers. If you add one counting number with another
counting number, your answer will always be a counting number. Hence,
under addition, natural numbers are closed. Meaning no new element comes
as a product that does not belong to the set we are dealing with [No new
element].
•
There exists exactly one Identity Element. Meaning, if the Set is a group, all
the elements of the set share one common Identity element.
•
Each element in the set will have its Own Inverse. Meaning, no two or more
elements will have the same inverse.
•
The Set must fulfil the Associative property.
In order for the system to be declared a
group, it must pass all four tests.
If it fails one or more test, it is not a group
10
EXAMPLE 1 Show that the system belonging to Set A given below is a group
Solution →
Step 1: The Set A = {a, s, d, f}
The elements with the smaller font are the results of the operation of the elements of
Set A. No new element is present in the solution. All solution belongs to Set A
hence, the Set A is closed.
Step 2: There exists exactly one Identity element
Let A denote an element and I denote an identity element.
It follows that A ∗ I = A
a∗ f = a
s∗ f = s
d∗ f =d
f∗f = f
[There exists exactly one identity element → Identity element is f ]
Step 3: Let A denote an element. It follows that element A operated by its inverse
will give the identity element as the product.
a ∗d = f
A ∗ A −1 = I
s ∗s= f
d ∗a = f
f∗f = f
•
•
•
•
The inverse of a is d
The inverse of s is s
The inverse of d is a
The inverse of f is f
Each element has its own inverse
11
Step 4: In this step, we will have to prove that associative property is valid for this
set.
a ∗ (b ∗ c) = (a ∗ b) ∗ c
a∗a = c∗c
It passes the associative test.
b=b
Conclusion
Since the system is closed under the operation *, there exist exactly one identity
element [d], each element has its own inverse and it is associative in nature, the
system (S *) is a group.
Note: In addition, if the system is commutative, it is called commutative group or
Abelian group.
Not all binary operations are given in a table. Some are expressed as an equation or
formula.
EXAMPLE 2
A binary operation is defined as a ∗ b = 2a − 4b. If a = 2 and b = 1 , evaluate a ∗ b .
Substitute in the equation a ∗ b = 2 a − 4b the values of a and b respectively.
a ∗ b = 2a − 4b
a ∗ b = 2 ( 2 ) − 4 (1)
a *b = 4 − 4
a *b = 0
If a ∗ b = −2 a × −b and a = −1 and b = −2 evaluate a ∗ b
EXAMPLE 3
a ∗ b = −2a × −b
−1∗ 3 = −2 ( −1) × − ( −2 )
= 2× 2 =
4
EXAMPLE 4
If a ∗ b ∗ c = −a 2 × −b 2 − c 0⋅5 and a = −1 and b = −2 c = 25,
evaluate a ∗ b ∗ c
a ∗ b ∗ c = − ( −1) × − ( −2 ) − ( 25 )
2
2
0⋅5
= −1 − 4 − 5 =
12
−10
EXAMPLE 5
If a ∗ b = a − b and a = −1 and b = −2 evaluate a ∗ b
a ∗ b = −1 − −2 = − 1 + 2 =
EXAMPLE 6
1
Given below is a ⊕ 5 modulus operation table.
⊕5
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
1
2
3
4
0
2
3
4
0
1
3
4
0
1
2
4
0
1
2
3
In row 3 column 6: 1 + 4 = 5 but you see in the table that the answer is zero. This is
because the answer cannot be greater or equal to 5 because this is a ⊕ 5 modulus
operation table. When the sum is either equal or greater than 5 in a ⊕ 5 modulus
operation table, you subtraction 5 from the sum until the final sum is less than 5.
Since 1 + 4 = 5, we will subtract 5 from the sum [5]. 5 – 5 = 0. This applies to all
sums greater or equal to 5.
EXAMPLE 7
Given below is a ⊗5 modulus operation table.
⊗5
1
2
3
4
1
2
3
4
1
2
3
4
2
4
1
3
3
1
4
2
4
3
2
1
In row 5 column 5: 4 x 4 = 16 but in the table the answer is 1. The answer cannot be
greater or equal to 5 because this is a ⊗5 modulus operation table.
When the product is either equal or greater than 5 in a ⊗5 modulus operation table,
you subtraction 5 from the sum until the final sum is less than 5.
4 × 4 = 16 16 − 5 = 11 11 − 5 = 6 6 − 5 = 1
13
EXAMPLE 8
Given below is a ∗4 modulus operation table. Note: * represents a binary operation.
∗4
1
2
3
1
2
3
1
2
3
2
3
1
3
2
1
•
∗4 Modulus operation is closed because all 9 products belong to the original
set {1, 2 and 3}.
•
There is only one identity element and that element is 1.
•
The inverse of 1 is 1, 2 is 3 and 3 is 3. Each element has its own inverse.
•
It has associative property:
(1∗ 2 ) ∗ 3 = 1∗ ( 2 ∗ 3) → 2 ∗ 3 = 1∗1 → 1 = 1
Since it passes all four conditions, the above ∗4 Modulus operation set is a group.
EXAMPLE 9
*
W
X
Y
Z
W
Y
Z
W
X
X
Z
W
X
Y
Y
W
X
Y
Z
Z
X
Y
Z
W
a. Name the identity element.
Identity element is Y
b. Find the inverse of Z.
X
c. Evaluate W * ( X * Y)
W*X
Z
14
Yes, a man may say, you have faith, and I have works: show me your faith without
your works, and I will show you my faith by my works (New Testament | James 2:18).
Exercise 1A
1. Show that the system belonging to Set S given below is a group.
2. A binary operation is defined as a ∗ b = 4 a − 3b.
evaluate a ∗ b .
If
a = 2 and b = 1 ,
3. If a ∗ b = − a × 2b and a = −1 and b = 3 evaluate a ∗ b
4. An operation is defined as a b = 4a 4 − 2b. Evaluate −2 −10.
5. If a • b = a × −2b and a = 100, b = −21, find a • b.
6. A binary operation ∗ on natural numbers is defined as a ∗ b =
( 2a + 2 ) .
( 2b − 2 )
Evaluate 3 ∗ 2.
7. A
binary operation ʘ is defined on a set of
a
a b = where a, b ∈ I . Explain why 5 ʘ 3 is not closed.
b
integers
as
8. Give two reasons why the set of odd integers under addition is not a group.
9. The set Z n = {0,1, 2, 3..., n − 1} for n ≥ 1 is a group under addition modulo n.
Write out a complete operation table for Z 4 .
10. Suppose the table given below is a group table. Fill in the blank entries.
15
11. Suppose the table given below is a group table. Fill in the blank entries.
12. The table given below shows the results of set S = {1, 2, 3, 4} under the
operation *.
A.
B.
C.
D.
Find the value of c ∗ ( d ∗ b )
Name the identity element.
What is the inverse of c?
What is the inverse of d
13. For the binary operation ∗ on the set S, if z ∗ ( x ∗ c ) = ( z ∗ x ) ∗ c is true for all
values of z, x, and c on S, what can you conclude about the operation ∗ ?
14. The operation given below is for the set {a, s, d } . If a is the identity element,
a.
b.
c.
Find the values for 1, 2, 3, 4 and 5.
Find the inverse of a.
Is the operation a group? Give a reason.
15. Show that {1, 2, 3} under multiplication modulo 4 is not a group but that
{1, 2, 3, 4} under multiplication modulo 5 is a group.
16
LAWS OF INDICES
Law 1
1 
→ ( a ) = an
n

Law 2
2 
→

a
( )
n
eg .12
a
4 
→ b = ax − b
x
Law 4
a
5 
→ − b = ax b
x
eg .14
x−a yb
6 
→ −b = a
y
x
−c
( )
( )
xa
−c
9 
→
•
x =x
a
x− 4
a
b
2
2
=
26 64
=
34 81
x − ac y bc
= −bc = ac
y
x
−c
( )
( )
b
= 26 = 64
2 x4
=
= 2 x4
1
( )
( )
−2
22
 22 
eg.17  3  =
5 
53
Law 9
2
2
23
 23 
eg .16  2  =
3 
32
x 
8 
→ b =
y 
yb
Law 8
2
2−3 32 9
eg.15 −2 = 2 =
3
2
4
Law 6
a
( )
23
2
2x− 4
=
= 2x− 4
4
1
x
eg .13
Law 5
c
= 23 = 8
42 16
4
eg .11   = 2 =
3
9
3
= a mn
 xa 
x ac
Law 7
7 
→  b  = bc
y
y 
3
2
x
xa
 = a
y
y
Law 3
3 
→ am
( 2)
eg.10
−2
−2
22×− 2 2− 4 56
= 3 ×− 2 − 6 = 4
5
5
2
eg .18
3
x =x
4
Law has been extended to enhance understanding
17
4
3
EXAMPLE 19
2− 4
Write − 3 as a fraction with positive power.
5
2− 4
53
→ 4
5− 3
2
When a base is raised to a power and its position changes in the fraction, meaning it is
either moved from numerator to denominator or from denominator to numerator, the
power sign of the power also changes.
EXAMPLE 20
 23 
Write  2 
3 
•
−2
as a fraction with positive power.
When a power is raised to another power, multiply both powers.
−2
( )
( )
23
2 
 2 = 2
3 
3
3
−2
−2
3 ×− 2
( 2)
=
2 ×− 2
( 3)
2− 6 34
= −4 = 6
3
2
Rule for changing from surd to base index form
x
numerator
denominator
=
denominator
x numerator
Note: Root takes place of the denominator
EXAMPLE 21
Write
5
x7
in base index form.
solution :
18
5
x7 = x
7
5
EXAMPLE 22
Write
x in base index form.
x means square root of x. The standard rule is
b
a
b
x =x .
a
For x , a = 1 [ x 1 = x ] and b = 2. You do not see root 2 in x because the smallest
root is square root. When a radical is a square root, you won’t see 2 in the place of
root. However, if the root is anything else besides square root, the radical must show
3
that root. In example 18, the root is 3 and hence
x 4 is read as cubic root of x
1
raised to the power 4.
EXAMPLE 23
•
Simplify
x ×
3
x5
Change the surds into base index form before performing multiplication.
x ×
x
x = x2
x is read as square root of x .
1
2
3
x5
5
3
× x = x
1 5
+
2 3
→ x
3 +10
6
→ x
13
6
→
6
x13
Example 24
2x + 2
Simplify
4
o
o
2x + 2
2 x × 22
is same as
4
4
2 x × 22
Perform possible cancellations
= 2x
4
2x + 2
2 x × 22
2x × 4
2x × 4
→
→
→
→ Ans : 2 x
4
4
4
4
19
Simplify
Example 25
o
o
2x + 2
2x
2x + 2
2 x × 22
is same as
2x
2x
2 x × 22
Perform possible cancellations
2x
= 22 or 4
2x + 2
2 x × 22
2 x × 22
2
→
→
→
Ans
:
2
or 4
x
x
x
2
2
2
2x + 2 4
Simplify 2 x + 4 ÷ 4
x
x
Example 26
xx + 2 4
x x + 2 x4
x x × 22 x 4
÷
→ 2x+4 ×
→ 2x
×
x2 x + 4 x4
x
4
x × x4 4
x × 2
x
2
x2 x × x4
×
x
4
→
( )
( )
4
x
→
x2x
xx
→
2
x
( )
( )
( ) (4 )
x
44 × 16
44
164 x + 1
164 x ×16
→
→
→
2x
4
48 x
48 x
4
4
xx
1
2
→
164 x + 1
Simplify
48 x
Example 27
(4 )
( )
(x )
1
xx
x
x +−2 x
1
× 16
(4 )
4
→
−x
1
× 16
→
20
16
(4 )
4
x
→
x
4
1
16
44 x
−2 x
× 16
1
xx
Example 28
164 x + 1
Simplify
−1
48 x
164 x + 1
164 x + 1 1 16 4 x + 1 − 48 x
−
1
as
a
single
fraction
→
− →
.
48 x
48 x
1
48 x
164 x + 1 − 48 x 164 x ×161 − 48 x
o 164 x + 1 means 16 4 x × 161. Hence,
=
48 x
48 x
o Write
o If you would notice that 164 x can be written with base 4. 164 x = ( 42 ) = 48 x .
4x
( )
42
164 x × 161 − 48 x
=
48 x
o
4x
×161 − 48 x
48 x
=
48 x × 161 − 48 x
.
48 x
(
)
48 x × 161 − 1
48 x × 161 − 48 x
is
equal
to
48 x
48 x
48 x is a common factor. Hence,
48 x × (161 − 1)
o Perform all possible cancellations
48 x
(16 − 1) = 16 − 1 = 15 = 15
1
=
1
1
1
Example 29
44 x + 1
Simplify
−1
28 x
44 x + 1
44 x + 1 1
44 x + 1 − 28 x
−1 → 8x − →
28 x
2
1
28 x
( )
22
44 x × 41 − 28 x
→
28 x
4x
× 41 − 28 x
28 x
(
)
(2 )
2
→
(
21
× 41 − 28 x
28 x
)
28 x 41 − 1
28 x 41 − 1
28 x × 41 − 28 x
→
→
28 x
28 x
28 x
41 − 1
4 −1
3
→
→ = 3
1
1
1
4x
Exercise 1B
1. Write the following indices as fractions having positive powers
A. 4 y − 9
B. −2 x −1
C.
(y )
2
−4
D.
(7 b)
−2
2. Write the following fractions in the base index form.
A.
5
y −5
1
a
B.
C.
1
x −1
2
x y2
D.
3. Simplify the following.
72 x3 y 7
A.
4 x y5
150 a b 2
B.
4 a6 b2
 2 x −2 y 2 
C. 

 4x y 
−3
 2y 
D. 

 5 
−3
4. Write the following as a simplified fraction.
3 x− 4
A. 4 + 4
x
x
1 − y 2 x2 + x2 y
B.
÷
x +1
4x + 4
C.
4y
÷4y
x
5. Write as a fraction with positive power.
 22 
A.  2 
3 
−2
 x2 
B.  5 
y 
6. A. Simplify
2 x+3
8
7. A. Simplify
2 x +3 2 x
÷
16 16
8.
−2
 x2 
C.  − 5 
y 
B. Simplify
2 x +3
2x
B. Simplify
24 x + 1
Simplify 8 x − 1
2
)
(
1
2
10. Simplify 36x12
11. Simplify
1
2
(
9. Simplify 25x100
)
4 x +1 × 4
22 n + 5
22
−2
C. Simplify
3x+ 3 27
÷
16 16
2 x+3
8 × 2x
9

12. Simplify 
 16 
−
−
5
2
1
13. Simplify  
4
3
2
(
14.
If a = 2, b = −2 and c = 3, find the value of y = a −8b
15.
If a = 16, and b = 2, find the value of y = 8 a b −2
(
2
16. Evaluate
 ( 2ab )2 − ( 3a )3 

 if a = 2 and b = 3.
2


2
a
b


2
17. Evaluate
18.
19.
20.
 ( 2ab −2 )2 − ( 3a )3 

 if a = 2 and b = 16.


2a −2b


1
400 2
( 2500x )
1
27 3
(8x )
1
5 5
( 59,049x )
21. Simplify
22. Simplify
24 x −6
18 x 6
4 x −2 y
22 x −3 y −5
23
)
2
)
c
Surds
Rational numbers can be expressed as a fraction. Some roots of rational numbers
cannot be expressed a fraction. For example, 3 is a root of a rational number
that cannot be expressed as fraction. Even thought 3 is a rational number, 3 is
an irrational number. Irrational numbers which incorporates the radical sign
are called surds.
Rule 1
1 
→ a × b = ab
9 × 100 = 9 ×100 = 900 = 30
3 × 10 = 30
Rule 2
2 
→ a× b = a b
Rule 3
3 
→
2× 3 = 2 3
100
100
=
= 4 =2
25
25
a
a
=
b
b
10
=2
5
Rule 4
4 
→ a x ± b x = (a ± b) x
2 5 + 4 5 = ( 2 + 4) 5 = 6 5
24
Example 30
Simplify the following surds
12 = 4 × 3
162 = 2 × 81
18 = 9 × 2
= 2× 3
= 2 ×9
= 3× 2
= 2 3
= 9 2
= 3 2
Example 31
(
Expand and simplify
2+ 3
)(
2− 3
)
2× 2 = 4 =2
(
2+ 3
)(
2− 3
)
2×− 3 = − 6
3× 2 = 6
3 × − 3 = 9 = −3
2 − 6 + 6 − 3 = −1
Example 32
(
(
Expand and Simplify
5+ 2
)(
5− 2
)(
5− 2
)
)
5 × 5 = 25 = 5 →
2× 5 =
5+ 2
10 →
5 × − 2 = − 10
2 × − 2 = − 4 = −2
5 − 10 + 10 − 2 = 3
Example 33
Expand and simplify
2× 2 = 4 = 2 →
(
2+ 3
)(
(
2× 3 = 6 →
)
2+ 3
)(
3× 2 = 6 →
2 + 3 → 2+ 6 + 6 +3= 2 6 +5
25
2+ 3
)
3× 3 = 9 = 3
CONJUGATE OF SURDS
2 + 3 is a binomial surd because it contains
two terms. The first term is 2 and the second term is 3 . Whenever, the sign of
the coefficient of the second term of any binomial surd is changed, it is called
conjugate surd. The conjugate of 2 + 3 is 2 − 3 .
The conjugate of
− 11 − 7 is − 11 + 7 .
A Binomial surd contains two terms.
Note: a monomial surd does not have any conjugate because a monomial surd has
only one term. The change in sign for conjugates happens only to the second surd.
The conjugate of 3 is 3 . The conjugate of - 3 is - 3 . There is no change in
the sign because there is no second term.
RATIONALIZING THE DENOMINATOR
Rationalising the denominator means to make the denominator an integer, without
changing the value of the fraction. To this, we multiply both the numerator and the
denominator with the conjugate of the denominator.
EXAMPLE 34
5
.
7
Rational the denominator of
Solution
5
7
5× 7
35
×
=
=
=
7
7
7×7
49
35
7
EXAMPLE 35
3− 2
. The conjugate of the denominator is 3 − 2 .
3+ 2
Multiply both the numerator and denominator by the conjugate of the denominator.
Rational the denominator of
2− 5
3+ 2
×
3− 2
3− 2
6 − 4 − 15 + 10
=
9− 6+ 6− 4
=
6 − 2 − 15 + 10
= − 6 + 2 + 15 − 10
3−2
EXAMPLE 36
3
. The conjugate of the denominator is 1+ 2 .
1− 2
Multiply both the numerator and denominator by the conjugate.
Rationalise the denominator of
(
)
3 1+ 2
3
1+ 2
3+3 2
3+3 2 3+3 2
×
=
=
=
=
= −3 − 3 2
1− 2
−1
1− 2 1+ 2
1+ 2 − 2 − 4
1− 2 1+ 2
(
EXAMPLE 37
)(
)
Rationalise the denominator of
3+ 2
.
2
 3 + 2  2 
3× 2 + 2× 2
6+ 4
=
=

 
 =
2  2 
2× 2
4

26
6 +2
2
EXAMPLE 38
Rational the denominator of
3+ 2
4 3+ 2
 3 + 2  4 3 − 2 

 

4
3
2
4
3
2
+
−



=
=
(
) ( 3 × − 2 ) + ( 2 × 4 3) + ( 2 × − 2 )
3) + (4 3 × − 2 ) + ( 2 × 4 3) + ( 2 × − 2 )
3×4 3 +
(4
3×4
4 9 − 6 +4 6 − 4
12 + 3 6 − 2 10 + 3 6
=
=
48 − 2
46
16 9 − 4 6 + 4 6 − 4
EXAMPLE 39
2+ 2
1+ 2
Rational the denominator of
.
 2 + 2   1− 2 

 

1
+
2
1
−
2



( 2 ×1) + ( 2 × −
=
(1×1) + (1× −
=
) (
2)+(
2 +
) (
2 ×1) + (
2 ×1 +
)
2)
2 ×− 2
2 ×−
2−2 2 + 2 − 4 2− 2 −2 − 2
=
=
=
−
−
1
2
1
1− 2 + 2 − 4
27
2
Exercise 1C
Rationalize the denominator of following leaving your answer in surd form
1.
2.
3
1− 2
3
1− 2
3.
3
1+ 2
4.
2+ 2
2− 2
5.
5+ 2
3− 2
6.
4+ 2
3− 2
7.
1− 5
8− 3
Expand and simplify the following leaving your answer in surd form
8.
(
3 3− 3
)
(
10. − 3 + 5
9.
)(
5+ 3
)
3
(
5+ 3
(
11. − 2 + 3
)
)(
3+ 2
)
Simplify the following
12.
32
13.
45
14.
500
15.
4500
16.
60
17.
18. Rationalise each denominator
A.
13 − 2 2
3− 2
B.
3+ 2
3−2 2
C.
4 2
5− 2
D.
E.
3+ 2
3− 2
F.
3+ 2
2
G.
3+ 2
2 −2
H.
28
3+ 2
2
2
3− 2
8
LAWS OF LOGARITHM
Law 1 → log x a + log x b = log x a × b = log x ab
EXAMPLE 40
log 5 2 + log 5 8 = log 5 ( 2 × 8 ) = log 5 16
Law 2 → log x b − log x a = log x ( b ÷ a )
EXAMPLE 41
 10 
log 3 10 − log 3 5 = log 3   = log 3 2
 5 
Law 3 → log x a b = b log x a
EXAMPLE 42
log 6 7 2 = 2 log 6 7
Law 4 → x a = b → in Logarithm form is log x b = a
EXAMPLE 43
42 = 16 → Log 416 = 2
Note: Calculators are programmed with base 10. When you enter logarithm of any
number with base 10, you do not need to enter 10.
For log10 100 , in the calculator just enter log 100 and your answer will be 2.
x a = b → log x b = a
log 10 100 = 2 → 10 2 = 100 .
When you come across logarithm in the form such as log 1000 without its base
written, you can assume the base is 10.
29
EXAMPLE 44
LAW 1 with base 10
log x a + log x b = log x a × b = log x ab
let x = 10
log10 a + log10 b = log10 ( a × b ) = log10 ab
Since calculators are programmed with base 10, log10 a + log10 b = log10 ab can be
written as
log a + log b = log ( a × b ) = log ab
EXAMPLE 45
LAW 2 with base 10
a
log x b − log x a = log x  
change base from x to 10
b
b
b
log 10 b − log 10 a = log 10  → log b − log a = log 
a
a
b
log b − log a = log ( b ÷ a ) = log  
a
EXAMPLE 46
LAW 3 with base 10
Log x a b = b log x a
change base from x to 10
log10 a b = b log10 a → log a b = b log a
log a b = b log a
EXAMPLE 47
LAW 4 with base 10
x a = b → in log form is log x b = a
change base from x to 10
10 a = b → in log form is log10 b = a → log b = a
EXAMPLE 48
Write log 2 + log 3 as logarithm of a single number
log 2 + log 3 → apply law 1 → log 2 + log 3 = log ( 2 × 3) = log 6
30
EXAMPLE 49
Write log 5 + log 10 − log 25 as a logarithm of a single number
log 5 + log10 − log 25 → log ( 5 ×10 ) − log 25 = log 50 − log 25
 50 
Apply Law 2
log 50 − log 25 →
log   = log 2
 25 
EXAMPLE 50
Simplify
log 3 + log 4
log 10 − log 5
Let’s simplify the numerator first. log 3 + log 4 = log(3 × 4) = log12
 10 
Now simplify the denominator. log 10 − log 5 = log  = log 2
5
log 3 + log 4 log 12
It follows that
=
log 10 − log 5 log 2
EXAMPLE 51
Simplify
log 4 + log 9
log 36 − log 6
log 4 + log 9 = log(4 × 9) = log 36
 36 
Now simplify the denominator by applying law 2. log 36 − log 6 = log  = log 6
 6 
log 4 + log 9 log 36
It follows that
=
log 36 − log 6 log 6
Simplify the numerator first by applying law 1.
log 36
can be further simplified.
log 6
log 36 = log 6 2 = 2 log 6
[law 3]
log 4 + log 9 log 36 log 62 2 log 6 2 log 6
=
=
=
=
= 2
log 6
log 36 − log 6 log 6
log 6
log 6
EXAMPLE 52
If log 2 = x and log 3 = y , express log 12 in terms of x and y.
log 12 = log(4 × 3)
= log 4 + log 3
= log 2 2 + log 3
= 2 log 2 + log 3
= 2x + y
31
EXAMPLE 53
If x = log 2 and y = log 3 , calculate x + y
x + y = log 2 + log 3
= log ( 2 × 3)
= log 6
EXAMPLE 54
If x = log 2 and y = log 3 , calculate 2 x + 2 y
2 x + 2 y = 2 log 2 + 2 log 3
= log 22 + log 32
= log 4 + log 9
= log ( 4 × 9 )
= log 36
EXAMPLE 55
If x = log 6 and y = log 2 , calculate 2 x − 2 y
2 x − 2 y = 2 log 6 + 2 log 2
= log 62 + log 22
= log 36 + log 4
= log ( 36 ÷ 4 )
= log 9
EXAMPLE 56
If x = log 6 and y = log 2 , calculate x − y
x − 2 = log 6 − log 2 = log ( 6 ÷ 2 ) = log 3
NATURAL LOGARITHM Symbol → ln
ln e = 1 ln1 = 0 ln 0 = undefined
Law 1 → ln x y = y ln x
32
Example 57
ln 25 = 5 ln 2
Law 2 → ln a + ln b = ln ab
Example 58
Write ln 2 + ln 3 as a ln of a single number
ln a + ln b = ln ab
ln 2 + ln 3 = ln (2 × 3) = ln 6
a
Law 3 → ln a − ln b = ln  
b
Example 59
Write ln 30 − ln 5 as a ln of a single number
a
ln a − ln b = ln  
b
 20 
ln 20 − ln 4 = ln   = ln 5
 4 
EXAMPLE 60
100 = 10 x
Solve for x.
100 = 10 x
ln100 = ln10 x [take ln of both sides ]
ln100 = x ln10
ln100
=x
ln10
EXAMPLE 61
x=2
Solve for x.
10 = 100 x
10 = 100 x
ln10 = ln100 x [take ln of both sides ]
ln10 = x ln100
ln10
=x
ln100
x=
1
2
33
In example 62, example 61 is redone but this time solution is found by using
logarithm instead of natural logarithm. Both will give the same answer.
EXAMPLE 62
Solve for x.
10 = 100 x
10 = 100 x
log10 = log100 x [take log of both sides ]
log10 = x log100
log10
=x
log100
x=
1
2
EXAMPLE 63
8 = 2 x −1
Solve for x.
8 = 2 x −1
log 8 = log 2 x −1 [take log of both sides ]
log 8 = ( x − 1) log 2
log 8
= x −1
log 2
3 = x −1 → x = 3 +1 → x = 4
EXAMPLE 64
Solve for x.
8 = 2− x −1
8 = 2− x −1
log 8 = log 2− x −1 [take log of both sides ]
log 8 = ( − x − 1) log 2
log 8
= −x −1
log 2
3 = − x − 1 → x = −1 − 3 → x = −4
34
So teach us to number our days, that we may apply our hearts
unto wisdom. (Old Testament | Psalms 90:12)
Exercise 1D
1. Solve for x.
log 5 x = 4
2. Express
1
log 25 + 2 log 5 as log of a single number
2
3. Simplify
log 4 + log 2
log 8
4. Solve for x.
2 x = 256
5. Solve for x.
3x = 60
6. Write ln 10 + ln 2 as ln of a single number
7. Write 5 ln 5 + 2 ln 2 as a ln of a s ingle number
8. Solve for x.
23 x −1 = 100
9. Solve for y.
ey = 1
10. Solve for x.
128 = 2 x
1
11. If x = log 9 and y = log , calculate x − y
3
1
12. If x = log 9 and y = log , calculate x + y
3
13. If x = log 2 and y = log 3 , calculate 4 x − 2 y
14. If x = log 400 and y = log10 , calculate x − 2 y
14. Express log ( ab ) − log a as a single logarithm.
1
15. Express log   − log a as a single logarithm.
a
35
2
I will fetch my knowledge from afar, and will ascribe righteousness to
my Maker. (Old Testament | Job 36:3)
ALGEBRA
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
Expanding brackets
Factorization by factor method
Factorization by completing the square method
Long Division
Remainder theorem
Solutions of quadratic and cubic equations by factors
Quadratic equation and formula
Nature of roots
Simplification, multiplication and solving of rational
algebraic expressions
Solving inequalities
Subject of the formula
Solving equations simultaneously
Sigma Notation. Arithmetic and geometric Sequences
and series.
Sum to infinity
Matrices: addition, subtraction and multiplication
Inverse and determinant of 2 by 2 matrices.
Application of 2 by 2 inverse matrix
36
Expanding brackets and Factorization
Expand (x + 6)(x + 1) → This is in factorised form
EXAMPLE 65
(x + 6 )( x + 1 )
=
x 2 + 1x + 6 x + 6
=
x2 + 7x + 6
Two simple methods can be used to factorise all expanded factors. The general form
of quadratic equation is ax 2 + bx + c = 0. If a = 1, the method given in example 66
is recommended. If a ≠ 1, then the method given in example 67 and 68 is
recommended. Both methods have just a slight difference between them.
EXAMPLE 66
Factorise x 2 + 5 x + 4 . Note: a = 1.
This is a quadratic equation, meaning the highest power is 2.
The following steps are to be used to factorise quadratic equations.
• Find the factors of the variable raised to the power 2.
• Find the factors of the constant term.
• Multiply the factors and add the products to get the middle term.
• If you get the middle term by adding the products, you can conclude that the
factors you have chosen are correct.
• If you do not get the middle, try changing the factors.
x2 + 5x + 4
factors of x 2 factors of 4 multiply the factors
x → 4 ⇒ 4x
x → 1 ⇒ 1x
add the product of the factors → 4 x + 1x = 5 x
Since the products of the factors add up to the middle term, you can conclude that the
factors are correct.
x → 4 ( x + 4)
x →1
( x + 1)
x 2 + 5 x + 4 → in factored form
37
( x + 4 )( x + 1)
Factorise 2 x 2 + 8 x + 6
EXAMPLE 67
Note: a ≠ 1,
2 x2 + 8x + 6
factors of 2 x 2 factors of 6 multiply the factors
2x → 1 ⇒ 2x
x → 6 ⇒ 6x
add the product of the factors → 2 x + 6 x = 8 x → middle term
Since a ≠ 1, the final answer will be determined diagonally.
2x
1 ⇒ 2x
x
6 ⇒ 6x
Answer: ( 2 x + 6 )( x + 1)
Factorise 2 x 2 − x − 6
EXAMPLE 68
2 x2 − x − 6
factors of 2 x 2 factors of − 6 multiply the factors
2 x → −2 ⇒ −4 x
x → 3 ⇒ 3x
add the product of the factors → −4 x + 3 x = − x → middle term
Since a ≠ 1, the final answer will be determined diagonally.
2x
→ −2
⇒
− 4x
x
→ 3
⇒
3x
Answer: ( 2 x + 3)( x − 2 )
38
Completing the Square
This method is used to solve quadratic equations by replacing the quadratic
expression:
2
b

b
x + bx + c by  x +  + c −  
2

2
2
2
Factorise x 2 + 5 x + 4 = 0 .
EXAMPLE 69
•
Step 1: Separate the variable from constant. x 2 + 5 x = −4
•
Step 2: Divide the coefficient of the middle term by 2 and then square the
result and add it to both sides of the equation of step 1.
2
5
5
x + 5 x +   = −4 +  
2
2
2
2
•
Step 3: The left hand side of the equation in step 3 can be written as
2
b
5


x+  →x+ 
2
2


•
2
Step 4: The right hand side of the equation in step 3 is simplified
2
−4 25 −16 + 25 9
5
−4 +   =
+
=
=
1
4
4
4
2
•
Step 5: The equation of step 2 can now be written as:
2
2
2
5
9
5
5

x 2 + 5 x +   = −4 +   →  x +  =
2
4
2
2

•
Step 6: Recall recall x 2 − y 2 = ( x − y )( x + y )
2
5 9

x+  =
2
4

2
5 9

x+  −
2 4

2
5 3

 x +  − 
2 2

•
2
2
2
2
2
2
5 3 
5 3 
5 3

 x +  −   =  x + −  x + + 
2 2 
2 2 
2 2

5 3 
2 
8

→  x +  −   =  x +  x + 
2 2 
2 
2

2
5 3

 x+  −  =
2 2

( x + 1)( x + 4 )
Compare the solution of this example with example 66
39
LONG DIVISION
Long division: An algorithm for division by a number or more than a single digit that
proceeds by subtracting from the initial segment of the dividend the largest multiple
of the divisor less than that initial segment; this process is repeated for the successive
remainders augmented by the next digit of the dividend.
Recall division you learned in your primary school.
The same procedure will be used for division of polynomials
3
34
2 68 → 2 goes into 6, 3 times → 2 68 → 2 goes in to 8, 4 times → 2 68
−60
8
−60
8
−8
EXAMPLE 70
Evaluate x 4 x 2 + 9 x
Multiply the quotient , 4 x,
x will go into 4 x 2 ,
4 x times. .
4x
2
x 4x + 9x
with the divisor , x.
4x
→
2
x 4x + 9x
(
− 4x2 + 0
)
Multiply the quotient ,9,
Perform
x will go into 9 x,
subtraction
9 times.
4x
4x + 9
2
x 4 x + 9 x → x 4x2 + 9 x
(
)
− 4x + 0
9x
9x
− 4x + 0
2
(
2
with the divisor , x, and
perform subtraction
→
4x + 9
x 4x2 + 9 x
(
− 4x2 + 0
)
)
9x
− 9x
0
40
Quotient
Divisor Dividend
Rules for Long Division
1. The order of terms must be arranged in descending powers of x.
2. For all missing consecutive powers of x, fill it with zeros as the
coefficient.
x 4 − 3x − 7, → should be written as x 4 + 0 x 3 + 0 x 2 − 3x − 7
3. Each term in the quotient is found by dividing the first term of the
divisor into the first term of the new dividend.
4. Once a division is done, multiply the new quotient with the divider.
5. Subtract the term found in step 4 from step 3. This gives a new term.
FACTOR THEOREM
If
( x + a)
is a factor of ax 2 + bx + c, then x + a ax 2 + bx + c will have a
remainder that is exactly equal to zero.
Suppose that P ( x) is a polynomial and α is a constant.
Then ( x − α ) is a factor if and only if f (α ) = 0
In general :
EXAMPLE 71
Show that (x − 2 ) is a factor f ( x) = 2 x 2 − x − 6
x goes into 2 x 2 , 2 x times.
Multiply the quotient 2 x, with
→ the divider , x − 2. 2 x ( x − 2 ) = 2 x 2 − 4 x
2x
x − 2 2x − x − 6
2
2x
2 x2 − x − 6
x−2
(
− 2 x2 − 4 x + 0
x goes into 3 x, 3 times
Perform subtraction
x−2
2x
(
2
2x
−x − 6 → x − 2
− 2x − 4x + 0
2
)
3x − 6
2x
(
2
2x + 3
−x − 6
− 2x − 4x + 0
2
)
)
Perform subtraction
→ x−2
2x
2x + 3
−x − 6
(
− 2 x2 − 4 x + 0
3x − 6
( 3x − 6 )
2
3x − 6
−
( 3x − 6 )
0
41
)
Show that ( x − 2) is a factor of f ( x) = x 3 + 2 x 2 − 5 x − 6 and
hence, completely factorise f ( x) = x 3 + 2 x 2 − 5 x − 6 .
EXAMPLE 72
x2
x − 2 x + 2 x − 5 x − 6 → x will go into x , x times x − 2 x + 2 x − 5 x − 6
3
2
3
2
3
* Perform subtraction
∗ Multiply the quotient x 2
x2
x 3 + 2 x 2 − 5x − 6
with the divisor x − 2
x−2
x2
x − 2 x 3 + 2 x 2 − 5x − 6
(x
3
− 2x + 0 + 0
2
(
− x3 − 2x 2 + 0 + 0
)
)
4x 2 − 5x − 6
∗ Multiply the quotient 4 x with the
∗ x will go into 4 x 2 , 4 x times
divisor x − 2
x2 + 4 x
x3 + 2 x 2 − 5 x − 6
x−2
2
x 2 + 4x
x + 2x − 5x − 6
x−2
− ( x3 − 2 x 2 + 0 +0 )
3
2
(
− x3 − 2x 2 + 0 + 0
4 x − 5x − 6
2
)
4x 2 − 5x − 6
4 x 2 − 8x + 0
*Perform subtraction ∗
x will go into 3x, 3 times.
x2 + 4 x
x + 2 x − 5x − 6
x−2
3
(
− x − 2x + 0 + 0
3
2
x2 + 4 x + 3
x + 2 x − 5x − 6
x−2
2
)
3
(
− x3 − 2 x 2 + 0 + 0
4 x2 − 5x − 6
( −)
( −)
3x − 6
x−2
x 2 + 4x + 3
x + 2x 2 − 5x − 6
(
− x3 − 2x 2 + 0 + 0
4 x 2 − 5x − 6
(− ) 4 x 2 − 8 x + 0
3x − 6
(− )
3x − 6
)
)
4 x2 − 5 x − 6
4 x2 − 8x + 0
3
2
4 x2 − 8x + 0
3x − 6
3x − 6
Since R = 0, x − 2 is a factor of f ( x) = x 3 + 2 x 2 − 5 x − 6
and the quotient x 2 + 4 x + 3 is also a factor that can
be further factorise.
x 2 + 4x + 3
x →1
1x
x→3
3x 1x + 3x = 4 x ← middle term
(x + 1)(x + 3) = x 2 + 4 x + 3
(x − 2)(x + 1)(x + 3) = x 3 + 2 x 2 − 5x − 6
0
42
EXAMPLE 73
Divide x 3+ x 2 + 3 by x + 1
For division, all polynomial should be written in general form
→ ax3 + bx 2 + cx + d
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
x will go into x 3 , x 2 times
x + 1 x3 + x 2 + 0 x + 3 →
x2
x + 1 x3 + x 2 + 0 x + 3
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
multiply the quotient x 2 ,
by the divisor x + 1
x2
x + 1 x3 + x2 + 0 x + 3
x3 + x2
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Perform subtraction
x2
x3 + x 2 + 0 x + 3
x +1
(
− x3 + x 2 + 0 + 0
)
3
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
note : x can not go into 3, hence the remainder = 3
remainder
3 

R
=
3
is
written
as
R
3,
or
which
in
this
case
is

divisor
x + 1 

x + 1 x3 + x 2 + 0 x + 3 ⇒ answer ⇒ x 2 R 3 ⇔ x 2 +
3
x +1
The above example showed that x + 1 is not a factor of x 3 + x 2 + 3 because
x 3 + x 2 + 3 when divided by x + 1 gives remainder which is not zero.
• When the remainder is zero, the divisor is a factor.
• When the remainder is not equal to zero, the divisor is not a factor.
43
REMAINDER THEOREM
If ( x − a ) is a factor of f ( x) = ax 3 + bx 2 + cx + d , it follows that
x−a =0
x =0+a
x = a → f (a ) = 0.
In example 72, it is proved that ( x − 2 ) is a factor of f ( x) = x 3 + 2 x 2 − 5 x − 6 by the
process called factor theorem or long division method. In example 73, the remainder
theorem will be used to show that ( x − 2 ) is a factor of f ( x ) = x 3 + 2 x 2 − 5 x − 6
EXAMPLE 74
Show that ( x − 2 ) is a factor of f ( x ) = x 3 + 2 x 2 − 5 x − 6
If x − 2 is a factor of f ( x) = x 3 + 2 x 2 − 5 x − 6 then
x−2=0
x = 0+2
x=2 →
f (2) = 0
Complete prove is given below
x=2
f (2) = (2) 3 + 2(2) 2 − 5(2) − 6
= 8 + 2(4) − 10 − 6
= 8 + 8
− 10 − 6
= 0
Since the remainder is 0, ( x − 2 ) is a factor of f ( x) = x 3 + 2 x 2 − 5 x − 6 .
In general, the remainder theorem states that if P(x) is a polynomial and α is
a constant, then the remainder after division of P(x) by x - a is P(α).
Proof: Since x – α is polynomial of degree 1, the division theorem tells us that there
are unique polynomials Q(x) and R(x) such that
P(x) = (x – α)Q(x) + R(x) and either R(x) = 0 or deg R(x) = 0.
Thus R(x) is a zero or a nonzero constant that can be written a r, and so
P(x) =(x- α)Q(x) + r
When substituting x = α the result is P(α) = (α – α)Q(α) + r
Hence, the result is r = P(α), as required.
44
EXAMPLE 75
Find the values of a, b, c, and d, if x3 − x = a ( x − 2)3 + b( x − 2) 2 + c( x − 2) + d
let x = 2
23 − 2 = a (2 − 2)3 + b(2 − 2) 2 + c(2 − 2) + d
8−2 = d
d =6
Equate the coefficients of x3
x3 − x = a ( x − 2)3 + b( x − 2) 2 + c( x − 2) + d
a =1
let x = 3
33 − 3 = a (3 − 2)3 + b(3 − 2)2 + c(3 − 2) + d
27 − 3 = a (1)3 + b(1) 2 + c(1) + d
24 = a + b + c + d
Given a = 1
d =6
24 = 1 + b + c + 6
24 − 6 − 1 = b + c → 17 = b + c → c = 17 − b
let x = 0
03 − 0 = a (0 − 2)3 + b(0 − 2) 2 + c(0 − 2) + d
0 = −8a + 4b − 2c + d
0 = −8(1) + 4b − 2c + 6 → 0 = −8 + 4b − 2c + 6
0 = −2 + 4b − 2c → 0 = −1 + 2b − c → c = −1 + 2b
let c = c
c = 17 − b
−1 + 2b = 17 − b
2b + b = 17 + 1
c = 17 − 6
3b = 18 → b = 18 ÷ 3 → b = 6
A=1
B=6
C = 11
45
c = 11
D=6
EXAMPLE 76
(x + 2) is not a
Show that
factor of f ( x) = x 3 + 2 x 2 − 5 x − 6
f (2) = ( −2 ) + 2 ( −2 ) − 5 ( 2 ) − 6
3
x+2=0
2
f (2) = −8 + 2 ( 4 ) − 10 − 6
x = 0−2 →
f (2) = −8 + 8 − 10 − 6
x = −2
f (2) = −16
(x − 2) is not a
Since the remainder is not zero,
factor of f ( x) = x 3 + 2 x 2 − 5 x − 6
EXAMPLE 77
If x − 1 is a factor of f ( x) = 2 x 3 + kx 2 − x − 8, solve for the value of k .
x −1 = 0 → x = 0 +1 → x = 1
f (1) = 2 (1) + k (1) − (1) − 8 = 0
3
2
2 (1) + k (1) − (1) − 8 = 0
3
2
2 + k −1− 8 = 0
k −7 = 0
k =7
EXAMPLE 78
If x − 4 is a factor of f ( x) = x 3 + kx 2 − 2 x + 8, solve for the value of k .
x−4 = 0 → x = 0+4 → x = 4
f (4) = ( 4 ) + k ( 4 ) − 2 ( 4 ) + 8 = 0
3
( 4)
3
2
+ k ( 4) − 2 ( 4) + 8 = 0
2
64 + 16k − 8 + 8 = 0
16k + 64 = 0
16k = 0 − 64
16k = − 64
k = −4
46
EXAMPLE 79
A polynomial is given as f ( x) = 2 x3 + kx 2 − x − 8 . When f(x) is divided by (x +1) the
remainder is zero. Determine the value of k.
x +1 = 0
x = 0 − 1 x = −1
f ( x) = 2 x 3 + kx 2 − x − 8
f (1) = 2 ( −1) + k ( −1) − ( −1) − 8 = 0
3
2
2 ( −1) + k (1) − ( −1) − 8 = 0
− 2 + k +1− 8 = 0
k −9 = 0
k = 0+9
k =9
EXAMPLE 80
A polynomial is given as f ( x) = x 3 + kx 2 − 2 x + 8 . When f(x) is divided by
the remainder is 3. Determine the value of k.
x −1 = 0
x = 0 +1 x = 1
f ( x) = x3 + kx 2 − 2 x + 8
f (1) = (1) + k (1) − 2 (1) + 8 = 3
3
(1)
2
3
+ k (1) − 2 (1) + 8 = 3
2
1+ k − 2 + 8 = 3
k +7 =3
k = 3−7
k = −4
Compare this example with example 78
47
( x − 1)
Solutions of Quadratic and Cubic Equations by Factors
EXAMPLE 81
If
Solve for x.
( x − 1)( x − 4 ) = 0,
( x − 1)( x − 4 ) = 0
either
( x − 1) = 0
or
( x − 4) = 0
When one of the brackets is equal to zero,
the product will be 0.
Zero × any number = 0
Hence, ( x − 1)( x − 4 ) = 0.
( x − 1) = 0
or
( x − 4) = 0
x = 0 +1
x = 0+4
x =1
x=4
Solution → {1, 4}
Note: For the product to be equal to zero, at least one of the brackets must be equal to
zero. All the brackets can be equal to zero as well.
EXAMPLE 82
Expand and simplify
( x − 1)( x − 4 )
( x − 1)( x − 4 )
( x − 1)( x − 4 ) = ( x × x ) + ( x × −4 ) + ( −1× x ) + ( −1× −4 )
( x − 1)( x − 4 ) = x 2 + −4 x + −1x + 4
( x − 1)( x − 4 ) =
x2 − 5x + 4
48
QUADRATIC EQUATION AND FORMULA
Quadratic Equation
Quadratic Formula
f ( x) = ax 2 + bx + c
•
x=
The general expression of any quadratic equation is:
ax 2 + bx + c = 0
•
→ x2 +
b
c
x+ =0
a
a
Separate the variables from constants. Leave the variables on the left hand
side of the equation but remove the constant and take it to the right hand side
of the equation.
x2 +
•
a≠0
Divide both sides of the above equation by a.
ax 2 b
c 0
+ x+ =
a
a
a a
•
− b ± b 2 − 4ac
2a
b
c
x=−
a
a
Use the method of completing the square to factorize the above equation.
2
b
c  b 
 b 
x + x+  = − + 
a
a  2a 
 2a 
2
b ± b 2 − ac
x+
=
2a
2a
2
2
b 
−c b 2

+ 2
x+  =
2
a
a
4a


2
b 
−ac + b

x
+
=


2a 
4a 2

→
± b 2 − ac b
x=
−
2a
2a
± b 2 − ac
x=
2a
2
−b
−b ± b 2 − ac
x=
2a
b
−ac + b 2
x+
=±
2a
4a 2
49
Use the quadtratic formula to solve x 2 − 5 x + 4 = 0
EXAMPLE 83
x2 − 5x + 4 = 0
a = 1 b = −5 c = 2
( −5) − 4 (1)( 4 )
2 (1)
2
− −5±
−b ± b 2 − 4ac
x=
→ x=
2a
x=
5±3
2
→ x=
5± 9
2
5+3
5−3
or x =
2
2
8
2
x=
x=
2
2
x=
x=4
solution set : {1, 4}
x =1
Compare example 83 with examples 81 and 82.
EXAMPLE 84
x2 + 7 x + 6 = 0
Use the quadtratic formula to solve x 2 + 7 x + 6 = 0
a =1 b = 7 c = 6
−7 ±
−b ± b 2 − 4ac
x=
→ x=
2a
(7)
2
− 4 (1)( 6 )
2 (1)
→ x=
−7 ± 25
2
−7 + 5
−7 − 5
x=
2
2
−7 ± 5
−2
−12
x=
→ x=
or x =
2
2
2
x=
x = −1
x=−6
Solution : x = {−6, −1}
50
EXAMPLE 85
Example 85 shows factorisation method of solving quadratic equation.
x2 + 7x + 6 = 0
(x + 1)(x + 6) = 0
→ You may need to refer to examples 65 and 68
(x + 1)(x + 6) = 0
x +1 = 0
x+6=0
x = 0 −1
x = 0−6
x = −1
x = −6
Compare example 85 with example 84
EXAMPLE 86
Use the quadratic formula to solve x 2 + 2 x + 1 = 0 .
x2 + 2 x + 1 = 0
a =1 b = 2 c =1
−2 ±
−b ± b 2 − 4ac
x=
→ x=
2a
x=
x=
( 2 ) − 4 (1)(1)
2 (1)
2
−2 ± 4 − 4
−2 ± 0
→ x=
2
2
−2 ± 0
2
x=
−2 + 0
−2 + 0
or x =
2
2
x=
−2
2
x = −1
•
Repeated root {-1}
51
x=
−2
2
x = −1
Nature of Roots
b2 − 4ac → concludes the nature of the root / s
b 2 − 4ac = 0 → one repeated real root
b 2 − 4ac > 0 → two real roots
b 2 − 4ac < 0 → no real root
Show that x 2 + 2 x + 1 = 0 have one repeated root
EXAMPLE 87
x2 + 2x + 1 = 0 a = 1 b = 2 c = 1
b 2 − 4ac → ( 2 ) − 4 (1)(1) = 4 − 4 = 0
2
Since b 2 − 4ac = 0, there exist only one real root.
Example 86 confirms the solution of example 87
EXAMPLE 88
Show that x 2 + 7 x + 6 = 0 has two real roots.
x2 + 7 x + 6 = 0
a =1 b = 7 c = 6
b 2 − 4ac → ( 7 ) − 4 (1)( 6 )
2
b 2 − 4ac → 49 − 24
b 2 − 4ac = 25
Since b 2 − 4ac > 0, x 2 + 7 x + 6 = 0 has two real roots
Example 84 and 85 confirms the solution of example 88
EXAMPLE 89
Show that x 2 + 3 x + 7 has no real roots.
x 2 + 3x + 7
a =1 b = 3 c = 7
b 2 − 4ac → ( 3) − 4 (1)( 7 )
2
b 2 − 4ac → 9 − 28
b 2 − 4ac = −19
b 2 − 4ac < 0
Since b 2 − 4ac < 0, x 2 + 3 x + 7 has no real root
52
Use the quadratic formula to solve for x. 2 x 2 + 8 x + 6 = 0
EXAMPLE 90
2 x2 + 8x + 6 = 0
a =2 b=8 c =6
( 8)
−8 ±
−b ± b 2 − 4ac
x=
→x=
2a
2
− 4 ( 2 )( 6 )
2 ( 2)
→x=
−8 ± 64 − 48
4
x=
−8 ± 64 − 48
−8 ± 16
−8 ± 4
−8 ± 4
→x=
→x=
→x=
4
4
4
4
x=
−8 ± 4
−8 + 4
−8 − 4
→x=
or x =
4
4
4
x=
−4
−12
or x =
4
4
x = −1 or x = −3 Solution set → {−3, −1}
Compare this example 90 with example 67
Use the quadratic formula to solve for x. x 2 + 8 x + 4 = 0
Give your answer correct to 3 decimal places.
EXAMPLE 91
x2 + 8x + 4 = 0
a =1 b = 8 c = 4
−8 ±
−b ± b 2 − 4ac
→x=
x=
2a
( 8)
2
− 4 (1)( 4 )
2 (1)
→x=
x=
−8 ± 64 − 16
−8 ± 48
−8 ± 6 ⋅ 928
→x=
→x=
2
2
2
x=
−8 ± 6 ⋅ 928
−8 + 6 ⋅ 928
−8 − 6 ⋅ 928
→x=
or x =
2
2
2
x=
−8 ± 64 − 16
2
−1 ⋅ 072
−14 ⋅ 928
or x =
2
2
x = −0 ⋅ 536 or x = −7 ⋅ 464 Solution set → {−0 ⋅ 536, −7 ⋅ 464}
53
Honour thy father and thy mother: that you days may be long upon the land which the
LORD your God has given to you. (Old Testament | Exodus 20:12)
Exercise 2 A
1. Expand simplify the following
a. (x − 2 )( x + 2 )
d. (x − 2 )( x − 3)(x + 1)
b. (2 x − 2 )( x + 2 )
c. (x − 9 )(− x − 5)
e. x( x − 3)( x + 1)
2. Either factorise the following or use the quadratic formula to solve for x.
a. x 2 + 6 x + 9 = 0
b. x 2 + 5 x + 4 = 0
c. x 2 + 4 x + 4 = 0
d. 2 x 2 + 11x + 12 = 0
e. 4 x 2 + 19 x − 12 = 0 f. 3 x 2 − 19 x + 12 = 0
g. − x 2 − 2 x + 3
h. 2 x 2 − 98
i. 4 x 2 − 64
3. Use the long division to show that x + 1 is a factor of f ( x) = x 3 − 5 x 2 + 2 x + 8
and hence completely factorise f ( x) = x 3 − 5 x 2 + 2 x + 8 .
4. Use the remainder theorem to show that
x + 1 is a factor of f ( x) = x 3 − 5 x 2 + 2 x + 8 and hence, completely factorise f(x).
5. Use the long division to show that
x − 1 is not a factor of f ( x) = x 3 − 5 x 2 + 2 x + 8
6. Use the remainder theorem to show that
x − 1 is not a factor of f ( x) = x 3 − 5 x 2 + 2 x + 8
7. If x − 1 is a factor of f ( x) = 2 x 3 + kx 2 − x − 8, solve for the value of k .
8. If x − 4 is a factor of f ( x) = x3 + kx 2 + 2 x + 8, solve for the value of k
and find the other two factors.
9. If x − 3 is a factor of f ( x) = x3 + kx 2 − 9 x − 18 , solve for k and find the other
two factors
10. If x + 1 is a factor of f ( x) = x 3 − kx 2 + x + 6 , solve for k and find the other
two factors.
11. Use the quadratic formula to solve for x. 2 x 2 + 3 x − 2 = 0
12. Use the quadratic formula to solve for x. 3 x 2 − 3 x − 3 = 0
54
13. Solve for x. 12 x 2 − 320 x + 1600 = 0
14. Determine the nature of roots for x 2 + 7 x + 12 = 0
15. Factorise and solve for x. x 2 + 7 x + 12 = 0
16. Use the quadratic formula to solve for x. x 2 + 7 x + 12 = 0
17. Determine the nature of roots for x 2 − 16 = 0.
18. Factorise and solve for x. x 2 − 16 = 0.
19. Use the quadratic formula to solve for x. x 2 − 16 = 0.
20. Determine the nature of roots for x 2 + 4 x + 4
21. Determine the nature of roots for x 2 + 4 x + 9
22. Solve for x. x 2 − 4 x + 3 = 0
23. A cubic function is defined as f ( x ) = x3 − kx 2 + x + 6.
( x + 1) is a
Find the value of k if
factor of f ( x ) and hence, find the other two factors.
24. A cubic function is defined as f ( x ) = x 3 + x 2 − 17 x + 15.
Show that ( x − 1) is a
factor of f ( x ) and hence completely factorize f ( x ) .
25. Use the quadratic formula to solve the expression f ( x ) = 2 x 2 + 4 x − 2.
26. If the two roots of x 2 − kx + 1 = 0 are unequal, find the value/s of k.
27. If the two roots of x 2 − kx + 1 = 0 are equal, find the value/s of k.
28. If x 2 − kx + 1 = 0 has no roots, find the value/s of k.
29. If ( x − 2 ) is a factor of the function f ( x ) = x3 + 2 x 2 − kx − 6, find the value of k .
30. Show that x = −1 is a solution of f ( x ) = x3 + 5 x 2 + 8 x + 4 and hence find the other
solution/s.
31. Show that x = −1 is a solution of f ( x ) = x3 + x 2 − 4 x − 4 and hence find the other
solution/s.
32. Write the solutions of question 31 in the form of factors.
55
Solving Linear Equations Containing Fractions
•
•
•
•
•
•
Multiply the denominators
Multiply each numerator with the product of the denominators from the above step.
Perform all possible cancellation
Remove parenthesis
Combine like terms
Get the variable alone
EXAMPLE 94
x − 2 3 x −1
+ =
3
2
2
Solve for x.
*Multiply the denominators ⇒ 3 × 2 × 2 = 12
12 ( x − 2 ) 12 ( 3) 12 ( x − 1)
+
=
3
2
2
*Perform all possible cancellation ⇒ 4 ( x − 2 ) + 6(3) = 6( x − 1)
*Multiply the product '12 ' with the numerators ⇒
*Remove parenthesis ⇒ 4 x − 8 + 18 = 6 x − 6
*Combine like terms ⇒ −8 + 18 + 6 = 6 x − 4 x ⇒ 16 = 2 x
*Get the variable alone ⇒ 16 ÷ 2 = x
EXAMPLE 95
Solve for x.
x =8
x − 2 x − 4 x −1
+
=
3
4
5
Multiply the denominators to find the common denominator. This method is
recommended for those who find it difficult to identify lowest common denominator.
Sometimes there is no lowest common denominator. This question does not have a
lowest common denominator.
Hence, finding the common denominator by
multiplying all denominators is recommended. ⇒ 3 × 4 × 5 = 60
• Multiply the common denominator ‘60’
x − 2 x − 4 x − 1 60 ( x − 2 ) 60 ( x − 4 ) 60 ( x − 1)
+
=
→
+
=
3
4
5
3
4
5
• Perform all possible cancellation
20
60 ( x − 2 )
3
15
+
60 ( x − 4 )
4
12
=
60 ( x − 1)
5
→ 20 ( x − 2 ) + 15 ( x − 4 ) = 12 ( x − 1)
• Remove parenthesis
20 ( x − 2 ) + 15 ( x − 4 ) = 12 ( x − 1) → 20 x − 40 + 15 x − 60 = 12 x − 12
• Combine like terms
20 x − 40 + 15 x − 60 = 12 x − 12 → 20 x + 15 x − 12 x = −12 + 40 + 60
• Solve for x.
20 x + 15 x − 12 x = −12 + 40 + 60 → 23 x = 88 → x = 88 ÷ 23 → x = 3 ⋅ 826 ( 3dp )
56
EXAMPLE 96
2 ( x − 1)
3 × 2 ( x − 1)
Solve for x.
2 ( x − 1)
3
− ( x − 4) = 5
Same as
− ( x − 4 ) = 5 →
2 ( x − 1)
3
cd = 3 × 1×1 = 3
−
3
3 × 2 ( x − 1)
3
3( x − 4)
−
=
1
3( x − 4)
1
3
−
( x − 4)
1
=
5
1
3× 5
1
=
3× 5
1
2 ( x − 1) − 3 ( x − 4 ) = 3 × 5
2 x − 2 − 3 x + 12 = 15
2 x − 3x = 15 + 2 − 12
− 1x = 5 x = 5 ÷ −1 = −5
EXAMPLE 97
Solve for x.
x = −5
2
−1
( x + 3) − ( x + 2 ) =
5
2
2
−1 Same as 2 ( x + 3) ( x + 2 ) −1
x
+
3
−
x
+
2
=
→
−
=
(
) (
)
5
2
5
1
2
2 ( x + 3)
5
−
( x + 2 ) = −1
1
2
cd = 5 × 1× 2 = 10
10 × 2 ( x + 3) 10 ( x + 2 ) 10 × −1
−
=
5
1
2
2
10 × 2 ( x + 3) 10 ( x + 2 ) 510 × −1
−
=
1
5
2
4 ( x + 3) − 10 ( x + 2 ) = −5 → 4 x + 12 − 10 x − 20 = −5
4 x − 10 x = −5 − 12 + 20 → − 6 x = 3 → x = 3 ÷ −6 → x = −0 ⋅ 5
57
EXAMPLE 98
Solve for x.
3
2
=
x + 4 4 ( x − 5)
3
2
=
x + 4 4 ( x − 5)
3 × 4 ( x + 4 )( x − 5 )
x+4
cd = 4 ( x + 4 )( x − 5 )
2 × 4 ( x + 4 )( x − 5 )
=
3 × 4 ( x + 4 ) ( x − 5)
( x + 4)
4 ( x − 5)
=
2 × 4 ( x + 4 ) ( x − 5)
12 ( x − 5 ) = 2 ( x + 4 )
12 x − 60 = 2 x + 8
x = 68 ÷ 10
4 ( x − 5)
→
12 x − 2 x = 8 + 60
10 x = 68
x = 6⋅8
Let’s substitute x = 6 ⋅ 8 in the equation to verify that the answer of the above
example is true.
3
2
=
x + 4 4 ( x − 5)
3
2
3
2
=
→
=
6 ⋅ 8 + 4 4 ( 6 ⋅ 8 − 5)
10 ⋅ 8 4 (1 ⋅ 8 )
3
2
=
10 ⋅ 8 7 ⋅ 2
0 ⋅ 277... = 0 ⋅ 277...
You can see that both sides of the equation are same and hence, the answer x = 6 ⋅ 8 is
correct.
58
Solving Inequalities
x 2
>
4 5
Solve for x.
EXAMPLE 99
x −2
cd = 4 × 5 = 20
>
4 5
20 ( x )
4
>
5 x > −8
20 ( −2 )
5
→
Cancellation
5
x>
20 ( x )
4
4
>
20 ( −2 )
5
−8
5
−4 x 2
>
3
5
Solve for x.
EXAMPLE 100
−4 x 2
>
cd = 3 × 5 = 15
3
5
15 ( −4 x )
3
>
15 ( 2 )
5
15 ( −4 x )
5
→
3
15 ( 2 )
3
>
5 ( −4 x ) > 3 ( 2 ) → − 20 x > 6 → x <
5
6
−20
x<−
3
10
Whenever multiplying or diving by a negative number, the inequality signs changes.
EXAMPLE 101
Solve for x.
•
Solve for x.
x
≤5
−3
x
≤ 5 → x ≥ 5 × −3 →
−3
x ≥ −15
The sign for inequality changes because the multiplier is a negative
number.
59
EXAMPLE 102
3
3
6
−
÷
4
y
y
Simplify
→
3 3 6
− ÷
4 y y
3 3 y
3 3
18 − 12
6
1
− × → − →
→
→
4 y 6
4 6
24
24
4
Division is done first
EXAMPLE 103
Simplify
4x
12 y
− 4y ÷
7
7x
4x
12 y
− 4y ÷
7
7x
Same as
→
4 x 4 y 12 y
−
÷
7
1
7x
4 x 4 y 12 y
4x 4 y 7x
4 x 28 xy
−
÷
→
−
×
→
−
7
1
7x
7
1 12 y
7 12 y
4 x 7 28 x y
4 x 7 x 12 x − 49 x
−37 x
− 3
→
−
→
→
7
7
3
21
21
12 y
EXAMPLE 104
Simplify
3x
15 y
− 5y ÷
2
2x
3 x 5 y 15 y
3x 5 y 2 x
3 x 10 xy
3 x 210 x y
−
÷
→
− ×
→
−
→
− 3
2
1
2x
2
1 15 y
2 15 y
2
15 y
3x 2 x
9x − 4x
5x
−
→
→
2
3
6
6
EXAMPLE 105
Simplify
Use the reciprocal of
x2 − 4
x−2
÷ 2
x + 4 x + x − 12
x−2
and multiply.
2
x + x − 12
Factor and divide out common factors.
x 2 − 4 x 2 + x − 12
×
x+4
x−2
( x − 2 )( x + 2 ) × ( x + 4 )( x − 3)
( x + 4)
( x − 2)
( x − 2 ) ( x + 2 ) ( x + 4 ) ( x − 3)
×
= ( x + 2 ) × ( x − 3) = ( x + 2 )( x − 3)
( x + 4)
( x − 2)
60
or x 2 − x − 6
Subject of the Formula [Revision Form 5 work]
EXAMPLE 106 Make y the subject of the equation 9x + 3y – 9 = 0.
9x + 3y − 9 = 0
3y − 9 = 0 − 9x
3 y − 9 = −9 x
3 y = −9 x + 9
y=
−9 x + 9
3
y=
9 ( − x + 1)
3
3
y=
9 ( − x + 1)
3
y = 3 ( − x + 1)
y = −3 x + 3
EXAMPLE 107 Make x the subject of the equation y = 2 x + 1 − 4
y = 2x +1 − 4
y + 4 = 2x +1
Square root means to the power of
1
1
2
y + 4 = ( 2 x + 1) 2
Our aim is to make x the subject. To do this we need to make its power equal to
exactly 1. This can be done by multiplying the power with its reciprocal. Multiplying
the power with its reciprocal must be done on both sides of the equation.
1
2
→ reciprocal
2
1
1 2 2
× = =1
2 1 2
→
2
( y + 4)
2
1
2
1


= ( 2 x + 1)  →


1
2
( y + 4 ) 1 = ( 2 x + 1)
2
( y + 4 ) = ( 2 x + 1)
2
( y + 4) −1 = 2x
( y + 4)
2
−1
2
1
=x
EXAMPLE 108 Make x the subject of the equation
5
x
= 9 x5 = 9 × y
y
1
5
( y + 4)
x=
5
61
−1
2
x5
=9
y
1
5
 
x = 9 y  x  = (9 y )
 
5
1
2
x = (9 y )
1
5
When I consider thy heavens, the work of thy fingers, the moon and
the stars, which thou hast ordained; what is man, that thou art mindful
of him? (Old Testament | Psalms 8:3 - 4)
Exercise 2 B
1. Solve the following equations
A. Solve (4 x − 3) = 19 − (2 + 4 x)
C. Solve
B. Solve 3( x + 1) = 6 + (5 − 3 x)
1
(2 x − 3) = 2 − (2 − 2 x )
2
D. Solve
1
3
(9 x − 2) = 2 − (2 − 2 x)
3
4
2. Solve the following equations
A.
5
=2
3x + 4
B.
3. Solve for x.
2
= 20
x+2
4x − 2 x +1 x −1
+
=
3
2
5
5. Solve for x. 2 ( x − 1) +
7. Solve for x.
C.
3 ( x − 1)
5
=
3
4
x − 2 x −1 x −1
−
≥
3
2
4
9. Solve for x. −
3x x − 1
≥
2
4
5
= −2
3x + 1
D.
1
=3
x +1
4. Solve for x.
x + 1 x + 1 x −1
+
=
10
2
3
6. Solve for x.
x 2 ( x + 1) 1
+
=
3
5
4
8. Solve for x.
−2
4
≤
x − 1 2 ( x − 3)
10. Solve for x.
11. Simplify
x 2 − y 2 y 2 + ab
÷
4
4x + 4
12. Simplify
13. Simplify
x 2 − y 2 y 2 + xy
÷
4
4y − 4
14. Simplify
3x − x − 1
≥
5
4
2 y y y2
− ÷
5 3 9
x −1 x2 − 4 x + 3 2
÷
+
3x − 9
x −3
3
15. Make t the subject in the following formula for the equation v = u + at .
16. Make x the subject of the formula for the equation y = 5 x 6 − 3
3
17. Make x the subject of the formula for the equation y − 3 =
x5 − 1
2
18 . Make u the subject of the formula for the equation v 2 = u 2 − 2as
19. Make v the subject of the formula KE = ½ mv2
62
SIGMA NOTATION
The symbol ∑ means to add or in other words ‘sum of’. Every sigma notation has a
lower limit and an upper limit. The lower limit determines from where the addition
will begin and the upper limit determines where the addition will cease.
EXAMPLE 109
5
∑n
Evaluate
5
∑
→
n =1
n = 1 + 2 + 3 + 4 + 5 = 15
n = 1
EXAMPLE 110
5
∑ 2n = 2 ( 3 ) + 2 ( 4 ) + 2 ( 5 )
n =3
6
Evaluate
5
∑ 2n
∑ 2n = 6 + 8 + 10
n =3
n =3
5
∑ 2n =
24
n =3
EXAMPLE 111
3
3
Evaluate
∑ ( n − 1)
∑ ( n − 1) = (1 − 1) + ( 2 − 1) + ( 3 − 1)
n =1
n =1
= 0 +1+ 2 = 3
EXAMPLE 112
3
∑ ( n − 1) = (1 − 1) + ( 2 − 1) + ( 3 − 1)
2
2
n =1
3
Evaluate
∑ ( n − 1)
= ( 0 ) + (1) + ( 2 )
2
2
2
n =1
= 0 +1+ 4 = 5
63
2
2
SEQUENCE AND SERIES
Arithmetic Progression
Arithmetic progression is a sequence of numbers, each differing by a constant amount
from its predecessor. This constant amount is called the common difference.
Let’s study the sequence < 3, 6, 9, 12, 15, 18, 21, … >. Notice that each term differs
by 3 and hence the common difference [d] is 3.
< 3, 6, 9, 12, 15, 18, 21, … > can be represented by an arithmetic formula.
3 is the first term, 6 is the second term, 9 is the third term, 12 is the fourth term, 15 is
the fifth term, 18 is the sixth term, and 21 is the seventh term and so on.
First term is represented by t1 or a .
Common difference is represent by small
letter d and is found by subtracting previous term from the current term.
d = t2 − t1 d = t3 − t2 d = t4 − t3
d = tn − tn −1
To find the nth term of any arithmetic progression, you will need to use the formula
given below.
tn = a + ( n − 1) d
tn → nth term
a → first term
d → common differnce
EXAMPLE 113
Find the 8th term of the arithmetic progression < 3, 6, 9, 12, 15, 18, 21, … >.
Just by looking at it, you can tell the answer is 24. Let’s use the arithmetic nth term
formula.
< 3, 6, 9, 12, 15, 18, 21,... >
tn = a + ( n − 1) d
Common difference
d = t2 − t1
d = 6−3 = 3 →
d = t3 − t2
d = 9−6 = 3
t8 = 3 + ( 8 − 1) 3 [bedmas law]
t8 = 3 + ( 7 ) 3
t8 = 3 + 21
64
t8 = 24
EXAMPLE 114
Find the 1001st term of the arithmetic progression < 3, 9, 15, 21, 27, 33, … >.
a = 3 n = 1001 d = t2 − t1 = 9 − 3 = 6
d = t3 − t2 = 15 − 9 = 6
tn = a + ( n − 1) d
t1001 = 3 + (1001 − 1) 6
t1001 = 3 + (1000 ) 6
t1001 = 3 + 6000 → t1001 = 6003
EXAMPLE 115
Find the 10th term of the arithmetic progression < 2, 4, 6, 8, 10, 12, 14, 16, 18, … >.
Just by looking at it, you can tell the answer is 20. Let’s use the arithmetic nth term
formula.
< 2, 4, 6, 8, 10, 12, 14, 16, 18,... >
a=2
tn = a + ( n − 1) d
find t10
t10 = 2 + (10 − 1) 2 [bedmas law]
Common difference
→ t10 = 2 + ( 9 ) 2
d = t2 − t1
d = 4−2 = 2
t10 = 2 + 18
d = t3 − t 2
d = 6−4 = 2
t10 = 20
EXAMPLE 116
Find the 101st term of the arithmetic progression < 2, 4, 6, 8, 10, 12, 14, … >.
< 2, 4, 6, 8, 10, 12, 14, 16, 18,... >
a=2
find t101
Common difference
tn = a + ( n − 1) d
t101 = 2 + (101 − 1) 2 [bedmas law]
→ t101 = 2 + (100 ) 2
d = t2 − t1
d = 4−2 = 2
t101 = 2 + 200
d = t3 − t2
d = 6−4 = 2
t101 = 202
65
EXAMPLE 117 An arithmetic progression has 10th term equal 20 and 101st term
equal to 202. Find the common difference and the first term of the progression.
tn = a + ( n − 1) d
tn = a + ( n − 1) d
t10 →
t10 = a + (10 − 1) d = 20
t202 →
t10 = a + 9d = 20
t101 = a + (101 − 1) d = 202
t101 = a + 100d = 202
a + 100d = 202
a + 9d = 20
Now solve both equations simultaneously.
a + 9d = 20
a + 100d = 202
make a the subject of the formula
a + 9d = 20
a + 100d = 202
a = 20 − 9d
a = 202 − 100d
a = 20 − 9d
a=a
20 − 9d = 202 − 100d
→
− 9d + 100d = 202 − 20
91d = 182 → d = 182 ÷ 91
•
a = 20 − 9 ( 2 )
a = 20 − 18
d =2
a=2
Compare this example 117 with examples 115 and 116
EXAMPLE 118 An arithmetic progression has 6th term equal 33 and 1001st term
equal to 6003. Find the common difference and the first term of the progression.
tn = a + ( n − 1) d
t6 →
t6 = a + ( 6 − 1) d = 33
t6 = a + 5d = 33
tn = a + ( n − 1) d
t1001 →
t1001 = a + (1001 − 1) d = 6003
t1001 = a + 1000d = 6003
a + 1000d = 6003
a + 5d = 33
•
Now solve both equations simultaneously.
a + 5d = 33
a + 1000d = 6003
make a the subject of the formula
a + 5d = 33
a + 1000d = 6003
a = 33 − 5d
a = 6003 − 1000d
a=a
33 − 5d = 6003 − 1000d
−5d + 1000d = 6003 − 33
995d = 5970
d = 5970 ÷ 995 → d = 6
66
a = 33 − 5d
→
a = 33 − 5 ( 6 )
a = 33 − 30
a=3
•
Compare example 118 with example 114.
Example 119
An arithmetic progression has 4th term equal log16 and 10th term equal to log1024 .
Find the common difference and the first term of the progression.
t4
t10
tn = a + ( n − 1) d
t4 = a + ( 4 − 1) d = log16
→
t4 = a + 3d = log16
tn = a + ( n − 1) d
t10 = a + (10 − 1) d = log1024
a + 3d = log16
→
t10 = a + 9d = log1024
a + 9d = log1024
Now solve both equations simultaneously.
a + 3d = log16
a + 9d = log1024
make ' a ' the subject of the formula
67
a + 3d = log16
a + 9d = log1024
a = log16 − 3d
a = log1024 − 9d
a=a
a = log16 − 3d
log16 − 3d = log1024 − 9d
−3d + 9d = log1024 − log16
 1024 
6d = log 

 16 
a = log16 − 3log 2
a = log16 − log 23
→ a = log16 − log 8
log 64
6d = log 64 → d =
6
 16 
a = log  
8
a = log 2
6
d=
log 2
6 log 2
→ d=
6
6
d=
6 log 2
→ d = log 2
6
Arithmetic Series
The sum of the terms of an arithmetic progression, for example 2 + 4 + 6 + 8 + ….
The sum of the first n terms of such a series, whose first term is a and whose common
difference is d, is
Sn =
n
n
 2a + ( n − 1) d  or Sn =
2
2
(a + l )
EXAMPLE 120
Find the sum of the first six terms of the arithmetic series 2 + 4 + 6 + 8 + 10 + 12...
•
You can see that all six terms are present and hence the sum of the first six
terms is 2 + 4 + 6 + 8 + 10 + 12 = 42.
•
Now use the sum formulas to get the sum of the first six terms. d = 2 a = 2
68
Sn =
n
6
 2a + ( n − 1) d  → S6 =  2 ( 2 ) + ( 6 − 1) 2 
2
2
S6 =
6
6
 4 + ( 5 ) 2  → S6 = [ 4 + 10]
2
2
S6 =
6
[14] → S6 = 3[14] → S6 = 42
2
Since we already know the first term and the last term that is the 6th term, we can use
another sum formula to find the sum of the first 6 terms.
Sn =
n
(a + l )
2
n
(a + l )
2
6
S6 = ( 2 + 12 ) → S6 = 3 (14 ) → S6 = 42
2
Sn =
EXAMPLE 121
Find the sum of the first 100 terms of the arithmetic series 1 + 2 + 3 + 4 + 5 + 6 + 7 + ...
100
 2 (1) + (100 − 1)1
2 
100
 2 + (199 ) 
=
2 
= 50 [ 201]
S100 =
a =1
d = t2 − t1 = 2 − 1 = 1
d = t3 − t2 = 3 − 2 = 1
n  2a + ( n − 1) d 
Sn = 
2
→
S100
S100
S100 = 10, 050
EXAMPLE 122 Find the sum of the first 200 terms of the arithmetic series
4 + 8 + 12 + 16 + 20 + 24 + 28 + ...
69
200
 2 ( 4 ) + ( 200 − 1) 4 
2 
200
8 + (199 ) 4
=
2 
= 100 [804]
S200 =
a=4
d = t2 − t1 = 8 − 4 = 4
d = t3 − t2 = 12 − 8 = 4
n  2a + ( n − 1) d 
Sn = 
2
→
S200
S200
S200 = 80, 400
EXAMPLE 123
Find the sum of the first 20 terms of the arithmetic series < 5, 7, 9, 11, 13, 15, 17…>
d = t2 − t1 = 7 − 5 = 2
d = t3 − t 2 = 9 − 7 = 2
Sn =
20
 2 ( 5 ) + ( 20 − 1) 4 
2 


→ S 20 = 10 10 + (19 ) 4 
S 20 = 10 [86]
S 20 =
a=5
S 20 = 860
n
 2a + ( n − 1) d 
2
EXAMPLE 124
On the right hand side is an arithmetic progression. log 5, log 25, log125, log 625,...
•
Show that the sequence is indeed an arithmetic progression.
An arithmetic progression is one that has a common difference.
common difference.
Let’s find the
 25 
d = t2 − t1 = log 25 − log 5 = log   = log 5
 5 
 125 
d = t3 − t2 = log125 − log 25 = log 
 = log 5
 25 
•
Find the 5th term of this arithmetic progression.
Since we now know that the common difference is log 5, we can add log 5 to the
4th term and this will give us the 5th term.
70
log 625 + log 5 = log ( 625 × 5 ) = log 3125 = log 55 = 5log 5
•
Calculate the sum of the first ten terms.
Sn =
n
 2a + ( n − 1) d 
2
S10 =
10
 2 ( log 5 ) + (10 − 1) log 5
2
S10 =
10
[ 2 log 5 + 9 log 5]
2
S10 = 5 [ 2 log 5 + 9 log 5]
S10 = 5 [11log 5]
S10 = 55log 5
But after that faith is come, we are no longer under a schoolmaster.
(New Testament | Galatians 3:25)
Once you have gained knowledge and understand the concepts learned in this topic, you do
not need the guidance of your schoolmaster anymore. Knowledge itself is not adequate
enough to hep you to succeed in mathematics. Understanding the knowledge gained in
mathematics is important because it is the understanding of the knowledge that makes you
remember learned mathematical concepts and principles.
Exercise 2 C
5
1. Evaluate
∑n
n =1
5
2. Evaluate
∑ 2n
n =1
71
5
3. Evaluate
∑ 2n + 1
n =1
4. Find the 121st term of the arithmetic sequence < 3, 9, 15, 21, 27, 33, … >.
5. Find the sum of the first 2000 terms of the sequence < 4, 8, 12, 16, 20, 24, 28…>
6. An arithmetic series has t 21 = 63 and t101 = 303 . Find the common difference and
the first term.
7. An arithmetic series has t15 = 50 and t 60 = 185 . Find the first term and the
common difference.
8. Find the sum of the first 100 terms of natural numbers.
9. The sum of the first nine terms of an arithmetic progression is 75 and the 25th term
is also 75. Find the common difference and the sum of the first hundred terms.
10. Samuel opens a savings account in a bank. He begins with a deposit of $100. He
plans to increase his deposit $20 each week in his savings account.
A. Write down the amounts deposited during the first 7 months.
B. How much money would in Samuel’s savings account at the end of the first 3
years provided he does not withdraw any amount and is exempted from tax
and hence, receives no interest.
11. A writer plans to increase his pace of typing using computers by 1 word per
month. If currently he can type 30 words per minute, how may words per minute
would he be able to type at the end of first 10 months.
12. Simon was playing with cards and made layers of triangle by successively adding
them in a sequence. Row 1 contains 3 cards and forms 1 triangle. Row 2 contains 6
cards and forms 3 triangles. Row 3 contains 9 cards and forms 5 triangles. Row 4
contains 12 cards and forms 7 triangles. Row 5 contains 15 cards and forms 9
triangles.
72
A.
B.
C.
D.
Calculate the number of cards in row 100.
Calculate the number of triangles in row 100
Calculate the sum of the cards used to complete the first 100 rows
Calculate the sum of the triangles in the first 100 rows.
13. A sequence is given by log 2, log 4, log 8, log16, log 32,...
A. Show that the sequence is arithmetic.
B. Show that the 20th term of the sequence is 20 log 2
C. Show that the sum of the first 20 terms of the sequence is 210 log 2
14. A kindergarten student stacks up small cubes in the order shown below.
A.
B.
C.
D.
Calculate the number of cubes in the 40th row.
Calculate the sum of cubes in the first 40 rows.
Which row will have 103 cubes?
The student stacks altogether 5,353 cubes using the same sequence of
arrangement. How many rows does this represent?
15. Find the 300th term of the arithmetic series < -1, -3, -5, -7, -9 …>.
16. Find the sum of the first 300 terms of the series in question 15.
Geometric Progression
A sequence of numbers whose successive members differ by a constant multiplier and
this constant multiplier is commonly known as common ratio. In this book common
ratio is used instead of constant multiplier.
The ratio is found by dividing the current term with the previous term.
r=
t2
t1
r=
t3
t2
In general r =
tn
tn −1
To find the next term, multiply the current term with the common ratio.
EXAMPLE 125
73
Show that < 2, 4, 8, 16, 32 …> is a geometric progression.
If it is a geometric progression, then there must be a common ratio.
t2
t1
r=
4
2
r=2
r=
r=
r=
t3
t2
8
4
r=2
r=
t4
t3
16
8
r=2
r=
r=
let n = 5
tn
r=
tn −1
t5
t5−1
r=
32
r=2
16
It is proofed that < 2, 4, 8, 16, 32 …> is a geometric progression because it has
common ratio equal to 2.
EXAMPLE 126
Show that < 1, 2, 4, 8, 16, 32, …> is a geometric progression.
r=
t2
t1
r=
t3
t2
r=
t4
t3
r=
2
→ r=2
1
r=
4
→ r=2
2
r=
8
→ r=2
4
•
Since it has a common ratio equal to 2, it is a geometric progression.
The nth term of any geometric progression is found by the formula using below.
tn = ar n −1
a → first term
r → common ratio
EXAMPLE 127
•
Show that 1, 3, 9, 27, 81,... is a geometric progression.
74
n → term
•
r=
t2
t1
r=
t3
t2
r=
t4
t3
r=
3
→r =3
1
r=
9
→r =3
3
r=
27
→r =3
9
Find the 6th term of this geometric progression.
From the information provided, we know that the 5th term is 81 and the common ratio
is 3. Find the 6th term; we will need to multiply the 5th term with the common ratio.
Each term in a geometric progress is found by multiplying the common ratio with the
previous term.
tn = tn −1 × r
tn = tn −1 × r
•
t6 = t6 −1 × 3
t 6 = t5 × 3
t6 = 81× 3
t6 = 243
Find the 6th term using the general formula for finding any term in a geometric
progression.
tn = a
( )
= (3 )
t6 = 1 36 −1
(r )
n −1
→ t6
given → n = 6 a = 1 r = 3
5
t6 = 243
EXAMPLE 128
A geometric progression is given by 1000, 500, 250, 125, 62 ⋅ 5,...
•
Find the common ratio
r=
•
t2
500
1
→r =
→r =
t1
1000
2
r=
t3
250
1
→r=
→r=
t2
500
2
Calculate the 10th term of this geometric progression
75
tn = a
  1 10 −1 
t10 = 1000    
 2  


(r )
n −1
9
1
→ t10 = 1000  
1
2
given n = 10 a = 1000 r =
2
t10 = 1 ⋅ 953125
EXAMPLE 129
A geometric progression is given by 19683, 6561, 2187, 729,...
•
Find the common ration
r=
•
t2
6561
1
→r =
→r =
t1
19683
3
r=
t3
2187
1
→r =
→r =
t2
6561
3
Find the 10th term of this geometric progression.
tn = a
  1 10−1 
t10 = 19683    
 3  


(r )
n −1
1
→ t10 = 19683  
1
3
given → n = 10 a = 19683 r =
3
t10 = 1
9
Geometric Series
Geometric Series: A series whose terms form a geometric progression.
The sum of the first n terms of a Geometric Progression can be found by the formula
76
Sn =
(
)
a r n −1
r −1
S n → sum of first n terms
a → first term
•
r → common ratio
n → number of terms
Geometric series is written in the form:
a + ar + ar + ... + ar
2
n −1
∞
+ ... = ∑ ar k −1
k =1
•
The partial sum of the general term of the geometric sequence is expressed as:
S n = a + ar + ar 2 + ... + ar n −1
•
equation 1
Multiply both sides of the above equation by common ratio r.
rS n = ar + ar 2 + ar 3 + ... + ar n
•
equation 2
Now subtract equation 2 from equation 1 and solve for Sn.
Sn =
S n − rS n = a − ar n
(1 − r ) Sn = a (1 − r
n
)
→
Sn =
(
a 1− rn
)
1− r
(
)
a r n −1
r −1
for r < 1
for r > 1
Remember either formula will give correct solution. This book uses only first
formula. You can use the second formula to verify the answer. The second
formula is found by subtracting equation 1 from equation 2.
1, 2, 4, 8... is a geometric progression.
EXAMPLE 130
Just by inspection, the sum of the four terms is 1 + 2 + 4 + 8 = 15 .
Now let’s use the geometric sum formula.
77
t2
2
r=
r=2
t1
1
r=
Sn =
(
)
a r n −1
S4 =
r −1
(
)
1 24 − 1
2 −1
r=
S4 =
t3
4
r=
r=2 .
t2
2
1(16 − 1)
S4 =
1
1(15 )
1
S4 = 15
EXAMPLE 131
Find the sum of the first 20 terms of the geometric series < 8, 4, 2, 1, ½, ¼ …>
First find the common ratio. r =
t
t2
4
1
2
1
r= r=
r= 3 r= r=
t1
8
2
t2
4
2
Now use the geometric sum formula
Sn =
a
(r
n
)
−1
r −1
 1  20 
8   − 1
 2 

S 20 = 
1
−1
2
S 20 = 15 ⋅ 999984 ( 6dp )
Infinite Geometric Progression
a (1 − r n )
•
For r < 1, the partial sum of a geometric progress is: S n =
•
a
This formula can be rewritten as S n =
1− rn
1− r
•
When the value of n approaches infinity, rn approaches zero.
(
S∞ =
a
1− r∞
1− r
(
S∞ =
)
→
As n → ∞ r → 0
n
1− r
)
a
(1 − 0 )
1− r
S∞ =
a
1− r
SUM TO INFINITY
Geometric progression with ratio less than 1 [ r < 1 ], approaches a limit.
78
1 1 1
1
1
,
,
,
,
,
2 4 8 16 32
common ratio is ½ which is less then 1.
8, 4, 2, 1,
1
1
,
... is a geometric series and its
64 128
Let’s try to add as many terms as possible and check if the series approaches a limit.
8 + 4 = 12
8 + 4 + 2 = 14
8 + 4 + 2 + 1 = 15
8 + 4 + 2 +1+
1
1
= 15 = 15 ⋅ 5
2
2
8 + 4 + 2 +1+
1 1
3
+ = 15 = 15 ⋅ 75
2 4
4
8 + 4 + 2 +1+
1 1 1
7
+ + = 15 = 15 ⋅ 875
2 4 8
8
8 + 4 + 2 +1+
1 1 1 1
15
+ + + = 15 = 15 ⋅ 9375
2 4 8 16
16
8 + 4 + 2 +1+
1 1 1 1 1
31
+ + + +
= 15 = 15 ⋅ 96875
2 4 8 16 32
32
8 + 4 + 2 +1+
1 1 1 1 1
1
63
+ + + + +
= 15 = 15 ⋅ 984375
2 4 8 16 32 64
64
8 + 4 + 2 +1+
1 1 1 1 1
1
1
+ + + + + +
= 15 ⋅ 9921875
2 4 8 16 32 64 128
If you notice closely, from step 4, the sums are very slowly increasing. However, as
you come down the steps, the sums are slowly approaching a certain value. In this
case, it is approaching 16. A sum to infinity formula can be used to calculate a point
or limit beyond which, this series would not exceed or pass.
SUM TO INFINITY FORMULA
79
S∞ =
a
1− r
The symbol ∞ stands for infinity. a is the first term and r is the common ratio.
EXAMPLE 132
8, 4, 2, 1,
1
,
2
1 1
1
,
,
,
4 8 16
1
,
32
1
1
,
...
64 128
The above geometric progression is also found in the page 81. This time the sum to
infinity formula will be used to show that the series is indeed bounded by a limit.
a =8 r =
S∞ =
1
a
8
8
S∞ =
S∞ =
S∞ =
2
1− r
 1
1
1 − 
 
 2
2
8
1
2
16
→ S∞ = 8 ÷ → S ∞ = 8 × → S ∞ =
→ S∞ = 16
2
1
1
1
 
2
This means no matter how many elements of this series you continue to add, the sum
will approach 16 but, the sum will never be 16 and will never cross 16.
EXAMPLE 133 Given below is a geometric series.
160 + 80 + 40 + 20 + 10 + 5,...
•
Find the sum of the first 10 terms. .
r=
r=
•
t2
80
1
r=
r=
t1
160
2
Sn =
t3
40
1
r=
r=
t2
80
2
(
)
a r n −1
r −1
  1 10 
160    − 1
 2 



S10 =
1


 − 1
2 
S10 = 319 ⋅ 6875
Find the sum to infinity of
a = 160
1
r=
2
S∞ =
a
1− r
S∞ =
160
 1
1 − 
 2
EXAMPLE 134
80
S∞ =
160
1
 
2
S∞ = 320
An overweight man is placed on a control diet with a training program that will surely
remove all his excess body weight. His weight is monitored at the end of every 6th
month. At the end of the first 6th month, he lost 20kg of the excess weight. The rest
of the end of the every other 6th month, he lost 10% of his previous lost weight.
A.
B.
C.
D.
E.
How much excess body weight is removed in the second 6th month?
How much excess body weight is removed in the third 6th month?
How much excess body weight is removed in the fourth 6th month?
Find the common ration.
How much excess body weight did the person had altogether?
A. t2 = 20 × 0 ⋅1 = 2 kg B. t3 = 2 × 0 ⋅1 = 0 ⋅ 2 kg
D. r =
C. t4 = 0 ⋅ 2 × 0 ⋅1 = 0 ⋅ 02 kg
t3
0⋅2
0 ⋅ 02
t
r=
r = 0 ⋅1 r = 4 r =
r = 0 ⋅1
2
0⋅2
t2
t3
E. S ∞ =
20
2
a
S∞ =
S ∞ = 22 kg
1− r
1 − 0 ⋅1
9
EXAMPLE 135
In 2000 the total number of visitor arrival in a town was 20, 000. Statistical reports
indicate that the number has been increasing by 12% every year since then.
A. Find the visitor arrival for the first five years.
t1 = 20,000
t 2 = 20,000
t 2 → 20,000 × 0 ⋅ 12 = 2400
t 2 = 20,000 + 2400 = 22,400
t 3 → 22,400 × 0 ⋅ 12 = 2688
t 3 = 22,400 + 2688 = 25,088
t 4 → 25,088 × 0 ⋅ 12 = 3010 ⋅ 56
t 4 = 25088 + 3010 ⋅ 56 = 28098 ⋅ 56
t 5 → 28098 ⋅ 56 × 0 ⋅ 12 = 3371.8272
t 5 = 31470 ⋅ 3872
B. Find the common ratio.
t2
t1
r=
r=
22,400
20,000
r = 1 ⋅ 12
r=
t3
t2
r=
25088
22400
r = 1 ⋅ 12
C. Find the total number of visitor arrival in the town from beginning of 2000 and
at the end of 2007.
Sn =
S8 =
(
)
a r n −1
r −1
→ S8 =
(
)→S
20, 000 (1⋅12 ) − 1
8
1⋅12 − 1
8
=
20, 000 (1⋅ 475963176 )
0 ⋅12
29,519.26353
→ S8 = 245,993.8627 ≈ 245,994
0 ⋅12
Therefore I esteem all your precepts concerning all things to be right;
and I hate every false way. (Old Testament | Psalms 119:128)
81
Exercise 2 D
1. Inflation rate in the housing sector of Fiji is running at 11% due to past
repeated devaluation of Fiji dollar and coup d'états. At the end of the first year
of the inflation, the ground rent value of a standard 400 m 2 Native Land Trust
Board house block was $50 per year.
A. Calculate the rent of the house block at the end of the second year of the
inflation.
B. Calculate the rent of the house block at the end of the third year of the
inflation.
C. Calculate the rent of the house block at the end of the fourth year of the
inflation.
D. Calculate the common ratio.
E. Calculate the number of years it will take for the ground rental to exceed $300
per year.
2. Solomon opens an account in the ANZ bank by depositing $100. Bank gives
him 5% interest per annum.
A. Calculate the sum of the money in Solomon’s account at the end of the 2nd
year, 3rd year and 4th year.
B. Calculate the common ratio.
C. In what year will the sum of money in Solomon’s account will exceed $200,
provide he does not make any withdrawal and any further deposits.
3. Samuel won a lottery worth a million dollars. He has to pay his government
tax on this lottery money. He is given two options. The first option is that the
government will charge him 35% and he will have to pay the tax within seven
days of receiving the lottery money. The second option is that at the end of
the 1st year, he will give the government $100, 000. The years after, at the end
of each year he will pay his government the 50% of the amount he paid the
previous year.
A. Calculate the total amount he would have paid if he opted for option one.
B. Calculate the total amount he would have to pay over the years if he opted for
option two.
C. Which option do you perceive as the best option and explain with reasons.
4. In 1990 the total amount of tithing collected from the members of a particular
church was approximately $25, 000. It has been noticed that this church is
82
growing in this particular area of Fiji by ¼%. If the amount of tithing
collected each year in the same area increases by 16% per year, calculate the
total amount of tithing collect within the first 10 years.
5. The Missionary program of The Church of Jesus Christ of the Latter Day
Saints indicates that recently the convert membership in the church has been
increasing at rate of 1% per year in Fiji. The Church of Jesus Christ of the
Latter Day Saints has been in Fiji for a while and is currently the fast growing
Church in Fiji as well. If by the end of year 2010, the Church had a sum of
4000 convert members, calculate the expected total population of the convert
members at the end of the 2017.
6. At the end of the year 2005, the costal area of the Wainividio village in Navua
has recorded lose of 0 ⋅ 002% of their costal land area per year. This is a
direct result of global warming. If the total costal area was 40, 000m 2 ,
calculate the total lose to the costal area by the end of the year 2009.
7. A geometric progression is given by 100, 50, 25, 12 ⋅ 5, 6 ⋅ 25,...
A. Find the common ratio.
B. Find the 10th term.
C. Find the sum of the first 10 terms
D. Find the sum of the first 50 terms
E. Find the sum of the first 100 terms
F. Find the sum to infinity.
8. In 2005 the total number of patron arrival in the House of the Lord, the LDS
Temple at the junction of the Lakeba street and Princess Road was 8, 000.
Statistical reports indicate that the number has been increasing by 2% every
year since then.
A. If it is projected that this statistical report would apply for the next 10 years,
find the expected visitor arrival for the first five years.
B. Find the common ratio.
C. How many patrons are expected to visit this temple by the end of the year
2020?
83
MATRIX
A matrix is a rectangular array of numbers:
 a11
a
 21
 
 am1
a1n 
a2 n 


amn 
...
...
a12
a22
am2
All the numbers in the array are called the entries or elements of the matrix. If a
matrix has m number of rows and n number of columns, it can be said that its size
is m by n. It is written m × n .
c1 c2 c3 c4
↓ ↓ ↓ ↓ →
r1 →  0 4 5 6 
r2 → 1 2 4 7 
r3 → 9 2 9 3 
c → Column r → Row
Size = 3 × 4
r represents rows and c represents columns. The dimension or the size of the
matrix is given as number of rows by the number of the column. The matrix in
this example has, there 3 rows and 4 columns. The size can be written as either
3 by 4 or 3 × 4.
SCALAR MULTIPLICATION
a
A =  11
a 21
•
a12 
a 22 
a
kA = k  11
a 21
a12 
a 22 
 ka
kA =  11
ka21
ka12 
ka22 
The scalar is multiplied with every element inside the matrix.
EXAMPLE 136
1 2 
1 2  1( 2 ) 2 ( 2 ) 
2 4
A=
→
2
A
=
2
=
→
2
A
=

3 4   3 2 4 2 
6 8 
( )
3 4 

  ( )


84
ADDITION OF MATRICES
a
A =  11
 a21
•
a12 
a22 
b 
b
B =  11 12 
b21 b22 
a + b
A + B =  11 11
 a21 + b21
a12 + b12 
a22 + b22 
Addition of matrices is performed by adding corresponding rows and columns of both matrices.
EXAMPLE 137
1 2 
 4 3
1 + 4 2 + 3
5 5 
A=
B
=
A
+
B
=
A
+
B
=

 5 1
3 + 5 8 + 1 
8 9 
3 8 






PROPERTIES OF BASIC MATRIX OPERATIONS
Given below are properties of basic matrix operations and it provides some useful
rules for doing matrix arithmetic. All properties are valid for any matrices
A, B and C for which the indicated operations are defined and for any scalar k . In
other word, addition of matrices is commutative.
A + B = B + A Additive commutativity
EXAMPLE 138
In example 137, the operation A + B was done. In this example B + A is done.
1 2 
 4 3
 4 + 1 3 + 2
5 5 
A=
B
=
B
+
A
=
A
+
B
=

5 1
5 + 3 1 + 8 
8 9 
3 8 






SUBTRACTION OF MATRICES
a
A =  11
a 21
•
a12 
a 22 
b
B =  11
b21
a − b
A − B =  11 11
a 21 − b21
b12 
b22 
a12 − b12 
a 22 − b22 
Subtraction of matrices is performed by subtracting corresponding rows and columns of both matrices.
EXAMPLE 139
1 2 
 4 3
1 − 4 2 − 3
A=
B=
A− B = 



3 8 
 5 1
3 − 5 8 − 1 
85
 −3 −1
A+ B = 

 −2 7 
Determinant
Determinant of matrix A in short form is written as det (A).
a
If A =  11
a 21
a12 
,
a 22 
then det ( A) = (a11 × a 22 ) − (a 21 × a12 )
 4 5
EXAMPLE 140 Find the determinant of matrix A = 

 3 9
det ( A ) = ( 4 × 9 ) − ( 3 × 5 ) det ( A ) = ( 36 ) − (15 )
 −4
EXAMPLE 141 Find the determinant of matrix B = 
 −1
det ( A ) = 21
24 
6 
det ( B ) = ( −4 × 6 ) − ( −1× 24 ) det ( B ) = ( −24 ) − ( −24 )
det ( B ) = 0
 −1 −3
6 
EXAMPLE 142 Find the determinant of matrix B = 
 2
.
det ( B ) = ( −1× 6 ) − ( 2 × −3) det ( B ) = ( −6 ) − ( −6 )
EXAMPLE 143
1
2
Find the determinant of matrix B = 
3
 4
1 1 3 2
det ( B ) =  ×  −  × 
 2 3  4 3
1  6 
det ( B ) =   −  
 6   12 
det ( B ) =
→
det ( B ) =
86
12 − 36
72
−24
72
det ( B ) = 0
2
3

1
3 
→ det ( B ) = −
1
3
INVERSE OF 2 by 2 MATRICES
Steps for find the inverse of 2 by 2 matrices:
1. Find the determinant of the matrix
2. Divide 1 by the determinant. Let’s call the answer, scalar.
3. Interchange a11 with a 22 .
4. Change the sign of a 21 and a12 .
5. Multiply the scalar of step 2 with changes made to matrix by from steps 3 and 4.
 a11 a12  −1
 a22 −a12 
1
A=
 A = det A  −a
a
a
( )  21 a11 
 21 22 
 4 10 
EXAMPLE 144 Find the inverse of matrix A = 

2 6 
Step 1 → det ( A ) = ( 4 × 6 ) − ( 2 ×10 )
det ( A ) = ( 24 ) − ( 20 ) det ( A ) = 4
Step 2 →
1
4
 6 10 
Step 3 → 

2 4 
 6 −10 
Step 4 → 
4 
 −2
6
1  6 −10   4
Step 5 → 
=
4   −2
4  −2
 4

87
−10 
3
2
4 
−1
→A =
4 
 −1

 2
4 

−5 
2


1 

 −4 −2 

 2 2
EXAMPLE 145 Find the inverse of A = 
det ( A ) = ( −4 × 2 ) − ( 2 × −2 )
det ( A ) = −8 − −4
 2
 −4
1  2 2 
=
−4  −2 −4  
−2

 −4
→ det ( A ) = −8 + 4
2
−4 

−1
 → A =
−4 

−4 
 1
− 2


 1

 2
det ( A ) = −4
1
− 
2

 −0 ⋅ 5 −0 ⋅ 5
 or 
1 
 0⋅5


1 
1 1

 2 3
EXAMPLE 146 Find the inverse of A = 
det ( A) = (1× 3) − (1× 2 ) = 3 − 2 = 1
det ( A ) = 1
 3 −1
1  3 −1 1  3 −1
=
=
1
 −2 1 
det ( A)  −2 1  1  −2 1 


 3 −1
A−1 = 

 −2 1 
 2 −1
EXAMPLE 147 Find the inverse of A = 

 6 −3
det ( A ) = ( 2 × −3) − ( −1× 6 ) = ( −6 ) − ( −6 ) = 0
1  −3 1 
1  −3
=


det ( A )  −6 2 
0  −6
det ( A ) = 0
1
2 
1
is undefined . This means matrix A has no inverse.
0
88
MULTIPLICATON OF 2 BY 2 MATRICES
Two matrices may be multiplied if the number of column/s of the first matrix is equal
the number of the rows of the second matrix.
a
If A =  11
c21
a
AB =  11
c21
EXAMPLE 148
b12 
d 22 
e
B =  11
 g 21
b12   e11
d 22   g 21
f12 
then AB is equal to :
h22 
f12   a11e11 + b12 g 21
=
h22  c21e11 + d 22 g 21
1 2
 4 6
A=
B=


3 4
7 8 
Find AB
a11 f12 + b12 h22 
c21 f12 + d 22 h22 
( AB
means
A × B)
Matrices are multiplied row by column. Take the first row of the matrix A and
multiply with the first column of the matrix B.
Note: The first element of the first row of matrix A is multiplied by the first element
of the first column of matrix B. Once this operation is completed, the second element
of the first row of matrix A is multiplied with the second element of the first column
of matrix B. Row one × Column one r1 × c1 → (1 × 4 ) + (2 × 7 ) = 4 + 14 = 18
1 2  4 6 (1 × 4 ) + (2 × 7 ) _____________ 
3 4 7 8  =  ___________ _____________ 


 

Take the first row of matrix A and multiply with the second column of matrix B.
Row one × Column two r1 × c 2 → (1 × 6 ) + (2 × 8) = 6 + 16 = 22
1 2 4 6 (1 × 4 ) + (2 × 7 )
3 4 7 8 =  __________ _


 
(1 × 6) + (2 × 8)
__________ _ 
Take the second row of matrix A and multiply with the first column of matrix B
Row two × Column one r2 × c1 → (3 × 4 ) + (4 × 7 ) = 12 + 28 = 40
1 2 4 6  (1 × 4 ) + (2 × 7 )
3 4 7 8 =  (3 × 4 ) + (4 × 7 )


 
(1 × 6) + (2 × 8)
___________ 
Take the second row of matrix A and multiply with the second column of matrix B
Row two × Column two r2 × c 2 → (3 × 6 ) + (4 × 8) = 18 + 32 = 50
1 2 4 6  (1 × 4 ) + (2 × 7 )
3 4 7 8  =  (3 × 4 ) + (4 × 7 )


 
(1 × 6) + (2 × 8)
(3 × 6) + (4 × 8)
89
18 22
= 

40 50 
IDENTITY MATRIX
A matrix multiplied by its identity will give the same matrix back as the product.
A× I = A
1 0 0 
1 0 


I2 = 
 , I3 = 0 1 0  and I 4
0
1


0 0 1 
1
0

0

0
0
1
0
0
0
0
1
0
I 2 , I3 and I 4 are identical
0
0 
matrices of second order,
→
0
third order and fourth

1
order, respectively
A matrix whose identity entries are all 1 and whose other entries are all 0 is called
n × n identity matrix or identity matrix of n th order.
EXAMPLE 149
3 4
1 0
If A = 
and its identity be I = 

, show that A × I = A.
6 7 
0 1 
3 4 1 0 3 4
In other words 

=

6 7  0 1  6 7 
 3 4  1 0   ( 3 × 1) + ( 4 × 0 )
6 7  0 1  =  6 ×1 + 7 × 0
) ( )


 (
( 3 × 0 ) + ( 4 ×1)  3 + 0
=
( 6 × 0 ) + ( 7 ×1) 6 + 0
0 + 4 3 4
=
0 + 7  6 7 
EXAMPLE 150
1 2 
1 0 
If A = 
and its identity be I = 

 , show that A × I = A.
3 4 
0 1 
1 2 1 0 1 2 
In other words 

=

3 4   0 1  3 4 
1 2  1 0   (1×1) + ( 2 × 0 )
3 4   0 1  =  3 × 1 + 4 × 0
) ( )


 (
(1× 0 ) + ( 2 ×1)  1 + 0
=
( 3 × 0 ) + ( 4 ×1)  3 + 0
90
0 + 2  1 2 
=
0 + 4  3 4 
A × A−1 = I
Any matrix multiplied by its inverse will give a product which is the identity matrix.
Let A be a matrix and lets its inverse be A −1 . It follows that A × A −1 = I .
EXAMPLE 151
1 1
 3 − 1
In example 146, A = 
and A −1 = 

 . In this example it will be
 2 3
− 2 1 
shown that A × A −1 = I .
1 1  3 −1  (1× 3) + (1× −2 )
 2 3  −2 1  =  2 × 3 + 3 × −2
) (
)


 (
(1× −1) + (1 + 1) 
( 2 × −1) + ( 3 ×1)
 3 − 2 −1 + 1 
1 0 
=
=

0 1 
6 + −6 −2 + 3


EXAMPLE 152
 −4 −2 
 −0 ⋅ 5 −0 ⋅ 5
and A−1 = 
.
In example 145, A = 

1 
 2 2
 0⋅5
In this example it will be shown that A × A −1 = I .
 −4 −2   −0 ⋅ 5 −0 ⋅ 5 ( −4 × −0 ⋅ 5 ) + ( −2 × 0 ⋅ 5 )
=
 2 2   0⋅5
1   ( 2 × −0 ⋅ 5 ) + ( 2 × 0 ⋅ 5 )


 2 −1 2 − 2 
1 0 
=
= 


 −1 + 1 −1 + 2 
0 1 
91
( −4 × −0 ⋅ 5) + ( −2 ×1) 
( 2 × −0 ⋅ 5) + ( 2 ×1) 
POINTS TO NOTE:
•
•
When two square matrices are multiplied with each other, the answer will be a
square matrix.
When two matrices of difference size are multiplied with each other, the
answer have a size of the rows of the first matrix and the columns of the
second matrix, provide the number of columns of the first matrix is equal to
the number of rows of the second matrix.
C1 C2
C1
↓
↓
↓
r1 →  2 3  r1 →  1 


 
r2 →  1 4  r2 →  0 
2× 2
2 ×1
2 × 2 2 ×1
Same means
that both
Matrices can
multiplied
2 and 1 is the dimension of the product matrix.
C1
↓
C1
↓
r1 →  2  r1 →  1 
 
 
r2 →  3  r2 →  0 
2 ×1
2×
2 ×1
1 2 ×1
different means
that both
Matrices cannot
be multiplied
92
EXAMPLE 153
1 2 
5 
A=
B= 

3 4 
6 
Find AB
( AB
means A × B )
1 2  5  1× 5 + 2 × 6  5 + 12  17 
AB = 
  = 
=
= 
3 4   6  3 × 5 + 4 × 6  15 + 24  39 
EXAMPLE 154
1 2 
1 0 2 
A=
B=


3 4 
 0 3 5
2× 2
Find AB
2× 2
2× 2 2 ×3
Same means
that both
Matrices can
multiplied
2 × 3 is the dimension of the product matrix
1 2  1 0 2  1× 1 + 2 × 0 ________ ________  1
3 4  0 3 5 =  ________ ________ ________  = 


 
 



1 2  1 0 2  1× 1 + 2 × 0 1× 0 + 2 × 3 ________  1 6
3 4  0 3 5 =  ________ ________ ________  = 


 
 



1 2  1 0 2  1× 1 + 2 × 0 1× 0 + 2 × 3 1× 2 + 2 × 5 1 6 12 
3 4  0 3 5 =  ________ ________ ________  = 



 
 

1 2  1 0 2  1× 1 + 2 × 0 1× 0 + 2 × 3 1× 2 + 2 × 5 1 6 12 
3 4  0 3 5 = 3 ×1 + 4 × 0 ________ ________  = 3



 
 

1 2  1 0 2  1× 1 + 2 × 0 1× 0 + 2 × 3 1× 2 + 2 × 5 1 6 12 
3 4  0 3 5 = 3 ×1 + 4 × 0 3 × 0 + 4 × 3 _______  = 3 12 


 
 

1 2  1 0 2  1× 1 + 2 × 0 1× 0 + 2 × 3 1× 2 + 2 × 5  1 6 12 
3 4  0 3 5 = 3 ×1 + 4 × 0 3 × 0 + 4 × 3 3 × 2 + 4 × 5 = 3 12 26 


 
 

93
Application of Inverse Matrix for Solving Linear Equations
Example 155
Use inverse of matrix method to find the point of intersection
of the lines y = 2x + 4 and y = -x +1
Both equations are rewritten with some changes in the order of the equations.
y = 2x + 4
y = −x +1
y − 2x = 4
−2 x + y = 4
y + x =1
x + y =1
Now write the above equations in matrix form
 −2
_
−2 x + y = 4

→
 −2
x + y =1
1

1  x 4
=
_   y   _ 
1  x   4 
=
1  y   1 
 −2 1
Now find the inverse of the first matrix 
 . For convenience, we will name this
 1 1
matrix A.
 −2 1
A=
 det ( A ) = ( −2 × 1) − (1× 1) = −2 − 1 = −3
 1 1
 1 1
− 3 3 
1  1 −1
−1
−1
A =− 
A =

3  −1 −2 
 1 2
 3 3 
Recall from form 5 maths that if Sin x = y then x = Sin-1y. Like manner
 −2 1  x   4 
 1 1  y  = 1 

   
 1
−
 x  3
 y =  1
  
 3
→
 x
 −2 1  4 
 y  = inverse of  1 1 × 1 
 

  
1
 4 1
 3
− + 
−


 x
 x  3 
 x   −1 
3 4
3 3
  →   = 
 →  =  →  = 
2  1 
 y  4 + 2 
 y  6 
 y   2
3 
3 
 3
 3 
Hence, x = -1 and y = 2. The point of intersection is (-1, 2). Compare this example
with example 215.
94
Failure is instructive. The person who really thinks learns quite as
much from his failures as from his successes. John Dewey
EXERCISE 2E
0 −1
1 0 
0 ⋅ 5 0 
0 1 
2 0
 −1 0 
A=
B=
C=
D=
E=
F=






1 0 
0 1 
 0 0 ⋅ 5
1 0 
0 2
 0 −1
Matrices A, B, C, D, E and F are used for the questions 1 to 14
1. Evaluate 2A
2. Evaluate -2C
3. Evaluate –A
4. Evaluate 2A – 3B
5. Evaluate A – B
6. Evaluate BA
7. Evaluate AB
8. Evaluate AC
9. Evaluate CA
10. Evaluate CE
11. Find the determinant of Matrix A
12. Find the inverse of matrix A
13. Find the inverse of matrix D
14. Find 2A – 3B
15. Use inverse of matrix method to find the point of intersection of the lines
A. y = 4x + 4 and y = -x +1
B. y = x – 1 and y = 3x – 1
C. x – 2y = 4 and 2x – 3y = 8
D. x – 2y = 9 and x – 3y = 9
95
3
Give instruction to a wise man, and he will be yet wiser: teach a just
man, and he will increase in learning (Old Testament | Proverbs 9:9).
FUNCTIONS AND GRAPHS
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
Linear Graph
Quadratic Graph
Cubic Graph
Equation of a circle and graph
Exponential graphs
Transformation of graphs
Hyperbolic Graphs
96
LINEAR GRAPHS
The standard form of linear graphs is
y = mx + c. m → gradient c → y − Intercept
When you are asked to graph any function, you will be required to find both the x and
y intercepts. Given below are the steps that you would need to follow to successfully
graph a linear function.
•
•
•
•
Find the y-intercept. To find the y-intercept, make x = 0.
Find the x-intercept. To find the x-intercept, make y = 0.
Plot both the x and y intercepts on the Cartesian plane.
Join the points by drawing a straight line.
EXAMPLE 156
Graph y = x + 2.
To find the y − Intercept , To find the x − Intercept ,
make x = 0 and then solve make y = 0 and then solve
for y.
y = x+2
for y.
y = x+2
y = 0+2
y = 0+2
0= x+2
0−2 = x
y=2
−2= x
97
x = −2
QUADRATIC GRAPHS
Quadratic graphs are U shaped.
•
•
•
•
It has a turning point.
The x -value of the vertex of the turning point is found by adding the roots and
dividing the sum of the roots by 2.
Quadratic graphs are symmetrical.
The y-value is found by substituting the x-value.
All Quadratic graphs have a standard equation → y = ax 2 + bx + c.
EXAMPLE 157 Graph y = x 2 + 3x + 2.
y = x + 3x + 2.
To find y → Intercept ,
2
make x = 0.
y = ( 0) + 3( 0) + 2
2
y=2
y = x 2 + 3 x + 2.
x → Intercept , make y = 0.
0 = x 2 + 3x + 2
x →1
x→2
1x + 2 x = 3 x ← middle term
( x + 1)( x + 2 ) = 0 →
x = −1 x = −2
The method given below for finding turning point is only applicable to quadratic functions and its graphs.
Turning Point →
x=
−1 + −2 −3
2
=
= −1 ⋅ 5 y = ( −1 ⋅ 5 ) + 3 ( −1 ⋅ 5 ) + 2 → y = −0 ⋅ 25
2
2
TP → ( −1 ⋅ 5, −0 ⋅ 25 )
98
EXAMPLE 158 Graph y = x 2 + 4 x + 4.
y = x + 4 x + 4.
To find y → Intercept ,
make x = 0.
2
y = x 2 + 4 x + 4.
x → Intercept , make y = 0.
y = ( 0) + 4 (0) + 4
0 = x2 + 4 x + 4
x→2
x → 2 2 x + 2 x = 4 x ← middle term
y=4
( x + 2 )( x + 2 ) = 0 →
2
•
•
•
x = −2
x = −2
All repeated roots become turning points of the function they belong to.
Any root that is repeated, the y value of its turning point will always be zero.
Since x = −2 is a repeated root, it becomes the x value of the turning point
and the y value of the turning point is zero. Hence, the turning point is ( −2, 0 )
Given below is the method used in example 157 for finding coordinates of turning
point of any quadratic graph.
Turning Point →
x=
−2 + −2 − 4
2
=
= −2 y = ( −2 ) + 4 ( −2 ) + 4 → y = 0
2
2
TP → ( −2, 0 )
99
CUBIC GRAPHS
General Equation : y = ( x + a )( x + b)( x + c) y = ax3 + bx 2 + cx + d
y = ax 3 + bx 2 + cx + d
If a is positive
If a is negative
+ ax 3
− ax 3
EXAMPLE 159
The factorised form of f ( x ) = x 3 + 2 x 2 − 5 x − 6 is f ( x) = (x − 2)(x + 1)(x + 3) .
Sketch the graph of f ( x).
First find the x and y intercepts.
To find x-intercept, make y = 0.
To find y − intercept , make x = 0
f ( x ) = ( x − 2 )( x + 1)( x + 3)
f ( x) = ( x − 2 )( x + 1)( x + 3)
f ( 0 ) = ( 0 − 2 )( 0 + 1)( 0 + 3)
0 = ( x − 2 )( x + 1)( x + 3)
f ( 0 ) = ( −2 )(1)( 3)
x − 2 = 0 x +1 = 0 x + 3 = 0
x=2
•
f ( 0 ) = −6
y = −6
x = −1 x = −3
Always start to draw any quadratic and cubic graph from the right hand side
and then move to the left hand side.
At this point you are not required to find the turning points for cubic functions.
100
EXAMPLE 160 Sketch the graph of f ( x) = ( − x − 2 )( x + 1)( x + 3) .
First find the x and y intercepts.
To find x-intercept, make y = 0.
To find y − intercept , make x = 0
f ( x ) = ( − x − 2 )( x + 1)( x + 3)
f ( x) = ( − x − 2 )( x + 1)( x + 3)
f ( 0 ) = ( 0 − 2 )( 0 + 1)( 0 + 3)
0 = ( − x − 2 )( x + 1)( x + 3)
f ( 0 ) = ( −2 )(1)( 3)
−x − 2 = 0 x +1 = 0 x + 3 = 0
x = −2
f ( 0 ) = −6
x = −1 x = −3
101
y = −6
EXAMPLE 161
Sketch the graph of
First find the x and y intercepts
To find y − intercept ,
To find x-intercept, make y = 0
make x = 0
f ( x) = ( x − 2 )( x − 2 )( x + 3)
f ( x ) = ( x − 2 )( x − 2 )( x + 3)
0 = ( x − 2 )( x − 2 )( x + 3)
f ( 0 ) = ( 0 − 2 )( 0 − 2 )( 0 + 3)
x−2 = 0 x−2 = 0 x+3= 0
x=2
x=2
f ( x) = ( x − 2 )( x − 2 )( x + 3) .
f ( 0 ) = ( −2 )( −2 )( 3)
x = −3
f ( 0 ) = 12
y = 12
•
Since x = 2 is a repeated root, it becomes the x value of the turning point and
the y value is 0. For any repeated root, the y value of the turning point will
always be 0.
•
All repeated roots become turning points of the function they belong to.
102
Equation of a Circle
( x, y ) in a Cartesian plane that are equidistance from a
C ( h, k ) . If r is the fixed distance, then a point P ( x, y ) is on the circle if
A circle is the set of all points
fixed point
and only if
d (C, P ) =
( x − h) + ( y − k )
2
2
=r
The above equation of the circle is equivalently expressed as
( x − h) + ( y − k )
2
2
=r
2
This is called the standard form for the equation of the a circle with centre C(h, k) and
radius r.
( x − h) + ( y − k ) = r2
2
2
103
If we set h = 0 and k = 0, we obtain the standard form for the equation of a circle
with centre at the origin.
( x − h) + ( y − k )
2
2
( x − 0) + ( y − 0)
2
= r2
( x) + ( y)
2
2
2
= r2
= r2
EXAMPLE 162 Find the equation of the circle with centre (2, -1) and radius 4.
( x − h ) + ( y − k ) = r 2 ( h, k ) → center of the circle r → radius of the circle
( 2, −1) → center of the circle 4 → radius of the circle
2
2
2
2
2
2
( x − h ) + ( y − k ) = r 2 → ( x − 2 ) + ( y − −1) = 42 → ( x − 2 ) + ( y + 1) = 16
2
2
EXAMPLE 163 Find the equation of the circle with centre (-3, -1) and radius 9.
( x − h ) + ( y − k ) = r 2 ( h, k ) → center of the circle r → radius of the circle
( −3, −1) → center of the circle 4 → radius of the circle
2
2
2
2
2
2
( x − h ) + ( y − k ) = r 2 → ( x − −3) + ( y − −1) = 92 → ( x + 3) + ( y + 1) = 81
2
2
EXAMPLE 164 Find the equation of the circle with centre (0, 0) and radius 4.
( x − h) + ( y − k )
( 0, 0 ) → center
2
( x − h) + ( y − k )
2
2
2
= r2
( h, k ) → center
of the circle r → radius of the circle
of the circle 4 → radius of the circle
= r 2 → ( x − 0 ) + ( y − 0 ) = 42 → ( x ) + ( y ) = 16
2
2
104
2
2
Exponential Graphs
The best way of graphing exponential functions is to plot as many points as you can.
Graph y = 2 x
Example 165
x
y
•
•
-3
-1/8
-2
-¼
-1
-½
0
1
1
2
2
4
3
8
Domain: This is the set of x-values or the first element of any ordered pair.
For the above graph, the domain on the left hand side of the graph approaches
infinity and on the right hand side the domain of the graph also approaches
infinity but faster than domain on the left hand side. The fact that both sides
are approaching infinity tells us that all x-values are part of the domain.
Hence, we conclude that the domain of the above graph is the set of all xvalues. We write this as domain : x ∈ R. This means that the domain is an
element of real numbers and all real numbers are included in the domain.
Range: This is the set of y-values or the second element of any ordered pair.
The above graph, the range is above zero but not equal to zero. Hence, the
range of the above graph is Range : y > 0, y ∈ R. This means that all real
numbers greater than 0 are included in the range.
105
Example 166
1
Graph y =  
2
x
y
-3
8
x
-2
4
-1
2
0
1
1
½
2
¼
3
1/8
•
Domain: For the above graph, the domain on the left hand side of the graph
approaches infinity and on the right hand side the domain of the graph also
approaches infinity. Hence, we conclude that the domain of the above graph
is the set of all x-values. We write this as domain : x ∈ R.
•
Range: The above graph, the range is above zero but not equal to zero.
Hence, the range of the above graph is Range : y > 0, y ∈ R. This means that
all real numbers greater than 0 are included in the range.
106
Transformation of graphs
Any graph can be shifted in
•
•
•
vertical direction
horizontal direction
and the combination of both directions
Transformation in Vertical Directions
Let c be a constant, the graphs of the sum y = f ( x ) + c and the difference
y = f ( x ) − c can be obtained from the graph of y = f ( x ) by a vertical shift.
Transformation in Horizontal Directions
c
be
a
constant,
the
graphs
of
the
compositions
Let
y = f ( x + c ) and y = f ( x − c ) correspond to horizontal shifts in the graph of
y = f ( x).
Function
Graph
y = f ( x) + c
Graph of y = f ( x ) shifted up c units
y = f ( x) − c
Graph of y = f ( x ) shifted down c units
y = f ( x + c)
Graph of y = f ( x ) shifted to the left c units
y = f ( x − c)
Graph of y = f ( x ) shifted to the right c units
y = x 2 is a parabolic graph facing upwards.
y = ( x − 1) is a translation of the graph y = x 2 one unit to the right.
2
y = ( x + 1) is the translation of the graph y = x 2 one unit to the left.
2
y = x 2 + 1 is the translation of the graph y = x 2 one unit vertically upwards.
y = x 2 − 1 is the translation of the graph y = x 2 one unit vertically downwards.
y = ( x − 1) 2 + 2 is the translation of the graph y = x 2 one unit horizontally right
and 2 units vertically upwards.
y = ( x − 1) 2 − 2 is the translation of the graph y = x 2 one unit horizontally right
and 2 units vertically downwards.
y = ( x + 1) − 2 is the translation of the graph y = x 2 one unit horizontally
towards the left and 2 units vertically downwards.
2
y = ( x + 1) + 2 is the translation of the graph y = x 2 one unit horizontally
towards the left and 2 units vertically upwards.
2
107
Example 167
Sketch the graph of y = x 2 and y = ( x − 1) on the same pair of axis.
2
You can see that the graph of y = ( x − 1) is a translation of the graph y = x 2 one unit
to the right.
2
Example 168
Sketch the graph of y = x 2 and y = ( x + 1) on the same pair of axis.
2
You can see that the graph of y = ( x + 1) is a translation of the graph y = x 2 one unit
to the left.
2
108
Example 169
Sketch the graph of y = x 2 and y = x 2 + 1 on the same pair of axis.
y = x 2 + 1 is the translation of the graph y = x 2 one unit vertically upwards.
Example 170
Sketch the graph of y = x 2 and y = x 2 − 1 on the same pair of axis.
y = x 2 − 1 is the translation of the graph y = x 2 one unit vertically downwards.
109
Example 171
Sketch the graph of y = x 2 and y = ( x − 1)2 + 2 on the same pair of axis.
y = ( x − 1) + 2 is the translation of the graph y = x 2 one unit horizontally
towards the right and 2 units vertically upwards.
2
Example 172
Sketch the graph of y = x 2 and y = ( x − 1)2 − 2 on the same pair of axis.
110
y = ( x ± b) ± c
2
In General
x ± b = 0 x = ±b
if x = + b → move y = x 2 to the right
if x = −b → move y = x 2 to the left
± c if +c → move y = x 2 c units upwards
Example 173
Graph y = x 2
and
y = (x − 2)
2
and
y = ( x − 2) + 2 on the same pairs of axis
2
111
Horizontal Compression of Quadratic Graphs
Example 174 The graph y = 2x2 is the graph of y = x2 compressed horizontally.
Example 175 The graph y = 4x2 is the graph of y = x2 compressed horizontally.
112
Horizontal Stretching of Quadratic Graphs
Example 176
Compare the two graphs given below. The graph of
1 2
y = x is the graph of y = x 2 stretched horizontally.
2
Example 177
Given below are the graphs of y = x 2 , y = 2 x 2 and y =
113
1 2
x on one pair of axis.
2
HYPERBOLA
General Equation :
y=
Ax + B
Cx + D
Example 178
Ax
A
→ horizontal asymptote y =
Cx
C
Cx + D = 0 → Vertical asymptote
y=
Graph
2x + 3
x−2
Horizontal asymptote → y =
2x
x
y=2
Vertical asymptote → x − 2 = 0 x = 0 + 2 x = 2
y → intercept make x = 0
y=
2 (0) + 3
0−2
x → intercept make y = 0
2x + 3
→ 0 ( x − 2) = 2x + 3 → 0 = 2x + 3
3 → 0=
x−2
y=
−2
0 − 3 = 2 x → −3 = 2 x → x = −1 ⋅ 5
The graph does not have any point that lies on the asymptote. In other words, the
graph does not cross the asymptotes. As x → ∞, y → 2 . This means as x continues
to increase or approach infinity, y approaches to 1. The graph comes very close to 1
but does not cross and lie on y = 2.
114
Graph y =
Example 179
x −1
x +1
Horizontal asymptote → y =
x
x
y =1
Vertical asymptote → x + 1 = 0 x = 0 − 1 x = −1
y → intercept make x = 0
y=
( 0) −1
0 +1
y=
−1
y = −1
1
x → intercept make y = 0
0=
•
•
•
•
x −1
0
x +1
( x + 1) = x − 1
0 = x −1 0 + 1 = x 1 = x x = 1
The graph does not have any point that lies on the asymptote. In other
words, the graph does not cross the asymptotes. As x → + ∞, y → 1 .
This means as x continues to increase or approach positive infinity,
y approaches to 1.
The graph comes very close to 1 but does not
cross and lie on y = 1.
As x → − ∞, y → 1 . This means as x approaches negative infinity, y
approaches to 1. Hence, the limit as x approaches infinity, y =1.
When x approaches -1 from the negative side, y approaches to positive
infinity.
When x approaches -1 from the positive side, y approaches to negative
infinity.
115
Example 180
Graph y =
2x − 4
x +1
Horizontal asymptote → y =
2x
x
y=2
Vertical asymptote → x + 1 = 0 x = 0 − 1 x = −1
y → intercept make x = 0
y=
2 (0) − 4
•
•
•
•
0 +1
x → intercept make y = 0
2x − 4
→ 0 ( x + 1) = 2 x − 4 → 0
x +1
= 2x − 4 0 + 4 = 2x 4 = 2x x = 2
→
−4
y=
y = −4
1
0=
The graph does not have any point that lies on the asymptote. In other
words, the graph does not cross the asymptotes. As x → + ∞, y → 2 .
This means as x continues to increase or approach positive infinity,
The graph comes very close to 2 but does not
y approaches to 2.
cross or touch the line y = 2.
As x → − ∞, y → 2 . This means as x approaches negative infinity, y
approaches 2. Hence, the limit as x approaches infinity, y = 2.
When x approaches -1 from the negative side, y approaches to positive
infinity.
When x approaches -1 from the positive side, y approaches to negative
infinity.
116
Learning in education is like a race. Each race has a finishing point. The day we were
born, we became automatic participants of this race. This race is for everyone. Some
are running this race at a very fast pace and they have reached the outer space. Sweet
and eternal victory comes not to those who runs at their pace in this race with their own
learning rate.
Author of this Book
Exercise 3
Sketch the graphs of the function given blow.
1. y = x 2 + 2 x + 1
2. y = ( x − 1)( x − 1)( x + 2)
3. y = ( x + 1) ( x − 1)
2
4. y = x ( x − 2 )
2
5. y = x 2 ( x − 4)
6. y = 2 x
7. y = −2 x
8. y = 2− x
9. y = −2 − x
10. y =
3
x −1
11. y =
x−2
x −1
12. y =
2x −1
x +1
117
4
Wisdom is better than weapons of war: but one sinner destroys much
good (Old Testament | Ecclesiastes 9:18).
COORDINATE GEOMETRY
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
Distance between two points
Mid point formula
Gradient of straight line
Equation of straight line passing through two points
Collinear Points
Parallel lines
Perpendicular lines
118
DISTANCE BETWEEN TWO POINTS
If A(x1 , y1 ) and B( x 2 , y 2 ) are two points then the distance between these two points
is found by the formula given below.
d = ( x2 − x1 ) 2 + ( y2 − y1 ) 2
x = ( x 2 − x1 )
d 2 = x2 + y2
d 2 = ( x 2 − x1 ) + ( y 2 − y1 )
2
d =
2
( x 2 − x1 ) + ( y 2 − y1 )
2
y = ( y 2 − y1 )
2
EXAMPLE 181 Find the distance between the points A ( 2, 1) and B ( 6, 4 )
A( 2, 1) B( 6, 4)
d=
( x2 − x1 ) + ( y2 − y1 )
2
2
↓ ↓
↓ ↓
x1 y1
x2 y2 d = ( 6 − 2) + ( 4 −1) d = ( 4) + ( 3) d = 16 + 9 d = 5
2
2
119
2
2
EXAMPLE 182
Find the distance between the points A ( −2, 3) and B ( 4, 5 )
A ( −2, 3) and B ( 4, 5 )
A ( x1 , y1 ) and B ( x2 , y2 )
d 2 = ( x2 − x1 ) + ( y2 − y1 )
2
d=
( 4 − −2 ) + ( 5 − 3)
2
2
2
d=
( x2 − x1 ) + ( y2 − y1 )
2
( 6) + ( 2)
2
d=
2
→
2
d = 40
EXAMPLE 183
Find the distance between the points A ( −1, 3) and B ( −4, 5)
A ( −1, 3) and B ( −4, 5 )
A ( x1 , y1 ) and B ( x2 , y2 )
d 2 = ( x2 − x1 ) + ( y2 − y1 )
2
d=
( −4 − −1) + ( 5 − 3)
2
2
2
d=
( x2 − x1 ) + ( y2 − y1 )
2
( −3) + ( 2 )
2
d=
2
2
→ d = 13
EXAMPLE 184
Find the distance between the points A (1, 0 ) and B ( 2, −3)
A (1, 0 ) and B ( 2, −3)
A ( x1 , y1 ) and B ( x2 , y2 )
d 2 = ( x2 − x1 ) + ( y2 − y1 )
2
d=
( 2 − 1) + ( −3 − 0 )
2
2
2
d=
d=
120
( x2 − x1 ) + ( y2 − y1 )
2
(1) + ( −3)
2
2
→
2
d = 10
MID POINT FORMULA
The mid point of a straight line segment joining the points ( x1 , y1 ) and ( x 2 , y 2 ) is
given by:
 x1 + x2 y1 + y2 
,


2
2


EXAMPLE 185
Find the mid point of A ( 0, 6 ) and B ( −6, 0 ) .
A ( 0, 6 ) and B ( −6, 0 )
A ( x1 , y1 ) and B ( x2 , y2 )
 x1 + x2
,

2

 0 + −6
,

2
→ 
 −6
 ,
 2
y1 + y2 

2 
6+0

2 
6

2
( −3, 3)
121
EXAMPLE 186 Find the mid point of A ( 3, 6 ) and B ( −7, 2 ) .
A ( 3, 6 ) and B ( −7, 2 )
A ( x1 , y1 ) and B ( x2 , y2 )
 x1 + x2 y1 + y2 
,


2 
 2
 3 + −7 6 + 2 
→ 
,

2 
 2
 −4 8 
 ,  → ( −2, 4 )
 2 2
EXAMPLE 187 Find the mid point of A ( 0, 0 ) and B ( 4, 2 ) .
A ( 0, 0 ) and B ( 4, 2 )
A ( x1 , y1 ) and B ( x2 , y2 )
 x1 + x2 y1 + y2 
 0+4 0+2
4 2
,
→
,
→




 ,  → ( 2, 1)
2 
2 
 2
2 2
 2
EXAMPLE 188 Find the mid point of A ( −1, 0 ) and B ( 9, 8) .
A ( −1, 0 ) and B ( 9, 8 )
A ( x1 , y1 ) and B ( x2 , y2 )
 x1 + x2 y1 + y2 
 −1 + 9 0 + 8 
8 8
,
,

→ 
 →  ,  → ( 4, 4 )
2 
2 
 2
2 2
 2
EXAMPLE 289 Find the mid point of A ( 5, 0 ) and B ( 9, − 6 ) .
A ( 5, 0 ) and B ( 9, −6 )
 x1 + x2 y1 + y2 
,

→
2 
 2
A ( x1 , y1 ) and B ( x2 , y2 )
 5 + 9 0 + −6 
 14 −6 
,
→


 ,
 → ( 7, −3)
2 
 2
 2 2 
122
GRADIENT OF STRAIGHT LINE
Let A( x1 , y1 ) and B( x 2 , y 2 ) be any two points on a straight line. The slope or the
gradient of the line AB can be calculated by the formula given below.
m =
y 2 − y1
x 2 − x1
or
m = ta n θ
m → g r a d ie n t
EXAMPLE 190
Find the gradient of the line joining the points A ( 2, 1) and B ( 6, 4 ) in the diagram
given below.
y2 − y1
4 −1
3
m=
→ m=
→ m=
x2 − x1
6−2
4
Calculate the angle the line segment AB makes with the positive x-axis.
m = tan θ
3
m=
4
m = tan θ
3
→ 3
→ tan −1   = θ → θ = 53 ⋅130
= tan θ
4
4
123
EXAMPLE 191
Find the gradient of the line joining the points A ( 2, 3) and B ( 6, 5) .
m=
y2 − y1
5−3
2
1
→ m=
→ m= → m=
x2 − x1
6−2
4
2
Calculate the angle the line segment AB makes with the positive x-axis.
m = tan θ →
1
1
= tan θ → tan −1   = θ → θ ≈ 26 ⋅ 570
2
2
EXAMPLE 192
Find the gradient of the line joining the points A (1, 3) and B ( 6, 8) .
m=
y2 − y1
8−3
5
→ m=
→ m = → m =1
x2 − x1
6 −1
5
Calculate the angle the line segment AB makes with the positive x-axis.
m = tan θ → 1 = tan θ → tan −1 (1) = θ → θ = 450
EXAMPLE 193
Find the gradient of the line joining the points A ( 2, 9 ) and B ( 6, 9 )
m=
y2 − y1
9−9
0
→ m=
→ m= → m=0
x2 − x1
6−2
4
Calculate the angle the line segment AB makes with the positive x-axis.
m = tan θ → 0 = tan θ → tan −1 ( 0 ) = θ → θ = 00
124
EQUATION OF STRAIGHT LINE
Equation of a straight line passing through any two points lets say
A( x1 , y1 ) and B( x2 , y2 ) can be found by using the equations given below.
y = mx + c or y − y1 = m( x − x1 )
EXAMPLE 194 Find the equation of the line passing through the
points A (−1, 1) and B (1, 3)
The equation of a straight line can be found by either using:
y = mx + c
or
y − y1 = m( x − x1 )
A(−1, 1) and B(1, 3)
m=
y2 − y1
3 −1
2
→m=
→ m = → m =1
x2 − x1
1 − −1
2
m =1
y = mx + c 

→ y = 1x + c → y = x + c
Substitute the
x and y value
A(−1, 1) 
→y = x+c
y = x + c → 1 = −1 + c → 1 + 1 = c → 2 = c
y = x+c → y = x+2
125
EXAMPLE 195 Find the equation of the line passing through the
points A (1, 3) and B (2, 6) .
The equation a straight line can be found by either using:
y = mx + c
y − y1 = m( x − x1 )
or
Both will be used in this example to proof that both will yield the same answer.
m=
y2 − y1
6−3
3
→ m=
→ m= → m=3
x2 − x1
2 −1
1
y = mx + c
y = 3x + c → (1, 3) lies on the line. So x = 1 and y = 3 .
3 = 3 (1) + c
[ Substitute
x and y values to solve for c value
]
3 = 3+ c → 3−3 = c → c = 0
y = 3x + c → y = 3x + 0 → y = 3x
y − y1 = m ( x − x1 )
It has been established that m = 3 . Let’s call point A (1, 3) as ( x1 , y1 ) . This means
that x1 = 1 and y1 = 3 .
y − y1 = m ( x − x1 )
y − 3 = 3 ( x − 1)
y − 3 = 3x − 3
y = 3x − 3 + 3
y = 3x
[Both methods give the same answer]
126
EXAMPLE 196
Find the equation of the line passing through the points A(2, 4) and B(1, 2) .
m=
y2 − y1
2−4
−2
→ m=
→ m=
→ m=2
x2 − x1
1− 2
−1
y = mx + c
→ ( 2, 4 ) lies on the line. So x = 2 and y = 4 .
y = 2x + c
4 = 2 ( 2) + c
[ Substitute
x and y values to solve for c value
4 = 6 + c → 4 − 6 = c → c = −2
Answer
y = 2 x + c → y = 2 x + −2 
→
y = 2x − 2
EXAMPLE 197
Find the equation of the line passing through the points A(2, −4) and B(3, 2) .
A( 2, −4) and B( 3, 2 )
( x 1,
m=
y1)
( x 2,
y 2)
y2 − y1
2 − −4
6
→ m=
→ m= → m=6
x2 − x1
3− 2
1
y − y1 = m ( x − x1 ) → y − −4 = 6 ( x − 2 )
y + 4 = 6 x − 12
y = 6 x − 12 − 4
Answer

→ y = 6 x − 16
127
]
COLLINEAR POINTS
Collinear points lies on a single line or a single line passes through those points.
Two or more points are collinear then the following properties must be prevalent.
• All the points must lie on one single straight line
• The gradient between any two randomly chosen points will be same.
• The equation of any two randomly chosen points will be same.
The best method of proving that points are collinear is to find the gradient between
two points and compare the gradient between other points. Same gradient means the
points are collinear that is they lie on the same line.
EXAMPLE 198
Show that the following points are collinear.
A (1, 2 )
B(3, 4 )
Let’s find the gradient between the points A (1, 2) and
A (1, 2 ) → ( x1 , y1 )
B ( 3, 4 ) → ( x2 , y2 )
m=
C ( 5, 6 ) → ( x2 , y2 )
m=
Since the gradients are same, A (1, 2 )
B(3, 4)
4−2
2
y2 − y1
m=
m = m =1
x2 − x1
3 −1
2
Now let’s find the gradient between the points B(3, 4)
B ( 3, 4 ) → ( x 1 , y1 )
C (5, 6 )
C (5, 6)
6−4
2
y2 − y1
m=
m = m =1
x2 − x1
5−3
2
B(3, 4 )
128
C (5, 6 ) are collinear points.
EXAMPLE 199
Show that the following points are collinear.
A ( 0, −4 ) B (1, −2 ) C ( 2, 0 )
Let’s find the gradient between the points A
( 0,
A
( 0,
m=
−4 ) and B (1, −2 )
−4 ) → ( x1 , y1 ) and B (1, −2 ) → ( x2 , y2 )
y2 − y1
−2 − −4
2
→ m=
→ m=
→ m=2
x2 − x1
1− 0
1
Now let’s find the gradient between the points B (1, −2 ) C ( 2, 0 )
B (1, −2 ) → ( x 1 , y1 ) C
m=
( 2, 0 ) → ( x2 ,
y2 )
y2 − y1
0 − −2
2
→ m=
→ m= → m=2
x2 − x1
2 −1
1
Since the gradients are same, points A ( 0, −4 ) B (1, −2 ) C ( 2, 0 ) are collinear.
EXAMPLE 200
Show that the following points are collinear.
A ( −2, 10 ) B ( 0, 0 ) C (1, −5)
Let’s find the gradient between the points A ( −2, 10 ) and B ( 0, 0 )
A ( −2, 10 ) → ( x1 , y1 ) and B ( 0, 0 ) → ( x2 , y2 )
m=
y2 − y1
0 − 10
−10
→ m=
→ m=
→ m = −5
x2 − x1
0 − −2
2
Now let’s find the gradient between the points B ( 0, 0 ) C (1, −5)
B ( 0, 0 ) → ( x 1 , y1 ) C
m=
(1,
−5 ) → ( x2 , y2 )
y2 − y1
−5 − 0
−5
→ m=
→ m=
→ m = −5
x2 − x1
1− 0
1
Since the gradients are same, points A ( −2, 10 ) B ( 0, 0 ) C (1, −5) are collinear.
129
EXAMPLE 201 If the points A ( x, 10 ) , B ( 0, 0 ) and C (1, −5) , lie on the
same line, find the value of x.
Since the points are collinear; the gradient between the points A and B will be same as
the gradient between the point B and C.
•
Let’s first find the gradient between the points B and C.
B ( −1, −3) → ( x1 , y1 ) and C (1, 3) → ( x2 , y2 )
m=
•
y2 − y1
3 − −3
6
→ m=
→ m= → m=3
x2 − x1
1 − −1
2
Now us the gradient to solve for x.
A ( x, 6 ) → ( x1 , y1 ) and B ( −1, −3) → ( x2 , y2 )
m=
y2 − y1
−3 − 6
−9
→ 3=
→ 3=
→ 3 ( −1 − x ) = −9
x2 − x1
−1 − x
−1 − x
−3 − 3x = −9 → −3x = −9 + 3 → −3x = −6 → x = −6 ÷ −3 → x = 2
EXAMPLE 202 If the points A ( −1, −2 ) , B ( 0, y ) and C ( −2, 0 ) , find the
value of x.
Since the points are collinear; the gradient between the points A and B will be same as
the gradient between the point B and C and hence, the gradient between the points A
and C will be same.
•
Let’s first find the gradient between the points A and C.
A ( −1, −2 ) → ( x1 , y1 ) and C ( −2, 0 ) → ( x2 , y2 )
m=
•
y2 − y1
0 − −2
2
→ m=
→ m=
→ m = −2
x2 − x1
−2 − −1
−1
Now us the gradient to solve for y.
A ( −1, −2 ) → ( x1 , y1 ) and B ( 0, y ) → ( x2 , y2 )
m=
y2 − y1
y − −2
y+2
→ −2 =
→ −2 =
→ −2 = y + 2
x2 − x1
0 − −1
1
−2 = y + 2 → −2 − 2 = y → −4 = y → y = −4
130
Example 203
Now let us return to example 198. Instead of finding gradient
between points to show that 3 points are collinear, we will use
distance formula to demonstrate collinear points.
Use the distance method or distance formula to show that the points A(1, 2), B(3, 4)
and C(5, 6) are collinear points.
If points A, B and C are collinear then it follows that the sum of distance between the
points A to B and B to C is exactly equal to the distance between points A to C.
•
Firstly, find the distance between the points A to B.
A (1, 2 ) and B ( 3, 4 )
d=
( x2 − x1 ) + ( y2 − y1 )
A ( x1 , y1 ) and B ( x2 , y2 )
d=
( 3 − 1) + ( 4 − 2 )
d=
( 2) + ( 2)
d 2 = ( x2 − x1 ) + ( y2 − y1 )
2
•
2
2
2
2
2
2
→
d= 8
Secondly, find the distance between the points B to C.
B ( 3, 4 ) and C ( 5, 6 )
d=
( x2 − x1 ) + ( y2 − y1 )
B ( x1 , y1 ) and C ( x2 , y2 )
d=
( 5 − 3) + ( 6 − 4 )
d=
( 2) + ( 2)
d 2 = ( x2 − x1 ) + ( y2 − y1 )
2
•
2
2
2
2
2
→
d= 8
Thirdly, find the distance between the points A to C.
d=
( x2 − x1 ) + ( y2 − y1 )
B ( x1 , y1 ) and C ( x2 , y2 )
d=
( 5 − 1) + ( 6 − 2 )
d=
( 4) + ( 4)
d 2 = ( x2 − x1 ) + ( y2 − y1 )
2
2
2
2
2
2
→
2
2
d = 32
Now show that the sum of the distance between the points A to B and
B to C is equal to the distance between points A to C.
AB + BC = 8 + 8 = 2 8
•
2
2
A (1, 2 ) and C ( 5, 6 )
•
2
AC = 32 = 4 × 8 = 2 8
Since the sum of the distance between the points A to B and B to C is
exactly equal to the distance between the points A to C, points A, B
and C are collinear points.
131
EXAMPLE 204
Use the equation of a straight line passing through two points method to show that the
points A(1, 2), B(3, 4) and C(5, 6) are collinear points.
•
Let’s find the equation of the straight line joining the points A(1, 2)
and B(3, 4).
A(1, 2)
y −y
3− 2
1
m= 2 1 → m=
→ m = → m =1
B(2, 3)
x2 − x1
2 −1
1
y = mx + c
y = x+c
•
→ 2 = 1+ c → c = 1
y = x +1
Let’s find the equation of the straight line joining the points B(3, 4)
and C(5, 6).
B(3, 4)
m=
C (5, 6)
y2 − y1
6−4
2
→ m=
→ m = → m =1
x2 − x1
5−3
2
y = mx + c → y = x + c → 4 = 3 + c → c = 1
•
Let’s find the equation of the straight line joining the points A(1, 2)
and C(5, 6).
A(1, 2)
C (5, 6)
m=
y2 − y1
6−2
4
→ m=
→ m = → m =1
5 −1
4
x2 − x1
y = mx + c → y = x + c → 2 = 1 + c → c = 1
•
y = x +1
y = x +1
Since the equations of the straight line passing through the points A to
B, B to C and A to C are exactly the same, points A, B and C are
collinear.
Note: To show that three points are collinear, you can use either of the examples:
•
•
•
•
Example 198
Example 203
Example 204
You can also find the equation of the line passing through the points
AC and then show that point B lies on it. You select the most
convenient method for you.
132
Parallel Lines
Two or more lines are considered to be parallel if and only if they meet three
conditions given below.
•
•
•
Their gradients are equal. This means the gradient of the first line m1
is equal to the gradient of the second line m2 .
They do not intersect at any point
Their separation distance is constant.
m1 = m2
EXAMPLE 205
The graphs of the function y = 3 x + 3 and y = 3 x − 3 are parallel to each other
because both graphs have the same gradient. m = 3
133
EXAMPLE 206
Find the equation of the straight line parallel to the graph of the function
y = 2 x + 2 and passes through the point ( 3, 5 ) .
Step 1
Since the parallel line in this case is a straight line, the standard equation
is y = mx + c.
Step 2
Now find the gradient of this parallel line. Remember the gradient of two parallel
lines is always equal to each other.
The gradient of the line y = 2 x + 2 is → m1 = 2
m1 = m2 → m2 = 2
Step 3
Now substitute this gradient m = 2 into the standard equation y = mx + c.
y = mx + c → y = 2 x + c
Step 4
You are given the information that the parallel line passes through the
point ( 3, 5 ) . Substitute these values for x and y.
( 3, 5)
( x1 , y1 )
→
y = mx + c
5 = 6+c
→ 5 = 2 ( 3) + c →
→ c = −1
y = 2x + c
5−6 = c
Step 5
Since you have now found the value of c, substitute this into the main equation of
step 3.
final Answer
y = 2 x + c 
→ y = 2x −1
134
EXAMPLE 207
Find the equation of the straight line parallel to the graph of the function
y = 4 x + 1 and passes through the point (2, 6 ) .
Equation of the straight line is y = mx + c and since it is parallel to y = 4 x + 1, m = 4.
y = 4 x + c It passes through the point (2, 6 ).
x=2
y = 6.
6 = 4 x + 2 [Substitute the x and y-values]
6 = 4(2 ) + c
6 =8+c
6−8 = c
c = −2
y = 4x − 2
EXAMPLE 208
Find the equation of the straight line parallel to the graph of the function
y = x + 1 and passes through the point (2, 6 ) .
Equation of the straight line is y = mx + c and since it is parallel to y = x + 1 , m = 1.
y = x + c It passes through the point (2, 6 ).
x=2
y = 6.
6 = 2 + c [Substitute the x and y-values]
6 = 2+c → 6−2 = c → 6−2 = c → c = 4 → y = x+4
EXAMPLE 209
Find the equation of the straight line parallel to the graph of the function
y = − x + 1 and passes through the point (2, 6 ) .
Equation of the straight line is y = mx + c and since it is parallel
to y = − x + 1 , m = −1.
y = − x + c It passes through the point (2, 6).
x=2
y = 6.
6 = −2 + c [Substitute the x and y-values]
6 = −2 + c → 6 + 2 = c → c = 8 → y = − x + 8
EXAMPLE 210
Find the equation of the straight line parallel to the graph of the function
y = 6 x + 1 and passes through the point ( −2, 0 ) .
Equation of the straight line is y = mx + c and since it is parallel
to y = 6 x + 1 , m = 6.
y = 6 x + c It passes through the point ( −2, 0 ) . x = −2 y = 0 .
0 = 6 ( −2 ) + c [Substitute the x and y-values]
0 = −12 + c → 0 + 12 = c → c = 12 → y = 6 x + 12
135
PERPENDICULAR LINES
m1m2 = −1
Two lines are perpendicular if and only if they meet the following criteria:
•
•
The product of their gradient is negative one ( m1 × m 2 = −1 ).
They meet at right angles.
EXAMPLE 211
The graphs shown above are perpendicular to each other. The gradient of the line
1
y = 2 x + 2 is 2 and the gradient of the line y = − x − 2 is – ½.
2
Now multiply both gradients.
2×
−1
=
2
2 −1
×
=
1
2
−2
2
= − 1 . Since the product of the gradients of both
1
graphs is -1, y = 2 x + 2 is perpendicular to y = − x − 2 .
2
136
EXAMPLE 212
1
Find the equation of the straight line perpendicular to y = − x + 1 and passes
2
through the point ( 0, 2 ) .
Step 1
Since perpendicular lines are also straight, their standard equation is y = mx + c.
Step 2
Now find the gradient of this perpendicular line. Remember the product of two
perpendicular lines is always -1.
1
1
The gradient of the line y = − x + 1 is → m1 = −
2
2
−1m2 = −2
1
− × m2 = −1
m1 × m2 = −1 → 2
→
→ m2 = 2
−2
m2 =
−1m2 = −1× 2
−1
Step 3
Now substitute this gradient m = 2 into the standard equation y = mx + c.
y = mx + c → y = 2 x + c
Step 4
You are given the information that the perpendicular line passes through the
point ( 0, 2 ) . Substitute these values for x and y.
( 0, 2 ) → ( x1 , y1 )
→
y = mx + c
y = 2x + c
→
2 = 2 ( 0) + c
→ c=2
2 = 0+c
Step 5
Since you have now found the value of c, substitute this into the main equation of
step 3.
final Answer
y = 2 x + c 
→ y = 2x + 2
137
Example 213
Find the equation of the straight line perpendicular to
y = −2 x + 1 and passes through the point ( 4, 2 ) .
m1 × m2 = −1 → −2 × m2 = −1 → m2 =
y = mx + c → y =
−1
1
→ m2 =
−2
2
1
x
x+c → y = +c
2
2
( 4, 2 ) → ( x1 , y1 )
2 = 2+c
x
4
+c → 2= +c →
→ c=0
2−2 = c
2
2
y = mx + c → y =
x
x
x
c=0
answer
y = + c 
→ y = + 0 → y =
2
2
2
Example 214
Find the equation of the straight line perpendicular to
2 y = − 8 x + 2 and passes through the point ( 4, 2 ) .
2 y = − 8x + 2 → y =
−8 x 2
+ → y = − 4x + 1
2
2
m1 × m2 = −1 → − 4 × m2 = −1 → m2 =
y = mx + c → y =
−1
1
→ m2 =
−4
4
1
x
x+c → y = +c
4
4
( 4, 2 ) → ( x1 , y1 )
y = mx + c → y =
y=
2 = 1+ c
x
4
+c → 2= +c →
→ c =1
2 −1 = c
4
4
x
x
x
c =1
answer
+ c 

→ y = + 1 
→ y = +1
2
2
2
138
Solving Equations Simultaneously
At the point of intersection, both graphs have the same x and y values. Taking this
into consideration, we can solve both equations simultaneously by making y = y
because at any point of intersection, both graphs will have the same y value.
EXAMPLE 215
Find the point of intersection of the graphs y = 2 x + 4 and y = − x + 1
y = 2x + 4 y = − x + 1
At the point of intersection both graphs has the same y − value
y=y
−3
→ 2 x + x = 1 − 4 → 3 x = −3 → x =
→ x = −1
2x + 4 = −x +1
3
Substitute the x − value in either of the equations to find the y − value.
For this example, x substitution is done for both equations to confirm
to you that both equations will give same y − value.
y = 2x + 4
y = −x +1
y = 2 ( −1) + 4
y = − ( −1) + 1
y = −2 + 4
y = 1+1
y=2
y=2
Answer : {−1, 2}
We can either solve two equations simultaneously to find their point/s of
intersection or graphing both equations and then determining from the graphs
their point/s of intersection.
Compare example 215 with example 155
139
EXAMPLE 216
To solve the above graphs we will need to make sure that y is the subject in both
equations.
3 y = x + 1 and 2 y − 4 x − 4 = 0
3y = x +1 → y =
x +1
3
2 y − 4x − 4 = 0 → → 2 y = 4x + 4 → y = 2x + 2
y=
x +1
3
y = 2x + 2
y=y
x +1
x +1 = 6x + 6
x = 6x + 5
−5 x = 5
= 2x + 2
→
→
→
3
x = 6x + 6 −1
x − 6x = 5
x = −1
x + 1 = 3(2 x + 2)
y = 2 x + 2 → y = 2( −1) + 2 → y = −2 + 2 → y = 0
Point of intersection is (-1, 0)
140
EXAMPLE 217 Find the points of intersection of the graphs
y = x 2 − 4 and
y = x+2.
factorization
y=y
x → −3
x→ 2
x2 − 4 = x + 2
x2 − 4 − x − 2 = 0
→ −3 x + 2 x = − x ← middle term
x2 − x − 4 − 2 = 0
x2 − x − 6 = 0
( x − 3)( x + 2 ) = 0
x −3 = 0 x + 2 = 0
x=3
x = −2
Since there are two x values we can conclude that both graphs intersect at two
different points.
x =3 → y = x+2
x = −2 → y = x + 2
y = 3+ 2
y = −2 + 2
y = 5 → ( 3, 5 )
y = 0 → ( −2, 0 )
141
EXAMPLE 218 Find the points of intersection of the graphs
y = x 2 − 4 and y = x − 2
factorization
x → −2
y=y
x→1
x −4 = x−2
2
x −4− x+2 = 0
2
→
x − x−4+2 = 0
2
x −x−2=0
2
•
−2 x + 1x = − x ← middle term
( x − 2 )( x + 1) = 0
x−2=0
x +1 = 0
x = 0+2
x=2
x = 0 −1
x = −1
Since there are two x values we can conclude that both graphs intersect
at two different points.
x = 2 → y = x−2
x = −1 → y = x − 2
y = 2 − 2 = 0 → ( 2, 0 )
y = −1 − 2 = −3 → ( −1, − 3 )
142
EXAMPLE 219
Find the point of intersection of the graph
y = x 2 + 3x + 2 and
y = x + 1.
At the point of intersection, both graphs will have the same y-value. Hence, we can
solve both equations simultaneously.
Now solve x 2 + 2 x + 1 = 0
y=y
x → 1 1x
x 2 + 3x + 2 = x + 1
x 2 + 3x + 2 − x − 1 = 0
x 2 + 3x − x + 2 − 1 = 0
x2 + 2x + 1 = 0
→
x → 1 1x 1x + 1x = 2 x ←
middle
term
x 2 + 2 x + 1 = ( x + 1)( x + 1) = 0
x + 1 = 0 → x = 0 − 1 → x = −1
To find the y value, use any of the given equations.
y = x + 1 → y = −1 + 1 → y = 0
Point of intersection →
Given below is the graph of y = x 2 + 3x + 2 and
of intersection is represented by a thick dot.
143
( −1, 0 )
y = x + 1. In the graph, the point
EXAMPLE 220 Find the points of intersection of x 2 + y 2 = 52 and y = x + 1 .
x 2 + y 2 = 52 y = x + 1
x 2 + y 2 = 25
x 2 + ( x + 1) = 25
2
2
2
 Substituted y = x + 1 in x + y = 25
→ x 2 + x 2 + 2 x + 1 = 25 [ Expanded ]
2 x 2 + 2 x + 1 = 25 [ Simplified ]
To solve for x, take 25 to the left hand side so that the 
equation can be made equal to zero



Factorise → 2 x 2 + 2 x − 24 = 0
2 x 2 + 2 x + 1 − 25 = 0
2 x 2 + 2 x − 24 = 0
2 x 2 + 2 x − 24 = 0
→ 2x → 4
x → −6 8 x − 6 x = 2 x ← middle term
( 2 x − 6 )( x + 4 ) = 0
either 2 x − 6 = 0 or x + 4 = 0
2x − 6 = 0 2x = 0 + 6 2x = 6 x = 6 ÷ 2 x = 3
x +4 = 0 x = 0−4 x = −4
[ Substitute
the x − value into y = x + 1]
y = x +1
y = x +1
y = 3 +1
→ y=4
y = −4 + 1
→ y = −3 ⇒
Points of intersections
(3, 4) and (−4, −3)
The above functions are drawn below showing their points of intersections.
144
EXAMPLE 221 Find the points of intersection of x 2 + y 2 = 52 and y = x .
 Substitute y = x into x 2 + y 2 = 25
x 2 + y 2 = 52
→
2
y=x
x 2 + ( x ) = 25 → 2 x 2 = 25
Now solve for x
2 x 2 = 25
x 2 = 25 ÷ 2
→
x 2 = 12 ⋅ 5
x = ∓ 12 ⋅ 5
→ x = ±3 ⋅ 54
Substitute the x − value in the equation y = x
x = 3 ⋅ 54
y=x
y = 3 ⋅ 54
x = −3 ⋅ 54
y=x
y = −3 ⋅ 54
Points of intersections
(3 ⋅ 54, 3 ⋅ 54) and (−3 ⋅ 54, − 3 ⋅ 54)
145
1
x
x
4
EXAMPLE 222 Find the points of intersection of y = and y = .
y=
1
x
y=
1
x
4
Solve both equations simultaneously
y=y
1 x
2
1 1 → = → 1× 4 = x × x → 4 = x
x 4
= x
x 4
4 = x2 → x2 = 4 → x = ± 4 → x = ± 2
Substitude the x value into any of the equations
x=2
y=
1
1
→ y=
2
x
x = −2
y=
1
1
→ y=−
2
x
1
 1

points of intersection →  2,  and  −2, − 
 2

2
146
1
x
EXAMPLE 223 Find the points of intersection of y = and y = x .
y=
1
x
y=x
Solve both equations simultaneously
y=y
1
→ = x → 1 = x × x → 1 = x2
1
x
=x
x
1 = x2 → x2 = 1 → x = ± 1 → x = ±1
Substitude the x value into any of the equations
x =1
y=
1
→ y =1
1
x = −1
y=−
1
→ y = −1
1
points of intersection → (1, 1) and ( −1, − 1)
147
The glory of God is intelligence, or, in other words, light and truth.
(Doctrine and Covenants | Section 93:36)
Exercise 4
1. Find the point of intersection of the graphs y = x + 4 and y = − x + 2
2. Find the point of intersection of the graphs y = 2 x + 1 and y = − x + 10
3. Solve the equations 2 y = 4 x + 4 and y + 2 x − 10 = 0 simultaneously.
4. Find the point of intersection of the graphs 5 y = −2 x + 2 and 9 y − 4 x − 4 = 0
5. Find the point of intersection of the graphs 3 y − 2 x + 1 = 0 and 4 y − 4 x − 4 = 0
6. Find the point of intersection of the graphs y = x + 1 and 6 y − 4 x − 2 = 0
7. Find the point/s of intersection of the graphs
y = x 2 + 3x + 2 and y = x + 2
8. Find the point/s of intersection of the graphs
y = x 2 + x and y = x + 1.
9. Find the point/s of intersection of the graphs
y = x 2 + 4 x + 4 and y = − x − 1.
10. Find the point of intersection of the graphs y − 3 =
3
3
( x + 1) and y − 3 = − ( x + 1)
2
2
11. Find the point/s of intersection of the graphs y = x 2 + 5 x and y = 4 x + 2
12. Find the point/s of intersection of the graphs y = x 2 + 2 x + 1 and y = x + 5
13. Find the point/s of intersection of the graphs x 2 + y 2 = 25 and y = x − 1.
14. Find the point/s of intersection of the graphs x 2 + y 2 = 4 and y = 3 x.
15. Find the point/s of intersection of the graphs x 2 + y 2 = 2 and y + x − 2 = 0.
16. Find the points of the intersection of the graphs y =
17. Find the point/s of intersection of the graphs y =
1
and y = x
x
1
and y = 2 x + 1
4x −1
18. Find the point/s of intersection of the graphs x 2 + y 2 = 8 and y = x.
148
5
Withhold not good from them to whom it is due, when it is in the
power of your hand to do it. (Old Testament | Proverbs 3:27)
TRIGONOMETRY
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
♣
♣
♣
Right angle triangle
Sine graphs
Cosine graphs
Solving trigonometric functions
Calculation of angles and sides for non right angle
triangle.
Sine and Cosine Rule
Arc Length
Area of segment
Area of sector
Area of non right angle triangle
149
Right Angle Triangle
Angle s and sides in any right angle triangle is found by using the soh, cah and toa
ruled learned in form 5.
soh
Sine θ =
cah
opposite
hypotenuse
Cosine θ =
toa
adjacent
hypotenuse
Tangent θ =
opposite
adjacent
EXAMPLE 224 Find the value of the angle θ in the diagram given below.
Sine θ =
opposite
hypotenuse
Sine θ =
3
 3
→ θ = Sin −1   → θ ≈ 36 ⋅ 870
5
5
or
Cosine θ =
adjacent
4
→ cos θ =
5
hypotenuse
4
 
θ = cos −1   → θ ≈ 36 ⋅ 87 0
5
Trigonometry Identity
sin θ =
3
4
and cos θ =
5
5
2
( sin θ )
2
9
3
and
=   → sin 2 θ =
25
5
2
( cos θ )
2
16
4
=   → cos 2 θ =
25
5
9
25
16
+ cos 2 θ =
25
sin 2 θ =
sin 2 θ + cos 2 θ =
9 16
25
+
→ sin 2 θ + cos 2 θ =
→
25 25
25
150
sin 2 θ + cos 2 θ = 1
SINE GRAPHS
y = ± A sin ( Bx ± C ) ± D
A is the amplitude
B is the number of periods in 360 0
Period =
360 0
B
C
C
is positive, move the y-axis to the right. If is
B
B
negative, move the y-axis to the left.
C is shift in the y-axis. If
D is the shift in the x-axis.
•
•
If D is positive, move the x-axis downward by D units.
If D is negative, move the x-axis upwards by D units.
EXAMPLE 225 Graph y = sin x
y = sin x
A =1
y = A sin ( Bx + C ) + D
0
B = 1 → Period =
151
0
360
360
=
= 3600
B
1
C =0
D=0
EXAMPLE 226
Graph y = 2 sin x
y = 2sin x
y = A sin ( Bx + C ) + D
A=2
B =1
3600
B
3600
=
= 3600
1
Period =
C =0
D=0
EXAMPLE 227 Graph y = 3 sin 3x
y = 3sin 3x
y = A sin ( Bx + C ) + D
A=3
B=3
3600 3600
Period =
=
= 1200
B
3
C =0
D=0
EXERCISE 5 A
In a few minutes, a computer can make a mistake so great that it would take many
men many months to equal it.
Merle L. Meacham
(0 ≤ x ≤ 360 )
(0 ≤ x ≤ 360 )
y = 2sin 2 x. ( 0 ≤ x ≤ 360 )
y = 2sin 3 x. ( 0 ≤ x ≤ 360 )
y = 2sin 9 x. ( 0 ≤ x ≤ 360 )
1. Sketch the graph of y = 4 sin x.
2. Sketch the graph of y = 4 sin 4 x.
3. Sketch the graph of
4. Sketch the graph of
5. Sketch the graph of
0
0
0
0
0
0
0
0
0
0
152
EXAMPLE 228 Sketch the graph of y = 3 sin (x + 30 0 )
(
y = 3sin x + 300
)
y = A sin ( Bx + C ) + K
A = 3 B =1
C = 300
Period =
3600 3600
=
= 3600
B
1
C 300
=
= 300
B
1
Shift y − axis 300 to the right
D=0
y = 3 sin (x + 30 0 ) Let’s first graph y = 3 sin x first
Now graph y = 3 sin (x + 30 0 ) . The graph of y = 3 sin x is shifted 30 0 to the left.
This is same as shifting y-axis to the right which is easier and is strongly
recommended.
153
Sketch the graph of y = 3 sin (2 x + 90 0 ) + 2
Example 229
y = 3sin 2 x
(
3600 3600
Period =
=
= 1800
A=3 B=2
B
2
This means one cycle is 1800
y = 3sin 2 x + 90
0
)
A=3
3600
B = 2 → Period =
= 1800
2
C 900
= 450
C = 90 → =
B
2
Move y − axis 450 towards right
0
Continued on the next page
154
Now graph y = 3 sin (2 x + 90 0 ) + 2
Since D = +2, the graph of y = 3 sin (2 x + 90 0 ) is shifted 2 units upwards. This is
same as shifting the x-axis 2 units downward. Shifting of the x-axis is recommended
because it is easier.
Negative sine graphs
They are inverted sine graphs or it is a reflection along the x axis of its positive graph
EXERCISE 5 B
No guts, no glory!
1. Sketch the graph of y = sin ( x + 450 )
2. Sketch the graph of
3. Sketch the graph of
4. Sketch the graph of
5. Sketch the graph of
( 0 ≤ x ≤ 360 )
y = − sin ( x + 45 ) ( 0 ≤ x ≤ 360 )
y = 2sin ( x − 45 ) − 2 ( 0 ≤ x ≤ 360 )
y = −2sin ( 2 x − 90 ) + 2 ( 0 ≤ x ≤ 360 )
π

(0 ≤ x ≤ 360 )
y = 3 sin  x −  + 2
4
0
0
0
0
0
0
0

0
0
0
0
0
0

π

6. Sketch the graph of y = −3 sin  x − 
(0 0 ≤ x ≤ 360 0 )
4


0
7. Sketch the graph of y = −2 sin (x + 90 ) − 2
(0 0 ≤ x ≤ 360 0 )
155
COSINE GRAPHS
y = A cos ( Bx + C ) + D
A is the amplitude
B is the number of periods in 360 0
Period =
360 0
B
C
C
is positive, move the y-axis to the right. If is
B
B
negative, move the y-axis to the left.
C is shift in the y-axis. If
D is the shift in the x-axis. If D is positive, move the x-axis downward by D units. If
D is negative, move the x-axis upwards by D units.
EXAMPLE 230 Graph y = cos x
EXAMPLE 231 Graph y = 2 cos 2 x
y = 2 cos 2 x
A=2
B=2
3600
2
Period = 1800
Period =
156
EXAMPLE 232
A=2
B=2
Period =
C = 900
Graph y = 2 cos(2 x + 90 0 )
B 3600
=
= 1800
2
2
Shift y − axis to the right by 450
0
C 90
=
= 450
B
2
EXAMPLE 233 Graph y = 2 cos(2 x + 90 0 ) + 2
When the graph of y = 2 cos(2 x + 90 0 ) in example 233 is moved 2 units upwards the
resulting graph has the equation y = 2 cos(2 x + 90 0 ) + 2 . Since D = + 2, the graph of
y = 2 cos(2 x + 90 0 ) is shifted 2 units upwards. This is same as shifting the x-axis 2
units downward. Shifting of the x-axis is recommended because it is easier.
157
EXERCISE 5 C
Take a break but do not apply brake!
(0
1. Sketch the graph of y = 3cos ( x )
2. Sketch the graph of y = 2 cos ( 2 x )
≤ x ≤ 3600 )
0
(0
≤ x ≤ 3600 )
0
3. Sketch the graph of y = 2 cos ( 2 x − 900 )
4. Sketch the graph of y = −2 cos ( x )
(0
5. Sketch the graph of y = −2 cos ( 2 x )
(0
0
≤ x ≤ 3600 )
6. Sketch the graph of y = −2 cos ( 2 x − 900 )
7. Sketch the graph of y = 2 cos ( x + 2700 )
8. Sketch the graph of y = cos ( x + 900 )
≤ x ≤ 3600 )
≤ x ≤ 3600 )
0
(0
0
(0
(0
(0
0
9. Sketch the graph of y = 2 cos ( x + 900 ) − 2
0
≤ x ≤ 3600 )
0
≤ x ≤ 3600 )
≤ x ≤ 3600 )
(0
10. Sketch the graph of y = 2 cos(x − 90 0 ) + 2
(0
π

11. Sketch the graph of y = 3 cos x −  + 2
4

(0
π

12. Sketch the graph of y = −3 cos x − 
4

13. Sketch the graph of y = −2 cos(x + 90 0 ) − 2
158
(0
0
≤ x ≤ 3600 )
0
≤ x ≤ 360 0 )
0
0
≤ x ≤ 360 0
≤ x ≤ 360 0
(0
0
)
)
≤ x ≤ 360 0 )
SOLVING TRIGONOMETRICS
EXAMPLE 234
Solve 2 sin x = 1
(0
0
≤ x ≤ 360 0 )
This book will employ two methods of solving trigonometry equations. The two
methods are graphical and formula.
Graphical Method
1
1
2 sin x = 1 → sin x =   → x = sin −1   → x = 300
2
2
You can see from the sin x graph that y = 1 cuts the graph at two different points.
One at 30 0 and the other at 150 0 . So the solution set for 2 sin x = 1 is { 30 0 , 150 0 }
Quadrant Method
159
EXAMPLE 235
Solve 2 cos x = 1 ( 00 ≤ x ≤ 3600 )
2 cos x = 1 → cos x =
1
1
→ x = cos −1   → x = 600
2
2
{
Solution set → 600 ,3000
160
}
Formula Method-Sine Equation
For sine equations only
n
( Bx + C ) = (1800 ) n + ( −1) α
A → Coeffiecient α → angle
Example 236
Let’s verify the answers obtained in example 234 using the formula method of solving
trigonometric equations.
1
1
→ x = sin −1  
2
2
2sin x = 1 → sin x =
x = 300 → α = 300
(
)
Bx + c = 1800 n + ( −1) α
n
( )
( )
x = (180 ) ( 0 ) + ( −1) ( 30 )
x = 1800 n + ( −1) α
n
let n = 0 →
0
0
0
x = 300
( )
()
x = (180 ) (1) + ( −1) ( 30 )
x = 1800 n + ( −1) α
n
0
let n = 1 →
1
0
x = 1800 − 300
x = 1500
If we let n = 2, the answer will lie outside the boundary given in the question. Hence,
the solution set is {300 ,1500 }
161
Solve for sin 2 x = 0 ⋅ 5
Example 237
(0
0
≤ x ≤ 360 0 )
sin 2 x = 0 ⋅ 5 → 2 x = sin −1 ( 0 ⋅ 5 ) → 2 x = 300 → α = 300
note : ( Bx + c ) = α
2x = α
→ do not solve for x at this stage
2 x = α = 300
(
)
Bx + c = 1800 n + ( −1) α
(
)
n
2 x = 1800 n + ( −1) 300
n
let n = 0
2 x = 0 + (1) 300 → 2 x = 300 → x = 150 → Now you solve for x.
let n = 1
(
)
(1)
)
( 2)
)
( 3)
2 x = 1800 (1) + ( −1) 300 → 2 x = 1800 + ( −1) 300 → 2 x = 1500 → x = 750
let n = 2
(
2 x = 1800 ( 2 ) + ( −1) 300 → 2 x = 3600 + 300 → 2 x = 3900 → x = 1950
let n = 3
(
2 x = 1800 ( 3) + ( −1) 300 → 2 x = 5400 − 300 → 2 x = 5100 → x = 2550
Solution set
{ 15 , 75 , 195 , 255 }
0
0
0
0
Note that when n = 4, x = 375 0 . This lies outside of the boundary 0 0 ≤ x ≤ 360 0
So there is no need to continue after n = 3.
162
Example 238
This is a verification of example 237. This time graphical method is used. Graphical
method takes time and accuracy. Hence, formula method is recommended because it
takes less time and once you have mastered the formulas, you will surely get the
correct answers.
See carefully and you’ll notice that the line y = ½ cuts the graph of sin 2x at 4
different points.
Solution set
{ 15 , 75 , 195 , 255 }
0
0
0
0
Solve for sin x = 0 ⋅ 5 ( 00 ≤ x ≤ 3600 )
Example 239
sin x = 0 ⋅ 5 → x = sin −1 ( 0 ⋅ 5 ) → x = 300 → α = 300
note : ( Bx + c ) = α
(
)
→ Bx + c = 1800 n + ( −1) α
x =α
n
x = α = 300
(
)
x = 1800 n + ( −1) 300
n
let n = 1
let n = 0
x = 0 + (1) 30 → x = 30
0
(
)
(1)
x = 1800 (1) + ( −1) 300
0
x = 1800 − 300 → x = 1500
Solution set → {300 ,1500 }
When n ≥ 2 , the solution is outside the boundary of ( 00 ≤ x ≤ 3600 )
163
Example 240
Solve for sin ( 2 x − 180 ) = 0 ⋅ 5 ( 00 ≤ x ≤ 3600 )
(
)
sin 2 x − 1800 = 0 ⋅ 5 → 2 x − 1800 = sin −1 ( 0 ⋅ 5 ) → 2 x − 1800 = 300 → α = 300
note : ( Bx + c ) = α
2 x − 1800 = α
→ do not solve for x at this stage
2 x − 180 = α = 30
0
(
)
0
Bx + c = 1800 n + ( −1) α
(
n
)
2 x − 1800 = 1800 n + ( −1) 300
n
let n = 0
2 x − 1800 = 0 + (1) 300 → 2 x − 1800 = 300
2 x = 300 + 1800 → 2 x = 2100 → x = 2100 ÷ 2 → x = 1050
let n = 1
(
)
(1)
2 x − 1800 = 1800 (1) + ( −1) 300 → 2 x − 1800 = 1800 − 300
2 x − 1800 = 1500 → 2 x = 1500 + 180 → 2 x = 3300 → x = 3300 ÷ 2 → x = 1650
let n = 2
(
2 x − 1800 = 1800
) ( 2 ) + ( −1)( ) 30
2
0
→ 2 x − 1800 = 3600 + 300
2 x − 1800 = 3900 → 2 x = 3900 + 1800 → 2 x = 5700 → x = 5700 ÷ 2 → x = 2850
let n = 3
(
)
( 3)
2 x − 1800 = 1800 ( 3) + ( −1) 300 → 2 x − 1800 = 5400 − 300
2 x − 1800 = 5100 → 2 x = 5100 + 1800 → 2 x = 6900 → x = 6900 ÷ 2 → x = 3450
{
Solution set → 1050 ,1650 , 2850 ,3450
164
}
Formula Method-Cosine Equation
Cosine equations only
( Bx + c ) = ( 3600 ) n ± α
Solve 2 cos x = 1 ( 00 ≤ x ≤ 3600 )
Example 241
In this example, confirmation of the answers in example 233 will be done using the
formula method.
2 cos x = 1 → cos x =
1
1
→ x = cos −1   → x = 600
2
2
( Bx + c ) = ( 3600 ) n ± α
(
let n = 0
(
)
→ ( Bx + c ) = 3600 n ± 600
let n = 1
)
(
x = 3600 ( 0 ) ± 600 → x = 600
{
Solution set 600 , 3000
)
x = 3600 (1) ± 600 → x = 3000
}
When n = 0, x = −60 0 and 60 0 . When n = 1, x = 300 0 and 420 0 .
− 60 0 and 420 0 are not included because they are out of the boundary 0 0 ≤ x ≤ 360 0
Graphical Method
The line y = ½ cuts the graph of cos x at two different points. The points
are 60 0 and 300 0 .
165
Solve 2 cos(2 x + 90 0 ) = 1
EXAMPLE 242
(
)
(
)
2 cos 2 x + 900 = 1 → cos 2 x + 900 =
( 2 x + 90 ) = cos
0
−1
(0
0
≤ x ≤ 3600 )
1
1
→ 2 x + 900 = cos −1  
2
2
(
)
1
0
0
0
  → 2 x + 90 = 60 → α = 60
2
(
)
( Bx + c ) = ( 3600 ) n ± α
(
)
2 x + 900 = 3600 n ± 600
(
)
let n = 1 → 2 x + 900 = 3600 (1) ± 600 → 2 x + 900 = 3600 ± 600
2 x + 900 = 3600 + 600
2 x + 900 = 3600 − 600
2 x + 900 = 4200
2 x + 900 = 3000
2 x = 4200 − 900
2 x = 3000 − 900
2 x = 3300
2 x = 2100
x = 3300 ÷ 2
x = 2100 ÷ 2
x = 1650
x = 1050
(
let n = 2 → 2 x + 900 = 3600
) ( 2) ± 60
0
→ 2 x + 900 = 7200 ± 600
2 x + 900 = 7200 + 600
2 x + 900 = 7200 − 600
2 x + 900 = 7800
2 x + 900 = 6600
2 x = 7800 − 900
2 x = 6600 − 900
2 x = 6900
2 x = 5700
x = 6900 ÷ 2
x = 5700 ÷ 2
x = 3450
x = 2850
Solution set {1050 , 1650 , 2850 , 3450 }
166
Formula Method-Tangent Equation
Tangent
Bx + c = (180 ) n + α
(0
Solve tan 3x = 1
Example 243
0
≤ x ≤ 180 0 )
tan 3 x = 1 → 3 x = tan −1 (1) → 3x = α = 450
(
)
(
)
Bx + C = 1800 n + α → 3x = 1800 n + 450
Let n = 0
Let n = 1
Let n = 2
( )
3x = (180 ) ( 0 ) + 45
( )
3 x = (180 ) (1) + 45
( )
3x = (180 ) ( 2 ) + 45
3x = 0 + 450
3 x = 1800 + 450
3x = 3600 + 450
3x = 450
3 x = 2250
3x = 4050
x = 450 ÷ 3
x = 2250 ÷ 3
x = 4050 ÷ 3
x = 150
x = 750
x = 1350
Let n = 3
Let n = 4
( )
3x = (180 ) ( 3) + 45
3 x = 1800 n + 450
3x = 1800 n + 450
0
3 x = 1800 n + 450
0
3x = 1800 n + 450
0
0
0
0
( )
3 x = (180 ) ( 4 ) + 45
0
3x = 1800 n + 450
0
0
3x = 5400 + 450
3 x = 7200 + 450
3x = 5850
3 x = 7650
x = 5850 ÷ 3
x = 7650 ÷ 3
x = 1950
x = 2550 → outside the boundary
{
Solution set → 150 , 750 ,1350 ,1950
167
}
0
Solve tan ( x + 450 ) = 1 ( 00 ≤ x ≤ 1800 )
Example 244
(
)
(
)
(
)
tan x + 450 = 1 → x + 450 = tan −1 (1) → x + 450 = α = 450
(
)
(
) (
)
Bx + C = 1800 n + α → x + 450 = 1800 n + 450
Let n = 0
Let n = 1
( x + 45 ) = (180 ) n + 45
( x + 45 ) = (180 ) ( 0 ) + 45
( x + 45 ) = 0 + 45
( x + 45 ) = 45
( x + 45 ) = (180 ) n + 45
( x + 45 ) = (180 ) (1) + 45
( x + 45 ) = 180 + 45
( x + 45 ) = 225
x = 450 − 450
x = 2250 − 450
x = 00
x = 1800
0
0
0
0
0
0
0
0
0
0
{
Solution set → 00 ,1800
0
0
0
0
0
0
0
0
0
0
0
}
Solve tan ( x + 300 ) = 1 ( 00 ≤ x ≤ 1800 )
Example 245
(
)
(
)
(
)
tan x + 300 = 1 → x + 300 = tan −1 (1) → x + 300 = α = 450
(
)
(
) (
)
Bx + C = 1800 n + α → x + 300 = 1800 n + 450
Let n = 0
Let n = 1
( x + 30 ) = (180 ) n + 45
( x + 30 ) = (180 ) ( 0) + 45
( x + 30 ) = 0 + 45
( x + 30 ) = 45
( x + 30 ) = (180 ) n + 45
( x + 30 ) = (180 ) (1) + 45
( x + 30 ) = 180 + 45
( x + 30 ) = 225
x = 450 − 300
x = 2250 − 300
x = 150
x = 1950 → outside the boundary
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
{ }
Solution set → 150
168
0
0
0
SINE RULE
a
b
c
=
=
Sin A Sin B SinC
or
Sin A Sin B Sin C
=
=
a
b
c
Not all triangles are right angle triangles where soh, cah and toa can be used. For
triangles that are not right angles, sine and cosine rule must be used.
Capital letters denote the angles while the corresponding same smaller letter denotes
the opposite side.
Example 246
Find the value of the side marked a and b. [diagram not to scale]
a
b
c
=
=
sin A sin B sin C
a
b
c
=
=
sin A sin B sin C
a
b
4⋅6
=
=
0
sin 32
sin B sin 260
5 ⋅ 56
b
4⋅6
=
=
0
sin 32
sin B sin 260
a
4⋅6
=
0
sin 32
sin 260
5 ⋅ 56
b
=
0
sin 32
sin1220
a=
4⋅6
× sin 320
0
sin 26
5 ⋅ 56
× sin1220 = b
0
sin 32
a = 5 ⋅ 56 ( 2dp )
b = 8 ⋅ 9 (1dp )
169
Example 247
Example 248
Find the value of the angle B
Find the value of the angle A
Sin A Sin B Sin C
=
=
a
b
c
Sin A Sin B Sin 920
=
=
4
5
b
0
Sin A Sin 92
=
4
5
 Sin 920 
Sin A = 
× 4
5


Sin A
=
a
Sin A
=
a
Sin B
=
4
Sin B Sin C
=
b
c
Sin B Sin 800
=
4
7 ⋅ 88
0
Sin 80
7 ⋅ 88
 Sin 800 
Sin B = 
×4
 7 ⋅ 88 
Sin A = 0 ⋅ 7999512661
Sin B = 0 ⋅ 4999
A = Sin −1 ( 0 ⋅ 799512661)
B = Sin −1 ( 0 ⋅ 4999 )
A = 53 ⋅ 080
B = 29 ⋅ 990
Students please remember that the next chapter will show how to use the cosine rule
to find the value of angles and measurement of sides. You are required to correctly
determine the right equation to use to solve such problems. Given below is the
criterion that requires the use of sine rule.
Sine Rule:
•
•
If you are asked to solve for length of a side: any two angles and side
length is given.
If you are asked to solve for an angle: any other angle and the side
lengths opposite these angles are given.
Cosine Rule:
• If you are asked to find a side length: two other side lengths and the
included angle are given.
• If you are asked to find angle: three sides are given.
It is recommended that you know and understand the above criterion. However, if
your cognitive and IQ is not that powerful and remembering the above criterion is
difficult, I suggest that you start with the Sine rule and if the sine rule does not work
than you can be assured that the cosine rule will definitely work. This suggestion is a
worthwhile one and I only recommend this to make mathematics easier. When you
face difficulties in mathematics, there is always a smart way that we can all use to
enjoy mathematics despite the vast difference in our cognitive abilities.
170
COSINE RULE
a 2 = b 2 + c 2 − 2bcCos A
b 2 = a 2 + c 2 − 2acCosB
c 2 = a 2 + b 2 − 2abCosC
Example 249
Example 250
Find the value of the angle A.
Find the value of the side marked a.
a 2 = b 2 + c 2 − 2bcCosA
4 2 = 5 2 + 6 2 − 2(5)(6 )CosA
16 = 25 + 36 − 60 Cos A
16 = 61 − 60 Cos A
16 − 61 = −60 Cos A
− 45 = −60 Cos A
a 2 = b 2 + c 2 − 2bc CosA
− 45
= Cos A ⇒ 0 ⋅ 75 = Cos A
− 60
Cos −1 (0 ⋅ 75) = A ⇒ A = 41 ⋅ 410
a 2 = 8 ⋅ 2215
a 2 = 5 2 + 7 2 − 2(5)(7 )Cos 20 0
a 2 = 25 + 49 − 70 Cos 20 0
a 2 = 74 − 70Cos 20 0
a 2 = 74 − 65 ⋅ 7785
a = 8.2215
a = 2 ⋅ 87
Recall the suggestion given on the previous page. So let’s begin both examples with
the sine rule.
Sin A Sin B Sin C
=
=
a
b
c
a
b
c
=
=
Sin A Sin B Sin C
Sin A Sin B Sin C
=
=
4
5
6
a
5
7
=
=
0
Sin B Sin C
Sin 20
You see in both examples, you can not make any further progress using the Sine rule.
Hence you can be assured that the Cosine rule will work and this is all ready done.
171
If I rest, I rust.
Martin Luther King
EXERCISE 5
1. Use the Sine Rule to find the value of the side marked a.
2. Use the Sine rule to find the value of the angle B
3. For the triangles given below use either the Sine or Cosine rule to find the marked
angles and sides.
4. Find the length of the sides marked x and y.
172
Circular Measure
The sum of the angles in any circle is 360 0 .
The distance around the circle is called
perimeter and since entire perimeter is
circular, it is commonly known as the
circumference. C = 2 π r
Of all two dimensional figures having the
same perimeter, the circle is known to have
the greatest area.
A = π r2
Note: when θ = 2π , S is the
circumference C. Hence, the length
of S ∝ θ .
S
θ
=
C 2π
θ
×C
2π
θ 2π R
S=
×
2π
1
S = Rθ
S=
2π = 3600 → π = 1800
3600 = 2π
π radians = 1800
1800 = π
1 radian =
1800
10 =
π
π
1800
Example 251 Convert 450 to radians.
1800 = π → 10 =
π
1800
450 =
π  450  π

 = radians
1800  1  4
Hence, 450 in radians is
173
π
4
C = 2π R
in circular measure is
π 

→1800
Example 252
π
Convert 600 into radians.
π
600
→ 60 =
×
→ 600 = 1 ⋅ 047 radians
1 =
0
0
180
180
1
0
0
ARC LENGTH
EXAMPLE 253 Find the arc length (S) given that that the radius is 5 cm and the
angle theta equal to 30 degrees.
S = Rθ → R = 5cm
S = 5 cm × 0 ⋅ 0 ⋅ 5235988
π
300
×
θ = 30 ⇒
1800 1
0
→
S = 2 ⋅ 62 cm
300 = 0 ⋅ 5235988 radian
EXAMPLE 254 Find the arc length (S) given that that the radius is 2 cm and the
angle theta equal to 20 degrees.
S = Rθ
π
200
200 = 0 ⋅ 35 radian
R = 2 cm θ = 20 ⇒
×
0
180
1
0
S = 2 cm × 0 ⋅ 35 S = 0 ⋅ 7cm
EXAMPLE 255 Find the arc length (S) given that that the radius is 10 cm and
the angle theta equal to 60 degrees.
S = Rθ → R = 10 cm
θ = 600 ⇒
π
180
×
0
600
1
600 = 1 ⋅ 04719755 radian
174
S = 10 cm ×1 ⋅ 04719755
→
S = 10 ⋅ 47 cm
AREA OF A SECTOR
A sector is portion of a circle. A sector is enclosed by two radiuses and an arc that is
part of the circumference.
Area =
1 2
Rθ
2
Given that θ = 20 0 and R = 2 cm , find the area of the shaded
region which is the area of the sector of a circle with radius equal to 2 cm.
Example 256
A=
1 2
R θ → θ = 0 ⋅ 35 radian
2
A=
1 2
( 2 ) ( 0 ⋅ 35) → A = 0 ⋅ 7cm2
2
Example 257
[ from
example 254]
In this example, the solution of exampled 256 is verified.
Area of the circle = π r 2
A = π ( 2)
2
A = 4π
200
200
1
20 → is
part of the circle →
→
0
0
360
360
18
0
1
1 4π
4π
2π
of 4π →
×
→
→
= 0 ⋅ 7cm 2
18
18 1
18
9
175
EXAMPLE 258
Find the area of the sector.
1 2
Rθ
θ = 35 0
2
1
2
A = (8 ⋅ 9 ) (0 ⋅ 61)
2
A = 24 ⋅ 2 cm 2
A=
θ=
π
180 0
×
35 0
1
θ = 0 ⋅ 61
Example 259
In this example the solution of example 258 is verified using the ratio method.
Area of the circle = π r 2
A = π (8 ⋅ 9 )
2
A = 79 ⋅ 21π
350
350
7
35 → is
→
→
part
of
the
circle
3600
3600
72
0
7
7 79 ⋅ 21π
554 ⋅ 47π
×
→
of 79 ⋅ 21π →
72
72
1
72
A = 24 ⋅19 cm2 = 24 ⋅ 2cm2
176
AREA OF A TRIANGLE
Area =
1
ab sin C
2
Area =
1
ac sin B
2
1
Area = bc sin A
2
EXAMPLE 260
Find the area of the shaded region. The curve AB is part of a circle with r = 10 cm.
1
a b Sin C
2
1
A = (10cm )(10cm ) Sin 32 0
2
A = 26 ⋅ 50 cm 2
A=
EXAMPLE 261
Find the area of the shaded region. The curve AB is part of a circle with r = 6 cm.
1
ab sin C
2
1
Area = ( 6cm )( 6cm ) sin 300
2
Area =
Area = 9cm 2
EXAMPLE 262
Find the area of the shaded region. The curve AB is part of a circle with r = 12 cm.
1
ab sin C
2
1
Area = (12cm )(12cm ) sin 300
2
Area =
Area = 36cm 2
177
AREA OF A SEGMENT
From the previous page you have learned that the area of any non-right angle triangle
1
can be found by using the formula A = a b Sin C . You have also seen that both a
2
and b have the same value because they are part of a circle with a common radius.
Hence, sides a and b are radius of a circle.
A=
1
abSin C If a and b are radius of a circle then a = r and b = r.
2
Hence, A =
1
1
1
abSinC can be written as A = ( r )( r ) SinC → A = r 2 SinC
2
2
2
Area of Segment = Area of the Sector – Area of the triangle
1 2
1
r θ − r 2 Sin C
2
2
1
Area of segment = r 2 (θ − Sin C )
2
∠C = θ
1
Area of segment = r 2 (θ − Sin θ )
2
Area of Segment =
EXAMPLE 263 Find the area of the sector
1
Areasector = r 2θ
2
A=
θ = 300 ⇒
π
180
× 30 ⇒ θ = 0 ⋅ 524
1
2
(10 ) ( 0 ⋅ 524 ) → A ≈ 26 ⋅ 2 cm2
2
EXAMPLE 264 Find the area of the triangle.
1
a b Sin C
2
1
Area of ∆ = (10 )(10 )Sin 30 0
2
Area of ∆ = 25 cm 2
Area of ∆ =
Example 265 Find the area of the shaded region.
Area of segment = Area of sec tor − Area of ∆
Area of segment = 26 ⋅ 18 cm 2 − 25 cm 2
Area of segment = 1 ⋅ 18 cm 2
178
EXAMPLE 266 Find the area of the shaded segment
First find the area of the sector
A=
1 2
Rθ
2
θ=
π
60 π
60π
×
→
→ radian
1 180
180
3
A=
1
2 π
( 20 )   → A = 209 ⋅ 4395 cm2
2
3
Now find the area of the triangle ABC
1
ab sin C
2
1
Area = ( 20cm )( 20cm ) sin 600
2
Area =
Area = 173 ⋅ 2050808cm 2
Area of segment = Area of sector – Area of triangle
209 ⋅ 4395102 cm 2 − 173 ⋅ 2050808cm 2 ≈ 36 ⋅ 23 cm 2
179
Nothing worthwhile comes easily….Work, continuous work and hard work, is the
only way to accomplish results that last. [Hamilton Holt]
EXERCISE 5
Figure 1
Figure 2
1. For this question refer to figure 1. 0 is the centre of the circle and diameter of
the circle is 12 cm.
A. Find the area of the sector.
B. Find the area of the triangle
C. Find the area of the shaded segment.
2. For this question refer to figure 2. 0 is the centre of the circle with the radius
of the length 20 cm.
A. Find the length of the line AB which is labelled as x.
B. Find the area of the sector
C. Find the area of the triangle
D. Find the area of the shaded region
3. The triangle 0BC is an equilateral triangle with sides 4 cm and the radius of the
circle is 2 cm.
A. Find the area of the ∆0 AD
B. Find the area of the sector
C. Find the area of the ∆0 BC
D. Find the area of the shaded region.
E. Find the length of the arc AD.
F. Find the perimeter of the ABCD.
180
6
If my people, which are called by my name, shall humble themselves,
and pray, and seek my face, and turn from their wicked ways; then will
I hear from heaven, and will forgive their sin, and will heal their land.
(Old Testament | 2 Chronicles 7:14)
PROBABILITY
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
Probability of using fair coins
Probability dealing with dice
Mutually exclusive events
Independent events
Probability tree and sample space
Percentages and probability
Vann diagrams
181
Definition of Probability
•
Probability is the measure of how certain it is for an event will occur.
Probability =
Number of favourable outcomes
Total number of all possible outcomes
A coin has two sides and all fair coins will have head on one side and tail on the other
side. Since a coin had two sides, the total number of possible outcome is 2. If you
toss a coin and would like to now the probability of getting head on the coin, then you
divide the number of favourable outcome with the total number of all possible
outcome.
The favourable outcome in the above case is head and there is only one head on any
given fair coin and hence, the number of favourable outcome is 1. Since a coin has
two sides, the total number of all possible outcomes is 2.
Probability of head =
1
2
A fair die has 6 sides with a number on each side and none of the numbers are
repeated. Some dice have dots instead of numbers. Suppose six of the students from
your Form including you participate in a fair die tossing activity. You are claiming
that when the dice is tossed; number 5 will appear on top of the dice. The other
members are claiming that either 3 or 4 will appear on the dice.
EXAMPLE 267 Find the probability that your claim will come true?
For you the favourable outcome is number 5. The number of favourable outcome is
only 1 because a fair die has number 5 on only one of its sides. The numbers of all
possible outcomes are {1, 2, 3, 4, 5, and 6}. Altogether there are 6 possible outcomes.
P (5) =
Example 268
1
6
Find the probability that your classmates claim will come true?
Your classmate’s favourable outcomes are either 3 or 4. The number to favourable
outcome is 2. The number of all possible outcomes is {1, 2, 3, 4, 5, and 6}.
Altogether there are 6 possible outcomes.
P(3 or 4) =
2
6
⇒
P(3 or 4) =
1
3
Since number of favourable outcomes ≤ total number of all possible outcomes, 0 ≤ p ≤ 1.
182
EXAMPLE 269
A box contains five marbles with number 2 printed on it, three marbles with number 3
printed on it and two marbles with number 4 printed on it. A marble is randomly
picked from the box and after viewing, it is replaced inside the same box.
What is the probability that a marble picked will have number 4 printed on it?
Two marbles have number 4 printed on it. All together there are 10 marbles.
P(4 printed on the marble) =
2
1
⇒ When simplified
10
5
EXAMPLE 270
A box contains 7 marbles. Two are coloured black with a smiling face; three are
coloured white with a smiling face, one coloured white with a dot and one coloured
completely in black. A marble is randomly picked from the box and after viewing, it
is replaced inside the same box.
What is the probability that a marble picked will be colour white with a dot?
P ( white with a dot ) =
1
7
What is the probability that a marble picked will be colour black with a smiling face?
P (black with a smiling face) =
2
7
What is the probability that a marble picked will not be colour black with a smiling
face?
P ( not colour black with a smiling face) =
183
5
7
MUTUALLY EXCLUSTIVE EVENTS
P ( A ∪ B ) = P ( A ) + P ( B ) − P( A ∩ B)
Look at your mathematics teacher. Someone in your class claimed that your
mathematics teacher is a female. Let’s call this claim event one (E1). You claim that
the mathematics teacher is a male. Now let us call this event two (E2). E1 and E2
cannot happen at the same time.
E1 and E2 are called mutually exclusive events.
Mutually exclusive events cannot happen at the same time. Either E1 or E2 can
happen but not both. In other words, your mathematics teacher is either a male or a
female but not both.
Since E1 and E2 are mutually exclusive events P( E1 ∪ E 2 ) = P( E1 ) + P( E 2 ) . It is
commonly written as P ( A ∪ B ) = P ( A) + P (B ) .
EXAMPLE 271
A fair dice is tossed once. What is the probability that either 1 or 6 will be shown on
the topside of the dice.
In one toss it is impossible to have both 1 and 6. Hence, this event is mutually
exclusive. Let A be 1 and B be 6 on the dice. It follows that P ( A ∪ B ) = P ( A) + P (B ) .
P( A) =
1
6
P( B) =
1
6
P( A ∪ B) = P( A) + P( B)
P( A ∪ B) =
1 1
2
1
+ → P ( A ∪ B) = → P ( A ∪ B ) =
6 6
6
3
Either and or are two words that can also help you establish that the events involved
are mutually exclusive.
EXAMPLE 272
A coin is tossed once. Find the probability of getting a head or a tail.
P ( H ∪ T ) = P ( H ) + P (T )
P (H ∪T ) =
1 1
+
2 2
P (H ∪T ) =1
184
Example 273
The probability that a Fiji citizen does not have a Fijian licence to drive in Fiji
is 0 ⋅19 . The probability that a Fiji citizen is registered under the regulations of the
Land Transport Authority of Fiji to drive passenger carrying vehicles is 0 ⋅ 001 . What
is the probability that Fiji citizen chosen at random being either passenger carrying
vehicle driver or not having a Fijian driving licence?
We know that no one cannot be a certified passenger carrying vehicle driver and at
the same time does not have drivers licence. They cannot happen at the same time.
Hence, these two events are mutually exclusive; meaning that they cannot occur at the
same time. The probability of either of them occurring is the sum of their
probabilities.
P ( A ∪ B ) = P ( A) + P ( B )
either
P ( A ∪ B ) = 0 ⋅19 + 0 ⋅ 001 = 0 ⋅191
either
Complementary Events
Properties of complementary events
•
•
They are mutually exclusive
Every possible outcome of a trial belongs in one of the events.
This means that if we join both events together, the result will be the complete sample
space of the event.
If A and B are complementary events, it follows that
•
P ( A ∪ B) = 1
•
P ( A ∪ B ) = P ( A) + P ( B )
•
P ( A ∪ B ) − P ( A) = P ( B )
1 − P ( A) = P ( B )
185
INDEPENDENT EVENTS
P ( A ∩ B ) = P ( A) × P ( B )
Two events lets say event A and event B, are independent if the outcome of one event
does not have any effect on the outcome of the second event. You say your prayers
before retiring to your bed every night for a wonderful night sleep and the King of the
Kingdom of Tonga drinks kava in the night so that he can sleep well in the night.
Both of these events are independent because you saying your prayers in the night
before retiring to your bed for a wonderful sleep in the night do in no way influence
the King of the Kingdom of Tonga to drink kava in the night to sleep well.
Independent events can happen at the same time because the outcome of the one event
has no influence on the outcome of the other event. Because independent events can
happen at the same time, they do have something in common. That something in
common is the probability of both events happening at the same time.
In mutually exclusives events you learned that two events are mutually exclusive if
both of them cannot happen at the same time. If you toss a coin once, you cannot get
a head and a tail at the same time. In other words, the outcome of the first event
affects the outcome of the second event.
Let A and B be two independent events. Since independent events can happen at the
same time without interference among events, they have something in common. That
something in common is represented by P( A ∩ B ) = P( A) × P(B ) .
Following key words can help you determine that the probability situation is indeed
independent.
[Both, and, in a row, all]
EXAMPLE 274
The probability that during cyclone season in Fiji it rains on Mondays is ¼ and the
probability that it rains on Fridays is ½. What is the probability that it will rain on
both Mondays and Fridays during cyclone season.
1 1
1
P ( A ∩ B ) = P ( A) × P ( B ) → P ( A ∩ B ) = × → P ( A ∩ B ) =
4 2
8
EXAMPLE 275
Two coins are tossed once. What is the probability that both coins will have head on
the surface facing upwards?
The probability of head on the first coin is ½ and the probability of head on the
second coin is also ½.
1 1
1
P ( A ∩ B ) = P ( A) × P ( B ) → P ( A ∩ B ) = × → P ( A ∩ B ) =
2 2
4
186
EXAMPLE 276
The probability that Navua soccer team wins against Ba soccer team in most crucial
games is ½. What is the probability that Navua beats Ba in 3 most crucial games in a
row?
1 1 1 1
× × =
2 2 2 8
PROBABILITY TREE and SAMPLE SPACE
A probability tree diagram is very helpful for listing sample space.
Tossing one coin
Tossing two coins
Sample space
HH
HT
Sample space is {H, T}
TH
TT
Carefully study the sample space for tossing two coins. Recall from example 301,
you were asked to find the probability to getting both heads for tossing two coins.
Now looking at the sample space for tossing two coins together, there are altogether 4
total possible outcomes but only 1 double head is seen in the sample space. Earlier,
we have agreed that the probability is given by
P robability =
Number of favourable outcomes
.
Total number of all possible outcomes
Number of favourable outcomes
. Now the
Sample Space
favourable outcome that we are looking for is double head [HH] which happens only
once. So the probability of getting both heads [HH] is ¼.
Now we can also state that P robability =
EXAMPLE 277
Find the probability of getting double tail when tossing two coins?
{(H, H), (H, T), (T, H), (T, T)} is the sample space of the tossing two coins.
Double tail means tail and a tail. Double tail appears only once in the sample space
and the possible outcomes in the sample space is 4.
So the probability of getting a double tail is ¼.
187
EXAMPLE 278
Construct a probability tree for tossing three coins and write down the sample space.
SAMPLE SPACE
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Both style of sample space
is acceptable but for
examination purpose, it is
recommended that you use
the method of listing the
sample space as shown
below.
SAMPLE SPACE
{(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
EXAMPLE 279
Three coins are tossed at the same time once only. Find the probability of getting all
heads?
{(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}. From the space, we
can conclude that all head (HHH) appears only once. Altogether in the sample space
1
there are 8 total possible outcomes. So the probability is .
8
EXAMPLE 280
Tossing three coins in example 278 are independent events because having head
appearing on the first coin does not influence the outcome of the second coin in any
way. Since the events are independent, we can multiply the probabilities.
1 1 1 1
P( A ∩ B ∩ C ) = × × = . You can see the answers of example 279 and 280 are
2 2 2 8
exactly same. There were only 3 coins and you are asked to find the probability of
getting all heads in example 279. All heads mean 3 head in a row and also a head
and a head and a head. Thus, the probabilities are multiplied.
EXAMPLE 281
What is the probability of getting at least 2 heads when three coins are tossed
simultaneously once only?
{(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)} At least 2 heads
means 2 and more heads. From the sample space at least 2 heads is 4. The total
possible number of outcome is 8. The answer is ½.
188
Sample Space for Two Dice
Let’s assume that on the first die the outcome is 1. On the second die, you can get a 1
as well. (1, 1) is part of the sample space.
Assume again that on the first die the outcome is 1. On the second die, you can get 2.
Hence, (1, 2) is part of the sample space as well. On the second die you can also get
3, 4, 5 or 6. {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)} are part of the same space.
Now assume that on the first die the outcome is 2. On the second die, you can get 1
and hence, (2, 1) is part of the sample space. Our assumption for the first die is 1, 2,
3, 4, 5 or 6 and on the second die we can get 1, 2, 3, 4, 5 or 6.
The complete sample space for tossing two dice together is given below.
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
EXAMPLE 282 What is the probability of getting 1 on both dice?
From the sample space you can see that (1, 1) only happens once and there are
1
altogether 36 total possible outcomes. The probability of getting 1 on both dice is .
36
EXAMPLE 283
In example 282, you were asked about the probability of getting 1 on both dice.
Getting 1 on both dice is same 1 in a row and also 1 and a 1.
The outcome of the first event in no way affects the outcome of the second event.
Hence, they are independent events and independent events can be multiplied.
1
Now the probability of getting 1 on the first die is
and the probability of getting 1
6
1
on the second die is also .
6
1 1
1
P ( A ∩ B ) = P ( A) × P ( B ) → P ( A ∩ B ) = × → P ( A ∩ B ) =
6 6
36
189
EXAMPLE 284
List the sample space for tossing a coin and a die.
{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
Example 285
When a fair coin and die is tossed together, what is the probability 3 will appear on
the die and head on the coin?
The sample space in example 310 shows that (H, 3) appears once only. The total
possible number of outcome is 12. Hence, the probability of number 3 appearing on
1
the die and head appearing on the coin is
.
12
PERCENTAGES AND PROBABILITIES
A probability lets say ½ can be converted to percentage by multiplying with 100.
½ x 100 = 50%. In the same manner, percentage can be expressed as a probability by
diving it with 100. 50% divided by 100 = ½.
EXAMPLE 286
In a city, 40% of the population eats boiled taro, 55% eats curried taro and 25% eats
both boiled and curried taro.
If a person is randomly selected from the same population, what is probability that the
person eats boiled taro only?
40% → P = 0 ⋅ 4 55% → P = 0 ⋅ 55
25% → P = 0 ⋅ 25
From the Vann diagram we can conclude
the 15% of the population eats boiled taro
15
= 0 ⋅ 15
only. The probability is
100
What is the probability that person chosen at random eats curried taro only?
From the Vann diagram we can conclude the 30% of the population eats curried taro
30
only. The probability is
= 0⋅3.
100
190
Example 287
In a class 30% of the students speak Hindi and 60% speaks Fijian while 20% speaks
both in Fijian and Hindi.
If a student is chosen at random from the same class, what is the probability that a
student speaks Fijian only?
From the Vann diagram, it can be concluded that 40% of the students speaks Fijian
40
only. Converting 40% into probability is
= 0⋅4 .
100
Example 288
In a town 45% of the population attends Methodist church, 70% attends The Church
of Jesus Christ of Latter Day Saints and 20% attends both churches.
If a person is chosen at random from the same town, what is the probability that the
person attends The Church of Jesus Christ of Latter Day Saints only?
For such type of questions, drawing a Vann diagram is recommended.
From the Vann diagram, it can be concluded that 50% of the population attends The
Church of Jesus Christ of the Latter Day Saints only. Converting 50% into
50
probability is
= 0⋅5 .
100
191
EXAMPLE 289
Komal and Nolene are two sisters. Both of them wake up 6 in the morning and take
separate routes to do 3 km walk every morning as part of their get fit program. On
any given morning, the probability that Komal arrives home before 7 am is 0 ⋅ 9 , the
probability that Nolene arrives home before 7 am is 0 ⋅ 8 and the probability that at
least one of them arrives home before 7 am is 0 ⋅ 92 . Find the probability that on any
given day,
A. both sisters arrive home before 7 am.
B. only Nolene arrives home before 7 am
C. neither of them arrives home before 7 am
The recommended procedure of solving this kind of problems is to perform some
preliminary calculations. Let Komal represent event A and Nolene represent event B.
Given P( A) = 0 ⋅ 9, P ( B ) = 0 ⋅ 8, P( A ∪ B) = 0 ⋅ 92
A. Both sisters arrive home before 7 am.
Given P( A) = 0 ⋅ 9, P ( B ) = 0 ⋅ 8, P( A ∪ B) = 0 ⋅ 92
P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ∩ B )
0 ⋅ 92 = 0 ⋅ 9 + 0 ⋅ 8 − P( A ∩ B )
0 ⋅ 92 = 1 ⋅ 7 − P( A ∩ B )
P( A ∩ B) = 1 ⋅ 7 − 0 ⋅ 92
P( A ∩ B) = 0 ⋅ 78
Both sisters → P ( A ∩ B )
Before doing the rest of the questions, it is recommended that you draw a Vann
diagram.
B. Only Nolene arrives home before 7 am
From the Vann diagram, the probability that only Nolene arrives home before
7 am is 0 ⋅ 02 .
C. Neither of them arrives home before 7 am.
Either of them ⇒ P( A ∪ B) = P( A) + P(B ) = 0 ⋅ 92
Neither of them = (P( A ∪ B )) = 1 − 0 ⋅ 92 = 0 ⋅ 08
C
192
[GIVEN IN THE QUESTION ]
EXAMPLE 290
The probability of Fiji winning against Wales in a 15 a side rugby game is ¼ and the
probability of Fiji winning against Canada is ½. If the probability of Fiji wining at
3
least one of the games is , what is the probability of winning both?
5
Let’s call the game against Wales’s event A and the game against Canada Event B.
3
P(A) = ¼
P(B) = ½
P(A U B) =
5
At least one means one or more. More can be all. Thus, at least one would be union
of both probabilities i.e. P (A U B).
Earlier you have learned that both means
event A and event B. You also learned
that in such case you would need to
multiply the probabilities to get P( A ∩ B) .
All those are correct. But when you are
given P( A ∪ B) , you will have to use the
formula
P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B )
3
=
5
3
=
5
1 1
+ − P ( A ∩ B)
4 2
3
− P ( A ∩ B)
4
3 3
P ( A ∩ B) = −
4 5
3
P ( A ∩ B) =
20
P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B )
EXAMPLE 291
Two dice are tossed together.
A. Find the probability of getting a 4 on the first die.
B. Find the probability of getting sum of 10 on both dice.
C. Find the probability of getting either 4 on the first die or a sum of 10
on both dice.
A. (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
B. (4, 6), (5, 5), (6, 4)
6 1
=
36 6
3
1
=
36 12
C.
P ( A ∪ B ) = P ( A) + ( B ) − P ( A ∩ B )
P ( A ∪ B) =
1 1 1 1 
+ − × 
6 12  6 12 
P ( A ∪ B) =
1 1 1
17
+ −
→ P ( A ∪ B) =
6 12 72
72
193
EXAMPLE 292
Two rugby ball kickers are trying to kick penalties from 50 meters at different angles.
Let’s call the kick takers A and B. The probability that A is successful from any
angle is ¼, and the probability that B is successful from any angle is ½. What is the
probability that either by A or B is successful from any angle?
Either means to add the probabilities i.e. P( A ∪ B ) = P( A) + P(B ) .
From the information give it can be concluded that A being successful or unsuccessful
in no way influences kicker B. The outcome of one kicker is not affected by the
outcome of the other. Since this is an independent event it can be concluded
that P( A ∩ B ) = P( A) × P(B ) .
Now let us answer the question. What is the probability that either by A or B is
successful from any angle?
Given P( A) =
1
4
P (B ) =
1
2
P ( A ∪ B ) = P ( A ) + P (B ) − P ( A ∩ B )
1 1 1 1
+ − × 
4 2 4 2
1 1 1
P( A ∪ B ) = + −
4 2 8
P( A ∪ B ) = 0 ⋅ 625
P( A ∪ B ) =
EXAMPLE 293
Samuel has a small chicken farm. In his farm he has 6 roosters and 4 hens. One day
Peter visited him so he went to his farm and randomly picked a chicken and
slaughtered it for his friend Peter. A week later Simon visited him and he picked
another chicken at random and slaughtered it for his friend Simon.
What is the probability that both chickens picked at random were roosters?
This is an independent event. This means that the random pick of the first one does
not affect the random selection of the second one. There are 10 chickens and the
6
probability of picking a rooster for slaughtering in the first pick is . Since a rooster
10
is already picked, there are now 5 roosters left and 4 hens. Altogether there are 9
5
chickens left. So the probability of a rooster in the second pick is . Now the
9
6 5 30 3 1
1
probability of picking both roosters is
× =
= =
Answer
10 9 90 9 3
3
194
Problems worthy of attack prove their worth by hitting back.
Piet Hein, “Problems,” Grooks (1966)
EXERCISE 6
1. What is the probability of getting a score of 7 when tossing two dice?
2. What is the probability of getting a score of 5 or 11 when tossing two dice?
3. A newly wed young couple wants to have three male kids with 10 years.
Adoption and twins or triplets are ruled out. What is the probability that the
couple will have 3 male kids?
4. Two gentle ladies are punching a helpless man who is doing is best to duck all
the shoots because he is currently under bound over. The probability that the
first gentle lady lands a successful upper cut is ¼ and the probability that the
second gentle lady lands a successful below the belt power shoot is ¾. What
is the probability that any target is hit by either the first gentle lady or the
second gentle lady?
5. Four fair dice are tossed. What is the probability getting 1 on the 1st die, 2 on
the second die, 3 on the 3rd die and 6 on the 4th die?
6. Five coins are tossed at once. What is the probability that outcome will be 4
heads.
7. Two sisters were asked to think about any natural number between 0 and 100
and write them on a piece of paper and pass it to their mother on individual
basis. If their numbers do not differ by more than 10, they will watch movies
for ½ day. However, if their numbers do differ by more than 10, they will
have to do house work for ½ day.
A. What is the probability that both will be watch ½ day movie on any given day
B. What is the probability that both will be doing ½ day house work?
8. In a country 50% of the population likes to curry chicken and 40% likes
vegetarian meals while 30% likes both curry chicken and vegetarian meals.
A person is chosen at random from the same population.
A. What is the probability that a person randomly chosen likes vegetarian meals
only?
B. If a sample of 200 is taken randomly from the population, how many are
expected to like vegetarian meals only.
9. Read example 293 and answer the following questions.
A. What is the probability that both chickens are hen?
B. What is the probability that the first one is a rooster and the second one is a
hen?
195
7
For if you forgive men their trespasses, your Heavenly Father will also
forgive you: But if you forgive not men their trespasses, neither will
your Father forgive your trespasses. (New Testament | Matthew 6:14 - 15)
STATISTICS
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
♣
Mean: Average
Median
Mode
Upper and lower quartiles
Range and inter quartile range
Standard deviation
Interpreting Graphs
Frequency and cumulative frequency
196
MEAN: AVERAGE SCORE
x=
Sum of marks
Total number of marks
Example 294
Find the mean of the numbers {1, 3, 5, 7, 9, 11}
x=
Sum of marks
1 + 3 + 5 + 7 + 9 + 11
→ x=
→ x=6
Total number of marks
6
MEDIAN: THE MIDDLE SCORE WHEN ALL THE SCORES ARE ARRANGED
EITHER IN THE ACCENDING OR DECENDING ORDER
EXAMPLE 295
Find the median of the scores {1, 4, 8, 2, 2, 10, 6}
First arrange the scores in the ascending order. {1, 2, 2, 4, 6, 8, 10}
The middle score is the median. {1, 2, 2, 4, 6, 8, 10}
Answer: Median is 4
EXAMPLE 296
Find the median of the scores {4, 5, 7, 3, 6, 5, 8, 10}
First arrange the scores in the ascending order. {3, 4, 5, 5, 6, 7, 8, 10}
{3, 4, 5, 5, 6, 7, 8, 10} → Not possible to get a middle number directly. When you
come across such cases, add the two most middle numbers and than divide by 2. The
two most middle numbers are 5 and 6.
5 + 6 11
=
= 5⋅5
Answer: Median is 5 ⋅ 5
2
2
MODE: THE MOST COMMON SCORE OR THE SCORE THAT APPEARS THE
MOST NUMBER OF TIMES
EXAMPLE 297
The mode of the scores {1, 3, 3, 4, 4, 6, 6, 6} is 6 because the score 6 happens three
times, more than all other events.
197
RANGE: HIGHEST SCORE MINUS THE LOWEST SCORE
EXAMPLE 298
Find the range of the scores {1, 3, 4, 10, 4}
Highest score is 10 and the lowest score is 1. 10-1= 9. Answer: RANGE = 9
QUARTILES: THREE VALUES OF A VARIABLE THAT DIVIDES ITS
DISTRIBUTION IN FOUR INTERVALS WITH EQUAL PROBABILITY; THE
25TH, 50TH OR 75TH PERCENTILES. THE 25TH AND THE 75TH PERCENTILES
ARE RESPECTIVELY THE LOWER AND THE UPPER QUARTILE.
When finding the lower and the upper quartiles, the scores must be arrange in an
ascending order. Before you can successfully find both quartiles, you will need to
first find the median.
EXAMPLE 299
Find the lower and upper quartiles and inter-quartile range of the scores given below.
{1, 3, 9, 4, 8, 2, 12, 4, 6}
Rearrange the scores in ascending order {1, 2, 3, 4, 4, 6, 8, 9, 12}
{1, 2, 3, 4, 4, 6, 8, 9, 12} median is 4.
Lower quartile is the median of the lower numbers after the median
2+3 5
= = 2⋅5
Answer: Lower Quartile = 2 ⋅ 5
{1, 2, 3, 4}
{1, 2, 3, 4}
2
2
The upper quartile is the median of the scores after median.
{6, 8, 9, 12}
{6, 8, 9, 12}
8 + 9 17
=
= 8⋅5
2
2
Answer: Upper Quartile = 8 ⋅ 5
Inter-quartile range = upper quartile – lower
Inter quartile range = 8 ⋅ 5 − 2 ⋅ 5
=6
EXERCISE 26
It is better to wear out than to rust out.
Bishop Richard Cumberland
Find the mean, median, mode, range, lower quartile, upper quartile and inter quartile
range of the following scores.
A. {1, 3 ,5, 5, 7, 7, 8, 8, 9}
B. {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
198
STANDARD DEVIATION
EXAMPLE 300
x
2
f
fx
4
8
x−x
2–6=-4
4
3
12
4–6=-2
6
2
12
6–6=0
8
3
24
8–6=2
10
4
40
10 – 6 = 4
∑f
x=
= 16
(x − x )
(− 4) =16
(− 2)2 = 4
(0)2 = 0
(2)2 = 4
(4)2 =16
2
∑ fx = 96
∑ fx ⇒ x = 96 ⇒ x = 6
16
∑f
s2 =
∑ f (x − x )
∑f
(
2
2
s2 =
152
16
Standard deviation is the square root of the variance. σ = s 2
)
f x−x 2
4 × 16 = 64
3 × 4 = 12
2×0 = 0
3 × 4 = 12
4 × 16 = 64
∑ f (x − x )
2
= 152
s 2 = 9 ⋅ 5 The v ariance is 9 ⋅ 5
σ = 9 ⋅ 5 σ = 3 ⋅ 08
The above is one way of finding standard deviation. The other way is inputting data in a
calculator and let the calculator perform analysis. If you are using Casio fx-82 MS calculator,
use the procedure given below.
{2, 2, 2, 2, 4, 4, 4, 6, 6, 8, 8, 8, 10, 10, 10, 10}
1. Press shift
2. Press mode
3. Press 1 to clear all existing data [ On the screen you should see Stat
Clear]
4. Press the = button
5. Press AC red button [On the screen you should see 0]
6. Press 2 then press M+ button
7. Press 2 then press M+ button
8. Press 2 then press M+ button
9. Press 2 then press M+ button
10. Press 4 then press M+ button
11. Press 4 then press M+ button
12. Press 4 then press M+ button
13. Press 6 then press M+ button
14. Press 6 then press M+ button
15. Press 8 then press M+ button
16. Press 8 then press M+ button
17. Press 8 then press M+ button
18. Press 10 then press M+ button
19. Press 10 then press M+ button
20. Press 10 then press M+ button
21. Press 10 then press M+ button
22. On the screen you should see 16. This means that altogether 16 data
entry has been inputted in your calculator
23. Press AC button
24. Press shift and then press 2
25. Press 2 to get standard deviation
26. Press = sign button
27. On the screen you should see 3 ⋅ 08
199
INTERPRETING STATISTICAL GRAPHS
BAR GRAPH
EXAMPLE 301
Hours Wasted Drinking Grog
10
Hours
8
6
4
2
0
1
Mon
2
Tues
3
Wed
4
Thurs
5
Fri
6
Sat
[Survey conducted in 2002-2003: Fijian Education: Problems and Solutions]
[Sunday is not included in this graph due to restrictions of Sabbath day]
The diagram above is a bar graph. It shows the number of hours per day wasted
drinking grog on average by those grog addicted people in a village settlement.
A. Altogether how many hours is wasted per individual drinking grog?
2 + 2 + 3 + 4 + 5 + 9 = 25 hours
B. Find the mean hours of drinking grog per individual.
x=
25
6
x = 4 ⋅ 17 hours
C. Find the median number of hours spent drinking grog.
{2, 2, 3, 4, 5, 9}
3+ 4 7
= = 3 ⋅ 5 hours
2
2
D. Calculate the standard deviation of drinking grog per week.
Using the calculator as explained in example 206, calculate the standard deviation.
Answer: 2 ⋅ 41 hours
•
On the next page is the line graph of the graph given above on top of this page.
200
Line Graph
Hours
Hours Wasted Drink Grog
10
9
8
7
6
5
4
3
2
1
0
Series1
1
2
3
4
5
Mon Tues Wed Thurs Fri
6
Sat
The pie chart given below contains exactly the same information on the previous page
with the bar graph
Hours Wasted Drink Grog
1
2
3
4
5
6
Though Kava is declared by some people as a drug, studies have shown no sign of alcohol
present in Kava. However, excess consumption of Kava makes one feel dupe and slight loss
of control of physical body. But the person is fully aware of his surrounding and displays
greater measure of peace. Some people drink Kava to accomplish wise purpose. Many Fijian
traditional disputes include few bowls of kava consumption in a ceremony to resolve disputes
and it has worked so far. When Kava consumption is abused, negative effects in the society is
seen. Wise use of it enhances a society but abuse of it breaks a society. Today, Kava is
mostly abused and less used to accomplish a worthwhile purpose.
201
CUMULATIVE FREQUENCY: The frequency of occurrence of all values less
than each given value of a random variable, equal to the sum of the frequencies of
each value of the variable less than that given value.
EXAMPLE 302
Given below is a set of Mathematics marks for the short test that Samson has done in
a year.
{60, 60, 70, 70, 80, 80, 80, 90, 90, 100, 100, 100}
Now let’s display the above information in a table.
Marks
60
70
80
90
100
Frequency
2
2
3
2
3
∑ f = 12
Cumulative Frequency
2
2+2=
4
2+2+3=
7
2+2+3+2=
9
2 + 2 + 3 + 2 + 3= 12
The above information is displayed on a cumulative frequency graph given below.
202
Error is a hardy plant; it flourisheth in every soil.
Martin Farquhar Tupper, Proverbial Philosophy [1838-1842]
EXERCISE 7
1. Use the table as shown in example 300 to find the standard deviation of marks
given below
{1, 1, 2, 2, 2, 3, 4}
2. When 3 marks are added to all the marks in question 1, we will get the
following results: {4, 4, 5, 5, 5, 6, 7}. Using your calculator only, find the
mean and standard deviation of these new marks.
3. When 2 is multiplied to all the marks in question 1, we will get the following
results: {2, 2, 4, 4, 4, 6, 8}. Using your calculator only, find the mean and
standard deviation of these new marks.
4. From the results you have obtained from question 2 and 3, what conclusion
can you make about the means and the standard deviation when you increase
or decrease each mark by the same constant?
5. Sarawan Gandhi sat for a series of Mathematics test and his mean mark was
calculated to be 45. A week later he was asked to taken another similar test
and his new mean was 65. What is the mean of the combined two tests?
6. Varoon’s test 1 subject marks were analysed and the mean and standard
deviation of his marks were 52 and 6 respectively. He had his marks
rechecked and all his subject marks were increased by 4. What would happen
to his mean and standard deviation?
203
8
A woman when she is in travail hath sorrow, because her hour
is come: but as soon as she is delivered of the child, she
remembered no more the anguish, for joy that a man is born
into the world. (New Testament | John 16:21)
NORMAL DISTRIBUTION
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
Properties of normal distribution curve
Expected number
Z-score
Standard deviation
Mean
Probability
204
NORMAL PROBABILITY DISTRIBUTION
Properties of Normal Distribution Curve
1.
2.
3.
4.
5.
6.
The curve is symmetrical about the mean.
Mean, median and mode are identical.
The curve is smooth.
68% of all scores lie within the range µ ± σ
95% of all the scores lie within the range µ ± 2σ
99% of all scores lie within the range µ ± 3σ
•
•
•
Probably is 1 standard deviation
Very Probably is 2 standard deviation
Almost Certainly is 3 standard deviation
205
EXAMPLE 303
5000 students sat for a Mathematics test. Their mean mark is 60 with a standard
deviation of 10.
A. Find the Z score for the mark between the mean and 65
B. Find the probability that the mark will be between the mean and 65
C. How many students are expected to have their marks between the mean and
65?
A. Z =
Dist ance from the mean
st andard deviation
Z=
x−µ
σ
Z=
65 − 60
5
Z=
5
5
Z =1
B.
Using the Z score, you can find the probability. The probability that correspond to
Z = 1 is 0 ⋅ 3413
This means that the probability that students mark will be between the mean and 65 is
0 ⋅ 3413
Given → n = 5000
C.
P = 0 ⋅ 3413
Expected number = np ⇒ 5000 × 0 ⋅ 3413 → ≈ 1707 students
206
EXAMPLE 304
In a Fiji School Leaving Certification Examination, students in a certain school have a
mean mark of 55 and a standard deviation of 10. The marks are normally distributed.
A. What is the probability that students mark will be between 45 and 55?
B. If 454 students sat for the examination, how many are expected to have their
marks within 45 and 55?
C. What is the probability that the mark will be below 45
D. How many are expected to have their marks less then 45?
E. More than 70
F. If 30 students sat for the examination, how many are expected to score more
that 70 marks?
A. The mean is 55 so the mark is between the mean and 45. The shaded area is
the probability between 45 and 55. But we can not find the probability
directly so there is a need to calculate the Z-score. The Z-score will help us to
find the required probability
Z=
µ−x
55 − 45
10
distance from the mean
→ Z=
→ Z=
→ Z=
→ Z =1
σ
standard deviation
10
10
From example 302, part B, you found that when Z = 1, probability is 0 ⋅ 3413
B. Expectation = np → Expectation = 454 × 0 ⋅ 3413 → ≈ 155 students
C. Since the graph is symmetrical about the mean, each side of the graph have a
probability equal to ½. Between 45 and 55, the probability is 0 ⋅ 3413 .
Below 45 is 0 ⋅ 5 − 0 ⋅ 3413 = 0 ⋅ 1587
D. Expected number = np
Expected number = np → 100 × 0 ⋅1587 → 15 ⋅ 87 → 15 students or 16 students
Counting is done using natural/counting numbers only. It is wrong to say that the
answer is 15 ⋅ 87 students. We do not count people using decimal.
207
E.
Probability is found using the Z score.
Z=
70 − 55
distance from the mean
x−µ
→ Z=
→ Z=
σ
standard deviation
10
Z = 15 ÷ 10 → Z = 1 ⋅ 5
The Z-score always gives the probability between the mean and the x-value. Since the
curve is symmetrical about the mean, 50% of the area is on the right hand side from
the mean. When you convert 50% to probability, you divide 50% by 100. The
probability is ½ from the mean to the left hand side. The probability from the mean to
70 is 0 ⋅ 4332 . This is the probability of the non-shaded area. The probability of the
shaded area is 0 ⋅ 5 − 0 ⋅ 4332 = 0.0668 . The probability that the mark will be more
than 70 is 0.0668
Expected number = np
F.
Expected number = 30 × 0 ⋅ 0668
Expected Number = 2 students
208
EXAMPLE 305
40 students sat for Physics test and their mean mark was 60 and the standard deviation
was 10. Find the probability that the students mark will be between 65 and 78.
You won’t be able to find the probability between 65 and 78 directly. You will need
to use the Z-score. However, Z-score gives the probability between the mean and the
x-value.
Step one. Find the probability between 60 and 65.
Step two. Find the probability between 60 and 78
Step three. Step two answer – Step one answer.
STEP ONE:
65 − 60
x−µ
Z=
Z=
σ
10
Z=
5
10
Z = 0⋅5
The probability between 60 and 65 is 0 ⋅ 1915
STEP TWO:
x−µ
78 − 60
18
Z=
Z = 1⋅ 8
σ
10
10
The probability between 60 and 78 is 0 ⋅ 4641
Z=
Z=
STEP THREE:
0 ⋅ 4641 − 0 ⋅ 1915 = 0 ⋅ 2726
The probability that a students mark will be between 65 and 78 is 0 ⋅ 2726
209
EXAMPLE 306
The Fiji Seventh Form Examination of a particular school had a mean of the
Mathematics mark equal to 60 with a standard deviation equal to 5.
Find the probability that a student’s mark will be more than 51.
Step One: Find probability between the mean 60 and the x-value 51
Step Two: Add 0 ⋅ 5 to the answer of step one.
STEP ONE:
Z-score helps us to find the probability between the mean and the x-value. Here the
mean is 60 and the x-value is 51.
Z=
µ−x
60 − 51
9
→ Z=
→ Z = → Z = 1⋅ 8
5
5
σ
The probability between the mean 60 and the xvalue 51 is 0 ⋅ 4641
STEP TWO:
The graph on the left of the mean has a total probability equal to 0 ⋅ 5 .
0 ⋅ 5 + 0 ⋅ 4641 = 0 ⋅ 9641
210
There’s a mighty big difference between good, sound reasons and reasons that sound
Burton Hillis
good.
EXERCISE 8
1. A long distance runner begins his training sharp 5 every morning. His mean
training time is 60 minutes with a standard deviation of 4 minutes.
A. What is the probability that the runner would train longer than the 60 minutes?
B. What is the probability that his training time will be within 60 to 68 minutes?
[Use the Z-score]
C. What is the probability that his training time will exceed 55 minutes?
2. A mineral water machine is regulated so that it discharges an average of 1000 ml
of per bottle. It is found that the volume of the water in the bottles is normally
distributed with a mean volume of 20 ml.
A. What fraction of the bottles filled with the mineral water will contain more than
1000 ml
Exactly 21,000 bottles are filled by mineral water each day by a company. Bottles
containing the lowest 10% are donated to Red Cross to help flood victims.
B. What is the minimum amount of fill that guarantees the packaging of bottles at
the end of the day?
C. How many bottles would overflow if 1040 ml bottles are used for filling?
3. Birth weights of students in a boarding school were taken before they were allowed
to board in the school dormitory. After one year residing in the school dormitory, the
student weights were retaken. It was found that all of the students gained weight.
The excess weight [weight gained] is normally distributed with a mean weight of 3 kg
and a standard deviation of ½ kg.
A. What is the probability that increase in the body weights would be less than 4
kg?
B. What percentage of the students would have increased in body weight fewer
than 4 and ½ kg?
C. What percentage of the students would have increased in body weight higher
tan 4 ½ kg?
D. If 100 students are selected randomly, what percentage of the students is
expected to have excess body weight between 4 and 5 kg?
211
9
Do not be afraid to say if something is wrong, you can
raise it to your supervisor or leaders and if your
supervisor does not do something, you can raise it to
the next level until you reach the top. (Counsel given to LDS
Teachers | Late President of The Church of Jesus Christ of the Latter Day
Saints: President Hinckley)
GEOMETRY
The following topics are discussed in this chapter:
♣ Application of 2 by 2 matrices to transformation
♣ Geometry of the plane based on the transformation:
Reflection
Rotation
Translation
Enlargement
Shear
♣ Properties of invariant points under the above
transformation.
♣ Combinations of transformation
♣ Determinant as scale factor for area
212
TRANSFORMATION-[Single matrix]
The points A(0, 0), B (2, 0) and C (2, 4) are points of a triangle as show below.
EXAMPLE 307
0 1
Transform the points A(0, 0), B(2, 0) and C (2, 4) using the matrix A = 
.
1 0
Find the coordinates of the points
A1, B1 and C1.
A
 0 1  0   0 
1

  =   → A = ( 0, 0 )
 1 0  0   0 
B
 0 1  2   0 
1

  =   → B = ( 0, 2 )
 1 0  0   2 
C
 0 1  2   4 
1

  =   → C = ( 4, 2 )
1
0
4
2

   
Find a matrix that will take A1 , B1 and C 1 back to points A, B and C .
The matrix that would do the opposite translation is the inverse the matrix that was
initially used for the first part of the operation.
0 1 
0 1 
1  0 −1
A=
A−1 = 
det ( A) = 0 − 1 det ( A) = −1 A−1 =




−1  −1 0 
1 0 
1 0 
Let’s check if it indeed does the opposite operation.
A1
B1
C1
0 1   0  0 
0 1  0   2
0 1   4   2 
1 0   0  = 0  A → ( 0, 0 ) 1 0   2  = 0  B → ( 2, 0 ) 1 0   2  =  4  C → ( 2, 4 )

   

   

   
An inverse matrix does the opposite operation.
213
Example 308
Given below is the diagram for square ABCD.
 0 −1
For the transformation given by N = 
 , find the coordinates of the image of
1 0 
rectangle ABCD. Label the images A ' B ' C ' D '.
 0 −1 2   ( 0 × 2 ) + ( −1× 2 )   0 + −2   −2 
=
 1 0  2  = 
 =   = A ' = ( −2, 2 )

   (1× 2 ) + ( 0 × 2 )   2 + 0   2 
 0 −1 4   ( 0 × 4 ) + ( −1× 2 )   0 + −2   −2 
=
 1 0  2  = 
 =   = B ' = ( −2, 4 )

   (1× 4 ) + ( 0 × 2 )   4 + 0   4 
 0 −1 4   ( 0 × 4 ) + ( −1× 4 )   0 + −4   −4 
=
 1 0  4  = 
 =   = C ' = ( − 4, 4 )

   (1× 4 ) + ( 0 × 4 )   4 + 0   4 
 0 −1 2   ( 0 × 2 ) + ( −1× 4 )   0 + −4   −4 
=
 1 0  4  = 
 =   = D ' = ( − 4, 2 )

   (1× 2 ) + ( 0 × 4 )   2 + 0   2 
Find a matrix that will map triangle A ' B ' C ' D ' back onto triangle ABCD.
 0 −1
1 0 1
N =
det ( N ) = ( 0 × 0 ) − (1× −1) = 1 


1  −1 0 
1 0 
214
 0 1
N −1 = 

 −1 0 
EXAMPLE 309
2 0
Transform the points A(0, 0 ), B(1, 0 ), C (1, 2 ) and D (0, 2 ) using the matrix P = 

0 2
P× A
 2 0  0 0
0 2 0 = 0 → A' = (0, 0 )

   
P×B
2 0 1 2
0 2 0 = 0 → B' = (2, 0 )

   
P×C
 2 0  1   2
0 2 2 = 4 → C ' = (2, 4 )

   
P× D
Area of rectangle ABCD is:
 2 0  0  0 
0 2 2 = 4 → D' = (0, 4 )

   
Area = L × W
= 1× 2
Area = 2 square unit
Area of rectangle A' B ' C ' D' is:
Area = L × W
= 2×4
Area = 8 square unit
Note: det(P ) = 4. Length scale
factor is det (P ) → 4 = 2 .
Length scale factor is 2. The length
of AB is 1 unit. The length of
A' B' = AB × sf .
A' B' = 1× 2 = 2 units .
Area of image = Area of object × Determ inant
Area of image = 2 × 4 = 8 square unt
215
EXAMPLE 310 Find the coordinates of the images of the points
2 0
A ( 2, 2 ) , B ( 4, 2 ) , C ( 4, 4 ) and D ( 2, 4 ) using the matrix P = 

0 2
 2 0  2  4
 2 0   4  8 
P× A: 
=
→
A
'
=
4,
4
P
×
B
:
(
)
   
 0 2   2  =  4  → B ' = ( 8, 4 )
0 2  2  4

   
 2 0   4  8
P×C : 
   =   → C ' = ( 8, 8)
 0 2  4  8
•
2 0 2 4
P×D : 
   =   → D ' = ( 4, 8 )
0 2   4  8 
Calculate the area of the square ABCD.
Area = L × W
= 2× 2
Area = 4 square unit
•
Calculate the length scale factor
det (P ) = 4. Length scale factor is det (P ) → 4 = 2 . Length scale factor is 2. The
length of AB is 2 units. The length of
A' B' = AB × sf .
A ' B ' = 2 × 2 = 4 units .
•
Calculate the area of image A ' B ' C ' D '.
Area of image = Area of object × Determ inant
Area of image = 4 × 4 = 8 square units
216
EXAMPLE 311 Find the coordinates of the images of the points
 0 −1
A ( 2, 2 ) , B ( 4, 2 ) , C ( 4, 4 ) and D ( 2, 4 ) using the matrix P = 

0 0 
P× A:
P× B :
0 −1  2   −2 
0 −1  4   −2 
1 0   2  =  2  → A ' = ( −2, 2 ) 1 0   2  =  4  → B ' = ( −2, 4 )

   

   
P×C :
P× D :
0 −1  4   − 4 
1 0   4  =  4  → C ' = ( −4, 4 )

   
•
•
0 −1  2   − 4 
1 0   4  =  2  → D ' = ( −4, 2 )

   
Analysing the position of the image in the above diagram tells us that the
matrix P is a rotational matrix with the centre of rotation at the origin (0, 0)
and the angle of rotation is 900 counter clockwise.
Rotational matrix : 900 counter clockwise
0 −1
0 0  →
Centre of rotation : (0, 0)


217
EXAMPLE 312 Find the coordinates of the images of the points
 −1 0 
A ( 2, 2 ) , B ( 4, 2 ) , C ( 4, 4 ) and D ( 2, 4 ) using the matrix P = 

 0 −1
P× A:
P× B :
 −1 0   2   −2 
 −1 0   4   −4 
 0 −1  2  =  −2  → A ' = ( −2, −2 )  0 −1  2  =  −2  → B ' = ( − 4, −2 )

   

   
P×C :
P× D :
 −1 0   4   − 4 
 0 −1  4  =  − 4  → C ' = ( − 4, − 4 )

   
•
 −1 0   2   − 2 
 0 −1  4  =  −4  → D ' = ( −2, − 4 )

   
Matrix P is a rotational matrix with the centre of rotation at the origin (0, 0)
and the angle of rotation is 900.
218
EXAMPLE 313 Find the coordinates of the images of the points
 0 1
A ( 2, 2 ) , B ( 4, 2 ) , C ( 4, 4 ) and D ( 2, 4 ) using the matrix P = 

 −1 0 
P× A:
P× B :
 0 1  2  2 
 0 1 4  2 
 −1 0   2  =  −2  → A ' = ( 2, −2 )  −1 0   2  =  − 4  → B ' = ( 2, − 4 )

   

   
P×C :
 0 1  4  4
 −1 0   4  =  4  → C ' = ( 4, − 4 )

   
•
P× D :
 0 1 2  4 
 −1 0   4  =  −2  → D ' = ( 4, −2 )

   
Matrix P is a rotational matrix with centre of rotation (0, 0) and the angle of
rotation 2700 counter clockwise or 900 clockwise.
219
EXAMPLE 314 Find the coordinates of the images of the points
 −1 0 
A ( 2, 2 ) , B ( 4, 2 ) , C ( 4, 4 ) and D ( 2, 4 ) using the matrix P = 

 0 1
P× A:
P× B :
 −1 0   2   −2 
 −1 0   4   − 4 
 0 1   2  =  2  → A ' = ( −2, 2 )  0 1   2  =  2  → B ' = ( − 4, 2 )

   

   
P×C :
 −1 0   4   − 4 
 0 1   4  =  4  → C ' = ( − 4, 4 )

   
P× D :
 −1 0   2   4 
 0 1   4  =  −2  → D ' = ( −2, 4 )

   
 −1 0 
 0 1  → Reflection matrix on the y − axis or x = 0


220
EXAMPLE 315 Transforms the points A(0, 0), B(1, 0), C(1, 1) and D(0, 1)
1 2 
using matrix P = 

0 1 
P× A:
P× B :
1 2   0   0 
1 2  1  1 
=
→
A
=
'
0,
0
(
)
0 1  0 0 
0 1  0  = 0  → B ' = (1, 0 )

   

   
P×C :
P× D :
1 2  1 3
1 2   0   2 
0 1  1 = 1 → C ' = ( 3, 1) 0 1  1  = 1  → D ' = ( 2, 1)

   

   
The above is called shear transformation. In a shear transformation, all the points in
one line or plane remain fixed while all other points move parallel to the fixed line or
plane by a distance proportional to their distance from the fixed line or plane. In the
above example, a shear transformation of a square produces a parallelogram.
221
Double Transformation
EXAMPLE 316 The diagram below shows a line segment AB joining the points
A(2, 1) and B(2, 4).
 0 −1
The line segment is transformed using matrix M = 
.
 −1 0 
Find the coordinates of A ' and B ', the images of points A and B under
transformation by matrix M.
B'
 0 −1  2   −1 
 0 −1   2   −4 

   =   = A ' = ( −1, −2 ) 
   =   = −4, −2
)
 −1 0   1   −2 
 −1 0   4   −2  (
Find a matrix that will transform A ' and B ' back onto A and B.
 0 −1
−1  0 1   0 −1
−1
M =
 det ( m ) = ( 0 × 0 ) − ( −1× −1) = −1 → M = 
=

1  1 0   −1 0 
 −1 0 
An inverse matrix does the opposite operation. Evidence is shown on the next
page.
222
M −1
A'
 0 −1  −1   ( 0 × −1) + ( −1× −2 )   0 + 2   2 
=

  = 
 =   = A ( 2,1)
 −1 0   −2   ( −1× −1) + ( 0 × −2 )   1 + 0  1 
M −1
B'
 0 −1  −4   ( 0 × −4 ) + ( −1× −2 )   0 + 2   2 
=

  = 
 =   = B ( 2, 4 )
 −1 0   −2   ( −1× −4 ) + ( 0 × −2 )   4 + 0   4 
Hence, an inverse matrix does the opposite operation.
1 1
Transform the images A ' and B ' further by using matrix N = 
 to give the
0 2
points A " and B "
N
A'
N
B'
B ''
 1 1   −1   −3 
 1 1   −4   −6 
=
=
=
−
−
=
=
A
''
3,
4
(
)

   

   
 0 2   −2   −4 
 0 2   −2   −4  ( −6, −4 )
Find a matrix that will map the points A and B directly onto the points A " and B " .
 1 1  0 −1  (1× 0 ) + (1× −1)
NM = 

=
 0 2  −1 0   ( 0 × 0 ) + ( 2 × −1)
(1× −1) + (1× 0 )   0 + −1
=
( 0 × −1) + ( 2 × 0 )   0 + −2
Evidence is shown on the next page.
223
−1 + 0   −1 −1
= 

0 + 0   −2 0 
NM
A
A"
 −1 −1  2   ( −1× 2 ) + ( −1×1)   −2 + −1  −3 
=
=
=
=
 

  
  
 −2 0   1   ( −2 × 2 ) + ( 0 ×1)   −4 + 0   −4  ( −3, − 4 )
NM
B
B"
 −1 −1  2   ( −1× 2 ) + ( −1× 4 )   −2 + −4   −6 
=
=
=
=



 

  
 −2 0   4   ( −2 × 2 ) + ( 0 × 4 )   −4 + 0   −4  ( −6, − 4 )
•
Find a matrix that will map the points A " and B " back unto A and B.
 −1 −1 
NM = 
 This matrix maps points A and B directly onto A " and B "
 −2 0 
Inverse of matrix will map A " and B " back unto A and B.
 −1 −1 
NM = 
 det ( NM ) = ( −1× 0 ) − ( −1× −2 ) = −2
 −2 0 
( NM )
−1
1  0 1   0 −0 ⋅ 5 
=− 
= 

2  2 −1  −1 0 ⋅ 5 
Evidence is shown below
( NM )
−1
A"
 0 −0 ⋅ 5  −3   ( 0 × −3) + ( −0 ⋅ 5 × −4 )   0 + 2   2 
=

  = 
 =   = A ( 2,1)
 −1 0 ⋅ 5  −4   ( −1× −3) + ( 0 ⋅ 5 × −4 )   3 + −2   1 
( NM )
−1
B"
 0 −0 ⋅ 5  −6   ( 0 × −6 ) + ( −0 ⋅ 5 × −4 )   0 + 2   2 
=

  = 
 =   = B ( 2, 4 )

−
1
0
⋅
5
−
4
6
+
−
2
−
1
×
−
6
+
0
⋅
5
×
−
4
(
)
(
)

  
  4
 
224
Points to Note
Single Transformation
If A is a matrix that transforms triangle ABC to triangle A ' B ' C ' , then the inverse of
matrix A, A −1 is a matrix that will transform triangle A ' B ' C ' back onto triangle
ABC.
Double Transformation
If A is a matrix that transforms triangle ABC to triangle A ' B ' C ' and B is a matrix that
transforms triangle A ' B ' C ' (The image of triangle ABC) to triangle A '' B '' C '' , then
the following rules applies:
Rule one. Matrix BA will transform or map triangle ABC directly onto
triangle A '' B '' C ''
Rule two. The inverse of the Matrix BA will transform or map triangle A '' B '' C ''
directly on triangle ABC.
I prefer to call rule one as FB rule. F stands for the Finishing matrix and B stands for
Beginning matrix.
The first matrix that we begin transformation of the triangle ABC is matrix A. This is
called the beginning matrix. The second matrix that we use to transform triangle
A ' B ' C ' is matrix B and is called the finishing matrix because we finish
transformation using the second matrix. Hence, if we want to go directly from
triangle ABC onto triangle A '' B '' C '' by using a single matrix, we will use the FB rule
to find this single matrix. This means we will multiply the matrix that we finish
rule
transformation with the matrix that we began transformation. FB →
BA . Now
the inverse of the matrix found using the FB rule will map the triangle A '' B '' C ''
directly back onto triangle ABC.
225
10
A woman when she is in travail hath sorrow, because her hour
is come: but as soon as she is delivered of the child, she
remembered no more the anguish, for joy that a man is born
into the world. (New Testament | John 16:21)
DIFFERENTIATION
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
First derivative functions
First Principle
Application of first derivative
Tangents to curves at a particular point
Normal or perpendicular lines
Turning points
Minimum and maximum points
226
DEFINITION of DIFFERENTIATION
DIFFERENTIATION: The operation or process of determining the FIRST
DERIVATIVE of a function.
If f ( x) = ax
n
then first derivative of f ( x) is
f ' ( x) = nax n −1
EXAMPLE 317
Find derivative of f ( x) = 3 x using both methods
f ( x) = ax n
f ' ( x ) = n a x n −1
f (x) = 3x
f ( x ) = 3 x1
f ' ( x ) = (1 ) 3 x 1 − 1
f '(x) = 3x0
N o te : x 0 = 1
f ' ( x ) = 3 (1)
f '(x) = 3
227
f ( x) = ax n → f ' ( x) = nax n −1
Example 318
f ( x) = 8 x 2 find f ' ( x)
f ' ( x ) = ( 2 ) 8 x 2−1 → f ' ( x ) = ( 2 ) 8 x1
f ' ( x ) = 16 x1 → f ' ( x ) = 16 x
Example 319
f ( x ) = x8
Find f ' ( x )
f ( x ) = x 8 → f ' ( x ) = 8 x 8−1 → f ' ( x ) = 8 x 7
Example 320
f ( x ) = 2 x3
Find f ' ( x )
f ( x ) = 2 x 3 → f ' ( x ) = ( 3) 2 x 3−1 → f ' ( x ) = 6 x 2
Example 321
f ( x ) = 2 x6
Find f ' ( x )
f ( x ) = 2 x6
f ' ( x ) = ( 6 ) 2 x 6 −1 → f ' ( x ) = 12 x 5
228
Derivative of a constant is always equal to zero
Example 322
Find the derivative of f ( x) = 9
f ( x) = 9 → f ' ( x) = 0
Example 323
Find the derivative of f ( x) = 100
f ( x) = 100 → f ' ( x) = 0
Example 324
Find the derivative of f ( x ) = 3 x5 + 5
f ( x ) = 3 x 5 + 5 → f ' ( x ) = ( 5 ) 3 x 5−1 + 0 → f ' ( x ) = 15 x 4
Example 325
Find the derivative of f ( x ) = 3 x 4 + 5 x + 4
f ' ( x ) = 12 x 3 + 5 x 0
f ( x) = 3x 4 + 5 x + 4
f ' ( x ) = ( 4 ) 3 x 4 −1 + (1) 5 x1−1 + 0
→
f ' ( x ) = 12 x 3 + 5 x 0
Example 326
Find the derivative of f ( x) = 5 x 2 + 4 x
f ' ( x ) = 10 x1 + 4 x 0
x0 = 1
f ( x ) = 5 x 2 + 4 x1
f
f ' ( x ) = 12 x 3 + 5 (1)
f ' ( x ) = 12 x 3 + 5
f ( x) = 5 x 2 + 4 x
'
x0 = 1
( x ) = ( 2) 5x
2 −1
+ (1) 4 x
1−1
→
f ' ( x ) = 10 x1 + 4 (1)
f ' ( x ) = 10 x + 4
f ' ( x ) = 10 x1 + 4 x 0
229
•
When you are asked to find derivative functions that has variable in the
denominator, you will first need to move all such variables in the
numerator.
Example 327
5
x3
Find f ' ( x )
f ( x) =
→
f ( x) =
5
x3
f ( x ) = 5x
→
−3
f ' ( x ) = ( −3) 5 x −3−1
f ' ( x ) = −15 x −4
Example 328
2
f ' ( x ) = ( 3) 2 x 3−1
−3
x
→
→
'
2
f
x
=
6
x
3
(
)
f ( x) = 2x
2
x −3
Find f ' ( x )
f ( x) =
f ( x) =
Example 329
4 −1
4
x
(
)
f ' ( x) =
x4
f ( x) =
2
Find f ' ( x)
2
→
→ f ' ( x) = 2 x 3
4 x3
f ( x) =
2
'
Example 330
4 43 −1
f ( x) = x
3
'
f ( x) = x
3
'
f ( x) = 3 x 4
4
Find f ( x)
→
f ( x) = x
4
3
230
→
f ' ( x) =
4x
3
1
3
•
When you are asked to find the derivative of functions written in
factored form, you will first need to expand the brackets or find the
product of the factors.
Example 331
•
Find the first derivative of the function y = ( x + 2 )( x + 1) .
First you will need to expand this function or find the product of the
factors and have it written in a simplified form.
y = ( x + 2 )( x + 1) → y = x 2 + 3 x + 2
•
Now you can find the derivative of the above function. Here, instead
of the function named as f(x), it is named y. You will be finding the
derivation of y with respect to x because in equation x is the variable.
y = x 2 + 3 x1 + 2
dy
= 2 x 2−1 + (1) 3 x1−1 + 0
dx
dy
= 2 x1 + 3 x 0
dx
dy
= 2 x + 3 (1)
dx
dy
= 2x + 3
dx
•
•
Note: f ' ( x) is just one of the ways to represent a function. Some of
dy
the common ones are y ' and
.
dx
dy
means differentiating y with respect to x and d stands for
dx
derivative.
231
Second Derivative
The second derivative means to differentiate a function twice.
Example 332
Find the second derivative of f ( x ) = 5 x8 .
f ( x) = 5 x8
f ' ( x) = 40 x 7
f ' ( x) = ( 8 ) 5 x8−1 → f '' ( x) = ( 7 ) 40 x 7 −1
f ' ( x) = 40 x 7
f '' ( x) = 280 x 6
GRADIENT
Derivative of a function gives the gradient of the function at any particular point.
This means when you differentiate a function, the result is equal to the gradient of the
function at any point of the graph.
Rise Change in y − values y 2 − y1 ∆y
Recall from co-ordinate geometry m =
=
=
=
Run Change in x − values x 2 − x1 ∆x
Now read example 195. The gradient of the function y = 3 x is equal to 3 as shown
in example 195. The example uses co-ordinate geometry to show that the gradient of
the graph of y = 3 x is 3. Example 333 uses calculus to find the gradient of y = 3x.
Example 333
If you differentiate y = 3 x , the answer will be 3.
y = 3x → y = 3x1
dy
dy
dy
dy
= (1) 3x1−1 →
= 3x 0 →
= 3 (1) →
=3
dx
dx
dx
dx
It is now confirmed that the first derivative of a function gives the gradient at any
point.
m=
dy
dx
232
Tangent
You have learned that
o The derivative of any function represents the gradient at any point of a graph.
o The gradient at a particular point on the graph of a function is actually the
gradient of the tangent to the graph at that particular point.
Example 334
Find the gradient of the tangent to the curve y = x 2 at any point on the curve.
The gradient of the tangent to any curve is always the first derivative of the equation
dy
of the curve.
→ gradient
dx
y = x2 →
dy
dy
dy
= 2 x 2−1 →
= 2 x1 →
= 2x
dx
dx
dx
The gradient of the tangent to any point on the curve is 2 x.
If you were asked to find the equation of the tangent to the curve y = x 2 at the point
(1, 4), you will have to use the method given below.
y = mx + c
(1, 4) ⇒ x = 1 y = 4
y
=
2
x
+
c
m = 2(1) →
m = 2x
m=2
4 = 2(1) + c → 4 = 2 + c → 4 − 2 = c → c = 2 → y = 2 x + 2
233
EXAMPLE 335
Find the equation of the tangent to the curve y = 3 x 2 − 4 x − 1 at the point where x =1.
The definition of tangent is: A straight line that touches another curve but does not
cross but intersect at one point only.
Since the tangent is a straight line, we can write y = mx + c as an equation for the
tangent.
Let’s
first
find
the
gradient
of
the
tangent.
dy
= 6 x − 4 This means m = 6 x − 4 .
dx
Given x = 1
m = 6x − 4
→ m = 6 (1) − 4 → m = 6 − 4 → m = 2
y = mx + c
y = 2x + c
To find the value of c, we must know the value of x and y. The value of x is already
given as x = 1 . Since x = 1 is a value on the graph of y = 3 x 2 − 4 x − 1 , y must also
be a value of the same graph corresponding with x = 1 .
y = 3 x 2 − 4 x − 1 → y = 3 (1) − 4 (1) − 1 → y = −2 ⇒ (1, −2 ) is point on the graph.
2
Now we will us this point to find the value of c.
y = 2x + c
−2 = 2 (1) + c
→ −2 = 2 + c → −2 − 2 = c → −4 = c → c = −4
The equation of the tangent to the curve y = 3 x 2 − 4 x − 1 is y = 2 x − 4
234
EXAMPLE 336
In example 335, you have learned that the equation of the tangent to the
curve y = 3 x 2 − 4 x − 1 is y = 2 x − 4 .
In the diagram you can see that points A and B lie on the tangent line. Now recall
from co-ordinate geometry, equation of a straight line passing through two points.
See examples 190 to 197.
A(1, −2) x1 = 1 y1 = −2
B (2, 0) x2 = 2 y2 = 0
y − y1 = m( x − x1 )
y − −2 = m( x − 1)
y + 2 = m( x − 1) → m =
y2 − y1
0 − −2
→ m=
→ m=2
x2 − x1
2 −1
y + 2 = 2( x − 1)
y + 2 = 2x − 2
y = 2x − 4
235
EXAMPLE 337
Find the equation of the normal to the curve
where x = 1 .
y = 3 x 2 − 4 x − 1 at the point
y = 3x2 − 4 x − 1 x = 1
y = 3 (1) − 4 (1) − 1
2
y = −2 (1, −2)
dy
= 6 x − 4 → m = 6 x − 4 → m = 6 (1) − 4 → m = 2
dx
m = 2 is the gradient of the tangent. Gradient of the perpendicular line is m1 m2 = −1 .
m1m2 = −1 → 2m2 = −1 → m2 =
−1
2
Normal/perpendicular line is a straight line and the equation of straight line
is y = mx + c and from the above working, the gradient of this normal line is = - ½ .
1
y = − x + c passes through (1, −2)
2
−2 = −
1
1
1
3
1
3
(1) + c → −2 = − + c → −2 + = c → c = → y = − x +
2
2
2
2
2
2
EXAMPLE 338 Find the equation of the normal to the curve y = 4 x 2 − 8 x at
the point ( 2, 0 )
y = 4 x 2 − 8 x → gradient of tangent →
m = 8x − 8
dy
= 8x − 8
dx
m = 8 ( 2) − 8 = 8
Gradient of the tangent of the curve y = 4 x 2 − 8 x at the point ( 2, 0 ) is 8. The
gradient of the normal to the curve and tangent at the same point is found by the
formula learned in coordinate geometry.
m1m2 = −1
8 m2 = −1 → m = −
1
8
1
1
2
1
y = mx + c → y = − x + c → 0 = − ( 2 ) + c → 0 = − + c → c =
8
8
8
4
1
1
Equation of the normal line → y = − x +
4
4
236
TURNING POINT FOR QUADRATIC GRAPHS
Note: When gradient of tangent = 0, it is sometimes called horizontal tangent.
EXAMPLE 339
Find the turning point of the graph of the function y = x 2 + 4 x + 4.
At the turning point the gradient of the tangent is equal to zero. The gradient of the
tangent is the first derivative of the function.
y = x2 + 4 x + 4 →
dy
= 2x + 4
dx
The slope of the tangent at any point is 2 x + 4 . At the turning point the slope of the
tangent is equal to zero i.e. 2 x + 4 = 0 .
−4
→ x = −2
2
To find the y-value of the turning point, substitute the x-value in the equation
y = x 2 + 4 x + 4.
2 x + 4 = 0 → 2 x = 0 − 4 → 2 x = −4 → x =
y = x 2 + 4 x + 4 → y = ( −2 ) + 4 ( −2 ) + 4 → y = 4 − 8 + 4 → y = 0
2
The turning point of the graph of the function y = x 2 + 4 x + 4 is
( −2, 0 )
Note: When x < −2 , the graph is decreasing. When x > −2 , the graph is increasing.
When x = −2 , the graph is neither increasing nor decreasing. It is stationary and
hence it is called the stationary point as well. At any stationary point, the slope of the
tangent to the same curve will always be equal to zero. Note: we draw graphs from
the right hand side but read graphs from the left hand side.
237
TURNING POINTS FOR CUBIC GRAPHS
x3
EXAMPLE 340 Sketch the graph of the function y = + 2 x 2 + 3x
3
At the turning point the gradient of the tangent is equal to zero. The gradient of the
tangent is the first derivative of the function.
y=
x3
dy 3x3−1
dy
+ 2 x 2 + 3x →
=
+ ( 2 ) 2 x 2−1 + (1) 3x1−1 →
= x2 + 4 x + 3
3
dx
3
dx
The slope of the tangent at any point is x 2 + 4 x + 3 . At the turning point the slope of
the tangent is equal to zero i.e. x 2 + 4 x + 3 = 0 .
x 2 + 4 x + 3 = 0 → ( x + 1)( x + 3) = 0 → x = −1 and x = −3
To find the y-values of the turning point, substitute the x-values in the main equation.
( −1) + 2 −1 2 + 3 −1 → y = − 1 + 2 − 3 → y = −1 1
x3
y = + 2 x 2 + 3x → y =
( ) ( )
3
3
3
3
3
1

turning point →  −1, −1 
3

( −3) + 2 −3 2 + 3 −3 → y = −9 + 18 − 9 → y = 0
x3
y = + 2 x 2 + 3x → y =
( ) ( )
3
3
3
turning point → ( −3, 0 )
238
MAXIMUM & MINIMUM TURNING POINTS
You know that at all turning points
dy
= 0 . The two types of turning points are maximum and
dx
minimum turning points.
•
•
At the maximum turning point the second derivative is less than zero.
At the minimum turning point the second derivative is bigger than zero.
NATURE OF TURNING POINTS
f " ( x) < 0 ⇒ Maximum Turning Point
f "( x ) > 0 ⇒ Minimum Turning Point
1

In example 340, the turning points are  −1, −1  and ( −3, 0 )
3

and the x-values of the turning points are −1 and −3 .
Example 341
x3
y = + 2 x 2 + 3x → y ' = x 2 + 4 x + 3 → y " = 2 x + 4
3
y " = 2 x + 4 → y " = 2 ( −1) + 4 → y " = 2 ⇒ y " > 0 → x = −1 → Minimum
y " = 2 x + 4 → y " = 2 ( −3) + 4 → y " = −2 ⇒ y " < 0 → x = −3 → Maximum
239
EXAMPLE 342
Find the x-values of the turning points of the function y =
1 3
x − 2 x 2 + 3 x + 4 and
3
determine if these are maximum or minimum values.
1
dy
y = x3 − 2 x 2 + 3x + 4 →
= x2 − 4x + 3
3
dx
factorize
x 2 − 4 x + 3 = 0 
→ ( x − 1)( x − 3) = 0
x = 1 and x = 3
Now let’s make a rough sketch to using the x components of the turning points. This
technique gives a clear picture of what is expected and helps you to focus on the
question.
Just looking at the sketch, you know that
•
When you approach the graph from the left hand side towards the right
hand side, the maximum comes first and this corresponds to x = 1
because from left, 1 comes before 3.
•
When approaching the graph from the right hand side towards the left
hand side, minimum comes second and this corresponds to x = 3
because from left, 3 comes after 1.
Now let’s use the second derivative to show that the above logic is true.
f ( x) =
1 3
x − 2 x 2 + 3x + 4
3
f ' ( x) = x 2 − 4 x + 3
f " ( x) = 2 x − 4 2 (1) − 4 = −2 < 0 f "( x) < 0 x = 1 ⇒ maximum
f "( x) = 2 x − 4 2(3) − 4 = 2 > 0 f "( x) > 0 x = 3 ⇒ Minimum
240
It requires a very unusual mind to make an analysis of the obvious.
Alfred North Whitehead
EXERCISE 10
1. Find the first derivative of the following:
A. f ( x) =
3
x
B. f ( x) =
1
3
x
2
C. f ( x) = x 3
3x 4
4
E. f ( x) = 3x 4 − 4 x 3 + 4
F. f ( x ) =
(
I. f ( x) = (2 x − 2)(2 x + 2)
H. f ( x) = 2 x − 2 x 2
)
D. f ( x) =
x
2
G. f ( x) = −3 x
J. f ( x) = x ( x + 4)
2. Find the second derivative of the following.
A. f ( x) = (2 x − 2)(2 x + 2)
C. f ( x) =
x
2
D. f ( x ) =
B. f ( x) = 2 x ( x + 4)
3x 4
4
E. f ( x) = x 3
3. Find the equation of the tangent to the curve y = x 2 − 3 x at x = 3
4. Find the equation of the tangent to the curve y = x 2 − 3 x at x = 2
5. Find the equation of the tangent to the curve y = x 2 − 4 at x = 3
6. Find the equation of the normal to the tangent of the curve y = x 2 − 3 x at x = 3
7. Find the equation of the normal to the tangent of the curve y = x 2 − 3 x at x = 2
8. Find the equation of the line perpendicular to y = − 4 x 2 + 6 x at x = 1
9. Find the equation of the normal to the curve given by y = 4 x − x 2 at x = −1
10. Find the equation of the normal to the graph of y = x 3 at the poi nt where x = 1
241
KINEMATICS 1
Application of Differentiation
This part deals with some physics concepts. It is the study of motion along a straight
line path. Kinematics is applied mathematics.
Altogether you will learn four concepts and all concepts are in the table given below
with their symbols and unit of measurements.
Distance
Symbol
s
v
Velocity
a
Acceleration
Time
t
Units
m ⇒ metres
m
s
m
s2
s ⇒ seconds
•
When you differentiate the distance function, the resulting function
will always be a velocity function. s → v
•
When you differentiate a velocity function, the resulting function will
always be an acceleration function. v → a
•
When you differentiate distance function twice, the resulting function
will be an acceleration function. s → v → a
Distance and displacement in form 6 mathematics is seen to be used interchangeably.
However, they do not mean the same thing. Sometimes distance and displacement are
same but most of the time depending on the nature of the question, there is a vast
difference in both answers. Distance means how far an object has travelled altogether
and its direction is not considered. Displacement is how far an object is from its
starting point.
Given below are some important things that you must remember
•
Initial means beginning and at the beginning, time is equal to
zero [t = 0] .
•
At the maximum height, velocity is equal to zero.
•
If the distance travelled is quadratic in nature, then the time it takes for
the particle to go up will be equal to the time for the particle to come
down or reach the ground.
242
EXAMPLE 343
The distance travelled by a particle is represented by s(t ) = t 3 + 2t 2 + t + 2
where s is in metres and t is in seconds .
A. Find the formula for the velocity of the particle at any point on its path.
s (t ) = t 3 + 2t 2 + t + 2 → s '(t ) = 3t 2 + 4t + 1 → v (t ) = 3t 2 + 4t + 1
B. Find an expression for the acceleration of the particle
v(t ) = 3t 2 + 4t + 1 → v '(t ) = 6t + 4 → a (t ) = 6t + 4 m / s 2
EXAMPLE 344
An object is thrown vertically upwards into the air. Its vertical distance above the
ground is given by s ( t ) = 20t − 4t 2 metres .
A. Find and expression for the velocity of the object.
s ( t ) = 20t − 4t 2 metres → s ' ( t ) = 20 − 8t ⇒ v ( t ) = 20 − 8t m / s
B. Find the initial velocity of the object.
Initial velocity means the velocity of the object when time is zero seconds.
v ( t ) = 20 − 8t m / s → v ( 0 ) = 20 − 8 ( 0 ) m / s → v ( 0 ) = 20 m / s
C. Find the velocity when time is 1 second.
v ( t ) = 20 − 8t m / s → v (1) = 20 − 8 (1) m / s → v ( 0 ) = 20 − 8 m / s → v ( 0 ) = 12 m / s
D. At what time is the velocity equal to zero?
To find the time when the velocity is equal to zero, make the velocity equation
equal to zero and solve for t .
v ( t ) = 20 − 8t → 0 = 20 − 8t → 8t = 20 → t = 20 ÷ 8 → t = 2 ⋅ 5 seconds
E. Find the maximum height reached by the object.
At the maximum height the velocity is zero. When time is 2 ½ seconds, velocity
is zero and this is the time the object reaches the maximum height.
s ( t ) = 20t − 4t 2 metres → s ( 2 ⋅ 5 ) = 20 ( 2 ⋅ 5 ) − 4 ( 2 ⋅ 5 ) → s ( 2 ⋅ 5 ) = 25 metres
2
243
EXAMPLE 345
An object is thrown vertically upwards into the air. Its vertical distance above the
ground is given by s ( t ) = 160t − 0 ⋅ 2t 2 metres where t is time in seconds.
A. Find an expression for velocity
s ( t ) = 160t − 0 ⋅ 2t 2 metres → s ' ( t ) = 160 − 0 ⋅ 4t → v ( t ) = 160 − 0 ⋅ 4t m / s
B. Find the time when velocity is equal to zero.
v ( t ) = 160 − 0 ⋅ 4t → 0 = 160 − 0 ⋅ 4t → 0 ⋅ 4t = 160 → t = 160 ÷ 0 ⋅ 4 → t = 400 s
C. Find the time it takes to reach the maximum height.
At the maximum height, the velocity is equal to zero. Since when t = 400 seconds
is the time velocity is equal to zero, it is also the same time when the object
reaches it maximum height. The time taken to reach the maximum height is 400
seconds.
[It is a good idea to draw a sketch to help you better understand the question]
s' (t ) = 0 i.e. v (t ) = 0
t = 400 s
D. Find the maximum height.
The height formula as given is s ( t ) = 160t − 0 ⋅ 2t 2 metres . You know that when
time is equal to 400 seconds, the object reaches the maximum height. Substitute
the time in the distance/height formula
s ( t ) = 160 ( 400 ) − 0 ⋅ 2 ( 400 ) metres → s ( 400 ) = 64, 000 − 32, 000 → s ( 400 ) = 32, 000 m
2
E. Find the time it takes to return to its starting point or on the ground.
Since the quadratics graphs are symmetrical and it takes 400 seconds to reach to
its maximum height, it will take the same amount of time to return to is starting
point or on the ground. The time taken to return to its starting point is 800 s.
244
EXAMPLE 346
Movement of a particle is described by its distance formula s ( t ) =
t3 t2
+ + 2t + 1 .
3 2
[t is time in seconds and s is dist ance in metres]
A. Find an expression for the velocity of the particle.
s (t ) =
t3 t2
+ + 2t + 1 → s ' ( t ) = t 2 + t + 2 → v ( t ) = t 2 + t + 2 m / s
3 2
B. Find the initial velocity of the particle.
Initial velocity means when time is equal to zero. In Kinematics, initial means
time is equal to zero.
v (t ) = t 2 + t + 2 → v ( 0) = (0) + (0) + 2 → v (0) = 2 m / s
2
C. Find the acceleration of the system.
v ' → a → v ( t ) = t 2 + t + 2 → v ' ( t ) = 2t + 1 → a ( t ) = 2t + 1 m / s 2
D. Find the acceleration of the particle when t = 2 s .
a ( t ) = 2t + 1 → a ( 2 ) = 2 ( 2 ) + 1 → a ( 2 ) = 4 + 1 → a ( 2 ) = 5 m / s 2
EXAMPLE 347
The distance travelled by a particle is represented by s(t ) =
t3 2
+t
3
where s is in metres and t is in seconds .
Find the velocity of the particle after 1 second.
s(t ) =
t3 2
3t 2
3t 2
+ t → s '(t ) =
+ 2t → s ' ( t ) =
+ 2t → s ' ( t ) = t 2 + 2t
3
3
3
s ' ( t ) = t 2 + 2t → s ' ( t ) = v = t 2 + 2t → v = (1) + 2 (1) → v = 3 m / s
2
Find the acceleration of the particle after 1 second.
v = t 2 + 2t → v ' = a = 2t + 2 → a = 2 (1) + 2 → a = 4 m / s 2
245
Business Calculus-Agricultural Sector
Calculus is used in business to maximise revenue and minimise cost. Application of
calculus is used in agricultural areas as well. Those students who take agriculture at
school should take mathematics with a willing heart and mind in order to maximise
productivity with the limited resources available. The Government of Fiji has
injected millions of dollars in agriculture but the productivity level is unsatisfactory.
The examples given below will enlighten you and should act as a catalyst to help you
take mathematics with a willing heart and a willing mind.
EXAMPLE 348
Assume that you are a farmer and your farm is next to the river. You plant cabbage
and you supply your product to a local market. Your farm is not fenced and your
neighbour’s goats often damage the cabbages in your farm. You are a poor person
and you made a request to the Government of Fiji for 1000 m long fence.
You want to bind the farm with this 1000 m fence. One side of the farm does not
need to be bounded because it is protected by the riverbank. What maximum area can
be bounded with the 100m m long fence?
Since the length of the fence is 1000 meters, the perimeter will be also 1000 meters.
The fence forms a square with the riverbank. The perimeter will include only 3 sides.
Perimeter = x + x + y → P = 2 x + y → 1000 = 2 x + y → 1000 − 2 x = y → y = 1000 − 2 x
Now you know that the length of the side opposite the riverbank is (1000 − 2x ) m .
The length of the other two opposites and equal sides is x metres . Area enclosed by
the fence is found by multiply length with the width.
Area = lw ( l = x and w = 1000 − 2 x )
A = x (1000 − 2 x )
A = 1000 x − 2 x 2
[continued on next page]
246
Since you now you the area function, let’s sketch the graph of the area function
A = 1000 x − 2 x 2 → A ' = 1000 − 4 x
A' is the gradient of A at any poi nt . At the turning point the gradient is always equal
to zero. This concept will help us to find the x-value of the turning point.
A ' = 1000 − 4 x = 0 → 1000 − 4 x = 0 → 1000 = 4 x → 1000 ÷ 4 = x → x = 250
The x-value of the turning point is 250. This information tells us that at x = 250, the
graph turns and because the graph is positive and quadratic in nature, it will have a
maximum turning point. So when x = 250, we will get the maximum value.
Substitute x-value in the equation for the area to find the value of the maximum area.
A = 1000 x − 2 x 2
A = 1000 ( 250 ) − 2 ( 250 )
2
A = 250, 000 − 125, 000
Areamax = 125, 000 m 2
Example 349
Manasa wants to build a pen for his chickens. One of the sides as shown in the
diagram given below will not require fence because of the existing concrete wall.
He has 240 metres of fence given to him by his father to enclose thee three sides of
the pen. Find the maximum area he can enclose his farm with this 240 m long fence.
Perimeter = x + x + y → P = 2 x + y → 240 = 2 x + y → 240 − 2 x = y → y = 240 − 2 x
Area = lw ( l = x and w = 240 − 2 x ) → A = x ( 240 − 2 x ) → A = 240 x − 2 x 2
A ' = 240 − 4 x = 0 → 240 − 4 x = 0 → 240 = 4 x → 240 ÷ 4 = x → x = 60
A = 240 x − 2 x 2 → A = 240 ( 60 ) − 2 ( 60 ) → A = 14, 400 − 7, 200
2
Areamax = 7, 200 m 2
247
Business Calculus-Box Production
In this section, you will learn about the usage and application of calculus in box
production.
EXAMPLE 350
A carton and cardboard production company is given an order to make boxes. Each
box should have an open top and it is to be made from square piece of cardboard by
cutting a square out of each corner and turning up the sides. Given that the cardboard
measures 20 cm on a side, find the dimensions of the box that will give the maximum
volume. Find the maximum volume.
V = L ×W × H
V = ( 20 − 2 x )( 20 − 2 x ) x
V = ( 400 − 40 x − 40 x + 4 x 2 ) x
V = ( 400 − 80 x + 4 x
2
)x
V ' = 12 x 2 − 80 x + 400 = 0
12 x 2 − 160 x + 400 = 0
x = 10 and x = 3 ⋅ 33
V = 400 x − 80 x 2 + 4 x 3
V = 4 x3 − 80 x 2 + 400 x
Maximum point occurs when x = 3 ⋅ 3 cm .
H = 3 ⋅ 33 cm
L = 40 − 2 x ⇒ 40 − 2 ( 3 ⋅ 33)
L = 13 ⋅ 34 cm
W = 13 ⋅ 34
V = lwh → V = 13 ⋅ 34 × 13 ⋅ 34 × 3 ⋅ 33 → V ≈ 592 ⋅ 59 m3
248
Business Calculus-Economics
Profit P ( x ) = Total Revenue R ( x ) − Total Cost C ( x )
EXAMPLE 351
The cost of producing x units of black and white 50 page book is given by the
formula C ( x) = 20 + 2 x + 0 ⋅ 01x 2 . The selling price of each text book is $5.
A. Find the cost of producing 100 books.
2
C (100) = 20 + 2(100 ) + 0 ⋅ 01(100 )
C (100 ) = 20 + 200 + 100
C (100 ) = $320
B. Find the total income or revenue earned from selling 100 books.
Total Revenue = Cost of item × number of items sold
Total Revenue = $5 × 100
Total Revenue = $500
C. Using the answers of A and B, find the profit earned from the sales of 100
books.
Profit = Total Revenue − Total Cost
Profit = $500 − $320
Profit = $180
D. Derive a formula for profit in terms of x .
Profit = Total Revenue − Total Cost
Total Revenue = Cost of item × number of items
Total Revenue = $5 × x
Total Revenue = 5 x
Profit = Total Revenue − Total Cost
Profit = 5 x − (20 + 2 x + 0 ⋅ 01x 2 )
Profit = 5 x − 20 − 2 x − 0 ⋅ 01x 2
Profit = − 0 ⋅ 01x 2 + 3 x − 20
E. Use the profit formula to find the profit for selling 100 books.
Profit = − 0 ⋅ 01x 2 + 3x − 20
Profit = − 0 ⋅ 01(100 ) + 3(100 ) − 20 = $180
2
Answers of part C and E are same. Why?
EXAMPLE 352
249
The cost production of
C ( x ) = 100 + 2 x + 0 ⋅ 01x 2
a
certain
medicine
is
given
by
the
formula
A. If the medicine firm sells the medicine at $50 per litre, what is the firm’s
income when it sells x litres ?
If the firm sells x litres at $50 per litre, its income is $50 x
B. What will be the firm’s profit from selling x litres of medicine?
P rofit = Income − Cost
(
P rofit = 50 x − 100 + 2 x + 0 ⋅ 01x 2
)
Profit = 50 x − 100 − 2 x − 0 ⋅ 01x 2
P rofit = − 0 ⋅ 01x 2 + 50 x − 2 x − 100
P rofit = − 0 ⋅ 01x 2 + 48 x − 100
C. Find the number of litres of medicine the firm needs to make in order to
maximise profit.
P = − 0 ⋅ 01x 2 + 48 x − 100
P' = − 0 ⋅ 02 x + 48 = 0
48 = 0 ⋅ 02 x
48 ÷ 0 ⋅ 02 = x
x = 2400 litres
D. Find the firms maximum profit.
Substitute the x-value into the profit equation.
P = − 0 ⋅ 01x 2 + 48 x − 100
P = − 0 ⋅ 01(2400) + 48(2400) − 100
P = −57600 + 115200 − 100
2
P = $57, 500
Experience enables you to recognize a mistake when you make it again.
250
Franklin P. Jones
Use calculus to find the turning points of the graph of the functions given below.
1. y = x 2 − 4
2. y = x 2 + 3 x + 2
3. y = x 2 − 6 x + 5
4. y = ( x − 1)( x + 1)
When I was young I observed that nine out of every ten things I did were failures, so I
did ten times more work.
George Bernard Shaw
Sketch the graphs of the functions given below.
1. y = ( x − 2)( x + 2)( x − 1)
2. f ( x) = − (x 3 + 2 x 2 − 5 x − 6 )
3. y = ( − x + 1)( x + 1)( x − 3)
4. y = x( − x + 4)( x + 4)
5. y = x( x − 2) 2
1. Use the second derivate to find the minimum turning point of the following.
A. y = x 3 − 3 x + 1
C. y = ( x − 4) 2
B. f ( x) = − x 3 + x 2 + x + 2
D. f ( x) = x 3 + 3 x 2 − 24 x + 2
2. Sketch the following graphs of the function showing clearly both x and y intercepts
and the turning points.
A. y = ( x − 2)( x + 2)( x − 1)
B. f ( x) = x 3 − 6 x 2
3. Find the maximum turning point of the graph of the function
f ( x) = x 3 − 2 x 2 − 4 x − 3
1.
A.
B.
C.
The movement of a certain creature is given by the formula s (t ) = t 3 − 4t .
Find the velocity of the creature when t = 2 s econds .
Find the acceleration of the creature when t = 2 s econds
What is the acceleration of the particle when v (t ) = 2 ?
2. Velocity of an object is v (t ) = 4t 2 − 2t m / s. Find:
A. the time/s when velocity is zero
B. the value of acceleration when t = 3 s
C. the acceleration when velocity is zero.
D. The velocity when acceleration is zero.
251
1. A rectangular plot of land is fenced into three equal portions by two
dividing fences parallel to two sides as shown in the diagram given
below. If the total fence to be used is 8000 metres, find the
dimensions of the enclosed land that has the greatest area and find the
greatest area.
2. Steven wants to fence a piece of land and divide into two plots as
shown. One side of the piece of the land is a hedge of thick trees. If
Steven has 2700 metres of fence with which to enclose and divide the
land, what is the maximum are that he can enclose?
3. A 15 metres long flat iron tin is bent in the shape as shown. Once side
overlaps another. What is the maximum possible area of this shape?
1. A sheet of metal square having all its sides equal to 20 cm, has four equal squares
cut out at the corners as shown. All shaded pieces in the diagram are bent up along
the dotted lines to form a rectangular box. Calculate the dimensions of the box that
will give a maximum volume.
252
2. A sheet of metal has four equal squares cut out at the corners as shown. All
shaded pieces in the diagram are bent up along the dotted lines to form a rectangular
box. Calculate the dimensions of the box that will give a maximum volume and find
the maximum volume.
You cannot have the success without the failures. H.G. Hasler, The Observer
1. A company determines that in the production of x units of a commodity its
revenue and cost functions are, respectively,
R( x ) = −3 x 2 + 970 x
and C (x ) = 2 x 2 + 500 .
A. Derive a formula for profit in terms of x .
B. Find the number of units of commodity that needs to be sold
to maximise profit.
2. A company finds that its cost for producing x units of a commodity
is C ( x ) = 3 x 2 + 5 x = 10 . Find the cost for making the 21st unit.
253
3. Given that C ( x) = 2 x 3 − 21x 2 + 36 x + 1000 is a cost function, determine the
intervals(s) for which the cost is increasing.
4. A garment factory sells each top class traditional bula shirt for $25. The total
cost to the factory producing x shirts is given by the formula given below.
C ( x ) = 25 + 2 x + 0 ⋅ 01x 2
A. If the garment factory sells x number of shirts, what is the
factory’s income?
B. What will be the factory’s profit from selling x number of
shirts?
C. Find the number of shirts the factory needs to produce in
order to maximise profit.
D. Find the company’s maximum profit.
5. A company’s cost of producing x items of its product is given
by C (x ) = 30 x − 0 ⋅ 1x 2 .
a. What is the company’s cost of producing 100 items?
b. Find the number of items that needs to be produced to give maximum
cost per product.
c. After how many items would the cost per item begin to decrease?
254
11
A woman when she is in travail hath sorrow, because her hour
is come: but as soon as she is delivered of the child, she
remembered no more the anguish, for joy that a man is born
into the world. (New Testament | John 16:21)
INTEGRATION
The following topics are discussed in this chapter:
♣
♣
♣
♣
♣
♣
♣
Steps for integration
Evaluation of definite integrals by anti derivatives
Integral of polynomial functions
Application of antiderivative
Area between a function and the x- axis.
Area below x-axis
Finding area enclosed between two curves
255
Definition of Integral
ax( )
Integral → ∫ ax dx =
+ c
( n + 1)
n +1
n
n≠0
EXAMPLE 353
The derivative of y = 2 x 4 − 9 x + 7 is
dy
= 8x3 − 9
dx
dy
dy
= 8 x 3 − 9, find the value of y . The anti-derivative of
is y. Thus,
dx
dx
integration is also known as anti-derivative.
Given
∫ dy =∫ (8 x
8 x( 3+1)
y=
( 3 + 1)
3
)
− 9 dx
8x4
→ y=
− 9 x + c → y = 2 x4 − 9x + c
4
− 9x + c
Note: the derivative of a constant is equal to zero. The integral or the anti-derivative
of a constant is equal to the constant multiplied with the variable it is integrated with
respect to.
∫ a dy = ay + c
∫ a dx = ax + c
∫ −2 dx = −2 x + c
∫ 2 dx = 2 x + c
∫ dx ⇒ ∫ 1 dx = 1x + c = x + c
C is a constant. Additional information is required to find the value of C.
Example 354
∫ (3x
2
∫ ( 3x
)
2
+ 6 x − 9 dx →
Example 355
)
+ 6 x − 9 dx
( 2 +1)
(1+1)
3x
6x
+
( 2 + 1) ( 2 )
3 x3 6 x 2
+
− 9x + c
2
− 9( x) + c → 3
x3 + 3x 2 − 9 x + c
∫ ( x + 1) dx
2 x(1+1)
2
∫ ( 2 x + 7 ) dx → (1 + 1) + 7 ( x ) + c → x + x + c
256
∫(
Example 356
)
3
x 2 dx
∫ ( x ) dx → ∫ x
3
2
5
3
2
3
x
dx →
2
+1
3
2
+1
3
5
3
5
 
 3
x
+c
5
 
3
+c →
5
3
x
5
x 3
3x
÷ +c →
× +c →
+c
1 3
1 5
5
x2
dx
Example 357 ∫
4
x2
x ( 2+1)
x3
x3
dx
→
+
c
→
+
c
→
+c
∫4
4 ( 2 + 1)
4 ( 3)
12
x3 + 1
Example 358 ∫ 3 dx
x
x3 + 1
∫ x3 dx →
∫(
∫(x
)
x 0 + x −3 dx →
Example 359
3
)
+ 1 x −3 →
∫(
∫(x
)
3+−3
+ x −3 dx
)
1 + x −3 dx → 1x +
x −3+1
x −2
+c → x+
+c
−2
−2
( x − 1) 2
∫ x dx
( x − 1) 2
x2 − 2 x + 1
dx = ∫ x 2 − 2 x + 1
∫ x dx ⇒ ∫
x
(
)
x dx
5
∫(x
2
∫(x
2
)
− 2x +1
)
∫x
x dx
1
2
− 2 x + 1 x dx
5
2
3
2
− 2x + x
1
2
dx →
x2
3
+1
5
+1
2
−
2x 2
1
+1
+
3
+1
2
x2
→
x
7
2
7
2
−
2x
5
2
5
2
+
x
3
2
3
2
257
→
7
2
5
2
3
2
2x
4x
2x
−
+
+c
7
5
3
+1
1
+1
2
+c
EXERCISE 38
Think and you won’t sink. B.C. Forbes, Epigrams
Evaluate the following.
1.
∫ ( 3x + 4 ) dx
5.
∫x
3
2
2.
∫ ( 3x
2
)
+ 4 dx
1
dx
6.
∫x
2
3.
∫ ( 3x
2
)
+ 4 x dx
3
dx
7.
∫ 4x
6
dx
4.
∫ ( 3x
8.
∫
2
)
+ 4 x + 2 dx
dx
3
x7
EXERCISE 39
Work now or wince later.
B.C. Forbes, Epigrams
Find the integral of the following of the following.
 3x 4 + 2 x3 − 7 x 
1. ∫ ( 3 x 4 + 2 x3 − 7 x ) dx
2. ∫ 
 dx
x4


3.
 3x 4

3
∫  4 + 2 x − 7 x  dx
5.

1 4

x + 9 x 3 + 10 x 2 − 13 x + 6  dx
2

4.
∫  4 x
∫ 
6.
∫ 
7.
 4 x5 
∫  3 x7  dx


8.
 8x3 − 5 x 4 
∫  3 x  dx


9.
∫ (10u
10.
∫(
 3x − 5 
 dx
x3 
)
u du
5
−
3−2 x 
 dx
x 

258
x −3
)
2
dx
Definite integral of a constant
Definite Integral
b is known as upper bound [UB ]
b
∫ k dx
b
= k ∫ dx = kx
a
a
b
= k (b − a )
a
b ≥ a → a is known as lower bound [ LB ]
upper bound − lower bound
3
Example 360
∫ 2 dx
Evaluate
1
3
upper bound ⇒ ( 2(3) + c ) = ( 6 + c )
3
∫ 2 dx → ( 2 x + c )
Evaluate
lower bound ⇒ ( 2(1) + c ) = ( 2 + c )
1
1
upper bound − lower bound → ( 6 + c ) − ( 2 + c ) → 6 + c − 2 − c = 4
1
Example 361
Evaluate
∫ ( 3x
2
)
+ 1 dx
0
1
∫ ( 3x
Evaluate
2
1
)
+ 1 dx →
0
∫ (3x
2
)
+ 1 dx →
0
1
1
3 x 2+1
+ 1x → x 3 + x + c
0
0
2 +1
UB = (1) + (1) = 2 LB ⇒ ( 0 ) + ( 0 ) = 0 UB − LB = 6 − 2 = 4 →
3
3
1
∫ (3x
2
)
+ 1 dx = 4
0
2
Example 362 Evaluate
∫ ( 3x
2
)
− x + 1 dx
−2
2
∫(
2
)
3 x 2 − x + 1 dx =
−2
∫ ( 3x
2
)
− x1 + 1 dx
−2
2 +1
2
2
2
3 x ( ) x1+1
3x3 x 2
x2
−
+ 1x + c
→
− + x+c
→ x3 − + x + c
−2
−2
−2
3
2
2
( 2 + 1) 1 + 1
( 2) + 2 + c → 8 − 2 + 2 + c = 8 + c
x2
3
+ x + c → ( 2) −
2
2
2
UB = x 3 −
LB = ( −2 )
3
( −2 )
−
2
2
+ −2 + c → −8 − 2 − 2 + c = −12 + c
UB − LB → ( 8 + c ) − ( −12 + c ) → 8 + c + 12 − c = 20
2
∫ ( 3x
−2
259
2
)
− x + 1 dx = 20
EXERCISE 40
When I make a mistake it’s a beaut!
Fiorello LaGuardia
Evaluate the following definite integrals.
4
1.
∫ x dx
2. ∫ 2x dx
1
5.
3
5
3.
4
2
∫ x ( x + 1) dx
−1
4
∫ ( 3x − 4 ) dx
4.
6.
∫ (x
−3
2
2
)
+ 1 dx
1
7.
∫ − dx
−1
260
) dx
+ 2x + 1
x
1
0
3
∫
(x
7
8. ∫ ( x − 1)( x + 1) dx
6
Area between the x-axis and the Graph
EXAMPLE 363
Find the area of the shaded region.
The shaded region is the area between the graph and the x-axis.
The area of the shaded region can also be found by integration of the formula and the
bounds will take the x-values. The upper bound is 0 and the lower bound is -4.
0
∫ ( x + 4 ) dx
−4
0
∫(
−4
0
0
x(1+1)
x2
+ 4x + c
→
+ 4x + c
x + 4 dx →
−4
−4
1+1
2
1
)
(0)
UB =
2
2
+ 4 (0) + c = c
( −4 )
LB =
2
2
+ 4 ( −3) + c = − 4 + c
UB − LB → c − ( − 4 + c ) → c + 4 − c → 4 square unit
0
∫ ( x + 4 ) dx =
4
−4
261
EXAMPLE 364
Solution
•
Now let’s find the area of the shaded region using calculus.
1
∫ 3 dx
⇒
−2
3x + c
1
−2
UB ⇒ 2 (1) + c = 3 + c
LB ⇒ 3 ( −2 ) + c = −6 + c
UB − LB
3 + c − ( −6 + c ) → 3 + c + 6 − c → 9 square unit
1
∫ 3 dx = 9
−2
262
EXAMPLE 365
Find the area of the shaded region.
First exp and ( x − 1)( x − 3)
y = ( x −1)( x − 3)
y = x2 − 3x −1x + 3
y = x2 − 4x + 3
4
∫(
3
4
4
x3 4 x 2
x3
x − 4 x + 3 dx →
−
+ 3x + c →
− 2 x 2 + 3x + c
3
3
3
2
3
)
2
Upper Bound
3
x
− 2 x 2 + 3x + c
3
→
Lower Bound
3
x
− 2 x 2 + 3x + c
3
→
( 4)
3
3
( 3)
3
− 2 ( 4) + 3( 4) + c →
2
64
4
− 32 + 12 + c → + c
3
3
3
− 2 ( 3) + 3 ( 3) + c = 9 − 18 + 9 + c = c
2
4
4
4

UB − LB →  + c  − ( c ) → + c − c →
square unit
3
3
3

4
∫(x
2
)
− 4 x + 3 dx = 4
3
263
EXAMPLE 366
Calculate the area of the shaded region
Area =
1
bh
2
1
( 3)( 3)
2
9
A=
2
A = 4 ⋅ 5 square unit
A=
Let’s use integration to find the area of the shaded region.
Whenever the shaded region is below the x-axis, absolute value of the integral must
be taken.
3
3
x2
x
−
3
dx
→
−
3
x
+
c
(
)
∫0
0
2
( 3)
UB ⇒
2
( 0)
LB ⇒
2
− 3 ( 3) + c = −4 ⋅ 5 + c
2
2
− 3 ( 0) + c = c
UB − LB → − 4 ⋅ 5 + c − c → − 4 ⋅ 5
3
∫ ( x − 3) dx = − 4 ⋅ 5
0
You know that area can not be negative. Looking at the graph you can tell that the
area is below the x-axis. Whenever the area is below the x-axis, area will always be
negative. When the area is negative, we will have to take a positive integral of the
shaded region.
3
∫ ( x − 3) dx = 4 ⋅ 5 square unit
0
264
EXAMPLE 367 Calculate the area of the shaded area.
6
A = ∫ ( x − 3) dx →
3
(6)
UB =
2
− 3(6) + c = c
2
( 3)
LB =
6
x2
− 3x + c
3
2
2
− 3 ( 3) + c = − 4 ⋅ 5 + c
2
UB − LB → c − ( − 4 ⋅ 5 + c )
c + 4⋅5 − c → A = 4⋅5
----------------------------------------------------------------------------------
 x2
3
B = ∫ ( x − 3) dx →  − 3 x + c 
 2
0
0
3
( 3)
UB →
2
( 0)
2
− 3 ( 3) + c → −4 ⋅ 5 + c
LB →
2
2
− 3( 0) + c → c
B = UB − LB → B = −4 ⋅ 5 + c − c → B = −4 ⋅ 5 → B = 4 ⋅ 5
Total Area = A + B = 4 ⋅ 5 + 4 ⋅ 5 = 9 square unit
Areas that are below the x-axis should be integrated separately from areas above the
axis. If you integrate both areas simultaneously using a single integral only, the area
below the x-axis will cancel with the area above the x-axis.
6
∫ ( x − 3) dx ⇒
0
( 6)
UB ⇒
2
6
x2
− 3x + c
0
2
2
− 3 ( 6 ) + c = 18 − 18 + c = c
( 0)
LB ⇒
2
2
− 3( 0) + c = 0 − 0 + c = c
UB − LB ⇒ c − c = 0
You can see that negative area cancels with the positive area. All area below the xaxis should be always calculated separately from the areas above the x-axis.
265
EXAMPLE 368 Calculate the area of the shaded region.
The entire area is below the x-axis. Setting up of a single absolute integral is
recommended.
y = ( x − 1)( x + 1) → x 2 + 1x − 1x − 1
x2 + 0 x − 1 → x2 + 0 −1 → y = x2 − 1
--------------------------------------------------------------------------------------------1
1
∫ ( x − 1)( x + 1) dx →
∫(
−1
−1
(1)
UB ⇒
)
x 2 − 1 dx →
1
x3
− 1x + c
−1
3
3
2
− 1(1) + c = − + c
3
3
( −1)
LB ⇒
3
3
− 1( −1) + c =
2
+c
3
UB − LB
2
2

= − + c − + c
3
3

2 2
−4
4
= − − +c−c →
→ A = square unit
3 3
3
3
266
EXAMPLE 369 Find the area bounded by the graph of the function
f ( x ) = x 2 − 4 and the x-axis with the area within the boundary x = 1 and x = 3 .
3
(
)
A = ∫ x 2 − 4 dx
2
2
∫(x
B=
2
)
− 4 dx
1
TotalArea = A + B
3
(
)
= ∫ x − 4 dx +
2
2
A=∫
2
B=∫
)
2
( 3)
UB ⇒
(
2
x3
x − 4 dx ⇒
− 4x + c
1
3
2
2
)
)
3
x3
− 4x + c
2
3
( 2)
LB ⇒
3
3
8
16
− 4 ( 2) + c ⇒ − 8 + c = − + c
3
3
3
− 4 ( 3 ) + c ⇒ 9 − 12 + c = −3 + c
3
)
− 4 dx
(
1
(
2
3
x3
x − 4 dx ⇒
− 4x + c
2
3
2
3
∫(x
1
3
A = ∫ x 2 − 4 dx ⇒
2
16
 16

UB − LB ⇒ −3 + c −  − + c  = −3 + + c − c ⇒
3
 3

A=
7
3
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
2
B=
∫(x
2
)
− 4 dx ⇒
1
2
x3
− 4x + c
1
3
( 2)
3
UB ⇒
(1)
3
8
16
− 4 ( 2) + c = − 8 + c = − + c
3
3
3
LB ⇒
3
UB − LB
= −
16
16 11
5
5
 11 
+c −− +c = − + +c −c = − ⇒ B =
3
3
3
3
3
3


A+ B =
7 5 12
+ = ⇒ A + B = 4 squareunit
3 3 3
267
1
11
− 4 (1) + c = − 4 + c = − + c
3
3
EXAMPLE 370 Find the shaded area enclosed between the graphs of the
function y = x − 2 and y = x ( x − 2) ( x − 4) .
Expand
y = x ( x − 2) ( x − 4)
y = x ( x − 2) ( x − 4)
(
)
= x2 − 2x ( x − 4)
= x3 − 4 x 2 − 2 x 2 + 8 x
y = x3 − 6 x 2 + 8 x
Let ' s find area A first.
2
∫(
0
2
x 4 6 x3 8 x 2
x4
x − 6 x + 8 x dx =
−
+
+c =
− 2 x3 + 4 x 2 + c
0
4
3
2
4
3
UB ⇒
LB ⇒
)
2
( 2)
4
− 2 ( 2 ) + 4 ( 2 ) + c = 4 − 16 + 16 + c = 4 + c
3
4
( 0)
2
4
− 2 (0) + 4 ( 0) + c = 0 + c = c
3
4
UB − LB = 4 + c − c = 4
2
Area A = 4 square unit
Now find area B.
2
∫ ( x − 2 ) dx =
0
UB ⇒
( 2)
2
2
( 0)
2
x2
− 2x + c
0
2
− 2 ( 2 ) + c = 2 − 4 + c = −2 + c
2
− 2 (0) + c = 0 + c = c
2
UB − LB = −2 + c − c = −2 = 2
LB ⇒
Area B = 2 square unit
Total Area = A + B = 4 + 2 = 6 square unit
268
Area Enclosed Between Two Graphs
When you are asked to find the area enclosed between two graphs, the following
procedures and steps must be followed:
•
•
•
First, integrate the function of the upper graph. This should give the
area between the graph and the x-axis. If the required area is below the
x-axis, make sure to take the absolute value.
Second, integrate the function of the lower graph. If the required area
is below the x-axis, make sure to take the absolute value.
Subtract the answer of step two from step one. [Answer: step1- step 2]
EXAMPLE 371 Find the area enclosed between the graphs of the function
y = x and y = x 2
•
First find the area between the upper graph and the x-axis.
1
∫ x dx =
0
1
x2
+c
0
2
(1)
2
1
UB ⇒
+c = +c
2
2
LB ⇒
( 0)
2
2
+c = c
UB − LB
=
1
1
1
+c−c =
Area between y = x and the x − axis is square unit
2
2
2
269
•
Now find the area between the lower graph and the x-axis
Area between y = x 2 and the x − axis
1
2
∫ x dx =
0
1
x3
+c
0
3
(1)
3
1
UB ⇒
+c = +c
3
3
UB − LB =
LB ⇒
( 0)
3
3
1
1
+c−c =
3
3
Area between y = x 2 and the x − axis is
•
•
+c = c
1
square unit
3
Now perform the calculation of the shaded area that is enclosed
between the two graphs.
Area below the upper graph minus the area below the lower graph
UG → Upper Graph
LG → Lower Graph
Area enclosed between the two graphs is :
Area between the UG and the x : axis − Area between the LG and the x : axis
Area =
1 1
−
2 3
Area =
3− 2
6
Area =
1
square unit
6
270
EXAMPLE 372
Find
the
shaded
area
enclosed
functions y = x( x − 2)( x + 2) and y = 3 x − 6 .
between
the
graphs
of
• First find the area between the upper graph and the x-axis.
2
2
3x 2
3
x
−
6
dx
=
−
6
x
+
c
(
)
∫1
1
2
UB ⇒
3( 2)
LB ⇒
2
− 6 ( 2 ) + c = 6 − 12 + c = − 6 + c
2
3 (1)
2
2
− 6 (1) + c = 1 ⋅ 5 − 6 + c = − 4 ⋅ 5 + c
UB − LB
= − 6 + c − (− 4 ⋅5 + c) = − 6 + c + 4 ⋅5 − c
= − 6 + 4⋅5 + c − c
= −1 ⋅ 5
= 1⋅ 5 square unit
271
the
•
Now find the area between the lower graph and the x-axis.
Expand y = x( x − 2)( x + 2)
(
)
y = x 2 − 2 x ( x + 2 ) = x3 + 2 x 2 − 2 x 2 − 4 x
y = x3 − 4 x
2
∫(x
3
)
− 4 x dx =
1
2
x4 4 x2
x4
−
+c =
− 2 x2 + c
1
4
2
4
2
x4
− 2 x2 + c
1
4
( 2)
UB ⇒
4
(1)
LB ⇒
4
− 2 ( 2) + c = 4 − 8 + c = − 4 + c
2
4
4
− 2 (1) + c = 0 ⋅ 25 − 2 + c = −1⋅ 75 + c
2
UB − LB
= − 4 + c − ( −1 ⋅ 75 + c )
= − 4 + c + 1⋅ 75 − c
= −2 ⋅ 25
= 2 ⋅ 25 square unit
Shaded area
= Area between the UG and the x − axis − Area between the LG and the x − axis
= 1 ⋅ 5 − 2 ⋅ 25
= −0 ⋅ 75
Shaded area = 0 ⋅ 75 square unit
272
EXERCISE 41
It is the function creative men to perceive that relations between thoughts, or things,
or forms of expression that may seem utterly different, and to be able to combine
them into some new forms-the power to connect the seemingly unconnected.
William Plomer
Find the area of the shaded region/s.
273
the
area
of
the
region
bounded
6. Find
2
of y = x + 2 and y = x on [−1, 2] .
7. Find the area of the region bounded by the graphs of
8.
9.
274
by
the
graphs
KINEMATICS 2: Application of Integration
In Kinematics 1 you learned that the:
•
Derivative of the distance or displacement gives velocity or speed of a
particle in motion.
•
Derivative of the velocity formula gives acceleration of a particle in
motion.
In Kinematics 2 you will learn that the:
•
The derivative of distance formula gives velocity, its opposite; the
integral of velocity formula will give distance or displacement formula.
•
The derivative of velocity formula gives acceleration, its opposite; the
integral of the acceleration formula will give distance formula.
It is recommended that you revise examples 343 to 347 before as a perquisite
requirement of understanding, before proceeding with this section.
275
EXAMPLE 373
The acceleration of a particle is given by the formula a (t ) = 2t − 6 m / s 2 .
A. Find an expression the velocity of the particle given that when
t = 1 sec velocity is 0 m / s .
v ( t ) = ∫ a ( t ) dt
2t 2
2
∫ ( 2t − 6 ) dt = 2 − 6t + c = t − 6t + c
v ( t ) = t 2 − 6t + c
Given : When t = 1, v = 0
v (1) = (1) − 6 (1) + c = 0
2
1− 6 + c = 0 − 5 + c = 0 c = 0 + 5 c = 5
v ( t ) = t 2 − 6t + 5 m / s
B. Find the expression for distance formula given that when
t = 0 sec, s = 4 metres
s ( t ) = ∫ v ( t ) dt
∫(
t3 t2
t3 t2
t − 6t + 5 dt = − + 5t + c s ( t ) = − + 5t + c
3 2
3 2
2
)
( 0)
3
Given t = 0 sec, s = 4 metres
3
t3 t2
s ( t ) = − + 5t + 4 metres
3 2
276
( 0)
−
2
2
+ 5 ( 0) + c = 4 c = 4
EXAMPLE 374
A particle is moving in a straight path and its acceleration after t seconds is given
by a (t ) = 2t + 2 m / s 2 .
A. Given that the initial velocity of the particle is 1 m/s, obtain an equation for its
velocity after a time t seconds .
v(t ) = ∫ a (t ) dt
∫ ( 2t + 2 ) dt =
2t 2
+ 2 t + c = t 2+ 2 t + c
2
v(t ) = t 2 + 2 t + c
Initial velocity is 1 m / s. Initial means t = 0.
t = 0 v =1m/ s
v(t ) = t 2 + 2t + c
v(0) = ( 0 ) + 2 ( 0 ) + c = 1
2
c =1
v(t ) = t 2 + 2 t + 1 m / s
B. Given s ( 3) = 24 metres , obtain an equation for the distance travelled by the
particle.
s ( t ) = ∫ v ( t ) dt
v ( t ) = t 2 + 2t + 1 m / s
(
)
s ( t ) = ∫ t 2 + 2t + 1 dt
t 3 2t 2
s (t ) = +
+ 1t + c
3
2
t3 2
s (t ) = + t + t + c
3
33
s ( 3) = + 32 + 3 + c = 24
3
21 + c = 24
c = 24 − 21 c = 3
t3 2
s ( t ) = + t + t + 3 metres
3
277
Solution
Exercise 1 A
1.
2. 5
3. 6
4. 84
5. 4200
6. 4
8. Does not contain the identity element and fails the closure test.
278
9.
0
1
2
3
0
0
1
2
3
1
1
2
3
0
2
2
3
0
1
3
3
0
1
2
12. A. a
B. b
C. a
D. d
13. Associative
15. Under modulo 4, 2 don’t have an inverse. Under modulo 5, each element has an
inverse.
Exercise 1 B
1. A.
4
y9
2. A. 5 y 5
B. −
2
x
B. a −1
3. A. 18x 2 y 2 B.
8.
2
−1
24 x
11.
20.
1
y8
D.
C.
b
b
or
2
7
49
D. 2x −1 y −2
C. x
75
2a 5
3x 4 + 1
4. A.
x8
6. A. 2 x
B. 8
7. A. 8
C.
8x9
125
D.
3
y
8y 3
34
81
5. A. 4 or
2
16
y10
B. 4
x
C. 1
B. 3x
9. 5x 50
1
2
9x
10. 6x 6
18.
50x 200
22.
2 xy 6
11
19.
2x9
Exercise 1 C
1. −3 − 3 2
2. − 3 − 6
3. −3 + 3 2
279
4. 3 + 2 2
7.
5. 15 + 10 + 6 + 2
6. 4 3 + 4 2 + 6 + 2
8 + 3 − 40 − 15
5
8. 3 3 − 3
9. 15 + 3
12. 4 2
13. 3 5
10. 2
18. A. 5 + 2
11. 1
B. 3 3 + 2 6 + 3 2 + 4
Exercise 1 D
1. x = 625
2. 3log 5
3. 1
4. 8
5. 3 ⋅ 726833028 ≈ 3 ⋅ 73
6. ln 20
7. ln12,500
8. 2 ⋅ 547952063 ≈ 2 ⋅ 55
10. 7
11. log 27
12. log 3
14. log b
Exercise 2 A
1. A. x 2 − 4
B. 2 x 2 − 4
2. A. ( x + 3)( x + 3) = 0 x = −3
B. ( x + 1)( x + 4 ) = 0 x = {−4, −1}
3. ( x − 4 )( x − 2 ) = 0 x = {4, 2}
4.
x + 1 = 0 x = −1
f (−1) = ( −1) − 5 ( −1) + 2 ( −1) + 8
3
2
f (−1) = −1 − 5 − 2 + 8
f ( −1) = 0
7. k = 7
1

11.  , −2 
2


12. {1 ⋅ 62, −0 ⋅ 62}
13. {20, 6 ⋅ 67}
22. {1,3}
25.
{
}
2 − 1, − 2 − 1
Exercise 2 B
1. A. 1 ½
B.
4
3
2. A. – ½
280
B. −1 ⋅ 9
3. −
1
49
4. −3 ⋅ 5
12.
2 y 2 − 15
5y
13.
5
19. v =
or
x=
5
17
( x − y )( y − 1)
6
11.
14.
y
15. v = u + at → v −u = at →
16. x = ( y + 3) 6
10. x ≥ −
( x + y )( x − y )( x + 1)
y 2 + ab
2
9 ( x − 3)
v−u
v−u
=t → t =
a
a
( y + 3)
5
18. u = v 2 + 2as
2KE
m
Exercise 2 C
1. 15
2. 30
3. 35
4.
tn = a + ( n − 1) d a = 3 n = 101 d = 9 − 3 → d = 6
t121 = 3 + (121 − 1) 6
t121 = 3 + (120 ) 6
t121 = 3 + 720 t121 = 723
n  2a + ( n − 1) d 
Sn = 
2
5.
2000  2 ( 4 ) + ( 200 − 1) 4 
2000 8 + (199 ) 4 
S2000 =
S2000 =
S 2000 = 804, 000
2
2
7. a = 8 d = 3
11.
tn = 1 d = 1 a = 1 → tn = a + ( n − 1) d → t10 = 1 + (10 − 1)1
t10 = 1 + 9 = 10 → 30 + 10 = 40
Exercise 2 d
 −6 −6 
1. M × N = 
2. 0

 −3 −3 
 3 2
 − 5 5   −0 ⋅ 6 0 ⋅ 4 
5. 
 or 

 − 4 1   −0 ⋅ 8 0 ⋅ 2 


 5 5
3. No inverse
0 0
6. 

5 5
281
 1 −2 
9. 

 4 −3 
4. 5
6 6
12. M + 2M = 

 3 3
 3

16.  10
− 1

 6
1
− 
5

0 

4 
14. C × D = 

 −10 
 4 0
17. 

 −1 9 
18. 36
Exercise 3
5. A
5. B
282
 0 −6 
15. AC = 

 −5 −9 
 1

19.  4
 1

 36

0

1

9
16. A
283