Math book.sn4 by Iradukunda Prince Maxime

A Level

Mathematics Senior 4 Student's Book

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Published by East African Publishers Rwanda Ltd. Tabs Plaza, Kimironko Road, Opposite Kigali Institute of Education P.O. Box 5151, Kigali RWANDA. Email: eaep@eastafricanpublishers.com Website: www.eastafricanpublishers.com East African Educational Publishers Ltd. Brick Court, Mpaka Road/Woodvale Grove Westlands, P.O. Box 45314 Nairobi – 00100 KENYA. East African Educational Publishers Ltd C/O Gustro Ltd P.O. Box 9997, Kampala UGANDA. Ujuzi Books Ltd. P.O. Box 38260, Dar es Salaam Tanzania.

 EAEP 2013 First published 2013

ISBN 978-9966-25-950-9

Printed in Kenya by Printwell Industries Ltd. P.O. Box 5216-0506 Nairobi, Kenya

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Content Chapter 1

.Introduction to mathematical logic................................................................... 1

Chapter 2

.Binary operators and algebraic structures........................................................ 15

Chapter 3

The set of real numbers......................................................................................... 31

Chapter 4

Polynomials and rational functions................................................................... 47

Chapter 5

Equations, inequations and functions of the first degree.............................. 67

Chapter 6

Equations, inequations and functions of the second degree......................... 92

Chapter 7 Trigonometry......................................................................................................... 116 Chapter 8

Linear Algebra........................................................................................................ 164

Chapter 9

Euclidean vector space.......................................................................................... 205

Chapter 10 Descriptive statistics............................................................................................. 224 Model exam paper.....................................................................................................245

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Chapter 1

Introduction to mathematical logic 1.1 Objectives In this chapter you will learn to: • use correctly vocabularies and symbols of mathematical logic. • determine whether a compound proposition is a tautology, a contradiction or a contingent proposition by using the truth table. • determine the negation, the converse, the contrapositive of a proposition. • name usual tautologies (De morgan, double negation, ...). • state in words tautologies expressed in symbols. • determine the negation of quantifiers.

1.2. Key words Proposition Conjunction Equivalence Quantifier Converse

Truth table Disjunction Tautology Propositional function Contrapositive

Negation Implication Contradiction De morgan’s laws

1.3 Theory 1.3.1 Definitions • A logical proposition is a declarative statement which is either true or false. Questions, commands and expressions of a feeling are not logical propositions. • A proposition is represented by letters P,Q ,R, and so on. • The truth value of a proposition is the truth (T or 1) or falsity (F or O) of the proposition. • A propositional function is an expression containing at least one letter (variable) that becomes a proposition when the letter is replaced by a specific value. Example 1 State whether each of the following is a proposition, a propositional function or neither. (a) Kigali is the capital city of Rwanda. (b) 2 is a multiple of 3. 1

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(c) A ≥ 5. (d) Will you come today? (e) Mathematics is very easy. Solution (a) A proposition (b) A proposition (c) A propositional function (d) Neither a proposition, nor a propositional function (e) Neither a proposition, nor a propositional function 1.3.2 Negation of a proposition • Let P be a proposition.

The negation of proposition P is the proposition denoted and defined by: P is true when P is false and P is false when P is true.

• The truth table of the negation of P is shown below: P 1 0

P 0 1

P or

T F

P F T

The negation of P can also be denoted as P or P.

Example 2 State in words the negation of: (a) Kigali in the capital city of Rwanda. (b) 2 is a multiple of 3. Solution (a) Kigali is not the capital city of Rwanda. (b) 2 is not a multiple of 3. 1.3.3 Conjunction of two propositions Let P and Q be two propositions: The conjunction of propositions P and Q is the proposition true only if both propositions P and Q are true and false in all other cases (from the moment at least one of the two propositions P and Q is false). 2

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It is denoted by P ∧Q, sometimes, it is simply denoted by PQ. The number of possibilities in a truth table depends on the number of propositions in the compound proposition. For two propositions P and Q there are four possibilities: • P and Q are simultaneously true; • P and Q are simultaneously false; • P is true and Q is false; • P is false and Q is true; For three propositions P, Q and R there are eight possibilities. More generally, for n propositions P1, P2, ...., Pn, there are 2n possibilities. Below is the truth table of conjunction: P 1 1 O O

∧ 1 O O O

Q 1 O 1 O

p T T F F

∧ T F F F

Q T F T F

p T T F F

Q T F T F

P ∧Q T F F F

Example 3 Draw the truth table of P ∧ Q and P ∧ Q . Solution P

P∧ Q

P∧Q

P ˄Q

T T F F

T F T F

F F T T

F T F T

F F F T

T T F F

T F T F

T F F F

F T T T

1.3.4 Disjunction of two propositions Let P and Q be two propositions: The disjunction of propositions P or Q denoted by P ∨ Q and read P or Q is the proposition false only if both propositions are false and true if at least one of the two propositions P and Q is true.

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Below is the truth table of disjunction.

P 1 1 0 0

Q 1 0 1 0

P∨Q 1 1 1 0

P T T F F

Q T F T F

P∨Q T T T F

P 1 1 0 0

∨ 1 1 1 0

Q 1 0 1 0

P T T E E

∨ T T T F

Q T F T F

Example 4 Draw the truth table of P ∨ Q and P ∨ Q . Solution P

T T F F

T F T F

P F F T T

Q F T F T

P ∨ Q F T T T

P ∨Q

T T F F

T F T F

T T T F

P ∨Q F F F T

1.3.5 Implication (conditional proposition) Consider the following: “If a number is a multiple of 10 then the number is a multiple of 5”. This proposition is made of two parts: • hypothesis – a number is a multiple of 5 • conclusion– the number is a multiple of 10. This proposition is called an implication or a conditional proposition. The propositional connective is “ If .... then ....”, where the blank spaces are filled by P and Q respectively. The proposition “If P then Q” is denoted by “P ⇒ Q” and also read: • P implies Q; • P is sufficient condition for Q; • Q is necessary condition for P 4

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The hypothesis of an implication is called antecedent and the conclusion is called consequent. An implication is false only if the antecedent is true and the consequent is false. Thus, the truth table of P ⇒ Q is shown below. P 1 1 0 0

Q 1 0 1 0

P⇒Q 1 0 1 1

P 1 1 0 0

⇒ 1 0 1 1

Q 1 0 1 0

P T T F F

Q T F T F

P⇒Q T F T T

P T T F F

⇒ T F T T

Q T F T F

The implication obtained by interchanging the antecedent and the consequent of an implication is called the converse. Thus, the converse of P ⇒ Q is Q ⇒ P. The implication Q ⇒ P is called the contrapositive of P ⇒ Q. Example 5 Consider the conditional proposition: "If a number is a multiple of 10 then the number is a multiple of 5". (a) What is the truth value of the proposition? (b) Find the converse of the proposition. (c) What is the truth value of the converse? (d) Find the contrapositive of the proposition (e) What is the truth value of the contrapositive? Solutions (a) True (b) If a number is a multiple of 5 then the number is a multiple of 10. (c) False 5

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(d) If a number is not a multiple of 5 then the number is not a multiple of 10. (e) True 1.3.6 Equivalence (biconditional proposition) Let P and Q be two propositions. The equivalence of P and Q is the proposition denoted by P⇔ Q and read "P if and only if Q"; "P is equivalent to Q" The equivalence of P and Q is the proposition true when both P and Q are true or both P and Q are false. The biconditional proposition is obtained by connecting a conditional proposition and its converse by "and". Truth table

P 1 1 0 0 P 1 1 0 0

Q 1 0 1 0

P 1 0 0 1

1 0 0 1

Q 1 0 1 0

P T T F F

Q T F T F

P T F F T

T F F T

Q T F T F

Example 6 Consider the following: "An angle is right angle if and only if its measure is 90°". Use logical symbols to express the proposition, stating the meaning of each letter. Solution P: An angle is a right angle Q: The measure of an angle is 90° P⇔ Q . 6

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Equivalent propositions

Two propositions are said to be equivalent if and only if they have the same column value. Example 7 Show that the following propositions are equivalent: (a) P and P (double negation) (b) P∧ (Q ∧ R) and (P∧Q) ∧R Solution Considering the truth tables: (a)

T F

F T

P T F

We observe that P and P have identical column value (T,F). Therefore, P and P are equivalent. (b) P T T T T F F F F

Q T T F F T T F F

R T F T F T F T F

Q∧R T F F F T F F F

Since P (Q R) and (P propositions are equivalent.

P∧Q T T F F F F F F

P∧ (Q∧R) T F F F F F F F

(P ∧ Q) ∧ R T F F F F F F F

Q ) R have identical column value, the two

1.3.7 Tautologies A tautology is a compound proposition that is true for all possible truth values of its components. A compound proposition identically false for all possible truth values of its components is called a contradiction. 7

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A tautology is said to be a valid proposition, a contradiction is said to be an inconsistent proposition.

•

A proposition that is neither a tautology nor a contradiction is said to be a contingent proposition.

•

A proposition that is not a contradiction is said to be a consistent proposition

De morgan's laws The tautologies P Q ⇔ P Q and P Q ⇔ P Q are called De Morgan's laws; they are used to transform a conjunction to a disjunction and vice-versa.

Example 8 Show that De morgan's laws are tautologies Solution P

T T F F

T F T F

P F F T T

Q F T F T

P∧Q T F F F

P∧ Q F T T T

P ∨Q F T T T

P ∧ Q ⇔P ∨Q T T T T

Since the column value consists of "T" only the proposition is a tautology P

T T F F

T F T F

P F F T T

Q F T F T

P∨Q T T T F

P∨ Q F F F T

P ∧Q F F F T

P ∨ Q ⇔ P ∧Q T T T T

Since the column value consists of "T" only, the proposition is a tautology. The most often used tautologies are listed below. You should be familiar with them and acquire the ability to operate using them as reference. For any propositions P, Q and R: 1. 2.

Denial of a contradiction: P P Law of double negation: P ⇔P

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3. Law of excluded middle: P P 4. Negation of an implication. P → Q ⇔ P Q 5. Denial of the antecedent: P → (P → Q) 6. Contraposition: (P → Q) ⇔ ( Q → P ) 7. Modus ponens: [(P → Q) P] → Q 8. Modus tollens: [(P → Q) Q ] → P 9. The syllogism law (ex toto): [(P → Q) (Q → R)]→ (P → R) 10. Properties of logical operations: P ⇔ P (reflexive property of equivalence) (P ⇔ Q) ⇔ (Q ⇔ P) (symmetric property of equivalence) [(P ⇔ Q) (Q ⇔ R)] → (P ⇔ R) (transitive property of equivalence) [(P Q) R] ⇔ [P (Q R)] [(P Q ) R]⇔ [P (Q R)] (associative properties of conjunction and disjunction) (P Q) ⇔ (Q P) (P Q ) ⇔ (Q P) (commutative properties of conjunction and disjunction) [P (Q R)] ⇔ [(P Q ) (P R)] (distributive property of conjunction over disjunction). [P (Q R) ] ⇔ [(P Q ) (P R)] (distributive property of disjunction over conjunction) 1.3.8 Quantifiers •

A propositional function is an expression containing at least one variable. For example, x < 5.

For each value given to x, we obtain a proposition. If x = 3, then "3 < 5" is a proposition true.

If x = 9, then "9 < 5" is a proposition false. 9

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•

The set in which the values of the variable are chosen is called universal set, or the domain E.

If we intend to assign to the variable any value in the universal set, then we write ( x E); P(x), where P(x) is the propositional function.

Notice that P(x) is not a proposition but "( x); P(x)" is a proposition, since it is either true or false.

" " is the universal quantifier

If we intend to assign to the variable x in the propositional function P(x) only some values from the universal set E, then we write ( x E); P(x)

" ( x E); P(x)" is a proposition since it is either true or false

" " is the existential quantifier

For example, "x + 5 = 0" is not a proposition since the truth or falsity of "x + 5 = 0" depends on the value of x.

But ( x

•

Order of quantifiers

If an expression contains quantifiers of different types then the order in which the quantifiers are presented matters.

Thus "( x E) ( x ' E); P(x)" is not equivalent to "( x' E) ( x E); P(x)"

To understand this we can make a simple comparison:

"There is a Headmaster for all the members of a school" (1)

"For all member of a school there is a Headmaster" (2). Propositions (1) and (2) are quite different propositions. Proposition (1) is true, the Headmaster of a school is unique.

Proposition (2) is false.

It means that the number of headmasters of a school is equal to the number of students of the school, since everyone has his headmaster. Which is quite illogic.

•

Negation of a quantifier

As the negation connective " " crosses a quantifier, the quantifier changes from universal to existential and from existential to universal.

Thus "It is not true that there is a student in the class without uniform" is equivalent to "every body in the class is in uniform"

); x + 5 = 0" is a proposition.

( x E); P(x) ⇔ ( x E)

P(x).

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1.4 Practice 1.

Find the truth table of each of the following: (a) [(P Q ) ⇒ Q ] ⇒ P Q (b) [ P

Q) ⇒ Q] ⇔ Q

P, Q and R represent the statements

"He is a candidate", "He is preparing for examinations" and "He will pass" respectively. Find appropriate sentences which describe each of the following. (a) P Q

(b)

P R

(d)

R⇔Q

(e) (P ∧ R ) 3.

Use the truth table to show that (P Q ) ⇒ [(P Q ) (Q R )] R is a tautology

Show that (P Q )

Draw the truth table of the proposition (P Q ) ⇒ R Q

Determine, using a truth table whether [(PQ ) Q ] ⇔ (P Q ) is a contradiction or not.

Given the propositions

P Q is a contradiction

P :"John reads New Vision"

Q: "John reads Imvaho"

R: "John reads Uvugizi" Write each of the following propositions in symbolic form. (a)

John reads New Vision or Imvaho but not Uvugizi

(b) John reads New vision and Uvugizi, or he reads neither New vision nor Imvaho. (c)

It is not true that John reads New vision but not Uvugizi

(d) It is not true that John reads Uvugizi or Imvaho but not New vision. 8.

Show that the following propositions are equivalent (a) P ⇒ Q and P

(b) P ⇔ Q and (P ⇒ Q ) (Q ⇒ P) (c) P ⇔ Q and (P Q ) (P

Q) 11

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Given that P represents "Tonny is strong" and Q represents "Peter is weak" Write a compound statement in English represented by each of the following (a)

P Q P (b)

(d)

P ∧Q

(e) P ⇒ Q 10. Write the negation of: (a) x ≥ 2 (b) x = 2 and x ≠ 3 (c) x ≥ 3 or x ≤ – 9 11. Consider the propositional function: If x = 3 then x 2 = 9 (a) Write down the converse of the proposiional function. (b) Write down the contrapositive of the propositional function. 12. Write down the negation of: (a) ( x (b) (

) ( x' )( x

); (x + x' = 0) and (x' + x = 0) ); (x + = x ) and ( + x = x)

13. Complete (a) (x,y) = (a,b) ⇔ … (b) (x,y) ≠ (a,b) ⇔ … (c) ab = 0 ⇔ … (d) ab ≠ 0 ⇔ … 14. Consider the proposition. « If Munyeshuri is a student and is not working hard, then he will fail». Let M, H and S be propositions M: Munyeshuri is a student H: Munyeshuri is working hard S: Munyeshuri will succeed Express the sentence above in logical formula. 15. Find the converse of each of the following implications and its truth value. (a) If integer n is divisible by 8, then the integer n is divisible by 4. (b)

If triangle MNP is such that NP2 = MN2 + MP2, then triangle MNP is right angled at M.

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16. Consider the propositions:

A: Munyeshuri is a student

B. Munyeshuri is revising

C. Munyeshuri will succeed

Give appropriate sentences which describe the following logical formulas:

(a)

A B

Bâ&#x2C6;§ C (b) 17. Write down the negation of: (a)

If y is a factor of 24 then y is a factor of 12.

(b)

If Munyeshuri does not work hard then he will be discontinued or he will repeat.

18. Write down the contrapositive of the proposition. "If triangle XYZ is right angled at X, then YZ2 = XY2 + XZ2 ." 19. Name the tautology used in the following: D â&#x2021;&#x201D; x C and x D

(a)

x C

(b)

Ntaukuri lies when she says that she lies is equivalent to Ntaukuri tells the truth.

20. Consider the following:

" If you work hard, then you will succeed. And you are working hard". (a)

Express the statement in symbolic form, stating the meaning of each letter.

(b)

What conclusion can you draw from the statement?

(c)

Which law allows you to draw such a conclusion?

(d)

Express the law in symbolic form.

21. Consider the following: " If you had worked hard then you would have succeeded. And you failed." (a)

Express the statement in symbolic form, stating the meaning of each letter.

(b)

What conclusion can you draw from the statement?

(c)

Which law allows you to draw such a conclusion?

(d)

Express the law in symbolic form.

22. Consider the following statement: "If you have a job, then you have an income, and if you have an income, then you must pay taxes". 13

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(a)

Express the statement in symbolic form, stating the meaning of each letter.

(b)

What conclusion can you draw from the statement?

(c)

Which law allows you to draw such a conclusion?

(d)

Express the law in symbolic form.

23. The indirect reasoning is based on the principle which states that: «If the negation of an implication is false, then the original implication is true».

Use indirect reasoning to prove that: « If n 2 is an even integer, then n is an even integer»

24. Consider the following:

«Any negative integer is less than or equal to 0". (a)

What is the truth value of the proposition?

(b)

What is the propositional function?

(c)

What is the universal set?

(d)

Express the proposition using logical symbols.

25. Consider the proposition:

"For any natural number x, there exists a natural number y, such that x + y = 0" (a)

What is the truth value of the proposition?

(b)

Express the proposition using logical symbols.

(c)

Express the negation of the proposition in words and by using logical symbols.

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Chapter 2

Binary operators and algebraic structures 2.1 Objectives In this chapter, you will learn to determine if: •

a set is a group under a binary operator.

•

a set is a ring under two binary operators.

•

a set is a field under two binary operators.

•

a set is a vector space under addition.

2.2 Key words Binary operator Identity property Distributive property Subgroup Scalar multiplication

Closure property Inverse property Group Ring

Associative property Commutative property Vector space Field

2.3. Theory 2.3.1 Binary operators A binary operator or a binary operation combines two elements to give a unique third element. Examples of binary operators on real numbers include addition and multiplication. In the case of transformations of the plane, the normal binary operator is the composite, in the sense that one transformation follows another to give a single third transformation. For example, rox is the reflection in the x - axis, roy is the reflection in the y-axis. Therefore, in the central symmetry about the origin, then roy º rox= So : rox followed by roy. y m O

x rox(m)

roy[rox(m)]

Fig. 2.1 15

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A binary operator in a finite set can be given by a table. For example, let E = {a, b, c} and a b c

a a b c

b b c a

be defined on E by means of the following table.

c c a b

This table is constructed so that the value of a same column as b.

b is shown in the same row as a and the

The closure property A set S is closed under a binary operator

if, whenever a and b are in S, a

b is also in S.

The set of non-zero real numbers is closed under multiplication, but not under addition since the product of any two non-zero real numbers is another (unique) non-zero real number, but the sum of any two-non zero real numbers is not necessarily another nonzero real number. For example, the sum of two non-zero real numbers 5 and â&#x20AC;&#x201C;5 is 0 which is not a non-zero real number. The set of numbers of the form a.bc where a is a natural number, "â&#x20AC;˘" is the decimal point and b and c are natural numbers between o and 9 inclusive, is closed under addition but not under multiplication. Example 1 Show that each of the following sets is closed under addition. (a) S = {x

; k

; x = 3k}; the set of all mutiples of 3

(b) E = {x

; a

, b

;x=a+b 2}

Solution (a) Let x and y be any elements of S.

Then there exists k and l, integers, such that x = 3k and y = 3l

x + y = 3k + 3l

= 3 (k + l)

= 3s, where s = k + l integer.

Therefore x + y is in S, showing that S is closed under addition. (b) Let x and y be any elements of S.

Then there exists integers a and b such that x = a + b 2 , and there exists integers c and d such that y = c + d 2 .

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Now, x + y = (a + b 2 ) + (c + d 2 ) = (a + c) + (b + d) 2 = A + B 2 , where A = a + c and B = b + d are integers.

This shows that E is closed under addition.

The associative law Set S is associative under the binary operation a

if, for all a, b and c in S,

(b c) = (a b) c

For Example, the set is associative under both addition and multiplication and so is any subset of , but is not associative under either subtraction or division since; a + (b + c) = (a + b) + c for all real a, b and c a (bc) = (ab) c for all real a, b and c a – (b – c) (a – b) – c, in general and a: (b:c) (a:b):c, in general Example 2 Show that is associative under the binary operator all real numbers a and b.

where a

b = 2ab + a + b, for

Solution If a, b and c are real, a

c) = a

( 2bc+ b + c)

= 2a (2bc + b + c) + a + (2bc + b + c)

= 4 abc + 2 ab + 2ac + 2bc + a + b + c

Also (a

c = (2ab + a + b)

= 2 (2ab + a + b ) c + (2ab + a + b) + c

= 4 abc + 2ab + 2ac + 2bc + a + b + c

Therefore, a (b c) = (a

c and so

is associative under

If the binary operator is given by a table, it is necessary to check all the ordered triples in order to conclude whether the set is associative or not under the given operator. If n(E) = p, then there are p3 orderded triples. 17

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The identity element A set A under a binary operator said to contain an identity element e if, for all a in A, a e = a. For example, , Q, N and all have zero as the identity element with respect to addition and 1 as the identity element with respect to multiplication. The set of non-negative even numbers { 0, 2, 4, 6,....} has 0 as the identity under multiplication. The set of positive odd numbers {1, 3, 5, 7....} has 1 as the identity under multiplication but no identity under addition. Example 3 Find the identity element of with respect to the operation defined on R by a b = a + b – 2ab for all a, b in . Solution Let e be the identity element, if it exists. Then a e = a; a + e – 2ae = a e (1 – 2a) = 0; e = 0 or a= 1 2 Hence, if a e = a for all a in , then e = 0. We must now check that e = 0 is in fact the identity element. a o = a + o – 2a (o) = a and o a = o + a – 2(o)a = a Therefore 0 is the identity element.

Properties The identity element, if it exists, is unique. In fact , let e and e' be identity elements for the operation defined on a set A. Then e e' = e' since e' is an identity element. Hence e' = e and so the identity element is unique.

Inverse elements Let A be a set with binary operator and with an identity element e. The element a' in A is said to be an inverse of the element a in A if a a' = a' a = e. Also, if a' is an inverse of a then a is an inverse of a'. For example, the inverse of a real number a under addition is its opposite –a, since a + (–a) = (–a) + a = 0, where 0 is the identity element under addition. 18

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The inverse of the non-zero number a under multiplication is its reciprocal a–1 or 1a , since a.a–1 = a–1.a = 1, where 1 is the identity element under multiplication. ( 0 has no inverse under multiplication). Example 4 Find the inverse of the real number a with respect to the operation defined on by a b = ab + a + b

for all a, b in .

Solution First we must find the identity element. Let e be the identity element, if it exists. Then a e = a and so a e + a + e = a. Thus e(a + 1) = 0 which gives a = –1 or e = 0. But a e = a for all a in and so e = 0. We must now check that 0 is the identity element a o = a(o) + o = a; o a = o(a) + o + a = a. Therefore, o is indeed the identity. Now let a–1 be the inverse of a if it exists. Then a a–1 = e, a.a–1 + a + a–1 = 0; a–1 ( a + 1) = 0; – a–1 = a provided a ≠ –1.

a+1

Finally, we must check that this is indeed the inverse of a (a ≠ –1).

a a+1 –

a + a2 + a – a a+1

– 2

a a+1 –

= 0. a

–

Therefore, the inverse of a is a + 1 provided a ≠–1, if a = –1, there is no inverse. 2.3.2 Groups

Definition A group is a set of elements G together with a binary operator which satisfies the following properties: (i)

G is closed under :

Whenever a, b are in G, a b is in G. 19

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(ii)

G is associative under

(iii) There exists an identity element e in G such that for all a in G, a e = e a = a. (iv) Every element a in G has a corresponding inverse element a –1 in G such that a a–1= e and a–1 a = e .

For example, the set

of integers is a group under addition:

(i)

is closed under addition since the sum of any two integers is also an integer.

(ii)

is a subset of

and so must be associative under addition.

(iii) 0 is the identity. (iv) If a is an integer, then so is (–a) and a +(– a) = (–a) + a = o Therefore, (–a) is the inverse of a for all integers a. Hence the set

is a group under addition.

Properties of a group 1.

Cancellation law Let G be a group under . Let a, b, x be in G. If a x = b x then a = b Proof: a x = b x; (a x) x–1 = (b x) x–1 (closure property) a ( x x–1) = b (x x–1) ( associative property) a e = b e (inverse property). a = b (identity property) 2. The inverse of an element of a group is unique. Proof: Suppose a' and a'' are two elements of the group, both inverses of the element a. Then a a' = e (inverse property) and a a'' = e (inverse property) a' = a'' (cancellation law). Therefore the inverse of a is unique. 3. Let a, b, x be elements of a group G under , if a x = b, then x = a–1 b. Proof: a x = b; a–1 (a x) = a–1 b ( closure and inverse property) (a – 1 a) x = a – 1 b (associative property) e x = a –1 b (inverse property) x = a –1 b (identity property) 20

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If (G ) is a group, then for any x, y in G, ( x y)–1 = y–1 x–1 Proof: a x = h; (x y) (y – 1 x – 1)

= x y (y y – 1)] x –1 (associative property) = x e x – 1 (inverse property)

= (x

x–1 (identity property)

= e ( inverse property). Similarly, (y–1 x–1) ( x y) = e Therefore, ( x y)–1 = y–1 x–1. Note: Abelian group A group(G, ) is said to be abelian (commutative) if x y = y x, for all x, y in G,.

Subgroup Let (G, ) be a group. If H is a subset of G, and (H, ) is a group, then (H, ) is said to be a subgroup of (G, ). Every subgroup of G must contain the identity element of G. Thus any group G contains at least two subgroups {e} and G. Since these subgroups are obvious, we distinguish all other subgroups as proper. A subset H of a group (G, ) is a subgroup of G if and only if: (i) H is closed under ; (ii) The identity e of G is in H; (iii) For all a ∈ H, a–1 ∈H. There is no need to check the associative property since any subset of (G, ) must be associative under the operation .

Isomorphism of groups Let (G, )and (H,o) be groups. If there exists a bijection ø: G → H such that ø (g1)= h1, ø(g2) = h2 and ø (g1 g2) = h1oh, then groups (G, ) and (H, o) are said to be isomorphic. For example, (R, +) and (R+0,•) are isomorphic groups.Two finite groups are clearly isomorphic if their “group tables” have exactly the same structure. Example 5 Show that E = { a + b 2 ; a ∈ Z, b ∈ Z} is a subgroup of R under addition.

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Solution (i) E is closed under addition:

In fact, let x ∈ E and y ∈ E.

Then there exists a and b in Z such that x = a +b 2

x + y = (a + b 2 ) + (c + d 2 ) = (a + c ) + (b + d) 2 , where a + c and b + d are integers ( since the set of integers is closed under addition).

Therefore, x + y ∈ E, showing that E is closed under addition. (ii) The identity element 0 ∈ E in fact 0= 0 + 0 2 . (iii) Let x = x = a +b 2 ∈ E. Then – x = –(a + b 2 )

= (– a) + (–b) 2 = – x ∈ E

It follows that E is a subgroup of R.

2.3.3 Rings

Definition A ring ( A, +, •) is a set A together with two binary operators + and •, which are called addition and multiplication, defined on A such that the following properties are satisfied: (i)

(A,+) is an abelian group

(ii) Multiplication is associative (iii) Multiplication is distributive over addition, that is a(b+c) = ab + ac and ( a+b)c = ac + bc, for all a, b, c in A.

For example (

, +, •), ( , +, •), ( , +, •) are rings.

Properties If ( A, +, •) is a ring with additive identity o, then for any a, b, in A we have: (1) o.a = a.o = o

(2) a(–b) = (–a)b = –(ab)

(3) (–a) (–b) = ab

Infact: (1)

o.a = a.(o + o) = a.o + a.o Then by the cancellation law for the additive group (A,+), we have o = a . o

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Likewise, oa = (o + o)a

= oa + oa

Implies that o.a = o

This proves property (1)

(2) In order to understand the proof of property 2, you must remember that, by definition –(ab) is the element that when added to ab gives o. Thus to show that a(–b) = –(ab); a(–b)+ ab = o. By the distributive property, a(–b) + ab = a (–b + b) =a.o =o Since a.o = o by condition (on property 1) Likewise, (–a)b + ab = (–a+a)b = ob

(3) For property 3, note that (–a)(–b) = –[–(ab)]

By property 2, repeated, –[a(–b)] = –[–(ab)] and –[–(ab)] is the element that when added to –(ab), gives o. This is ab by defination of –(ab), and by the uniqueness of an inverse in a group.

Thus, (–a) (–b) =ab.

Integral domain

If a and b are two non-zero elements of a ring (A, + , •) such that ab = 0, then a and b are said to be the divisors of 0. A ring in which the multiplication is commutative is a commutative ring. A ring with a multiplicative identity is said to be a unitary ring. A unitary, commutative ring containing no divisors of 0 is said to be an integral domain. Thus ( , +, •) is an integral domian. An example of a ring with divisors of zero is the set m, constructed as follows:

of all residue classes, modulo

Two integers x and y are said to be congruent modulo m if and only if x - y is a multiple of m (in other words, the remainder of the division of x by m is equal to the remainder of the division of y by m). We write x ≡ y (mod m). It can be shown that the relation « – – – ≡ ––– (mod m)» is an equivalence relation. It partitions into classes, called residue classes modulo m. The set of all residue classes, modulo m is denoted by Addition and multiplication are defined in • • If x ∈ m and y ∈ m,

as follows:

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• • • then: x + y = x + y • • • and x y = x y • • • For example, 3 = {0 , 1 , 2 }, • where 0 = { ..., –6, –3, 0, 3, 6, ...} the remainder of the division of each of the numbers in this set by 3 is 0. • 1 = {..., –5, –2,1, 4, 7, ...}: the remainder of the division of each number of this set by 3 is 1. • 1 = {..., –4, –1, 2, 5, 8, ...}: the remainder of the division of each of the numbers in this set by 3 is 2. • • • • • 1+2=1+2 =3=0 • • • • • 2 ×2 =2+2 = 4 = 1 In general, (

, +, •) is a ring, but not an integral domain.

2.3.4 Fields

Definition A field ( F, +, •) is a ring such that (i) (F*, +) is an abelian group. (ii) (F*, •) is an abelian group, where F* is the set F –{o} o: identity element of addition. (iii) Multiplication is distributive over addition. Example 6 (Q, +, •) and ( , +, •) are fields. ( inverse property does not hold).

,+,•) is not a field, since (

*, •) is not a group (the

Properties Let (F, +, •) be a field. Then; (1) For all a ∈ F, a.o = o.a = o (2) For all a and b in F, ab = o implies a = o or b = o 24

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(3)

In any field F, for any non zero element a of F, for any real number b , there exists a unique element b such that ax + b = o. Infact, in the group (F,+), ax + b = o ⇔ ax = –b, and since the non zero element a has an inverse a–1(ax) = a–1(–b) ⇔ x = a–1(–b).

2.3.5 Vector spaces

Scalar multiplication Let E and K be two sets. A scalar multiplication or an external multiplication defined on E, in any mapping.

:K×E→ E (α, x)

α. x

Definition of a vector space A set V, whose members are u, v, w, ... , called vectors, is said to be a vector space over , or an -vector space if and only if V is an abelian group under addition together with a scalar multiplication that satisfies the following four properties: (αβ)u = α(βu);

(α +β)u = αu +βu;

α(u + v) =αu + αv

1. u = u

For example, consider the abelian group ( 2, +) = {(a,b);a ∈ , b ∈ } which consists of ordered pairs of real numbers under addition by components, that is (a, b) + (c, d) = (a + c, b + d).

Define scalar multiplication for scalars in by α (a,b) = (αa, αb). With these operations, ( 2, +) becomes an -vector space. The properties of a vector space are easily checked.

Properties (1) (2)

For any vectors u and v in a vector space V, a + u = a + v implies u = v .

x + u = v; x = v – u

= v + (– u).

(3) For any vector u in V, o.u = o. Infact, let ∈ ( + o) u = u + ou;

and u ∈ V

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But + o = . Therefore, ( + o) u = u; u + o u implies o.u = o; (4) For any scalar , o = o Infact, ( u + o) = u + o Since u + o = u, (u + o) = u u + o = u implies o = o

2.4 Practice 2.4.1 Binary operators 1.

Determine whether the set of all non zero real numbers is closed or not under: (a)

addition

(b) multiplication 2.

Determine whether or not: (a)

is closed under division

(b)

* is closed under multiplication.

Binary operator

is defined in E = {a, b, c} by

a b c (a)

a a b c

b c b c c a a b

Calculate: (a a) b and a (a b)

(b) Find x and y such that:

(i) x

b = b

(ii) y

b=a

Consider the operations

x y = LCM (x, y),

and defined in A = { 1, 2, 3, 4, 6, 12} by: x y = HCF (x, y)

(a) Draw the tables for the operations (b) Find the identity element for each operation (c)

Determine whether the inverse property is verified or not

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Find the inverse of the real number a with respect to operation defined on in each of the following:

(a) a

(b) a b = ab + 2a + 2b

b = 2a b + a + b

Binary operators and are defined in by a b = Inf (a,b) and a b = (a b) + 2, where Inf (a,b) denotes the smaller of a and b, or the common value if a = b. Calculate 4 7 and 6 6.

Given that A = {x, y}. Write down the elements in the power set of A and draw the table for union in the power set of A.

8. In , x + y =

x – 3xy + 5 2

and x y = x2 + 2xy + y – 8

Calculate: (a) 3 ⊥( 1 0) (b) (3 ⊥1) (3 ⊥ 0). 9.

Determine whether the binary operator defined in is commutative or not in each of the following cases:

(a) x y = x – y

(b)

y = x2 + y2

(d)

y = (x – y)2

10. Binary operator T is defined in the set of all natural numbers by x T y = xy + yx + 3:

(a)

Calculate: 23 and 32

(b)

Determine whether T is commutative or not.

11. Binary operator T is defined in the set of all real numbers by x T y = xy – 2.

Determine whether T is associative or not.

12. Determine the identity element, if it exists, for the binary operator T defined in by aTb = ab + 2a + 2b + 2. 13. Complete the following table so that the binary operator defined in

A = {a, b, c} contains an identity element. a a b c

c c a a a b

14. Binary operator is defined in the set of real numbers by x y = xy – (x + y) + 2, where x ≠ 1 and y ≠ 1. Find the identity element and the inverse of –1. 27

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15. Binary operators and T are defined in the carteian plane by: A B = I, Where I is the midpoint of line segment [AB]

ATB = J, where J is the symmetrical point of A about point B.

(a)

Is the operation

(i) commutative?

Is the operation T:

(b)

(i) commutative

(d)

(ii) associative T (ii) associative?

distributive over T?

Is T distributive over ?

2.4.2 Groups is defined in * by x

y = xy . Show that ( *, ) is an abelian

Operation group.

Let S be the set of all real numbers, except –1. Define on S by a b = a + b + ab (a)

Show that ( S, ) is a group.

(b) Find the solution of the equation 2 x 3 = 7 in S. 3.

Let * be the set of all real numbers, except o. Define where a is the absolute value of a.

on * by a

b = a b,

(a)

Determine whether is associative or not (b) Determine whether verifies the identity property or not.

(c)

In ( *, ) a group?

Let H = {1, –1}. Determine whether (H, multiplication) is a group or not.

E is the set of even integers. Detemine whether : (a)

E is a subgroup of ( , +) or not.

(b) E is closed under multiplication or not. 6.

Explain the properties used in solving in the set of real numbers, the equation 5x + 2 = 0.

The binary operator T is defined in –{1} = E by x T y = xy – x – y + 2.

(a)

Show that (E,T) is an abelian group.

(b)

Solve for x the equation 1 T x = 3

of integers is defined by S = {x ∈ Subset S of the set that is S is the set of all multiples of 4.

Show that (S, +) is a subgroup of ( , +).

,( k

); x = 4k},

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List three subgroups (by their tables) of the group defined by the table below: x y z t

x x y z t

y z t y z t z t x t x y x y z

10. Operation T is defined in the set of real numbers by a T b = a2 + b2 ( , T) is not a group. why? 2.4.3 Rings-fields-vector spaces 1.

Simplify ( x + d) (x2 – dx + d2)

Simplify (3√2 + 1) (3√4 – 3√2 + 1)

Prove that the set { a 3√4 + b3√2 +c , where a ∈ , b ∈ and c ∈ } is closed under multiplication. Find the inverse of 3√2 + 1 and also that of 3√2 – 1 .

Given vectors u = (3, –1, 2) ; v = (1, 1, –3) and w = (–1, 2, 1). Find: (a) 2u – v + 3w (b)

x such that u + 2 x – w = o

Solve the equation x2 – 5x + 6 = 0:

(a)

in the set of real numbers

(b)

in the set

Find the remainder of the division of 54 × 69 × 137 by 7.

Solve in

Find the remainder of the division of 1323 × 2741 by 8.

Read the following carefully.

the equation x3 – 2x2 – 3x = 0

Let x and y be any real numbers and Z the difference x – y.

We have x = y + z

Multiplying both sides by x – y, we obtain (x – y) x = (x – y) (y + z)

Expanding: x2 – yx = xy + xz – y2 – yz

Noticing that x2 – xy – xz = xy – y2 – yz, where x is a common factor on the left hand side and y is a common factor on the right side, we have,

x(x – y – z) = y (x – y – z). 29

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From the cancellation law, we obtain x = y. So we have proved that « any two real numbers are equal», which is not true.

Find the mistake in the proof above.

10. Find the divisors of zero in the following rings: (a) (

, + , •)

(b) (S, , ∩ ) where S is the power set of E = { a, b, c},

is the symmetric difference of sets,

∩ is the intersection of sets.

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Chapter 3

The set of real numbers 3.1 Objectives In this chapter you will: • revise what you have learnt about numbers. • simplify expressions using the laws of exponents, radicals and logarithms to base 10. • apply properties of absolute values. • solve simple logarithmic and exponential equations.

3.2 Key words nth root Rationalization

Surd Logarithm

Power Absolute value

Exponent Radical

Base

3.3 Theory 3.3.1 Number system: revision and extension You already know about the following sets of numbers: • Natural numbers = { 0, 1, 2, 3, ....} is closed under addition and multiplication. • The integers: = { ..., –3, –2, –1, 0, 1, 2, 3, .....} is closed under addition and multiplication.

Rational numbers A rational number is any number that can be written in the form ab , where b≠0, and as a whole a ∈ . The set of all rational numbers is denoted by . Thus = { ab ; a ∈ and b ∈ and b ≠ 0}. Common fractions are rational numbers, for example 3 . 4

Decimals which are terminating or have a recurring sequence of digits are rational numbers.

Real numbers The rational and irrational numbers together make up the set of real numbers, denoted by . 31

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An irrational number is any number that cannot be expressed in the form ab , where a ∈ , b ∈ and b ≠ 0. Some irrational numbers you know about are π, 2 , 5 , ... When expressed in decimal form, irrational numbers are non-terminating and nonrecurring. Example: π = 3.14115926535897932384626433832 ... 2 = 1.414213562373095048801688724097 ...

The sets , and

are all subsets of . In fact , ⊂ ⊂ ⊂ .

3.3.2 Algebraic fractions Remember that the denominator of a fraction, ab cannot be zero. x O

is undefined. We usually write this as xy , y ≠ o. We have:

a b

= a. 1b , b ≠ 0;

a b

+ dc = ad + bc ; b ≠ 0, d ≠ 0

a.c b d a b c d

4. 5. 6.

ac = bd ; b ≠ 0, d ≠ 0

= a . dc = ad , b ≠ o, c ≠ o d ≠ o b bc

If b ≠ o and c ≠ o then a.c = a b.c b

a b

–

= –a = – ba b

Example 1 Simplify (x–2) (x+1) + 3(x –2) , x ≠ 2 x–2

Solution (x–2) (x+1) + 3(x –2) x–2 (x–2) (x+1 + 3) x–2

= x + 4, x ≠ 2

3.3.3 Powers The rules used to manipulate exponential expressions should already be familiar to you. 32

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These rules may be summarized as follows 1.

Product rule: (a m)(an) = a m+n,

2. 3.

Quotient rule: a n = a m–n, a ≠ o Power of a power rule: (a m)n = amn

Power of product rule: (ab)m = a mb m

m am Power of a quotient rule: a = bm , b≠o b

For the quotient rule to hold when m = n, we must agree that a m = a m–m = a°, a≠o. Then it is natural to define: a° = 1 (a≠o). Now, 1 n = an = ao–m = a–m (a≠o), which gives the meaning to negative exponents. m

In the expression a = y, a is referred to as the base, n as the exponent or index and y as the power. Thus, in order to get a given power, we have to raise a certain base to a given exponent. When n is an integer, we say that an is an integral power. n

Example 2 –2 3 Express the following without brackets or negative exponents: 2x–3

Solution

2x–2 y–3

23 x–6

= y–9 = 8y6 x

3.3.4 nth roots: rational exponents In general, if n is an even positive integer, then in the real number system, (a) If a >0, a has two nth roots, one positive and one negative. (b) If a = 0, a has one nth root which is zero. (c) If a < 0, 0 has no nth root. If n is an odd positive integer (not 1) then in the real number system, (a) If a > 0, a has one nth root which is positive. (b) If a = 0, a has one nth root which is zero (c) If a < 0, a has one nth root which is negative If n is a positive integer other than 1 or 2 the radical sign n is used to indicate an nth root of a real number.

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If n = 2 just the radical “ 4 81 = 3 and

– 4 81

’’ is used. For example, 81 has two fourth roots denoted by

= –3, since (–3)4 = 81 and 34 = 81.

–4 has no square roots (in the set of real numbers) 3 – –8 has one cube root ( 8) = –2

When an nth root is irrational, it is referred to as a surd. For example 2 , 3 10 are surds. Numerical expressions containing surds are also referred to as surds. For example, 3 5

is a surd.

The same rules that hold for integral exponents also hold for rational exponents. These laws then also hold for surds. When simplifying surds, it is sometimes convenient to obtain rational exponents first. If the product rule for exponents is to hold, then 1

(a 2 ) (a 2 ) = a

1 1 2+ 2

= a1

=a 1

But we know that ( a ) ( a ) = a (a ≥ 0). Thus a 2 is a square root of a (provided a ≥0). 1

To avoid any confusion, take a 2 = a , a ≥ 0, that is a a 2 is taken to be the positive square 1

n root of a (a > 0). More generally, if n is a positive integer a 2 is defined by a n = a .

Note that if n is even , a must be a non-negative number, if n is odd, a may be any real number. Next we define a

m n

= a

1 m n

= na

a = n a which exists for all if n is odd, and exists for a≥0 if n is even. m

Example 3 1.

Evaluate :

(a) 27

5 3

(b)

0.064

–4

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Solve the following equations: 3

(a) 81–x = 42x + 3

(b) a 2 = 8

Solution 1.

5 3

(a) 27

5 3 3 3

= 35 = 243

(b) 0.064

= 42x + 3

= 22(2x + 3)

3 – 3x = 4x + 6

7x = –3

x = – 37 2

(1 – x)

23 – 3x = 24x + 6

–

4 3

–

–3

4 3

10 4

625 16 3

a 2 = 8 3 3 a 2 =2

a = 23

2 3

= 22 =4

= 16

4 3

= 52 4

(b)

–3

= 4–4.104

(a) 81–x

= 4 .10 3

= 4

= 16

–

4 3

2 3

3 2

= ± 43

= ± 64. 35

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Rationalization If a binomial surd is multiplied by its conjugate, the result is rational. We make use of these results when simplifying certain surd expressions by “rationalising the denominator”. Example 4 Express

3 3√2 –2 √3

with a rational denominator and in its simplest form.

Solution

3(3√2 + 2 √3 ) (3√2 – 2√3 ) (3√2 ) + 2√3 )

3(3 √2 + 2 √3 ) 18 – 12

3(3 √2 + 2 √3 6

√2 + 2√3 = 3 . 2

3.3.5 Logarithms

Definitions Each exponential expression has a corresponding logarithmic expression. log a x = y if and only if ay = x, where a > 0 and a ≠ 1. The logarithm, to base a, of x is the exponent to which base a is raised to obtain x. If a = 10, then loga x is denoted by log x and is said to be the decimal logarithm of x. Thus log x = y ⇔10y = x.

Characteristic and mantissa of a logarithm The logarithm of a number has two parts: •

The integral part before the decimal point is called the characteristic. This can be either positive or negative and it is the power of 10 when the number is written in standard form.

•

The decimal part after the decimal point is called the mantissa and it is always positive.

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Example 5 Find the values of x in each of the following cases (a) log2 32 = x

(b)

log3 x = 5

(b)

log3 x = 5 ⇔35 = x

(c)

logx 625 = 2

Solution (a) log2 32 = x ⇔ 2 x = 32 ⇔ 2x= 25

⇔ x = 243

⇔x=5

x = 25

(x > 0, x ≠ 1)

Properties of logarithms ( to base 10) 1.

log 1 = 0

log 10 = 1

log 1 = –log x, where x > 0

log ( xy) = log x + log y, where x > 0 and y > 0 more generally,

log (x1 x2 ....xn) = logx1 + logx2 +.....+ logxn, where x1> 0, x2 > 0...., x n >0.

log (xn)

= log (x.x.....x)

= log x + log x......+ log x

= n log x

Therefore, log (xn) = n log x

log ( y )

= log (x. y ), x > 0, y > 0

= log x + log y

= log x + log y–1

= log x – log y

Therefore, log xy = log x – log y

Note: (logx)n ≠n log x

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Example 6 Solve in the set of real numbers, the equations: (a) 2 log x = log (x + 6) (b) log (x + 3) = log (5 – x) Solution (a) 2log x = log (x + 6)

(b) log (x + 3) = log (5 – x)

Restrictions on x:

x > 0 and x + 6 > 0

x > 0 and x > –6;

x > 0 ...... (i)

2log x = log (x + 6); log x2 = log (x + 6)

x2 = x + 6

x2 – x – 6 = 0

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

⇔ x+3>0 5–x >0 x + 3 = 5 –x

– ⇔ x > 3 x< 2 2x = 5 – ⇔ 3 < x < 5 x=1

The solution set is S = {1}.

x+2 =0

or x – 3 = 0

Either x = –2, 0r x = 3 ...... (ii)

The only value to consider is x = 3

The solution set is S = {3}.

3.3.6 Absolute value

The one-to-one correspondence between the real numbers and the points on the number line is familiar to us all. Corresponding to each real number there is exactly one point on the line. Corresponding to each line on the line there is exactly one real number. 0 M A x 0 1 Intervals are particular subsets of the set of real numbers. If a and b are two real numbers such that a < b, then: 38

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Closed interval:

[a, b] = {x∈ ; a ≤ x ≤ b}: The set of all real numbers between a and b inclusive of both a and b. a

Open interval

]a, b[ = {x∈ ; a < x< b}: The set of all real numbers between a and b, exclusive of both a and b.

a 3.

Half closed intervals:

[a, b[ = {x ∈ ; a ≤ x<b} ]a, b] = {x ∈ ; a <x ≤ b} 4.

Half-lines

]–∞, a] ={x ∈ ; x ≤ a} –∞ ]–∞, a[ ={x ∈ ; x < a} [a, + ∞[ = {x ∈ ; a ≤ x}

∞

–

a a

]a, + ∞[ = {x ∈ ; a < x} 5.

+∞ +∞ +∞ +∞

= ]– ∞, + ∞[

Order properties 1.

If a and b are any two real numbers, then either a < b or b < a or a = b.

The sum and product of any two positive real numbers are both positive.

The opposite of a negative number is positive

If a < b then b – a > 0,

If a > b then a – b> 0 5. If a < b and c is any real number, then a + c < b + c that is we may add (or subtract) any real number to (or from) both sides of an inequality. 6. (i) If a < b and c > 0, then ac < bc

(ii) If a < b and c < 0, then ac > bc

That is, we may multiply (or divide) both sides of an inequality by a positive real number, but when we multiply (or divide) both sides of an inequality by a negative real number, we must change the direction of the inequality. 39

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If 0 < a < b then 0 < a2 < b2

7. (i)

(ii) If a < b < 0 then 0 < b2 < a2

That is we may square both sides of an inequality if both sides are positive. However, if both sides are negative we may square both sides but we must reverse the direction of the inequality and then both sides become positive.

If one side is positive and the other is negative we cannot square both sides

For example, –3 < 4 and (–3)2 > 42, but – 4 < 3 and (– 4)2 > 32

(i)

(ii) If a < b < 0 then 1 < 1 < 0.

That is, we may take the reciprocal of both sides of an inequality only if both sides have the same sign and, in each possible case, we must reverse the direction of the inequality.

1 If 0 < a < b then 0 < 1 < a b b

Example 7 Solve the inequality 2x + 1 ≤ 4x + 5 Solution 2x + 1 ≤ 4x + 5; 2x – 4x + 1 ≤ 4x + 5 – 4x: subtracting 4x from both sides.

–

2x + 1 ≤ 5; calculating

–

2x + 1 – 1 ≤ 5 –1: subtracting 1 from both sides

–

2x ≤ 4: calculating

2x 2

–

≥ –2 : dividing both sides by –2 (the direction of the inequality changes)

x ≥–2

Solution set s = [ –2, + ∞[.

Absolute value Sometimes, we order numbers according to their size. We denote the size or absolute value of a real number x by |x| = the distance from the origin to the point representing the number. Thus |x| = –x if x ≥ 0 |x|2 = x2 x if x < 0; 40

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Properties 1.

In general, if a is any positive real number, If |x| < a then –a < x < a.

Infact, |x|< a; |x|2 < a2; x2 < a2;

x2 – a2 < 0; (x –a) (x + a) < 0 x x–a x+a x2 – a2

∞

–

a – – – – 0 + + 0 – –

a 0 + 0

+∞ + + +

x2 – a2 < 0 if and only if –a < x < a 2.

If |x| > a then x > a or x < –a

The following relations are true for all real numbers a and b.

– (i) |– a| = |a| (ii) |a| ≤ a ≤ |a|

|a|

(iii) |ab| = |a|.|b| and | b | = |b|, where b ≠ 0

(iv) |a + b| ≤ |a| + |b| (v) |a – b| ≥ |a| – |b|

Example 8 Solve the inequality:

(a) |5x + 3| ≤ 2

(b) |5x + 4| > 12

Solution (a)

2 ≤ 5x + 3 ≤ 2

–

–2–3 ≤ 5x + 3 – 3 ≤ 2 –3; –5 ≤ 5 x < – 1

–

5≤ 5 ≤ 5 ≤ 1; 5 –

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–1 ≤ x ≤ – 1 5

–

The solution set is S = [–1, – 5 ]

(b) |5x + 4|> 12; 5x + 4 < –12 or 5x +4 > 12; 5x < –16 or 5x > 8; 8

x < – 16 or x > 5 5

The solution set s = ]–∞,– 16 [ ∪] 8 , + ∞[ . 5

3.4 Practice 3.4.1 Number system 1. Copy and complete the table by or . 5 1–2 5 –

4 –

10 7 3 8

State whether each of the following statements is true or false. If false, write down the correct satement: (a)

The product of two rational numbers is a rational number.

(b)

The difference of two rational numbers is always a rational number.

Simplify (x2 + 2xy + y2)½, x > 0, y > 0

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Express with a radical sign and simplify where possible: (a)

1 3 8

3 8

1 3

(b) a–1

(c)

3–1

(a + b)

1 10

3. Simplify: (a) 4.

(x ) (x y) .x (b) (xy)–2n y–m 2 m–1

–2m

(a ) (a b) (xy)–2n. b–m 4 n+1

–3m

(3x)–2 (c) 3x–2

2 3

Simplify, precising the condition for the simplification to be valid. (a)

4x2 – 1 (b) 4x2+ 2x

(c)

2(b – 2)2 – 2b + 4 b2 – 5b + 6

(a + b)2 + 4(a + b) a+b

5. Simplify: (a)

3ax + 2 – 2ax (3n)2 + 32n – 1 (b) x x–2 3a – 2a 9n

(c)

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

6. Simplify: x2m– 1 . y3.z1 – m (12a2 b6)2 (a) m – 3 2–m 2 –2m (b) 4 7 x .y z 8a b 7. Simplify (a) 8.

62x . 9x (b) (2a3)2 (4a4)3 27x . 8x. 2x

Express as one power and calculate, where possible,

(1)–3 .8 (b) 9 3. 3

(a) (–2)–12 . 2 9.

x–1

2x + 1

Find the value of x if.

(a) 32x – 4 = 9–x + 2 (b) 4x = 12(2x) – 27 43

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10. Find the value of x if:

(a)

2 3 ( 1 )3x – 1 (b) 4 – 9 =0 x

92x – 3 = 27

x 2

– 3x

3.4.3 nth roots 1. 2.

Find rational numbers a and b if ( a + b 2 ) ( 2 + 3 3 ) = 1 Find the values of rational numbers a and b for which:

(a) (a + b 3 ) (5 – 2 3 ) = 1 (b)

Solve for x the equation:

(a)

(b) 36x + 36x – = 97

2 + 4 (2– ) = 5 x 2

(a + 2 3 ) (4 – b 3 ) = 4

2 3

x 2

2 3

4. Simplify: 12 (81x4)3

(a) 5 5.

8x2 5 8x2

4 324

Simplify: (a)

27 3 . 49 2 . 16 4 (b) 5 2432 (b)

6. Express 6 5 and 4 3 with the same index. 7. Simplify:

1 x4 – 3 3 x + 3 x2 + 3 9 x for x = 4 3

8. Simplify: (a) ( 9.

3 4 3 4

) ( 5 16) 3 5 4 3 3125 ( 2 )

(b) 5

8 256 6 729 . 5 3125 3

Simpplify 1

(4x4) 2 8 +4 (a) 4 11 (b) 2 x .x. x 8 +4 10

7 15

10. Given that 5 x = 3 , find 14 x6 . 44

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3.4.4 Logarithms 1.

The pH of an aqueous solution is the opposite of the logarithm to base 10 of the H+ concentration in the solution, that is pH = – log [H+]. Find the H+ concentration of an aqueous solution with pH. (a) 2 (b) 11 2.

Solve for x the equations: 9x – 3 × 81(1– x) = 27x

(c)

242 1 – 1 = 243 3

(b) log (2x – 11) – log 2 = log 3 – log x

Simplify completely: (a)

(a)

3log 2 + log 24 – log 3 13 log 64 – log 128–1 2

(b)

1 log25 – log 5–1 log 625 + log 5 –1

Find, without using a calculator, the exact value of: 1

(a) log3 9 – log3 3 (b) log3 3 9 5.

Find the value of x if:

(a) log (3x + 4) – log (x – 2) = 3 log 2 (b) log (7x + 1) = log (x – 2) 6. If y = 3x2, find a linear expression connecting log x and log y. 7.

Write an expression equivalent to log y = 3 – 2 log x without using logarithm.

Given that log 2 = 0.30103, find:

(a) log 8 (b) log 2048 (c) log

125 (d) log25 2

Simplify.

log (2 + 2 + 2 ) + log (2 – 2 + 2 )

10. Find the value of x if: (a) log (x + 1) + log (x – 2) = log (3x – 1) (b) 2 log (x – 1) = log (3 – x) 45

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3.4.5 Absolute value 1.

Solve fox x:

(a) 2(3x + 2) > 5

(b)

1 – x ≥ 7

(c)

1 – 3(x + 2)≤10

Find the values of x that satisfy the inequality.

(a) 2(3x + 2) > 5

(b)

1–x≥7

1 – 3(x + 2) ≤ 10

(c)

Decide whether each of the following statements is true or false. If it is false, give an example to confirm that it is false.

(a) If x > 2 then x2 < 4

(b)

x2 > 4 then x > 2

1 (c) If x < y < 0 then 1 > y x

Write in interval notation.

(a)

E = {x ∈ ; 2 ≤ x ≤ 3 or 4 < x < 7}

(b)

E = {x ∈ ; |x| ≥ 5}

(c)

E = {x ∈ ; –3 ≤ 2x + 5 < 4}

Represent on a number line.

(a) [–3, 0]∪ [2, 3]

(b)

[–2, –1[ ∪[0, 1[

Express each of the following sets as intervals:

(a)

Solve, in the set of real numbers.

E = {x ∈ ; | x | ≥3}

(a) |3x – 2| <1

(b)

F = {x∈ ; |x – 3| <2}

(b)

| 2xx +–11 |<2

Find the complement, in the set of real numbers of:

(a)

E = {x ∈ ; x < –3 or x > 5}

(b)

F = { x ∈ ; –2 < x < 7}.

Show that for all real numbers x and y; α > 0, β > 0, α x + βy

If x < y then x < α + β < y. 10. Solve the inequalities

0< x+1 ≤ x .

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Chapter 4

Polynomials and rational functions 4.1 Objectives In this chapter you will learn to: •

factorize polynomials using different methods.

•

simplify algebraic fractions.

4.2 Key words Polynomial Remainder theorem Partial fractions

Leading coefficient Factor theorem

Degree Proper fraction

4.3 Theory 4.3.1 Definitions A polynomial in x, a variable, is an expression of the form P(x) = anxn + an–1 xn–1 + an – 2 xn–2 + …+ a2 x2 + a1x + a0 where n is a positive integer and a0 , a 1, a2 , …, an–1, an are constants The degree of the polynomial is n, the constant term is a0 If P(x) is a polynomial which p (x) = 0 for all x, then p(x) is called the zero polynomial.

Equality The polynomials P(x) = anxn + an–1 xn–1 + … + a1 x + a0 an≠ 0, and Q(x) = bm xm + bm – 1 xm–1 + … + b1 x + b0 are equal if and only if m = n and ai = bi for i = 0, 1, 2, …, n. Thus two polynominals are equal if and only if they have the same degree and all the corresponding coefficients are equal. 4.3.2 Operations on polynomials

Addition Addition of polynominals involves "collecting like terms" that is terms of the same degree. 47

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Example 1 Find the sum of the polynomial P(x) = 3x2 + x3 + 5 and Q (x) = 3x – 3x2 + 2 Solution P(x) + Q(x) = x3 + (3–3) x2 + 3x + (5 + 2) = x3 + 3x + 7 Let n be the degree of polynomial f(x), and m the degree of polynomial g(x). It can easily be shown that: If n < m, then f(x) + g(x) is of degree m. If m < n, then f(x) + g(x) is of degree n. If m = n, then either f(x) + g(x) is of degree 0, or the degree is less than or equal to n. If f(x) and g(x) are polynomials, then the degree of f(x) is less or equal to the greatest of the degrees m and n.

Multiplication Polynomials are multiplied in the usual way. Each term of the first polynomial is multiplied by each term of the second polynomial and the like terms are added. Example 2 Given that f(x) = 5x – 1 and g (x) = 4x3 + x – 5, Find f(x).g(x) =(f.g)(x) Solution (f.g) (x) = f (x). g (x) = (5x – 1) (4x3 + x – 5) = 20x4 + 5x2 – 25x – 4x3 – x + 5 = 20x 4 – 4x3 + 5x2 – 26x + 5 The product of two polynomials is the zero polynomial if at least one of the polynomials is the zero polynomial.

Property Let [X] denote the set of all polynomials in x, with real coefficients. Then ( [X] , + ,•) is an abelian unitary ring. In fact, (1) ( [X], +) is an abelian group: 48

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[X] is closed under addition: the sum of any two polynomial is a polynomial. Addition of polynomials is associative; The zero polynomial is the identity element. Any polynomial has an additive inverse (opposite). The opposite of polynomial P(x) = an xn + an – 1 xn – 1 + … + a1 x + a0 is – P(x) = – (an xn + a n – 1 x n – 1 + … + a1 x + a0)

= – a n x n – a n – 1 x n – 1 – … – a1 x – a 0

Addition of polynomial is commutative.

[x] is closed under multiplication: the product of any two polynomials is a polynomial; Multiplication of polynomials is associative and commutative The constant polynomial e(x) = 1 is the identify element.

Multiplication of polynomials is distributive over addition of polynomials:

For any polynomials f,g and h in [X]

[f.(g + h)] (x) = (f.g) (x) + (f.h) (x)

and [(f + g).h) (x) = (f.h) (x) + (g.h) (x).

Example 3 Use polynomials f(x) = 5x2 – x + 1, g (x) = 3x + 2 and h(x) = x3 – 2x + 5 to illustrate the distributive property. Solution f. (g + h) = (5x2 – x + 1) [(3x + 2) + (x3 – 2x + 5)]

= (5x2 – x + 1) (x3 + x + 7)

= 5 x2 (x3 + x + 7) – x (x3 + x + 7) + (x3 + x + 7)

= 5x5 + 5x3 + 35x2 – x4 – x2 – 7x + x3 + x + 7

= 5x5 – x4 + 6x3 + 34x2 – 6x + 7

(f.g) + (f.h) = (5x2 – x + 1) (3x + 2) + (5x2 – x + 1) (x3 – 2x + 5)

= 15x3 + 10x2 – 3x2 – 2x + 3x + 2 + 5x5 – 10x3 + 25x2 – x4 + 2x2 – 5x + x3 – 2x + 5

= 5x5 – x4 + 6x3 + 34 x2 – 6x + 7

We have [f.(g + h)] (x) = (f.g.) (x) + (f.h) (x) 49

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In the same way, we have [(f + g).h](x) = (f.h) (x) + (g.h) (x) The process of multiplying polynomials to obtain a polynomial is called expansion.

Division of polynomials We can divide one polynomial by another in a fashion similar to our method for long division of decimal numbers. Example 4 Divide 2x4 – 3x2 + x + 2 by x2 – 2x – 1 Solution 2x2 + 4x + 7 x2 – 2x – 1 2x4 – 3x2 + x + 2 2x4 – 4x3 – 2x2 4x3 – x2 + x 4x3 – 8x2 – 4x 7x2 + 5x + 2 7x2 – 14x – 7 19x + 9 2 The quotient is Q(x) = 2x + 4x + 7 and the remainder is R (x) = 19x + 9. Thus

2x4 – 3x3 + x + 2 x2 – 2x – 1

19x + 9

= 2x2 + 4x + 7 + x2 – 2x – 1

or 2x 4 – 3x3 + x + 2 = (2x2 + 4x + 7) (x2 – 2x – 1) + 19x + 9.

Remainder and factor theorems The remainder theorem states that: "When a polynomial P(x) is divided by x – a until the remainder R (x) is independent of x, then R(x) = P(a). Proof Let Q(x) be the quotient when P(x) is divided by x – a. Then by the division process, P(x) = (x – a) Q(x) + R(x) Putting x = a gives P(a) = R(a) Thus, the remainder is P(a) 50

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The factor theorem states that: "Polynominal P(x) has x – a as a factor if P(a)=0" Proof: 1.

If P(a) = 0 then when P(x) is divided by x – a, the remainder is zero (remainder theorem).

Therefore, x – a is a factor of P(x). 2. If x – a is a factor of P(x) then there exists a polynominal Q(x) such that P(x) = (x – a) Q (x) for all x. Putting x = a gives P(a) = 0 Example 5 Find the value of the real number a if P(x) = x3 + ax2 + 3x – 1 leaves a remainder of 1 when divided by x – 2. Solution When P(x) is divided by x – 2, the remainder is P(2) Therefore, P(2) = 1 Thus 8 + 4a + 6 – 1 = 1 giving a = –3. Contracted (Synthetic) division The following is a method which may be used to determine the quotient and the remainder when a polynominal is divided by a linear factor x – k. Then P (x ) = (x – k) Q(x) + P(k) (remainder theorem)

= (x – k) (bn – 1 xn–1 + bn–2 xn–2 + … + b1 x + b0) + P(k)

= bn–1 xn + (bn–2 – kbn–1) xn–1 + … + (b0 – kb)x + P(k) – kb0

Equating coefficients gives a0 = bm–1; am–1 = bn– 2 –kbn – 1; an–2 = bn–3– kbn–2, …, a2 = b1 – kb2; a1 = b0 – kb1; a0 =p(k) – kb0 Thus the coefficients of Q(x) can be found as follows: bn–1 = a0, bn–2 = an–1 + kbn–1 bn–3 = an–2 + kbn–2 51

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b1 = a2 + kb2 b0 = a1 + kb1 Finally, the remainder when p(x) is divided by x – k, P(k), can be found from p(k) = a0 + kb0. We set this work out in the following manner k

an – 1

an–2

kan

kan–1 + k2an

an–1 + k an

= bn–1 = bn–2

… an – 2 + kan–1+ k2an …

a0 kb0 P(k)

= bn – 3 …

The procedure is as follows: each element of the second row (except the first) is found by multiplying each element in the third row (in the column of the immediate left) by k. Each element of the third row is found by adding the correponding elements of the first and second rows. The remainder is the last element in the third row. Example 6 Find the quotient and remainder when 2x3 – 3x2 + x + 2 is divided by x – 2 Solution 2 2 2

–3 1 4 2 1 3

2 6 8

Therefore, the quotient is 2x2 + x + 3 and the remainder is 8.

Division by (x – k1) (x – k2) … (x – kp) Where k1, k2 ,…, kp are distinct real numbers It can be shown that polynomial P(x) is divisible by the product (x – k1) (x – k2) … (x – kp) if and only if P(k1) = … = P(kp) = 0>>. Note that if some of the factors are equal,the theorem does not hold.

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For example, (x – 1)3 is divisible by (x – 1)2 and (x –1)3, but (x –1)3 is not divisible by (x – 1)5 Let p(x) be a polynomial. The following statements are equivalent: (1) k is a solution of the equation p(x) = 0 (2) k is a root of the equation p(x) = 0 (3) k is a zero of the polynomial p(x) (4) x - k is a factor of the polynomial P(x)

Multiple zero If a linear factor of a polynomial is repeated, then the zero is repeated. A repeated zero is called a multiple zero. A multiple zero has a multiplicity equal to the number of times the zero occurs. Example 7 For the function f(x) = (x – 1)3 (x + 2) (x + 5)2, find multiple zeros and state the multiplicity of each zero. Solution Multiple zeros are 1 and –5 The multiplicity of 1 is 3, the multiplicity of –5 is 2. 4.3.3 Factorisation of a polynomial Factorisation is the reverse process of expansion. It consists of obtaining a product from a sum. Some methods of factorisation are discussed below:

Factorisation by common factors If the terms of a sum have common factors then the distributive property of multiplication over addition can be used to express the sum as a product. Example 8 Factorise: (a) 3a + 3b – 3c

(b) 5a2 b2 – 2a3b 53

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Solution (a) 3a + 3b – 3c = 3(a + b – c) (b) 5a2b2 – 2a3b = a2b(5b – 2a). Sometimes, it may be necessary to transform the expression in order to obtain a common factor, either by a proper grouping of terms or otherwise. Example 9 Factorise: ax + bx + ay + by Solution ax + bx + ay + by = (ax + bx) + (ay + by) = (a + b)x + (a + b) y = (a + b) (x + y).

Factorisation by using identity patterns We have: 1. (a – b) ( a + b) = a (a + b) – b(a + b)

= a2 + ab – ab – b2

= a2 – b2: quadratic difference of two squares

2. (a + b)2 = (a + b)(a + b)

= a (a + b) + b (a + b)

= a2 + ab + ab + b2

= a2 + 2ab + b2: perfect square trinomial

3. (a – b)2 = (a – b) (a – b)

= a (a – b) – b (a – b)

= a2 – ab – ab + b2

= a2 – 2ab + b2: perfect square trinomial

4. (a + b) (a2 – ab + b2) = a (a2 – ab + b2) + b (a2 – ab + b2)

= a3 – a2b + ab2 + a2b – ab2 + b3

= a3 + b3: sum of cubes

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5. (a – b) (a2 + ab + b2) = a (a2 + ab + b2) – b (a2 + ab + b2) = a3 + a2 b + ab2 – a2b – ab2 – b3 = a3 – b3 : difference of cubes. Example 10 Factorise the following expressions (a) 4a2 – 9b2 (b) a 4 + 2a2 + 1 (c) 4a2 – 12ab + 9b2 (d) 8x3 – y6 (e) a3 + 8 Solution (a) 4a2 – 9b2 = (2a – 3b) (2a + 3b) (b) a 4 + 2a2 + 1 = (a2 + 1)2 (c) 4a2 – 12ab + 9b2 = (2a – 3b)2 (d) 8x3 – y6 = (2x – y2) (4x2 + 2xy2 + y4) (e) a3 + 8 = (a + 2) (a2 – 2a + 4).

Factorisation of quadratic expresions Let P(x) = ax2 + bx + c, where a ≠ 0 Guess, if possible two numbers b1 and b2 such that b1 b2 = ac, then split ax2 + bx + c = ax2 + b1x + b2x + c and group properly the terms of obtain a common factor. Example 11 Factorise 2x2 – x – 1 Solution Two numbers with sum –1 and product –2 are –2 and 1. 2x2 – x – 1 = 2x2 – 2x + x – 1

= (2x2 – 2x) + (x – 1)

= 2x (x – 1) + (x – 1)

= (2x + 1) (x – 1) 55

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Completing the square Consider the following diagram. 1

x x

Fig. 4.1

The diagram represents a square in the centre with side x. It is surrounded by 4 rectangles, each with sides measuring x and 1. The area of each of the four rectangles surrounding the square is 1.x, such that the total area of the four rectangles is 4 (1.x) = 4x. The area of the whole figure is x2 + 4x. Complete the longer (outer) square by adding to the figure. Show the additional areas by shading them and write the area of each of these additions:

1 1

1 Fig. 4.2

The area of each of the additional figures is 1 × 1 = 1. The total area of the additional figures is 4 (1) = 4 Therefore (x + 2)2 = x2 + 4x + 4. If we wish to write x2 + 4x as an expression containing the perfect square x2 + 4x + 4, that is (x + 2)2, we need to "balance" the expression x2 + 4x + 4 by subtracting the term 4. So we write x2 + 4x = x2 + 4x + 4 – 4

= (x + 2)2 – 4.

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You may have noticed that the rule is to take half of the coefficient of x, square that number and add the result to the original expression to obtain a perfect square, and subtract the quantity added to show the original expression. Example 12 Factorise by completing the square the expression x 2 – 6x + 5 Solution x2 – 6x + 5 = x2 – 6x + 9 – 9 + 5 = (x2 – 6x + 9) – 4 = (x – 3)2 –22 = (x – 3 – 2) (x – 3 + 2) = (x – 5) (x – 1). In completing the square of ax2 + bx, we just make the coefficient of x2 appear to be 1. Example 13 (a) Complete the square of the expression 2x2 –7x + 3 and write it as; a (x – p)2 + q (b) Factorise 2x2 – 7x + 3. Solution 7

(a) 2x2 – 7x + 3 = 2 (x2 – 2 x) + 3

= 2(x2 – 2 x + 16 – 16 ) + 3

= 2 (x2 – 2 x + 16 ) + 3 – 8

= 2 (x – 4 )2 – 8

Therefore, a = 2, p = 4 , q = – 8 7

[ (

(b) 2(x – 4 )2 – 8 = 2 x – 4

( 2 x–

7 2 4

2 – 54

)]2 – ( 542 )2

) ( 2 x – 742

2 + 54

). 57

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Factorising a polynomial with integral coefficients Consider the polynomial P(x) = an xn + an – 1xn–1 + … + a1x + a0 where an , an – 1, … , a1, a0 are all integers, an ≠ 0 It can be shown that, <<

if an integral zero of P(x) exists then it must be a factor of the constant term a0>>

Example 14 Factorise in the set of real numbers, P(x) = 5 x3 + 4x2 + 6x + 7. Solution Try the factors of the constant term; 7: 1; -1; 7; –7: P(1) = 5 + 4 + 6 + 7 ≠ 0, so x – 1 is not a factor of p(x) P(–1) = –5 + 4 – 6 + 7 = 0, so x + 1 is a factor of P(x) P(x) = (x + 1) Q(x). Q(x) can be found from synthetic division of P(x) by x + 1 5 –1 5

4 –5 –1

6 1 7

7 –7 0

Q(x) = 5x2 – x + 7 Therefore, 5x3 + 4x2 + 6x + 7 = (x + 1) (5x2 – x + 7). Consider the polynomial P(x) = anxn + an–1xn – 1 + … + a1x + a0 , where an, … , a1, a0 are all integers, an≠ 0 α

If x = β (α and β are integers) is a rational root of the equation p(x) = 0, then x – β is a factor of p(x) so that α must be a factor of a0 and q must be a factor of a0 and β must be a factor of an.Thus if a rational zero of p(x) exists then it must be such that the numerator is a factor of an. Example 15 Factorise 6x3 + x2 + 3x + 2 in the set of real numbers. 58

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Solution The factors of the constant term are 1; –1; 2; –2 None of the factors above is a zero of P(x) = 6x3 + x2 + 3x + 2 Therefore, the polynomial has no integral zero. Trying for rational roots. By the rational root property, the only possible rational roots of the equation P(x) Factor of 2 have the form . So the possible rational roots are: Factor of 6 1 1 1 1 1 1 2 2 2;– 2; 3; 3; 6;–6; 3;–3

Testing each rational root; 1

P(– 2 ) = 6 (– 2 )3 + (– 2 )2 + 3 (– 2 ) + 2 3

=– 4 + 4 – 4 + 4

= 0. 1

So – 2 is a root Using synthetic division; 6 1

–2 6

–3

–2

6x3 + x2 + 3x + 2 = (2x + 1)(3x2 – x + 2). 4.3.4 Highest Common Factor of polynomials Let f(x) and g(x) be two non zero polynomials. The highest common factor (HCF) of f and g is a polynomial h(x), with 1 as leading coefficient, a factor common to both f(x) and g(x), the degree of h is less or equal to the degree of each of the two polynomials f(x) and g (x), the degree of h is maximum. Example 16 Find the highest common factor of polynomials f(x) = 2x4 + x3 –x2 – x – 1 and g(x) = 2x4 – 7x3 + 5x2 + 4 59

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Solution Factorising f(x) and g(x) f(x) = (x – 1) (2x2 + x + 1) g(x) = (x – 2)2 (2x2 + x + 1) The highest common factor of f and g is h (x) = x2 + 12 x + 12 . 4.3.5 Rational functions N(x)

A rational function is any function that can be expressed as f(x) = D(x) , where N(x) and D(x) are polynomials with no common factor. The domain of a rational function is the set of all real numbers x, except those for which D(x) = 0. Example 17 Find the domain of f(x) = 2xx 2––3x Solution Domain: Dom f = {x ∈ ; 2x2– x ≠ 0}

= {x ∈ ; x ≠ 0 and x ≠ 12 }

= ]– , 0[ ]0, 2 [ ] 2 , +

Simplification of rational functions A rational function is in its simplest form when the numerator and denominator are polynomials that have no common factor, except 1. For example x –x 1 , xx2 –+33 and 3x are in their simplest form. 2x – 6 2 But x2 x– x , 3x 2 – 9 are not in their simplest form, since x and x – x have a common factor x, 2x – 6 and 3x – 9 have a common factor x – 3.

You can simplify a fraction that is not in its simplest form, basing on the property if ma mb

a and m ≠ 0, b ≠ 0 then ma mb = b .

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Example 18 Simplify the following rational expressions, after stating any restrictions on the variable + 25 (a) f(x) = xx2++10x 9x + 20 2

4 + x3 – x2 – x – 1 (b) f(x) = 2x 2x4 – 7x2 + 5x2 + 4

Solution + 25 (a) f(x) = xx2++10x 9x + 20 2

(x + 5) (x + 5)

= (x + 5) (x + 4)

= xx ++ 54 , where x ≠ –5 and x ≠ –4

4 + x3 – x2 – x – 1 (b) f(x) = 2x 2x4 – 7x2 + 5x2 + 4

(x –1) (x + 1) (2x2 + x + 1)

= (x – 2) (x – 2) (2x2 + x + 1) (x – 1) (x + 1)

= (x – 2) (x – 2)

2 1 = (xx ––2) 2 , where x ≠ 2 and x ≠ 1 and x ≠ –1.

Multiplying and dividing rational functions Remember that: a c b. d

1. 2.

a b c d

ac = bd , where b ≠ 0 and d ≠ 0

= ba . dc , where b≠ 0, and d ≠ 0, c ≠ 0

Example 19 Find the domain of each of the following functions and simplify. 8x – 2 2x – 8 (a) f(x) = 4x + 4 . 4x – 1

(b) f(x) =

x2 x – 6x + 9 x x–3 2

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Solution (a) Domain: Dom f = {x ∈ ; 4x + 4 ≠ 0 and 4x – 1 ≠ 0} = {x ∈ ; x ≠ – 1 and x ≠ 14 }

– { –1; 14 }

8x – 2 2x – 8 f(x) = 4x + 4 . 4x – 1

(8x – 2) (2x – 8)

= (4x + 4) (4x – 1) 2 (4x – 1) 2(x – 4)

= 4(x + 1) (4x – 1)

= xx –– 41 where x ≠ 1 and x ≠ 4 (b) f(x) =

x2 x2 – 6x + 9 x x–3

Domain:

Dom f = {x ∈ ; x2 – 6x + 9 ≠ 0, x ≠ 0 and x – 3 ≠ 0 = { x ∈ ; x ≠ – 3 and x ≠ 0)

f(x) =

x2 x2 – 6x + 9 x x–3 x2

f(x) = (x –3)2 .

(x – 3) x

x. x . (x – 3)

= (x – 3) (x – 3)x

= x –x 3 where x ≠ 0 and x ≠ 3.

Adding and subtracting rational functions Remember that:

1. ba + bc = a b+ c , where b ≠ 0 62

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a b

c d=

a[ LCMb (b,d) ] + c[ LCMd(b,d) ]

2. + LCM (b,d) Where b ≠ 0, d ≠ 0, LCM: lowest common multiple Example 20 Express as a single fraction, after stating the restrictions on x 1

f (x) = x2 + 5x + 4 + 3x + 3 Solution Restrictions on x: x2 + 5x + 4 ≠ 0 and 3x + 3 ≠ 0; (x + 1) (x + 4) ≠ 0 and 3 (x + 1) ≠ 0; x ≠ – 1 and x ≠ – 4 1 x2 + 5x + 4

1(3)

5x(x + 4)

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

3 + 5x2 + 20x

= 3(x2 + 5x + 4) 5x2 + 20x + 3

Therefore, f(x) = 3(x2 + 5x + 4) .

4.4 Practice 4.4.1 Definitions and operations 1.

3. 4.

Express the polynomial as a perfect square: (a) x6 – 2x5 + 5x 4 – 2x3 + 2x2 + 4x + 1 (b) 4x6 + 4x 4 + 12x3 + x2 + 6x + 9 Determine p and g so that the polynomial is a perfect square (a) 4x4 – 12x3 + x2 + px + q (b) px 4 + 18x3 + qx2 – 6x + 1 Determine a polynomial of degree 4 if 1, 2, –1 and 3 are the zeros and the value of the polynomial at 0 is 12. Find the quotient and the remainder of the division of: (a) 4x2 – x3 + 7 + 2x5 – 6x by 4 – 3x – 2x2 (b) x6 – x7 + 3 – 6x + 5x2 + 2x 4 + 7x3 by 3x – x3 + 2 63

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Find the remainder when 3x5 – x2 + 1 is divided by x + 2 The expression 2x3 + ax2 + bx + 2 is exactly divisible by x + 2 and leaves a remainder of 12 on division by x–2. Calculate the values of a and b and factorise the expression completely. 7. Find a quadratic polynomial that leaves a remainder of 3 on division by x – 1 and a remainder of 12 on the division by x + 2. 8. When x4 + Kx2 + 4x + 2 is divided by x +3, the remainder is 8. Find the value of K 9. Find the remainder when P(x) = 2x3 –3x2 – 8x –3 is divided by: (a) x – 1 (b) x + 3 10. When the polynomial P(x) = x3 + a x2 – 3x + 4 is divided by x –3, the remainder obtained is twice the remainder obtained when the polynomial is divided by x – 2. Find the value of a. 5. 6.

11. A can consists of a cylinder with base radius r, a height of 10cm, and a hemispherical top. Write a polynomial to express the total volume of the can. 12. Given the polynomials P(x) = 5x – 8 and Q(x) = a(x – 2) + b(x – 1), find the values of real numbers a and b if P and Q are equal. 13. Given that the polynomials P(x) = 7x + 1 and Q(x) = a (x + 1) + b(2x – 1) are equal. Find the values of a and b. 14. A rectangular box is 3a units high, 2a –3 units wide and 2a + 3 units long. Express its volume as a polynomial in a. 15. A metal worker wants to make an open box from a sheet of metal of width 12cm and length 16cm by cutting equal squares of side x cm from each corner.

(a)

Write an expression for the length. width and height of the box.

(b)

Express the volume of the box as a polynomial in x.

4.4.2 Factorisation of polynomials 1.

Find the factors of x3 – 5x2 + 2x + 8

Three of the factors of x 4 + ax3 + bx2 + x + c are x, x + 1 and x – 1, find a, b and c.

Factorise completely the expression 3–7x + 5x2 – x3

When the polynomial P(x) is divided by x – 2 the remainder is 4, and when P(x) is divided by x – 3 the remainder is 7.

Find the remainder when P(x) is divided by (x – 2) (x –3)

Find a rational root of the equation P(x)=0. If P(x) = 8x3 + 4x2 – 34x + 15 and factorise completely P(x)= 8x3 + 4x2 – 34x + 15.

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Write apolynomial in standard form with.

(a)

–2, 3 and –3 as zeros

(b)

, 1 and –2 as zeros

Find the zeros of the polynomial and precise the multiplicity of each zero:

1 4

(a) P(x) = x 4 + 6x3 + 8x2 (b) P(x) = (2x – 1)3 (x + 1)2 (x – 5) 8.

Find a polynomial of degree 4, given that the zeros of the polynomial are:

(a)

1; 2; –1; –3 and the value of the polynomial at 0 is 12.

(b) 0; 2 – 2 , 1 and the value of the polynomial at 2 is 16. 9.

Factorise the expression πr2 + 2πrh. Hence, find the value od πr2 + 2πrh when π = 22 7 , r = 14, h = 43.

10.

(a)

Determine polynomials Q (x) and R(x) such that

x5 + 4x3 + 3x2 + 1 = (2x 2 – x + 5) Q(x) + R(x) where R(x) is of degree less than 3. (b) Find the quotient and the remainder of the division of 2x3 + 5x2 + 8x + 5 by x2 + 2x + 3. 11. When a polynomial P(x) is divided by 2x2 – 3x + 1, the quotient is Q(x) and the remainder is 5x – 4. When Q (x) is divided by x – 2, the remainder is –2. Show that x – 2 is a factor of (P) (x) 12. Find the value of a if polynomial P(x) = x3 – ax2 + x – 6 is divisible by x – a – 1. 13. Find the roots of the equation 2x3 – x2 + 2x – 1 = 0. 14. Determine a and b so that polynomial P(x) is a perfect square: (a) P(x) = 4x 4 – 12x3 + x2 + ax + b (b) P(x) = ax 4 + 18x3 + bx2 – 6x + 1 15. When a polynomial P(x) is divided by x + 1, the remainder is 4. When P(x) is divided by x – 2, the remainder is 1. Find the remainder when P(x) is divided by the product (x + 1) (x – 2). 4.4.3 Highest common factor-rational function 1.

Find the least common multiple of (a)

9x2 – 36 and 6x2 + 24x + 24

(b) 3x2 – 6x + 3 and x2 – x (c)

x2 – 1 and x2 – x 65

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Find the highest common factor of (a)

x3 + 2x + 3 and x 4 – x3 + 4x2 – x + 3

(b) 3x 4 + 2x3 + 4x2 + x + 2 and 6x5 – 2x 4 + 10 x3 – 5x2 + 5x – 2 3.

Calculate and simplify, after stating the restrictions on the variable: (a)

5x x –9

(b)

5y 10 2y + 4 – y2 + 2y

– x2 + 5x + 6

Calculate and simplify, after stating the restrictions on x: (a)

1 x–2

(b)

x x 1 – 3x + 1 +3x

3–x

– 2 + x + 4 – x2 18x2

+ 1 – 9x2

2 1 4 Let y = 1x + x + 1 – x – 1 + x2 – 1

(a)

Find the values of x for which the expression has no meaning

(b)

Express y as a single fraction in its simplified form.

Assume the denominator is not zero and simplify: x6 – 64

(a) x3 – 8 7.

4 3 2 Find the restrictions on x , then simplify 2x4 – 3x3 – 7x2 – 5x – 3 .

x – x – 2x – 9x – 9

8. Multiply 9.

x15 + 1

(b) x3 + 1

2x2 + 7x + 3 x2 – 16 by 2 x + 8x + 15 , state the restrictions on x and simplify. x–4

Express as a single fraction and state restrictions on x: 7x

(a) 5x2 – 125 – 3x +15 x

2x + 1

(b) 3x2 – 9x + 6 – 3x2 +3x – 6 x3 + 1

Ax + B

Cx + D

10. Given that (x2 + 4)2 = x2 + 4 + (x2 + 4)2 , find the values of A, B, C and D.

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Chapter 5

Equations, inequations and functions of the first degree 5.1 Objectives In this chapter you will learn to: •

solve equations and inequations of the first degree in one unknown.

•

solve and discuss parametric equations and inequations of the first degree in one unknown.

•

solve simultaneous equations in two unknowns.

•

solve simultaneous equations in three unknowns.

•

solve problems leading to equations.

•

represent graphically the solutions of inequalities.

5.2 Key words Equation Parametric equations Simultaneous equations Elimination method

Solution Inequalities Substitution method Cramer’s method

Root Inequation Combination method Homogeneous equations.

5.3 Theory 5.3.1 Equations of the first degree in one unknown

Definition An equation of the first degree is one unknown is any equation that can be expressed as ax + b = 0, where a and b are constants and a ≠ 0. Solving an equation of the first degree in one unknown Let ax + b = 0; then ax = –b.

If a ≠ 0 then the equation is of first degree and has a unique solution x = – a b

The solution set is s = {– a } If a = 0 and b ≠ 0 then the solution set is s = ø: the equation has no root. If a = 0, b = 0, then 0x = 0 which is true for any value of x. The solution set is S = . 67

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Example 1 Solve, in the set of real numbers, the equations; (a) 5(x – 1)= 2x – 7

x+2 3

(b)

+1= 2 – 4

(a) 5(x–1) = 2x –7 Check 2

5x – 5 = 2 –7 5 (– 3 –1) = 2 (– 3 ) – 7; 5x – 2x = –7+5

5( – 3 ) = – 3 – – 3

3x = –2

–3 = –3

x = – 3

(b)

x+2 3

+ 1 = 12 – 4x ;

4(x + 2) + 12 12

Check:

6 – 3x

= 12 4(x + 2) + 12 = 6 – 3x

–2 +2 1 3 +1= 2

– –2 4 ;

0 + 1 = 12 + 12 1 = 1[verified]

4x + 8 + 12 = 6 – 3x; 4x+20 = 6 –3x; 4x + 3x = 6 – 20 7x = –14 x = –2

3x + 3 = 3x + 7; 0x = 4 No value of x verifies the equation The solution set is S = ø

(d) 3(x + 1) = 3x +3;

3x +3 = 3x +3; 0x =0

Any real number verifies the equation.

Solution set S = . 68

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Parametric equations In the equation ax + b = 0, where at least one of the coefficients a and b depends on a free variable that can assume any value then the equation is said to be parametric. The solution of a parametric equation depends on the value given to the parameter. The discussion of a parametric equation consists in considering all the possibilities for the parameter and to each possibility finding the solution set. Example 2 Solve and discuss with respect to the values of parameter m the solutions of the equation. m2 x + m + 6 = 2m(x + m) Solution m2 x + m + 6 = 2mx + 2m2 m2 x –2mx = 2m2 – m – 6 m(m – 2)x = (2m+3) (m–2) If m ≠ 0 and m ≠ 2, then the equation is of the first degree and has a unique solution x = (2m +3) ( m–2) m(m–2)

Solution set S = {

2m + 3 m

}

If m = 0, the equation becomes 0.x = –6 No value of x verifies the equation. The solution set is S = ø If m = 2, the equation becomes 0.x = 0 Any real number x varifies the equation. The solution set is S = We have the following representation. m 2 0 –∞

S={

+∞

2m + 3 m

}S=ø {

2m + 3 m

}S=

S={

2m + 3 m

}

Problems leading to equations of the first degree in one unknown Example 3 1.

The area of a triangle is 18cm2 amd its base 6cm. Find the height of the triangle. 69

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Solution

Let x be the height of the triangle. 6x

Then 2 = 18; 3x = 18 2.

x=6

A father is 36 years old and his son is 10 years old. In how many years will the father be twice as old as his son.

Solution

Say in x years the father will be twice as old as his son. By that time, the father will be (36 + x) years old and his son (10 + x) years old

We have: 36 + x = 2(10 + x);

36 + x = 20 + 2x; 2x – x = 36 –20; x = 16

In 16 years time the father will be twice as old as his son

Check: 36 + 16 = 52

The son will be 10 + 16 = 26

The son will be 26 years old

We have 52 = 2(26).

5.3.2 Product and fractional equations reducible to equations of the first degree

Equations of the form A.B.C. = 0 The solution of an equation of the form A.B.C. = 0 is based on the zero product law: if a.b = 0 then a = 0 or b = 0. Example 4 Solve, in the set of real numbers, the equations; (a) x3 + 2x2 – x – 2 =0 (b) (x + 1)(x2 – 4) = 3 (x – 2)(x +1)

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Solution (a) x3 + 2x2 – x – 2 =0 x2(x + 2) – 1(x + 2) = 0 (x2– 1)(x + 2) = 0

Either (x – 1)(x + 1)(x + 2) = 0

Therefore x – 1 = 0 or x +1 = 0 or x + 2 = 0 x = 1 or x = –1 or x = –2

The solution set is S ={1, –1, –2}

Check: when x = 1, (1)3 + 2 (1)2 – (1) –2

=1+2–1–2

when x = –1, (–1)3 + 2(–1)2 – (–1) –2

= –1 + 2 + 1 – 2

when x = –2, (–2)3+2(–2)2 – (–2) –2

= –8 + 8 + 2 – 2

= 0.

The three are considered

Fractional equations These are equations that can be expressed as

N(x) D(x)

To solve such equations proceed as follows; 1.

Find the restriction on the unknown;

Transform the equation to get D(x) =0;

Solve the equation N(x) = 0 (from the fact that if b = 0 and b ≠ 0 then a = 0) Take into account the restriction to write the solution set of the equation.

N(x)

Example 5

Solve, in the set of real numbers, the equation

x– 2 x

+ x–2 = x2–2x 71

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Solution 1.

Restriction on x

x ≠ 0 and x – 2 ≠ 0 and x2 – 2x ≠ 0

that is x ≠ 0 and x ≠ 2

x–2 x

+ x–x = x2–2x ;

x–2 x

+ x – 2 – x(x – 2) = 0;

(x – 2)(x – 2) + 4x – 8 x(x–2)

x2 – 4x +4 + 4x – 8 = 0

x2 – 4 = 0

= 0;

(x – 2)(x + 2) = 0 Either x = 2 or x = –2

x = 2 is eliminated from the restriction on x. The only solution is x = –2

The solution set is S = {–2}.

Parametric equations Example 6 Solve in the set of real numbers and discuss with respect to parameter m, the solution of 2m

m+1

the equation x + 1 = x–3 . Solution Domain of the equation: x + 1 ≠ 0 and x – 3 ≠ 0 that is x ≠ –1 and x ≠ 3 The equation is equivalent to each of the following equations: 2mx +1

m+1

– x – 3 = 0;

2m(x–3) (x + 1)(x–3)

(m + 1)(x + 1)

– (x–3)(x+1) = 0;

2mx – 6m – mx –m – x – 1 = 0; 72

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mx – 7m – x – 1 = 0; mx – x = 7m + 1; (m – 1)x = 7m + 1 7m + 1

If m–1 ≠ 0, that is m ≠ 1, then x = m – 1 But x ≠ –1 and x ≠ 3 7m + 1

Then m – 1 ≠ –1 and 7m + 1 m–1

(1)

7m + 1 m–1

≠3

+1=0 7m + 1 + m – 1 m–1

m = 0 m = 0 is eliminated. (2)

7m + 1 m–1

–3=0

7m + 1 – 3m + 3 = 0 4m + 4 = 0

m = –1

Therefore, if m ∉ {–1, 0, 1} then the solution set is S = { m–1 }

If m = 1, the equation becomes: 0. x = 8: no real value of x verifies this equation. The solution set is S = ø

If m = 0, the solution set is S = ø

If m = –1, then the solution set is S = ø

The representation on a number line is as follows:

7m+1

–1

–∞ 7m + 1

0 7m + 1

S = { m–1 } S = ø { m–1 } S = ø

1 7m + 1

{ m–1 }

+∞ 7m + 1

S = ø { m–1 }

5.3.3 Inequations of the first degree in one unknown 1.

An inequation is said to be of the first degree if it can be expressed as ax + b < 0 (or ax + b ≤ 0), where a ≠ 0.

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Solving the inequation ax + b < 0 consists in determining the set S of real numbers x verifying the inequations. From the study of the sign of the expression f(x) = ax + b, we can determine on the number line the solution set.

If a = 0, then f (x) = b has the sign of b If a ≠ 0, then f (x) = ax + b is of first degree f(x) = ax + b b

= a(x + a ) b

x + a > 0 if and only if x > – a

x + a = 0 if and only if x = – a

x + a < 0 if and only if x < – a

But f(x) is the product of x + a and a Therefore; b

– a

–∞

x f(x)

sign of –a

+∞ sign of a

Example 7 Solve, in the set of real numbers, the inequations and represent the solution set on a number line. (a) 3x – 4 > 7x + 8 5

(b) 2x – 3 ≥ 3 – 2 (c) 2(x–3)≥ 2x +2 (d) 2(x–3) > 2x – 9 Solution (a) 3x – 4 > 7x + 8 –4x – 12 > 0 Let f(x) = –4x –12 74

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The sign of f(x) is shown in the table below:

f(x)

–3

–∞

+∞

+++++++ 0 ––––––––––

The solution set is S = ] – ∞, –3 [

–∞ Where the unwanted region is shaded 5

–3

+∞

(b) 2x – 3 ≥ 3 – 2 ; 2x – 3 ≥ 3 – 2 > 0; 5 3

x– 6 >0 5

7 10

–∞

f(x)

Let f(x) = 3 x – 6 +++++++ 0 ––––––––––

The solution set is

7 S = ] 10 ,+∞[

+∞

7 –∞ 10 Where the unwanted region is shaded.

+∞

–∞

f(x)

––––––––––––––––––––

Solution set S = ø the empty set

–∞

+∞

(d) 2(x –3) > 2x – 9 2x – 6 > 2x –9 0x + 3 > 0 75

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Let f(x) = 3

+∞

–∞

f(x)

+++++++++++++++++++++++++

Solution set S = ] – ∞, + ∞ [ –∞

+∞

Parametric inequations Example 8 Solve the inequation in the set of real numbers and discuss the solutions with respect to parameter m. 2mx + 3m ≤ m2 + 6x Solution 2mx + 3m ≤ m2 + 6x 2mx – 6x ≤ m2 – 3m (2m – 6)x ≤ m(m – 3) 2(m – 3)x ≤ (m – 3)m If m = 3, then 0.x ≤ 0: true for all x∈ The solution set is S = 3 x –∞

+∞

m–3 – 0 Let f(x) = 2(m – 3) x – m (m – 3)

If m<3, the sign of f(x) is shown below m x –∞ 2

m–3

+ +

–

Solution set S = [ 2 , + ∞[

If m > 3, the sign of f(x) is shown below x

+∞

f(x)

m 2

–∞ –

+∞ +

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Solution set S = ]– ∞, 2 ]

The number line for discussion is presented below m 3

–∞

+∞

S = [ 2 , + ∞[ S = R S = ] – ∞, 2 ]

5.3.4 Inequalities of the type AB≤0; AB≥0; AB<0; AB>0; A ≤0; A ≥ 0;

A < 0; A >0 B B

To solve the inequations above, study the sign using the sign rule for product and quotient. Example 9 Solve in the set of real numbers the inequations: (a) x4 + 2x3 + x –1 ≤ x3 + x2 +1 (b)

3x – 1 x2 – 1

≥

2 x+1

Solution (a) x 4 + 2x3 + x –1 – x3 – x2 –1 ≤ 0

x4 + x3 – x2 + x – 2 ≤ 0 1 1 1

1 1 2

–1 2 1

1 1 2

–2 2 0

x 4 + x3 – x2 + x – 2 = (x – 1)(x3 + 2x2 + x + 2)

= (x – 1)[x2(x + 2) + (x + 2)]

= (x – 1) (x2 + 1) (x + 2) x x–1 x 2+ 1

–∞ –2 1 +∞ –– – –– 0++ ++ + ++ +++ 77

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x+2 ( x – 1 ) ( x 2+ 1 ) ( x + 2 )

– 0 ++ ++ + +0 –– 0 ++

The solution set is S = [–2, 1]

(b)

3x – 1 x2 – 1

3x – 1– 2(x –1) (x – 1) (x + 1)

≥0

x+1 (x – 1) (x +1)

≥0

1 x–1

–

2 x+1

≥0

≥ 0; x ≠ –1

x –∞ 1 +∞ 1 ++++++++ x–1 –– 0 ++++ 1 – –||++++ x–1

The solution set is S = ]1, +∞ [.

5.3.5 Simultaneous equations

Definitions and equivalence principles Consider the equation in two unknowns 2x + y = 4........................(1) A solution of this equation is an ordered pair of values, one for x and another for y, that verify the equation. Making y the subject of the formula, y = 4 – 2x. To each value of x corresponds a value of y. Thus the equation has an infinity of solutions. If we consider two such equations, we obtain Simultaneous equations of the first degree in two unknowns. Thus, simultaneous equations of the first degree in two unknowns are of the type; ax + by = c a′x + b′y = c′ The solution of simultaneous equations is based on the following principles: 78

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If we make one of the unknowns the subject of the formula in one equation and we replace this unknown by its value in the other equation, we obtain an equivalent system.

When an equation is replaced by another equation obtained by adding to the equation to replace, side by side, other equations, we obtain an equivalent system. x–y=3 x=3+y For example, ⇔ 2x + y = 5 2(3 + y ) + y = 5

x – y = 3 2x + y = 5

x–y=3 2x + y + x – y = 5 + 3

Solving two simultaneous equations of the first degree in two unknowns Method 1: Substitution method Example 10 Solve the simultaneous equations; 2x+ y = 5 ........................... (1) 3x – 4y = 2 ........................ (2) Solution Making y the subject of equation (1) gives; y = 5 – 2x, and replacing in equation (2) y = 5 – 2x 3x – 4(5–2x) = 2

y = 5 – 2x 11x = 22 hence x = 2 y = 5 – 2x x =2

Therefore, y = 5 – 2(2) = 1. The solution is (2,1)

Check: 4 + 1 = 5

6 – 4 = 2.

Method 2: Elimination When the values of two unknowns are needed, we may use the elimination method. The aim of this method is to eliminate one of the unknowns by either adding or subtracting the two equations. 79

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Example 11 Solve the simultaneous equations 2x + 5y = 9 ...................... (1) 3x – 4y = 2 ....................... (2)

Solution For the coefficients of x to be opposite, multiply equation (1) by –3 and equation (2) by 2: –6x – 15y = –27 6x – 8y = 4

(1) (2)

Adding, side by side, gives –23 y = –23;

y = 1. The system is equivalent to

3x – 4y = 2 y = 1

3x – 4 = 2 x = 2

x=2 y=1

The solution of the system is (2, 1).

Method 3: Cramer’s rule ax + by = c a′x + b′y = c′

a b = ab′ – a′b ≠ 0 a′ b′

If D =

then the system is a Cramer’s system and has a unique solution. x = Dx ; y = Dy , D

where Dx =

c b c′ b′

and Dy =

a c a′ c′

a b = 0, the system is not a Cramer’s system. It may possess no solution a′ b′ or an infinity of solutions.

If D =

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Example 12 Solve the simultaneous equations 2x – y = 3 5x + 3y = 2 Solution D=

2 5

Dx = x=

–1 = 6 + 5 = 11 ≠ 0: the system is a Cramer’s system and has a unique solution 3 3 2

–1 2 = 9 + 2 = 11; = Dy = 3 5

Dx D

3 2

= –11

Dy = 11 = 1; y = D = 11 = –1 11

The solution of the system is (1, –1). Method 4: Graphical method We have seen that the equation ax + by = c has an infinity of solutions, the ordered pairs (x, c – a x). If you plot these points (two points suffice) and join them you obtain a b

straight line. ax + by = c The solution of the system a′x + b′y=c′ , if it exists is the ordered pair, coordinates of the

intersection point of the two lines. Example 13 –y=4 Solve, graphically, the system 2x x + y =2 Solution

Let (l1): 2x – y = 4 and (l2): x + y =2 (l1) (l2)

x 0 2 0 2

y –4 0 2 0

Point (0,–4) (2, 0) (0, 2) (2,0) 81

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2x – y = 4 1

–

0 –

x x + y =2

The intersection point is (2, 0): solution of the system. Solving three simultaneous equations of the first degree in three unknowns.

They are of the type: ax + by + cz = d aʹx + bʹy + cʹz =dʹ a˝x + b˝y + c˝z = d˝ Method 1: Substitution method Example 14 Solve the simultaneous equations: (a) x + y = 4 5x – y = 2 x + 2y + 2z = 5

(1) (2) (3)

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x – 2y + z = – 4 3x + y –2z = –5 (b) x + 2y + 2z = 5

(1) (2) (3)

Solution (a) Equations (1) and (2) do not contain unknown z. We can solve them simultaneously to find the values of x and y. Replacing x and y by their values in (3), we find z:

(1) + (2): 6x = 6

x = 1

1 + y = 4; y = 3; we substitute for x = 1 and y = 3 in equation (3).

Equation (3) becomes;

1 + 2(3) – z = 5

z=1+6–5=2

The solution is (1, 3, 2)

Check (1): 1 +3 = 4

(2): 5 – 3 = 2

(3): 1 + 6 = 5.

The equations can be arranged as follows:

x + y + = 4 ×1 5x – y = 2 ×1 x + 2y + 2z = 5

x = 1 5 – y = 2 1 + 2y – z = 5

6x = 6 5x – y = 2 x + 2y – z = 5

x = 1 y = 3 1 + 6 – z = 5

x = 1 y=3 z=2

Solution x = 1, y = 3, z = 2 or (1, 3, 2) (b)

x – 2y + z = –4 (1) 3x + y – 2z = –3 (2) x + 2y + 2z = 5 (3)

Making y the subject of equation (1); x = –4 + 2y – z Substituting x by its value in equations (2) and (3); x = –4 –z + 2y (4) 7y – 5z = 9 4y + z = 9

(5) (6)

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Equations (5) and (6) do not contain x. Solving them simultaniously to obtain y and z, gives y = 2, z = 1 Substituting y and z by their values in (4): x = –4 –1 + 4 = –1 The solution of the system is x = –1, y = 2, z = 1 or (–1, 2,1) Check: (1) –1 – 4 + 1= –4

(2) –3 + 2 – 2 = –3 (3) –1 + 4 + 2 = 5

The equations can be arranged as follows: x – 2y + z= –4 3x + y – 2z = –3 x + 2y + 2z = 5

⇔

x = –4 + 2y – z –12 + 6y – 3z + y – 2z = –3 –4 + 2y – z + 2y + 2z = 5

⇔

x = –4 + 2y – z 27 y = 54 z = 9 – 4y

⇔

x = –4 + 4 – z y=2 z=9–8

⇔

x = –4 + 4 – 1 y=2 z=1

⇔

x = –1 y=2 z=1

Solution: x = –1, y = 2, z = 1 or (–1, 2, 1). Method 2: Linear combination (addition method) Example 15 Solve the simultaneous equations; 2x + 3y – z = –6 (1)

3x + 2y + 4z = 7 (2) x – 2y + 5z = 15 (3)

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Solution Eliminating y between (1) and (2), by multiplying side by side (1) by –2, and (2) by 3, and adding side by side, gives; 5x + 14z = 33 We obtain equivalent system: 5x +14z = 33 (4) 4x + 9z = 22 (5) x – 2y + 5z = 15 (3)

Eliminating x between equations (4) and (5), by multiplying side by side (4) by 4, (5) by –5, and then adding side by side: 11z = 22 to get z = 2 Substituting z by 2 in (5) gives; 4x + 18 = 22; x = 1 Substituting x by 1 and z by 2 in equation (2), gives;

1 – 2y + 10 = 15; y = –2

The solution is x = 1, y = –2, z = 2 Check: (1): 2 – 6 – 2 = –6

(2): 3 – 4 + 8 = 7 (3): 1 + 4 + 10 = 15

The equations can be arranged as follows:

2x + 3y – z = –6 × –2 3x – 2y + 4z = 7 × 3 x – 2y + 5z = 15

5x + 14z = 33 x + 9z = 22 x – 2y +5z = 15

⇔

11z = 22 z=2 4x + 9z = 22 ⇔ 4x + 18 = 22 x – 2y + 5z = 15 x – 2y + 5z = 15

⇔

x = 1 z = 2 ⇔ 1 – 2y + 10 = 15

×4 × –5

x=1 z=2 y = –2

Solution: x = 1, y = –2, z = 2 or (1, –2, 2) ax + by = 0

ax + by + cz = 0

Note: Systems a′x + b′y = 0 and a′x + b′y + c′z = 0

are said to be homogeneous.

a′′x + b′′y + c′′z = 0

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They have obvious solutions x = 0, y = 0, z = 0. 5.3.6 Simultaneous inequations Example 16 Solve graphically the simultaneous inequations: (a)

x–y>0 2x + 3y < 10

(b)

4x + 3y –1< 0 3x – 2y + 12 > 0 x + 5y + 4 > 0

Solution (a)

Graphing the lines x – y = 0 and 2x + 3y = 10 y

x–y=0

2x + 3y = 10

The unwanted region is shaded.

(b) Graphing the lines 4x + 3y – 1 = 0, 3x – 2y + 12 = 0 and

x + 5y + 4 = 0

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x + 5y – 4 = 0 (1, –1) (–4, 0) 3x – 2y + 12 = 0 (–4, 0) (–2,3)

4x + 3y – 1 = 0 (–2, 3) (1, –1)

The unwanted region is shaded.

5.4 Practice 5.4.1 Equations of the first degree in one unknown 1.

Solve, in the set of real numbers, the equations:

x+x =x+1 (a) 2 4 x 12 + = x (b) 3 2

Solve, in the set of real numbers, the equations:

(a) 24x + 3 = 13 – x

(b)

x + 2 + x – 7 = 2x – 5 3

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2 (–3 – 3x) + 2x – 1 = 5 + x (d) 5(2x – 6) – 10 (x – 3) = 0

(c)

Solve, in the set of real numbers, the equations:

(a) 5x (x + 2) = (x – 1)(x +2) (b) (c) (3x + 2) (x – 1) = 3(x – 1)(x + 11 )

7(x–5) = (x – 2)(5 – x)

Solve and discuss with respect to parameter m:

(a) mx – 5m = 3x + 4m –2 5.

(b)

8mx – 3m = 7x – 4

Solve and discuss with respect to parameter m:

mx – 3 + 2mx – 5 = x – 7 – x – 7 (a) 2

(b) 3(m + 2)x – 5 = 3x + 7(m + 2) 6.

Solve, in the set of real numbers, the equations: 2

(a) (3x + 1)2 = 9(x – 3 )2

(b) 4x2 + 20x + 25 = 0

(d)

x2 – 9x = 0

Solve, in the set of real numbers, the equations: 2x – 1

–

2x + 5

4 – 2x

x +4 (a) = x + 3

(b) x – 1 + x + 1 = x2 – 1

3x – 5 = x (c)

(d)

x +3

x +4

5x – 7

Solve and discuss, with respect to parameter m: m–3

(a) x – 2 = 6 9.

2x2 –1 1 – x +1 –x +1 = 1

(b)

m–5 3 x –3 =x –2 ß

Solve for x and discuss with respect to the values of parameter m:

(a) 2mx + m – 1 = 0 (b) (m2 – 1) x – m – 1 = 0 10. Solve for x, and discuss with respect to the values of parameter m, each of the following equations. m

3m – 4

(a) x – 1 = x – 2

(b)

1 – mx

m+x

m – x = 1 + mx

5.4.2 Inequations 1.

Solve, in the set of real numbers, the inequations

(a) 3x – 2 > 2x + 7

(b)

5x – 2 – x ≥ 7x– 8

(d)

x – 8 < 7x + 14

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Solve, in the set of real numbers, the inequations:

3x – 1 > x + 5 + x (a) 2 2 x

x x (c) – > – 2 2 3 6

3(x – 2) – x < 2x + 1

(d)

x 2

– (x – 1) >

2–x 2

Solve, in the set of real numbers, the inequations: 1

(a) 4x – 1 < x + 2 3

3x – 2

(b)

5x + 2 2 + 1

x–1

3x + 2 1 < 2 – (x + 1) 2

(b) 7 + (d)

4x x – 5 2

< 10 + 1

Solve, in the set of real numbers, the inequations:

(a) x2 + x – 12 > 0

(b)

x3 – 4x ≤ 0 x+3

x+1

Solve, in the set of real numbers, the inequations:

(x – 1) ( x + 1) ≤ 1 (a) x+3 (16 – x ) – ( x + 4) ≤ 0 (c) 4 2 2

x –4

(b)

3x – 1 > 2 x +3

(d)

2x + 6 < –7 x–3

Discuss, with respect to parameter m, the solutions of the inequations:

(a) (m + 1) x + 2m – 2 > 4x + 5 7.

(m + 1) x – 1 ≥ 1 x–m

(b)

Discuss, with respect to parameter m, the solutions of the inequations:

(a) 3mx + 2x + 5 <5 (x – 2)

(b)

mx + 3x > m + 5 2x + 6

Solve in the set of real numbers, the inquation x – 3 ≤ – 7

Solve for x and discuss with respect to the values of parameter m, the solutions mx – 1

of the inequation x – m < 1 10. Solve in the set of real numbers, the inequations: (a) x (2x – 3) (1 – 3x) <0

(b)

x2 – 4

x – x2 – x + 1 ≥ 0 3

5.4.3 Simultaneous equations 1.

Solve the simultaneous equations by substitution

x + 3y = –5 (a) 2x + 5y = –11

(b)

3x – 2y = 5 2x – 5y = 7 89

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Solve the simultaneous equations by Cramer’s rule:

x + 2y = 4 (a) 5x – 2y = 8 3.

3x – 2y + 5z = 19 (b) –8x + 6y + 3z = –3 2x + y – 6z + 20 = 0

(b)

x + y – 2z = 1 2x – y + 3z = 1 3x + y – z = 2

Solve, by any method, the system:

x – 2y + z = –4 (a) –4x + y – 2z = 1 2x + 2y – z = 10 6.

Solve, by linear combination (addition):

3x – 5y + 4z = 5 (a) 7x + 2y – 3z = 2 4x + 3y – z = 7 5.

4x + 3y + 3= 0 8x + 5y + 9 = 0

Solve, by substitution

x + 5y – 5z = 2 (a) 3x – 2y + 2z = 6 – 2x + 3y – 5z = –6 4.

(b)

2x + 3y – z = –6 3x + 2y + 4z = 7 x – 2y + 5z = 15

The total cost of a meal and a bottle of soda is 1250 FRW. The meal costs four times the cost of a bottle of soda. Express the problem using simultaneous linear equations, stating the meaning of each letter. Find the cost of a bottle of soda and the cost of the meal. Gitera has 19 coins in her pocket some of them are 50 FRW and the rest are 100 FRW coins. The total value of the coins is 1550 FRW. Find the number of 50 FRW coins, and the number of 100 FRW coins she has. Binen bought one kilogram of potatoes, one kilogram of rice and one kilogram of bananas for a total of 770 FRW. Lero bought, at the same place, 4kg of potatoes, 3kg of rice and 4kg of bananas for a total of 2580 FRW. Pira bought 16 kg of potatoes, 9kg of rice and 4kg of bananas for a total of 7020 FRW. Find the price of 1 kg of potatoes, the price of 1 kg of rice and the price of 1 kg of bananas.

5.4.4 Simultaneous inequations Solve graphically and shade the wanted region: 1.

x + 2y + 1< 0 x–y–4>0

x + 4y + 2 ≥ 0 3x – 12y – 12 ≤ 0

–

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y≥5 y+x–2≤0

x–4<0 y>0 x–y+2>0

Each week you must do a minimum of 18 hours of homework. Participation in sports requires at least 12 hours per week. You do not have more than 35 hours per week to devote to these activities.

(a) (b)

Write a system of inequations for this situation. Graph and solve the system.

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Chapter 6

Equations, inequations and functions of the second degree 6.1 Ojectives In this chapter you will: •

solve quadratic equations in one unknown:

•

solve and discuss parametric equations of second degree.

•

solve inequations of the second degree.

•

solve and discuss quadratic inequations.

•

study the sign of quadratic expressions.

•

solve problem leading to quadratic equations.

•

graph a quadratic function.

6.2 Key words Equation Solution Roots Parametric equations Quadratic equation Maximum Minimum Axis of symmetry Intercepts Vertex Standard form Parabola Extraneous solution

6.3 Theory 6.3.1 Quadratic equations Quadratic equations in one unknown are equations of degree , which means that the highest power of the variable will be . All quadratic equations can be written in the form ax2 + bx + c = 0, where a ≠ o; this is said to be the standard form of the equation. The solutions are also said to be the roots of the equation or the zeros of the function f(x) = ax2 + bx + c = 0. Solving a quadratic equation

Let ax2 + bx + c = 0, a ≠ 0 The equation is equivalent to each of the following equations; a(x2 + b x + ac ) = 0, a

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x2 + b x + ac = 0, a

c b2 b2 x2 + ba x + 4a2 – 4a2 + a = 0 b2 – 4ac 4a2

(x + 2a )2 –

Let Δ = b2 - 4ac Case 1: Δ > 0 b

Then (x + 2a )2 – ( 2aΔ )2 = 0, This is factorisable using the identity of a2 – b2= (a + b) (a – b) b b thus; (x + 2a – 2aΔ )(x + 2a + 2aΔ ) = 0, b b Δ x + 2a – 2a = 0 or x + 2a + 2aΔ = 0,

either x = –b + Δ or x = –b – Δ where Δ = b2 – 4ac 2a

Case 2: Δ = 0 b 2 ) = 0, (x + 2a b x + 2a = 0, b

x1 = x2 = – 2a Case 3: Δ < 0

If Δ is negative and ax2 + bx + c ≠ 0 for all values of x:

the equation ax2 + bx + c = 0 has no real roots.

The quantity Δ = b2 – 4ac is called the discriminant of the equation.

Therefore, the number of the roots of the equation ax2 + bx + c = 0, a ≠ 0, in the set of real numbers, depends on the sign of the discriminant.

If Δ > 0 then the equation ax 2 + bx + c = 0 has two distinct roots.

If Δ = 0 then the equation ax2 + bx + c = 0 has one root (or a repeated root) and ax2 + bx + c = 0 is a perfect square

If Δ < 0 then the equation ax2 + bx + c = 0 has no real roots, that is the equation has zero roots. 93

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Example 1 Solve, in the set of real numbers, the equation (a) 2x2 + 7x + 3 = 0

(b)

x2 + 2x + 3 = 0

Solution (a) Δ = 49 – 24 = 25 > 0

1 −7 + 5 4 =− 2 −7 − 5 x2 = 4 = –3

Solution set S = {− 12 , −3}

x1 =

(b) Δ = 4 − 12 = −8 < 0

The equation has no real roots.

The solution set is s = φ

c) 4x2 − 4x + 1 = 0

Δ = 16 − 16 = 0

x1 = x2 = 48 = 12

Solution set S = { 12 }

Note If a and c are of opposite signs, then − 4ac ≥ , 0 and b2 − 4ac ≥, 0: we can conclude that the equation has two distinct roots. If b is even, that is b = 2b′, 2

then −b ± Δ = −2b′ ± 4b′ − 4 ac 2a

2(−b′ ± b′ 2− ac) 2a

= −b′ + Δ′ , where Δ′ = b2 − ac a

Properties of the roots of a quadratic equation 1.

Sum and product of the roots Let ax2 + bx + c = 0, where a ≠ 0, Δ ≥ 0 If x1 and x 2 are the roots,

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The sum, S = x1 + x2 =

−b + Δ 2a

−b − Δ 2a

= −2b 2a

=− a

The product; P = x1 x2 = (

= b 4a−2Δ

= 4ab 4a2

= ac

−b + Δ 2a

)(

−b − Δ 2a

)

b2 − (b2 − 4ac) 4a2

Conclusion: Without actually calculating the roots of a quadratic equation we can write down their sum S and their product P. c S = − ba and P = a

Some important problems

Example 2 (a) Find a quadratic equation whose roots

and are given.

Solution The equation (x − ) (x − ) = 0 has x2 − ( + )x +

and as roots and is such that

(b) Find two numbers whose sum S and product P are given. Solution The required numbers are the possible roots of the equation x2 − Sx + P = 0 2 2 If S2 − 4P > 0, the required numbers are S + S − 4P and S − S − 4P

If S − 4P = 0, the required numbers are 2

S 2

and

S 2

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If S2 − 4P < 0, these are no numbers with such sum and product Discussion of the equation ax2 + bx + c = 0 If P < 0, then the equation has two distinct roots of opposite signs. Let x1 be the negative root and x2 the positive root. If S < 0, then | x1| > | x2|

If S = 0, the | x1| = | x2|

If S > 0, the | x1| < | x 2|

If P = 0, so c = 0, the equation has two roots (Δ = b2): 0 and S If P > 0 and if the roots exist, then they are of the same sign. If Δ ≥ 0, the equation has two distinct roots of the same sign as S

If Δ = 0, the equation has two roots S2 and S2 (repeated roots) If Δ < 0, then the equation has no real roots. The discussion can be summarised as follows: P<0

S<0

S=0 S>0

S<0

P=0

S=0 S>0

Δ<0 P>0

Δ=0

S<0

S>0

S<0

S>0

Δ>0

Example 3 Consider the equation in x: (m + 1)x2 – (m + 3)x + 3 − m = 0. (a)

Solve the equation for m = −1

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(b) Assume m ≠ −1. Find the set of values of m for which:

(i) the equation has two roots of opposite signs.

(ii) the equation has two distinct positive roots.

Solution (a) If m = −1, the equation becomes −2x + 4 = 0. It has a unique root; x = 2. (b) Suppose m ≠ −1. The discriminant is Δ = 5m2 − 2m − 3

= (m − 1) (5m + 3)

The equation has two roots if and only if Δ > 0, that is (m − 1) (5m + 3) > 0 m Δ = (m − 1)(5m + 3)

−3

− ∞ 5 1 +∞ + 0−−0+ −3

Δ > 0 if and only if m ∈] −∞, 5 [U] 1, + ∞ [ To determine the sign of these roots, we need the signs of the product P = 3m−+m1 and m+3 the sum S = m + 1 (i) The roots are of opposite signs if and only if P < 0 We have: P < 0 ⇔ m∈] −∞, −1[U]3, + ∞[

Therefore, the equation has two roots of opposite signs if and only if

m∈] −∞, −1 [U]3, + ∞ [ (ii) The roots are all positive if and only if P > 0 and S > 0

We have:

P > 0 ⇔ m∈] −1, 3[ S > 0 ⇔ m∈] −∞, −3[U] −1, + ∞ [

Therefore, the equation has two distinct positive roots if and only if 2

m∈] −1, − 5 [U]1, 3 [ .

6.3.2 Quadratic functions

Definition A quadratic function is any function of the type; f: → x

f (x) = ax2 + bx + c , a ≠ 0. 97

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Graphing function f(x) = x2 Table of values x y

−2 4

1 2

−1

−

1 4

1 2

1 4

4 3 2 1 x –3

–2

–1

A parabola The graph of f(x) = ax2 + bx + c , a ≠ 0 is obtained from the graph of y = x2 by applying transformations: f(x) = ax2 + bx + c b 2 Δ = a(x + 2a ) − 4a

y = x2

(1)

b 2 y = (x + 2a )

(2)

b 2 y = a(x + 2a )

(3)

b 2 Δ y = a(x + 2a ) − 4a

b 2 b The graph of y = (x + 2a ) is the graph of y = x2 translated 2a units horizontally in the

(a)

b (b) positive x – direction if 2a is negative.

b 2 Δ b 2 ) – 4a is obtained from the graph of y = (x + 2a ) by The graph of y = a(x + 2a

b is positive negative x – direction if 2a

b 2 multiplying the y-coordinate of each point on the graph y = (x + 2a ) by a

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Δ −Δ b 2 b 2 The graph of y = a(x + 2a ) – 4a is the graph of y = a(x + 2a ) translated 4a units vertically in the:

−Δ

positive y − direction if 4a is positive

(a)

−Δ

(b) negative y − direction if 4a is negative The curve y = ax2 + bx + c , a ≠ 0, has the following characteristics; Δ

b vertex V(− 2a , − 4a )

(a) (b)

the vertex is a minimum if a > 0,

(c)

the curve is concave up, if a > 0

(d)

The vertex is a maximum if a < 0, and the curve is concave down

(e)

b The axis of symmetry is x = − 2a

The y − intercept is (0, c)

(f)

(g) The x − intercept has the following properties;

−b + Δ (i) If Δ > 0, there are two x − intercepts: ( −b − Δ , 0) and ( 2a , 0) 2a

−b

(ii) If Δ = 0, there is only one x − intercept ( 2a , 0): the x − axis is a tangent to the parabola. (iii) If Δ < 0, these is no x − intercept: the parabola does not cut the x − axis. Example 4 1.

Explain how to sketch the graph of y = −4x2 + 8x + 3 from the graph of y = x2 by applying transformations and state whether the parabola y = −4x2 + 8x + 3 is a concave up or concave down.

Solution

− 4x2 + 8x + 3 = –4(x2 − 2x) + 3

= − 4 [(x − 1)2 − 1] +3

= − 4 (x − 1)2 + 7

To transform the graph of y = x2 to the graph of y = −4x2 + 8x + 3 the steps are as follows; y = x2

(1)

(x − 1)2

(2)

4(x − 1)2

(3)

− 4(x − 1)2

(4)

y = − 4(x − 1)2 + 7 99

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(i)

horizontal translation 1 unit in the positive x − direction

(ii) the y − coordinates of each point is multiplied by 4, the x − coordinates is unchanged (iii) reflection in the x − axis (iv) vertical translation of 7 units in the positive y − direction y 4

x 0

Fig 6.1 (i) y 0

Fig 6.2 (ii) 1

Fig 6.3 (iii) y 7 3

−4

Fig. 6.4 (iv)

The parabola is concave down

x 1

Fig 6.5 (v) y = −4x2 + 8x + 3

Given the parabola y = − 2 x2 + 2x, (a) State whether the parabola is concave up or concave down. (b) Find the coordinates of the vertex and state the nature of the vertex. (c)

Find the equation of the axis of symmetry.

(d) Find the intercept of the parabola. Solution 1 (a) a = − 2 < 0

Therefore, the parabola is concave down.

(b) Vertex V (2, 2) is a maximum. (c)

Axis: x = 2

(d) Intercepts: (0, 0) and (4, 0). 100

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Zero and signs of y = ax2 + bx + c, a ≠ 0 −b + Δ If Δ > 0, the parabola cuts the x − axis at two distinct points ( −b − Δ , 0) and ( 2a , 0). 2a

The function has two zeros, y has the sign of a for x < x1 and for x < x2 (assume x1 < x2)

y has the sign of −a for x1 <x < x2

Δ> 0

Δ< 0

Δ= 0

Fig. 6.6

Fig 6.7

Fig 6.8

Δ>0 x

−∞

x − x1

−

x − x2

−

(x − x1 )(x − x2)

−

+ +

y = a(x − x1)(x − x2) sign of a

+∞

x2 +

sign of −a

sign of a

Fig 6.9 Δ=0 x (x − x1)2 y = a(x − x1)2

−∞ + sign of a

+∞

sign of a

Fig 6.10 101

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Δ < 0: y is of the sign of a x y

−∞

+∞ sign of a

Fig 6.11 6.3.3. Applications of quadratics

Factorisation of ax2 + bx + c, a 0 If Δ > 0 then ax2 + bx + c = a(x − x1)(x − x2), where x1 and x2 are the roots of the quadratic equation. If Δ = 0 then ax2 + bx + c = a(x − x1)2 If Δ > 0 then ax2 + bx + c cannot be factored in a set of real numbers. Example 5 Use the quadratic formula to factorise in : (a) 3x2 − 3x − 6 (b) −x2 + 2x − 1 (c) 2x2 + x + 1 Solution (a) 3(x2 − x − 2)

Δ = (−1)2 − 4(1)(−2)

= 9 = 32

x1 = 1 −2 3 = − 1, x2 = 1 +2 3 = 2, Therefore, 3x2 − 3x − 6 = 3(x + 1)(x−2)

(b) −x2 + 2x − 1

= − (−x2 − 2x + 1)

Δ = (−2)2 − 4(1) = 0

x1 = x 2 = 2 = 1

Therefore, −x2 + 2x − 1 = − (x − 1)2 102

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=1−8 =−7< 0

2x2 + x + 1 cannot be factorised in the set of real numbers. Inequations ax2 + bx + c ≤ 0, ax2 + bx + c < 0, ax2 + bx + c ≥ 0, ax2 + bx + c > 0, a ≠ 0 To solve such inequations, we study the sign of ax2 + bx + c, and we take into account the given sign. Example 6 Solve, in the set of real numbers, the inequation −x2 + 2x + 3 > 0. Solution Studying the sign of y = −x2 + x + 3 Δ′ = 1 + 3 = 4 x1 = −1 − 2 = 3, x2 = −1 + 2 = −1 −1

x y

−1

−∞ −

−1 0 +

3 0

+∞ −

Solutions: −1 < x < 3 Solution set: ]−1, 3[ . Inequations of the type A.B.C ≥≤ 0, where A, B, C are of the first degree or quadratics

To solve such inequations, we study the signs of A,B and C. Example 7 Solve, in the set of real numbers, the inequation (x − 1)(x2 + x − 6)(2x2 + x + 1)≥ 0 Solution We study the sign of each factor: x − 1= 0; x2 + x − 6 = 0; 2x2 + x + 1 = 0 x = 1; x = –3; x = 2 103

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−∞

−3

x−1

−

–

x2 + x − 6

–

−

2x2 + x + 1 + +

− 0

−

Product

1 2

+∞

Solution set S = [−3, 1] U [2, + ∞[ Fig. 6.12

A.B.C Inequations of the form D.E ≤≥ 0

Where A, B, C, D, E are of first degree or quadratic. A.B.C

We study the sign of D.E Example 8

2 + x + 1) ≤0 Solve, in the set of real numbers, the inequation (x − 1)(2x 2

(x + x − 6)

Solution Studying the sign of

(x − 1)(2x2 + x + 1) (x2 + x − 6)

−∞

x−1

−

2x2 + x + 1

x2 + x − 6

(x − 1)(2x2 + x + 1) (x2 + x − 6)

−3 − – −

1 2

+∞

−

+ + +

+ 0

−

Solution set S = ]−∞, −3[U[1, 2[ .

Parametric equations To discuss a quadratic parametric equation whose coefficients depend of a parameter, we study the sign of Δ, P and S.

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Example 9 Discuss with respect to parameter m the equation x 2 − 2(m − 2)x + m = 0. Solution 1.

Δ′ = (m − 2)2 − m = m2 − 5 m + 4 = (m − 1)(m − 4) Δ′ = 0 if m = 1 or m = 4 Δ′ > 0 if m < 1 or m > 4 Δ′ < 0 if 1 < m < 4 Product P = m Sum S = 2(m − 2) S = 0 if m = 2, S > 0 if m > 2, S < 0 if m < 2.

The conclusions are shown in the table below m 0 1 2 4

Δ′

+ + + 0 − − − 0 +

− 0 + + + + + + +

− − − − + 0 + + +

Conclusion 2 roots of opposite signs x1 < 0 < x2; | x1 | > | x2 | 1 root (zero), 1 root is negative 2 negative roots 1 repeated negative root No roots 1 repeated positive root 2 positive roots

Comparing a number to the roots of a quadratic equation To compare a number λ to the roots x1 and x2 (x1 < x2) of the equation ax2 + bx + c = 0, a ≠ 0, without solving the equation, let f(x) = ax2 + bx + c.

If f(λ) and a have opposite signs, then x1 < λ < x2 If f(λ) and a have same signs, no conclusion can be drawn without calculating Δ. If Δ < 0, the problem is meaningless b b If Δ = 0, then x1 = x2 = − 2a . It is easy to compare λ to x1 = x2 = − 2a .

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b If Δ > 0 and − 2a > λ then λ < x1 < x2 b If Δ > 0 and − 2a < λ then x1 < x2 < λ

Example 10 Compare: (a) − 1 to the roots of the equation 2x2 − 3x − 4 = 0 1

(b) 2 to the roots of the equation −x2 + x + 3 = 0 Solution (a) Let f(x) = 2x2 − 3x − 4 f(−1) = 2 + 3 − 4 = 1 (sign of 2) 3

b = 4 > −1 Δ = 4 > 0; − 2a

Therefore, −1< x1 < x2 (b) Let f(x) = −x2 + x + 3 1

(opposite sign of − 1) f( 2 ) = 4 + 2 + 3 = 13 4 1

Therefore, x1 < 2 < x2 6.3.4 Equations reducible to quadratic equations

Solving equations by using substitution There are equations where an algebraic expression is repeated in the equation. This algebraic expression can be replaced by a single variable. The replacement reduces the original equation to a quadratic equation that can easily be solved. The following are some forms of the disguised quadratic equations. (i)

ax 4 + bx2 + c = 0, a ≠ 0

(ii) ax 4 + bx3 + cx2 + bx + a = 0, a ≠ 0 To solve the equation ax 4 + bx2 + c = 0, let y = x2 The equation becomes ay2 + by + c = 0 106

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Example 11 Solve, in the set of real numbers, the equation x 4 − 5x2 − 36 = 0. Solution Let x2 = y. The equation becomes: y2 − 5y − 36 = 0 Δ = (−5)2 − 4(1)(−36) = 169 = 132 y1 = 5 +213 = 9 y2 = 5 − 13 = − 4 2

But x2 = y For y2 = − 4, the equation becomes x2 = − 4: no value of x verifies the equation For y = 9, x2 = 9; (x − 3)(x + 3) = 0; x = 3, x = − 3 The solution set is S = {3; − 3}. Equations of the form ax4 + bx3 + cx2 + bx + a = 0. Observe that the terms equidistant from the middle value have equal coefficients. ax4 + bx3 + cx2 + bx + a = 0 (ax4 + a) +(bx3 + bx) + cx2 = 0 a(x2 + 12 ) + b(x + 1 ) + c = 0; x x a[(x + 1 )2 − 2] + b(x + x a[(x + 1 )2 − 2] + b(x + x Let x + 1 = y x

1 ) + c = 0; x 1 ) + (c − 2a) = 0; x

The equation becomes ay2 + by + c − 2a = 0.

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Example 12 Solve, in the set of real numbers, the equation 6x 4 − 5x3 − 38x2 – 5x + 6 = 0 Solution Let x + 1 = y. The equation becomes; x

6y − 5y + (−38 − 12) = 0 2

6y2 − 5y − 50 = 0 5

y1 = 10 and y2 = 2 3

For y = 10 , the equation becomes; 3

x + 1 = 10 that is 3x2 − 10x + 3 = 0 3

x1 = 3 and x2 = 3

For y = − 5 , the equation becomes; 2

x + 1 = − 5 that is 2x2 + 5x + 2 = 0 2

x3 = −2 and x 4 = − 2

The solution set is S = {3; 3 ; − 2; − 2 }. Solving equations of the type ax3 + bx2 + bx + a = 0; ax3 + bx2 − bx − a = 0; ax4 + bx3 − bx − a = 0 These equations are solved by factorising the left hand side. Example 13 Solve in the set of real numbers, the equations: (a) 2x3 − 3x2 − 3x + 2 = 0 (b) 2x 4 + 5x3 − 5x − 2 = 0 Solution (a) 2x3 − 3x2 − 3x + 2 = 0 (2x3 + 2)–(3x2 + 3x ) = 0 2(x3 + 1) −3x(x + 1) = 0 108

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(x + 1)(2x2 − 2x + 2 − 3x ) = 0 (x + 1)(2x2 − 5x + 2) = 0 Either x + 1= 0 or 2x2 − 5x + 2 = 0 1

x = −1 or x = 2 or x = 2 1

Solution set: S = {−1; 2; 2 } 2x 4 + 5x2 − 5x − 2 = 0

(b)

2(x 4 − 1)+ 5x(x2 − 1) = 0 2(x2 − 1) (x2 + 1) + 5x(x2 − 1) = 0 (x2 − 1)(2x2 + 2 + 5x) = 0 x2 − 1 = 0 or 2x2 + 2 + 5x = 0 1

x = 1; x = −1; x = − 2; x = − 2 1

Solution set: S = {1; −1; 2; − 2 }.

Irrational equations Simple irrational equations are equations containing the unknown as radicand To solve such equations, eliminate the radical sign by equating the two sides, keeping in mind the following rules: (i)

represents a positive number and has a meaning only if A ≥ 0

(ii) The equation A = B can be replaced by A 2 = B2 (consider only values of the unknown for which A and B are of the same sign). In particular, the solution of the equation c = D are the solution of equation C = D2, where D ≥ 0. When solving an irrational equation, eliminate extraneous roots (roots derived from an original equation but it is not a solution of the original equation) Example 14 Solve in the set of real numbers the equation; (a) 2x + 3 + x = 6 (b) 2x − 1 + x = 2

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Solution (a) 2x + 3 + x = 6

2x + 3 = 6 − x

Squaring both sides; 2x + 3 = 36 − 12x + x2 x2 − 14x + 33 = 0 x = 11; x = 3 Check: 2(11)+ 3 + 11 = 6 5 + 11 ≠ 6; x = 11 is eliminated For x = 3, 6 + 3 + 3 = 6 3+3=6 Solution set S = {3} (b)

2x − 1

+ x =2

Squaring both sides; 2x − 1 + 2 x(3x − 1) + x = 4 2 x(2x − 1) = 5 − 3x Squaring both sides; 4x(2x − 1) = 25 − 30x + 9x2 x2 − 26x + 25 = 0 x = 1; x1 = 25 Check: For x = 25, 20 − 1 + 25 = 2

7 + 5 ≠ 2; x = 25 is eliminated.

For x = 1, 2 − 1 + 1 = 2 1+1=2 Solution set S = {1}. 6.3.5 Problems leading to quadratic equations The procedure for solving a problem consists of the following steps: 1. Understand the context of the problem 2. Set up the model 3. Solve the model mathematically 4. Translate back to the context. 110

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Example 15 A small rectangular vegetable garden is enlarged by increasing the length by 3 meters and the width by 1 meter. The area of the garden is now three times larger than that of the original garden. Determine the original dimensions of the garden, if its area was 6m2. Solution Let x represent the length of the original garden and let y represent the width of the original garden 8

Then xy = 6; a = y The area of the garden: (x + 3)(y + 1) Area of the new garden is (x + 3)(y + 1) = 3xy xy + x + 3y + 3 = 3xy

3y + x + 3 − 2xy = 0 Since xy = 6, then x = y and substituting in 3y + x + 3 – 2 xy = 0 6 6 gives: 3y + y + 3 − 2y ( y ) = 0 3y2 − 9y + 6 = 0 y2 − 3y + 2 = 0 (y − 2)(y − 1) = 0 y − 2 = 0 or y = 1 y = 2 or y = 1 x = 3 or x = 6 Hence , the length of the original garden was 3m and the width was 2m, or the length was 6m and the width was 1m

6.4 Practice 6.4.1 Quadratic equations 1.

Find the roots of the equations: (a) 3y2 = 2y

(b)

7x + 5x2 = 0

Solve, in the set of real numbers: (a) x + 6x2 − 11= 0 (b) 2 2 m + 1 = m2 111

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Solve, in the set of real numbers, the equations: (a) 2x2 − 3x − 4 = 0

(b)

2x2 − 3x + 4 = 0

Find a quadratic equation in x whose roots are −3 and 7

5. x1 and x2 are the roots of the equation 3x2 + 18x − 123 = 0. Calculate x1 + x2 and x1 x 2 6. x1 and x2 are the roots of the equation ax 2 + bx + c = 0, a ≠ 0, express x12 + x22 in terms of a, b, and c. 7. 8. 9.

Given that 35x2 − 11x − 6 = 0, x1 and x2 are the roots, find x 2 + 12 . x2 1

1 If 3x2 − 6x + 8 = 0 has roots α and β, find the equation whose roots are 1 and β . α Given that the roots of x2 + px + q = 0 are α and β, form a quadratic equation 1 whose roots are 1 and β . α

10. If α2 and β2 are the roots of x2 − 21x + 4 = 0 and α, β are positive. (a)

1 find: αβ; α + β; the equation with roots 12 and β2 α

1 (b) if α + β = 5 and αβ = 2, calculate 1 + β and determine the values of m α 1 and n such that x2 + mx + n = 0 has roots 1 and β α

11. Determine the value of m so that the equation (m – 1) x 2 – 2 (m + 3) x + m = 0 has only one root. 12. Determine the set of real values of m for which the equation mx2 – (2m + 1) x + m – 1 = 0 has no real roots. 13. If the sum of the roots of the equation (a + 1) x2 + (2a + 3)x + (3a + 4) = 0 is –1, find the product of the roots of the equation. 14. Without calculating the roots of the equation x 2 – 4x – 21 = 0, calculate the value of the expression a 2 b3 + a3b2 . 15. a and b are the roots of the equation x 2 – 2 (m + 1) x + m2 = 0.

(a)

(b)

determine the set of values of m if the equation has two district roots. without calculating a and b, determine the value of a – 2 + b – 2 in 2a + 1 2b + 1 terms of m.

6.4.2 Quadratic functions 1.

Graph on the same figure: (a) y = x2 and y = −x + 6

(b)

y = x2 and y = 2 x + 2

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2. Let f(x) = x2 + 4x + k, k∈ (a)

Find the value of k such that the parabola passes through the origin

(b) Find the coordinate of the vertex. 3.

Find the equation of the parabola with vertex at point (1, −5) which cuts the y−axis at (0, 4)

4. Given f(x) = 2x2 − x − 3, find; (a)

Whether the curve is concave up or concave down.

(b)

The coordinates of the vertex and state its nature.

(c)

The equations of the axis of symmetry.

(d) The intercepts. 5.

Explain how to obtain the graph y = x 2 − 2x from y = x2 step by step.

Parabola (P) is given by f(x) = – x2 – 2x + 3 (a)

Determine whether (P) is concave up or concave down.

(b) Find, by its coordinates, the turning point and state its nature. (c)

Find the interval where function f is:

(i) increasing

(ii) decreasing

(d) Find the range of function f. (e)

Find the intercepts of the parabola.

(f)

Sketch the parabola in the Cartesian plane.

Given the parabola (P): y = –x2 – 4x + 12,

(a)

Find the values of m, n and r such that y = m (x – n)2 + r.

(b)

Hence, find the vertex and state its nture.

(c)

Find the intercepts of y.

(d) Describe the transformations to apply to the graph of y = x 2 to obtain the parabol (P). 8.

Given the parbola (P): y = 2x2 – 8x + 8,

(a)

Determine whether (P) is concave up or concave down.

(b)

Find, by its coordinates, the turning point and state its nature.

(c)

Find the equation of the axis of symmetry.

(d)

Find the intercepts of the parabola.

(e)

Find the range of function y = f(x). 113

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6.4.3 Inequations 1. Solve: 2 (a) 2x 2− 9x − 5 ≤ 0.

x − 2x

(b)

(−x2 − x − 1)(2x2 − 3x + 1) ≤ 0.

2 − 6x − 9) > 0. (c) −5x(−x 2

x − x − 20

2. Solve: (a) −2x2 + x − 7 > 0. (b) x2 − 8x + 7 > 0. (c) –9x2 + 6x − 1 ≤ 0. 3.

Solve, in the set of real numbers, the inequations:

(a) x2 ≤ 4x + 12. 4.

(b)

x2 – 5x > –6.

Solve, in the set of real numbers, the inequations.

(a) x 4 ≥ x. (b) x 4 + x3 – 5x2 + x – 6 < 0. 5.

Solve, in the set of real numbers, the inequations:

4x + 5 ≥ 3x. (a) x+2

(x – 1)2 (x2 + x + 3) (b) 2 5x – 6 – x

Solve, in the set of real numbers, the inequations:

(a) –2x2 + 5x + 3 < 2x + 1. (b) x + 1 + x + 2 ≥ 3. 7.

Discuss the solutions of the inequation x 2 – (m – 2) x + 3 (m – 5) ≤ 0, where m is a parameter.

Determine the values of m for which 2 is less than both roots of the equation

(m + 4) x2 – 2 (m – 2) x + m – 4 = 0.

Miscellaneous questions 1.

Solve the equation: 5x2 − 6x + 2

2. a and b are solutions of the equation x 2 − 15x + 25 = 0. Without calculating the roots, find the value of (2a2− 5)(2b −25) . a + 3ab + b

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3. Solve: (a)

2x 4 − 5x2 − 12 = 0

(b) 9x 4 − 37x2 + 4 = 0 (c) x 4 − 6x2 + 5 = 0 4. Solve: (a) x 4 + 9 − 10x2 = 0 (b) x 4 − 16 = 13x2 (c) x 4 − 5x2 + 4 = 0 5. Solve: (a) 2x 4 − 7x3 + 10x2 − 7x + 2 = 0 (b) x 4 + x3 − 4x2 + x + 1 = 0 6. Solve: 2 (a) 2xx +−11 = − 1 + x +1 1

(b) x − 2 − x + 3 = − 1 7.

Solve the equations: 1

(a) x − 2 + x − 5 = 10 (b)

x+1 x−2

x−2

+ x+2 +2=0

8. Solve: (a) 2x + 17 = x + 1 (b) x − 7 = x + 2 − 3 9.

10. Find value of m for which the equation mx2−(m − 1)x + 2m − 2 = 0 has two real roots.

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Chapter 7

Trigonometry 7.1 Objectives In this chapter you will: •

express the angles in different units of measurements.

•

represent a given angle in a cartesian plane.

•

locate the angle in a trigonometric circle.

•

define sine, cosine, tangent and cotangent of any oriented angle.

•

verify trigonometric identities using fundamental formula.

•

represent graphically the sine, cosine, and tangent of an angle.

•

represent graphically the trigonometric functions by using coordinates.

•

apply trigonometric ratios to solve problems related to the right-angled triangle and any other triangle.

•

use addition formulae, double angle (duplication) formulae and Simpson’s formulae to simplify the trigonometric expressions.

•

solve problems related to trigonometric equations and inequations.

7.2 Key words Standard position Initial side Unit circle Cosine of Ɵ Intercepted arc Radian Cosine function Tangent of Ɵ Secant Cotangent Trigonometric ratios for a right triangle

Terminal side Coterminal angles Sine of Ɵ Central angle Sine function Cine curve Tangent function Cosecant Trigonometric identity Law of sines and law of cosines.

7.3 Theory 7.3.1 Angle and its measurements

Definition An angle is formed by two rays drawn from a common point. One of the rays of an angle is the initial side and the other is the terminal side. The size of an angle is the amount of rotation from the initial side to the terminal side. 116

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If the rotation is in counter clockwise direction, the angle is positive. If the rotation is in clockwise direction, the angle is negative. V Vertex

Terminal side

Vertex

Initial side

Terminal side

Initial side

Fig. 7.1 Standard position

An angle is in standard position when the vertex is at the origin and one ray is on the positive x-axis. The ray on the x-axis is the initial side of the angle, the other ray is the terminal side of the angle.

initial side x

Units of angle Degree

terminal side

Fig. 7.2

The angle formed by rotating the initial side about the vertex in the counterclockwise direction exactly once until it coincides with itself is said to measure 360 degrees. Thus, 1 one degree equals 360 revolution. We write 1 revolution = 360 degrees, 1° = 1 360

revolution. Submultiples of degree are minutes and seconds.

1° = 60′ ; 1′ = 60′′

An angle of, say 25 degrees 38 minutes 7 seconds is written 25° 38′ 07 ′′. Example 1 Find the measure of the angle below: y

Solution x

20º

The angle measures 20° more than a straight angle of 180°

Fig. 7.3 Since 180 + 20 = 200, the measure of the angle is 200º. 117

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Two angles in standard position are coterminal angles if they have the same terminal side. y

Angles that have measures of 135º and –225º are coterminal 135º –

225º

Fig. 7.4 Radian A central angle of a circle is an angle with a vertex at the center of a circle. An intercepted arc is the portion of the circle with end points on the sides of the central angle and remaining points within the interior of the angle.

1 rad.

r r

Fig. 7.6

Intercepted arc central angle

Fig. 7.5

When a central angle intercepts an arc that has the same length as a radius of the circle, the measure of the angle is defined to be one radian. Like degrees, radians measure the amount of rotation from the initial side to the terminal side of an angle.

Because the circumference of a circle is 2πr, there are 2π radians in any circle. Since 2π radians = 360º, therefore π radians = 180º. You can use a proportion such as r = r rad to convert between degrees and radians, or, in short: d = where π π 180 d: number of degrees are r: number of radians. dº 180

To convert degrees to radians, multiply by multiply by

180º . π radians

π radians 180º

. To convert radians to degrees,

When the terminal side of an angle lies on a coordinate axis, we say that the angle is quadrantal. Thus, angle of 0°, 90° 180°, 270° are quadrantal.

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Example 2 (a) Find the degree measure of an angle of – 3π radians. 4

Solution –

3π radians 4

180º 180º = 3π radians. multiply by π radians π radians 4

45 180º simplify = – 3π radians. π radians 4 1 = – 135º

An angle of – 3π measures – 135º. 4

(b) Find the radian measure of an angle of 27º Solution 27º = 27º.

π radians 180º

multiply by

π radians 180º

3 π = 27º. radians simplify 180º 20 3π radians. 20 3π An angle of 27º measures radians. 20 =

Activity 7.1 Try the following: 1.

Convert to degrees:

11π (a) 7π rad (b) rad 9

Convert to radians:

(a)

12°

(b)

105°

3. Convert:

(a)

50° 06′ 21′′ to a decimal in degrees.

(b)

21.256° to degrees, minutes, seconds form. 119

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Find the measure of an angle between 0° and 360° coterminal with:

(a)

In which quadrant or on which axis, does the terminal side of each of the following angles lies.

(a)

415°

150°

(b)

–225°

(b)

270°

7.3.2 Unit circle and the six trigonometric ratios The unit circle has a radius of 1 unit and its centre at the origin of the coordinate plane. You can use the symbol Ɵ for the measure of an angle in standard position. The six trigonometric ratios

Cosine and sine of an angle Suppose an angle in standard position has measure Ɵ. The cosine of Ɵ (cos Ɵ) is the x-coordinate of the point at which the terminal side of the angle intersects the unit circle. The sine of Ɵ (sin Ɵ) is the y-coordinate. Example 3 Find the cosine and sine of 60º. 1

–

P (cos 60º, sin 60º)

60º

b C

Fig. 7.7

Solution As the figure suggests, the x-coordinates of point P is In the triangle OCP, from the Pythagorean theorem, 1 3 = . Therefore b = 23 . 4 4 It follows that sin 60° = 23 .

, so cos 60º = 2

( 21 )

or 0.5.

+ b2 = 12;

b2 = 1 –

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More generally, in a 30°, 60°, 90° triangle, the length of the side opposite the 30° is half the length of the hypotenuse. The length of the adjacent side is 23 times the length of the hypotenuse.

The tangent The sine and cosine of an angle derive from the coordinates of a point on the unit circle. The tangent of an angle derives from the coordinates of a point on a line tangent to the unit circle. y x=1 For an angle Ɵ in standard 1 T (1, tan Ɵ) position, the tangent of Ɵ is the yThe line containing coordinate of the point where the P(cos Ɵ, sinƟ) the terminal side of line containing the terminal side Ɵ Ɵ interscts the line x of the angle intersects the tangent 1 x =1 at T 0 line x = 1. Fig. 7.8

Exercise Draw a unit circle (use 4 cm as one unit). Draw an angle of 225° in standard position and show on the figure, sin 225°, cos 225° and tan 225°.

Cosecant, secant and cotangent From the basic trigonometric ratios (sine, cosine and tangent), three other trigonometric ratios are related to the reciprocals of the basic trigonometric ratios. 1

The cosecant (csc) : csc Ɵ = sin Ɵ 1

The secant (sec) : sec Ɵ = cos Ɵ 1

The cotangent (cot): cot Ɵ = tan Ɵ Their domains do not include the real numbers Ɵ that make a denominator zero.

Activity 7.2 1.

On a unit circle, show an angle of 225°. Show sin 225°, cos 225° and tan 225°.

Given the numbers:

–

14 13 ; 35 ; –1.982; –0. 976; ; 2000 and – list those representing: 2 13 14

(a)

the sine of an angle 121

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(b) the cotangent of an angle (c)

the secant of an angle

(a)

Given that cos Ɵ =

(b)

State the sign of the cosine and sine in each of the following angles:

(a)

135°

(b)

205°

(c)

307°

(d)

29°

, find sec Ɵ. 16 Given that tan Ɵ = 2, find cot Ɵ.

7.3.3 Trigonometric identities The reciprocal identities

An equation is an equality true only for some values of the unknown, in general, but an identity is an equality true for all values of the unknown, except, possibly for the values making the expression undefined. 1

and cot Ɵ = tan Ɵ are known as The equalities csc Ɵ = 1 , sec Ɵ = 1 sin Ɵ cos Ɵ the reciprocal identities, where sec Ɵ, csc Ɵ (or cosec Ɵ) and cot Ɵ stand respectively for secant of angle Ɵ , cosecant of angle Ɵ and cotangent of angle Ɵ. Tangent identities

You can derive some other identities from the unit circle definitions of sine, cosine and tangent. In the diagram below, P (cos Ɵ, sin Ɵ) and Q (1, tan Ɵ). The slope (or gradient) of line PQ, which is the same as the slope of line OQ, or OP, since point O(0,0) lies on sin Ɵ – 0

sin Ɵ

line PQ, is cos Ɵ – 0 = sin Ɵ . On the other hand, using points P and Q, the slope is tan Ɵ – 0 = tan Ɵ 1–0

(cos Ɵ, sin Ɵ ) 1

y x=1

sin Ɵ

So, tan Ɵ = cos Ɵ

Ɵ 1

–

(transitive property of equality)

Using the reciprocal identity cot Ɵ = tan Ɵ , and the tangent identity tan Ɵ =

sin Ɵ cos Ɵ

, you can derive

–

Q (1, tan Ɵ)

Fig. 7.9

the cotangent identity cot Ɵ = cos Ɵ as follows: sin Ɵ

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cot Ɵ = tan Ɵ : reciprocal identity

= sin1 Ɵ : substitute cos Ɵ sin Ɵ

= cos Ɵ : divide sin Ɵ cos Ɵ The equalities tan Ɵ = cos Ɵ and cot Ɵ = sin Ɵ are called, respectively, tangent identity and cotangent identity.

Pythagorean identities You can derive another identity from the definitions of cos Ɵ and sin Ɵ. The ordered pair (cosƟ, sinƟ) is a point on the unit circle, and for any point (x, y) on the unit circle, x2 + y2 = 1. So, cos2Ɵ + sin2Ɵ = 1 . This is one of the three pythagorean identities. Note: To verify an identity, you should transform one side of the equation until it is the same as the other side. This eliminates the possibility of introducing errors that can be caused by squaring both sides of an equation or multiplying both sides of an equation by an expression that equals 0. These are the errors that can introduce extraneous roots when solving equations. It is sometimes helpful to write all the trigonometric ratios in terms of sine and cosine. Example 4 Verify the pythagorean identity 1 + tan2Ɵ = sec2Ɵ. Solution 1 + tan2Ɵ = 1 +

sinƟ cos Ɵ

tangent identity

2 = 1 + sin 2Ɵ simplify cos Ɵ 2 2 = cos2Ɵ + sin Ɵ write the fractions with common denominators cos Ɵ cos2Ɵ

cos2Ɵ + sin2Ɵ add cos2Ɵ

1 pythagorean identity; = sec2Ɵ reciprocal identity cos2Ɵ 123

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The left side of the equation has been transformed into the right side. Therefore the equation is an identity. We have the following identities: • Reciprocal identities 1

cscƟ = sin Ɵ

secƟ = cos Ɵ

cotƟ = tan Ɵ

• Tangent cotangent identities sin Ɵ cotƟ = cosƟ tanƟ = cot Ɵ sin Ɵ • Pythagorean identities

cos2Ɵ + sin2Ɵ = 1

1 + tan2Ɵ = sec2Ɵ

1 + cot2Ɵ = csc2Ɵ

Exercise 1. Determine the signs of the trigonometric ratios of Ɵ if Ɵ lies in quadrant: (a) I (b) II (c) III (d) IV. 2. Use the trigonometric identities to find the missing trigonometric ratios if: (a) tan Ɵ = 2 and Ɵ lies in quadrant III

(b)

cos Ɵ = 3 and Ɵ lies in quadrant IV.

There are many trigonometric identities in addition to the identities named in the summary above. Example 5 Verify the identity tan2Ɵ – sin2Ɵ = tan2Ɵ sin2Ɵ . Solution tan2Ɵ – sin2Ɵ =

sinƟ cos Ɵ

– sin2Ɵ

tangent identity

2 = sin 2Ɵ – sin2Ɵ

simplify

cos Ɵ

2 2 2 Ɵ = sin 2Ɵ – sin Ɵ cos subtract 2

cos Ɵ

sin2Ɵ (1 – cos2Ɵ) cos2Ɵ

factor

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2 2 Ɵ) = sin Ɵ(sin 2

pythagorean identity

cos Ɵ

sin2Ɵ

= cos2Ɵ sin2Ɵ rewrite the fraction = tan2Ɵ sin2Ɵ

tangent identity

Activity 7.3 1.

Determine the signs of trigonometric ratios for an angle in:

(a)

quadrant I

(b)

qudrant II

(c)

quadrant III

(c)

quadrant IV

Find the sine, cosine and cotangent og angle Ɵ in each of the following cases:

(a)

tan Ɵ = 2 and 180° < Ɵ < 270°

(b)

tan Ɵ = –2 2 and 270° < Ɵ < 360°

Given that sin Ɵ = 3 and 90° < Ɵ < 180°, find cos Ɵ, csc Ɵ, tan Ɵ, cot Ɵ and sec Ɵ.

Show that:

tan2 Ɵ

= sin2 Ɵ (b) sec2 – tan2 Ɵ = 1 (a) 1 + tan2Ɵ 7.3.4 Calculations involving the trigonometric ratios Complementary angles

Two angles α and β are said to be complementary if α + β = 90°, that is β = 90° – α. If α and 90° – α are in standard position, their terminal sides cut the unit circle at points M and M′ symmetrical about line y = x. y M′

S′ S o

We have cos α = oc ; sin α = os cos β = oc ′ sin β = os ′ M (cos α, sin α)

M β

α c′

Since M and M′ are symmetrical about the line y = x , M′ (sin α, cos α) But also M′ (cos β, sin β) It follows that:

Fig. 7.10 125

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cos (90° – α) = sin α

cos ( 2 – α) = sin α

sin (90° – α) = cos α

sin ( 2 – α) = cos α

tan (90° – α) = cot α

or tan ( π – α) = cot α 2 π cot ( 2 – α) = tan α

cot (90° – α) = tan α sec (90° – α) = cosec α

sec ( 2 – α) = cosec α

cosec (90° – α) = sec α

cosec ( 2 – α) = sec α Example 6 π

(a) Simplify cos ( 2 – x) – 2 sin x. (b) Without using a calculator, find the exact value of

2 sin2 63° – 4 cos2 41° + 2 sin2 27° – 4 cos2 49°.

Solution π

(a) Cos ( 2 – x ) – 2 sin x = sin x – 2 sin x = – sin x (b) 2 sin2 63° – 4 cos2 41° + 2 sin2 27° – 4 cos2 49°

= (2 sin2 63° + 2 sin2 27°) – (4 cos2 41° + 4cos2 49°)

= 2(sin2 63° + cos2 63°) – 4(sin2 49° + cos2 49°)

= 2(1) –4(1)

= –2

Trigonometric ratios of common angles:

0° cos

sin

tan

30°

45°

3 2

2 2

1 2 3 3

2 2

60°

90° 0

1 2 3 2 3

1 _

It is advisable to commit these values to memory so as to use them as reference.

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Example 7 Find, without using a calculator, the exact value of : (a) 3 sin 45° – 4 tan 30° (b) 4 cos 60° + 3 cot 30° Solution (a) 3 sin 45 – 4 tan 30° = 3 ( 22 ) – 4 ( 33 ) =

9 2 –8 3 6

(b) 4 cos 60° + 3 cot 30° = 4 ( 2 ) + 3 ( 3 ) = 2 + 3 3 Supplementary angles

Two angles α and β are sid to be supplementary if α + β = 180°, that is β = 180° – α. If α and 180° – α are in standard position, their terminal sides cut the unit circle at points M and M′ symmetrical about the y - axis. y

We have cos α = oc , sin α os , cos β = oc′ , sin β = os′ .

S′ S

M′

β C′

M α

If M (x, y) and M′ (x′, y′) are symmetrical about the y- axis, then x = x′ and y = y′. From M (cos α, sin α), M′ (cos β, sin β), cos β = – cos α and sin β = sin α.

Fig. 7.11 cos (180° – α) = – cos α sin (180° – α) = sin α tan (180° – α) = – tan α cot (180° – α) = – cot α sec (180° – α) = – sec α cosec (180° – α) = cosec α

But β = 180° – α. Therefore, cos (π – α) = – cos α sin (π – α) = sin α tan (π – α) = – tan α or cot (π – α) = – cot α sec (π – α) = – sec α cosec (π – α) = cosec α

Example 8 Find, without using a calculator, the six trigonometric ratios of 120°.

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Solution sin 120° = sin (180° – 60°)

= sin 60°

= 23 cos 120° = cos (180° – 60°).

= – cos 60° –1

= 2 tan 120° = – tan 60° = –

cot 120° = – cot 60° = – 33 sec 120° = –2

2 3

cosec 120° = 3 . Opposite angles

Angles α and – α are opposite angles. Their terminal sides cut the unit circle at points M and M′ symmetrical about the x axis.

O S′

We have cos α = oc , sin α = os , cos β = oc ′ , sin β = os ′ .

S α β

Since M and M′ are symmetrical about the x- axis, if M

C C′

(x , y) and M′ (x′, y′), then x′ = x and y′ = –y. But M (cos α, sin β) and M′ (cos β, sin β). Therefore, cos β = cos β

M′ Fig. 7.12

and sin β = – sin α.

So sin ( – α) = – sin α cos ( – α) = cos α tan ( – α) = – tan α cot (– α) = – cot α sec (– α) = sec α cosec (– α) = – cosec α

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Example 9 Find, without using a calculator, the six trigonometric ratios of – 30°. Solution 1

sin (–30°) = – sin 30° = – 2 cos (–30°) = cos 30° = 23 tan (–30°) = – tan 30° = ‑ 33 cot (–30°) = – cot 30° = – 3 2 3

sec (–30°) = sec 30° = 3 coses (–30°) = –2. Coterminal angles

Consider the quadrantal angles 0°, 90°, 180°, 270° and 360°, and A, B, A′, B′ the points, on the unit circle, associated to the angles. Then, Point A B A′ B′

Angle 0° = 0 rad π

90° = 2 rad 180° = π rad 3π

270° = 2 rad

General expression 360°k or 2 kπ π

90° + 360°k or 2 + 2kπ 180° = π rad or π + 2kπ 3π

270° + 360°k or 2 + 2kπ

where k ∈ .

In general, if α is a particular angle, 0° ≤ α < 360°, and β is the general expression of the angle, then β = α + 360°k, where k ∈ . This shows that the remainder of the division of β by 360 is α , and k is the quotient. It follows that: sin (α + 360°k) = sin α

sin (α + 2k π) = sin α

cos (α + 360°k) = cos α

cos (α + 2k π) = cos α

tan (α + 360°k) = tan α

or tan (α + 2k π) = tan α

cot (α + 360°k) = cot α

cot (α + 2k π) = cot α

sec (α + 360°k) = sec α

sec (α + 2k π) = sec α

cosec (α + 360°k) = cosec α

cosec (α + 2k π) = cosec α

where k ∈ . Note; tan (α + 180°k) = tan α; cot (α + 180°k) = cot α 129

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Example 10 Find, without using a calculator, the exact value of: 17 π

49 π

(a) sin 4 , cos 6 (b) the six trigonometric ratios of 315° Solution 17 π

(a) sin 4

16 π

= sin ( 4 + 4 ) π

= sin [ 4 + 2 (2) π]

= sin 4

= 22 49 π

cos 6

48 π

= cos ( 6 + 6 ) π

= cos ( 4 +8π)

= cos 6

= 23 (b) sin 315° = sin (–45° + 360°)

= – sin 45° = – 22

cos 315° = cos (–45° + 360°)

= cos 45°

= 22 tan 315° = – tan 45° = –1

cot 315° = – cot 45° = –1

sec 315° = sec 45° = 2

cosec 315° = – cosec 45° = – 2 .

Reference angle The reference angle of angle θ is the acute angle α formed by the terminal side of θ and either the positive x axis, or the negative x – axis.. It is usually easier to find the reference angle of θ by making a quick sketch of θ in standard position. 130

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The reference angle theorem states that: « If θ is an angle, and α its reference angle, then: sin θ = ± sin α ; cos θ = ± cos α ; tan θ = ± tan α ; cot θ = ± cot α ; sec θ = ± sec α; cosec θ = ± cos α . Where, the sign + or – depends on the quadrant in which θ lies» Example 11 Find the reference angle, then evaluate the trigonometric ratio: (a) cos 150°

(b)

tan (–45°)

(c)

sin 210°

Solution (a) The reference angle of 150° is 30°. Therefore cos 150° = ± cos 30°.

Since 150° lies in quadrant II, cos 150° < 0.

Therefore, cos 150° = – cos 30° = – 23 .

(b) The reference angle of – 45° is 45°.

Therefore, tan (–45°) = ± tan 45°.

Since – 45° lies in quadrant IV, tan (–45°) <0.

Therefore, tan (–45°) =– tan 45° = – 1.

Therefore, sin 210° = ± sin 30°.

Since 210° lies in quadrant III, sin 210° < 0, therefore, sin 210° = – sin 30° = – 2 .

Activity 7.4 1.

Show that:

(a)

sin (π + x ) = – sin x

(b)

cos ( π + x) = – cos x

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Show that:

(a)

sin ( 2 + x ) = cos x

(b)

cos ( 2 + x ) = – sin x

3. 4.

Calculate, without using a calculator, the exact value of: (a) sin 240°; tan 240° (b) sin 330°; cos 330°; tan 330° Evaluate, without using a calculator:

sin 300° – cos 990°

(a) cos 2520° – tan 675°

(b)

(cos 18040π ) 2 + (sin 2020π )2 9 9

5. Simplify:

(a)

7π

2 cos ( 2 – x) + sin ( 4 + x ) – 4 sin (6π + x ) sin (90° + x) cos (–x) cot (180° – x)

(b) cos(360° + x) cos (180° + x) tan (90° – x) 7.3.5 Triangles

Right triangle A triangle in which one angle is a right angle (90°) is called a right triangle. The side opposite the right angle is called the hypotenuse, and the remaining two sides are called the legs of the triangle. A hypotenuse c B

From Geometry, we have: A + B = 90° and a2 + b2 = c2 (Pythagorean formula)

b The trigonometric ratios for a right triangle are the six different ratios of the sides of a right triangle. These ratios do not depend on the size of the right triangle. They depend C only on the measures of the acute angles in the triangle.

Fig. 7.13 For the acute angles, the unit circle definition of sine is equivalent to the definition of sine for right triangles. Using the unit circle, sin Ɵ = y- coordinate of p = PQ

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y B

1 Ć&#x;

sin A = leg opposite to angle A

hypotenuse

BC AB

Fig. 7.14

Since triangles APQ and ABC are similar triangles,

So, sinĆ&#x; = PQ = 1 = PA = AB = sin A. More generally, in any right triangle,

sine of acute = Hypotenuse ;

cosine of acute = Hypotenuse ,

Opposite tangent of acute = Adjacent .

PQ PA

BC AB

Opposite

Adjacent

Example 12 5 In triangle ABC, angle C is a right angle and sin A = . Find cos A, cot A and sin B in 13 fraction and in decimal form. Solution Use triangle ABC to calculate the ratios; Calculate the ratios;

cos A =

length of leg adjacent to A length of hypotenuse length of leg adjacent to A

cot A = length of leg opposite A sin B =

= c = 13 = 0.9231 = a = 5 = 2.4

length of leg adjacent to B b length of hypotenuse = c

= 13 = 0.9231.

Example 13 In triangle DEF, angle at vertex F is a right angle, f = 13 and e = 5. Find the angle at vertex D to the tenth of a degree. 133

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Solution

Draw a diagram. E 13 F

Fig. 7.15

Use a cosine ratio: 5 cos D = 13

D = 67.38°; use a calculator

To the nearest tenth of a degree, D = 67.4º.

Solving a right triangle Solving a right triangle is to determine the lengths of all three sides and the measures of all three angles. Different possibilities are summarised in the table below. Given Hypotenuse C An acute angle B

Problem Formula Acute angle A legs a and b B = 90° – A a = c cos B; b = c sin B

One leg b and

Acute angle

an acute angle B

A Hypotenuse C

A = 90° – B ; c = sin B a = b cot B

the other leg a Hypotenuse C and one leg b Acute angles A , B the other leg a

sin B = c A = 90° – B a = c2 – b2

The two legs a and b

Acute angles A and B Hypotenuse

tan B = a A = 90° – B c = a2 + b2

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Example 14 1.

Calculate the size of angle at vertex A in each of the following cases:

(a) C

(b) A

A C B B 5 8 Fig. 7.16 2.

(a)

Calculate the length of the opposite side QR: Q

(b)

42°

Fig. 7.17

Calculate the length of adjacent side XY: X

35°

6 Z

Solution

(a)

(b) tan A = 5 ; A = 38.7° tan 42° = QR ; QR = 6 tan 42°;

(b)

tan 35° = XY ; XY = tan 35° = 8.57.

Fig. 7.18 8

tan A = 3 ; A = 69.4°

QR = 5.40 6

Elevation and depression The angle of elevation is the angle above the horizontal through which a line of view is raised. 135

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Object α

Horizontal

Observer's eye

Fig. 7.19

α : Angle of elevation. The angle of depression is the angle below the horizontal through which a line of view is lowered. Observer's eye

Horizontal

Object

Fig. 7.20

α : Angle of depression Example 15 A vertical tower AB of height 40 metres is observed from two points C and D in the same horizontal plane as B, the foot of the tower. The points B, C and D lie in a straight line and BC = CD. Given that the angle of elevation of A from D is 60°, calculate: (a) the distance of C from the foot of the tower; (b) the angle of elevation of A from C. Solution A

40 D

60°

Fig. 7.21

From the trigonometric ratio: Tangent of acute =

Opposite Adjacent

We have: tan 60° =

40 Adjacent

Adjacent = BD =

40 tan 60°

3 = 40 3

3 (a) The distance of C from the foot of the tower is 20 3

11.5 metres

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(b) tan c =

= 40 ×

20 3 3

3 20 3

=2 3

The angle of elevation of A from C is 73.8°.

Exercise Try the following. 1.

An eagle, 200 metres above the ground level, is observed from two points x metres apart at ground level. From one point, which is due South of the eagle, the angle of elevation is 30°, and from the other point, which is due East of the eagle, the angle of elevation is 40°. Calculate the value of x.

An observer notes that the angle of elevation of the top of a tower is 30° from a point A, while at a point B, 40 meters nearer to the tower, the angle of elevation is 60°. What is the height of the tower?

Oblique triangles An oblique triangle is any triangle that is not a right angled triangle. Law of sines

The formula for the area k of a triangle is k = 2 bh In any oblique triangle ABC with side length a, b, and c, h = c sin A. B

A C Fig. 7.22 b Therefore k = 1 bh = 1 bc sin A 2

Similarly k = 1 ac sin B and k = 1 ab sin C

The area of a triangle is a constant. 137

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Hence 1 bc sin A = 1 ac sin B = 1 ab sin C

Multiply all through by 2 gives sin A = sin B = sin C

abc

This is the law of sines and can be used in two ways namely to get the missing measurements of angles and sides and also to get area. Example 16 B

Find the area of the triangle in

the figure shown. 10 m

31º

A C 5m Fig. 7.23

Solution

In the triangle, b = 5, c = 10 and A = 31º

k = 1 bc sin A = 1 (5)(10) sin 31º 2

= 12.9

The area is about 12.9 m2.

Example 17 In triangle ABC, a = BC = 10cm, A = 30° and B = 40°. Find the side AB = c. Solution Given that the angle sum of a triangle is 180°; C = 110°. Also c

sin C

a ; sin A

C = a sin C = 10 sin 110° = 18.8 cm. sin A

sin 30°

138

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Law of cosines Since you cannot use the law of sines to find the missing measures in the triangles shown in the figure below, another formula is needed.

? ? 4

65°

4 6

Fig. 7.24 Fig. 7.25 In Figure 7.17, oblique triangle ABC has height h. Let AD = x and AB = c – x.

In ADC, b2 = x2 + h2 and cos A = b or x = b cos A C b

x c–x A B Fig. 7.26 D In triangle CBD:

a2 = (c – x)2 + h2

pythagorean theorem

= c2 – 2cx + x2+h2

square the binomial

= c – 2cx + b

substitute b2 for x2 + h2

= c2 – 2cb cos A + b2 substitute b cos A for x = b2 + c2 – 2bc cos A commutative property of addition Similarly b2 = a2 + c2 – 2ac cos B and c2 = a2 + b2 – 2ab cos C These equations relate the length of a side of any triangle to the measure of the opposite angle. It applies to any of the three sides and is called the law of cosines. a2 = b2 + c2 – 2bc Cos A b2 = a2 + c2 – 2ac Cos B c2 = a2 + b2 – 2ab Cos C

Example 18 Find the measure of angle C in the triangle below. Round your answer to the nearest tenth of a degree. 139

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Solution Choose the form of the law of cosines that contains angle c.

c 2

122 = 52 + 82 – 2(5)(8) cos C substitute

144 = 25 + 64 – 80 cos C simplify

= a2 + b2 – 2ab cos C

55 = –80 cos C combine like terms – 55 = cos C solve for cos c 80 B – cos–1 55 = magnitude of angle C solve for magnitude 80 of angle C Magnitude of angle C ≈ 133.4º use a calculator

A 12 8 5

Fig. 7.27

Note When the triangle is obtuse the cosine of the obtuse angle is negative Example 19 Given that triangle ABC is such that AB = 10cm, BC = 5cm and AC = 6 find the measure of x (i) cos C in the form y where x and y are integers (ii) angle in radian

(iii) angle C in degrees Solution (i) Using the law of cosines c2 = a2 + b2 – 2ab cos C and substituting for a = 5, b = 6 and C = 10 we have 100 = 25 + 36 – 60 cos C

Thus cos C = – 49 60

(ii) C = 2.256 radian from the calculator (iii) C = 129.2º Note: Heron's formula If the three sides of a triangle are known, Heron's formula can be used to find the area of the triangle. 140

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Heron's formula states that. «The area of a triangle with sides a, b and c is

, where = a + b + c

Area = s(s – a)(s – b) (s – c)

Example 20 Calculate the area of a triangle whose sides are of lengths 4 cm, 5 cm, and 7 cm. Solution s= 4+5+7 =8 2 Heron's formula gives Area = 8(8 –4) (8 – 5) (8 – 7)

=4 6

The area is 4 6 cm2

Notice that in any triangle, the sum of the lengths of any two sides is greater than the third side. Inclination and depression The angle of elevation (inclination) is the angle above the horizontal through which a line of view is raised. The angle of depression is the angle below the horizontal through which a line of view is lowered. observer's eye angle of depression

object

horizontal

object

angle of elevation observer's eye

Fig. 7.28

horizontal N P

Bearings

In navigation, the direction of bearing from a point O of an object at point P means the positive angle measured clockwise from the north (N) to the ray, as shown in the figure aside.

θ O

Fig. 7.29 141

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Example 21 A straight trail with uniform inclination leads from a hotel, elevation 5000 metres to a scenic overlook, elevation 12 000 metres. The length of the trial is 14 000m. What is the inclination of the trial? Solution S

Given: AH = 5000 m

A′

A′S = 12 000 m HS = 14 000m

Fig. 7.30

The angle at vertex H of triangle HOS is such that sin H = OS = 7 000 = 1 HS 14 000 2 The angle is H = 30° Therefore, the inclinalion of the trail is 30°.

Activity 7.5 1.

An eagle which is soaring at 200 meters above the ground level, is observed from two points that are x metres apart at ground level. From one point which is due South of the eagle, the angle of elevation is 30° and from the other point, which is due East of the eagle, the angle of elevation is 40°. Calculate the value of x.

An observer notes that the angle of elevation of the top of a tower is 30° from a point A. While, at a point B, 40 metres nearer to the tower, the angle of elevation is 60°. Find the height of the tower.

In triangle ABC, AB = x , AC = 2x – 1 and BC = 2x + 1. Show that x > 2.

Find the area of a triangular flower bed which has one angle of 30° and the sides about this angle of lengths 10m and 12m.

7.3.6 Angle identities (transformation formulas) In a right triangle, the acute angles are complementary. So A + B = 90º and B = 90º – A, where A and B are the measures of the acute angles 142

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B c

C b

Fig. 7.30

Sin (90º – A) = sin B

A and B are complementary angles

= b

definition of sine in a right triangle

definition of cosine in a right triangle

= cos A

Exercise Try the following

Given that sin x = 2 < x < π, find without using a calculator: (a) cos x and sin(x + π)

(b)

tan (x + 2π) π

In the complementary identities, any angle Ɵ is subtracted from 2 . There are also identities for subtracting any two angles. It is convenient to start with an identity for finding the cosine of the difference of two angles. In Fig. 7. 31, angles A, B and A-B are shown. Use the distance formula to find the square of the distance between P and Q. (PQ)2 = (x1 – x2)2 + (y1 – y2)2

= (cos A – cos B)2 + (sin A – sin B)2

= cos2A – 2cos A cos B + cos2B + sin2A – 2sinA sinB + sin2B

= 2 – 2cos Acos b – 2 sinAsinB

Use the pythagorean identity

sin2Ɵ + cos2Ɵ = 1

143

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1 y P(cos A, sin A)

Q(cos B, sin B) A–B A

–

x 1

Fig. 7.31 Now use the law of cosines to find (PQ)2 in triangle POQ. (PQ)2 = (PO)2 + (QO)2 – 2 (PO)(QO) cos(A–B) = 12 + 12 – 2(1)(1)cos(A-B)

= 2 – 2 cos (A-B)

The transitive property for equality tells you that the two expressions for (PQ)2 are equal. 2 – 2 cos (A-B) = 2 – 2 cosA cosB – 2sinA sinB –

2cos(A-B)

= –2cosA cosB –2sinA sinB

subtract 2 from each side

cos(A-B)

= cosA cosB + sinA sinB

divide each side by –2

You can also derive an identity for sin (A-B). Then you can use the tangent identity to derive an identity for tan (A-B). We have; sin (A-B)

= sin A cosB – cosA sinB

cos (A–B)

= cos A cosB + sinA sinB

tan (A-B)

= 1 + tan A tan B

tan A – tan B

Example 22 Find the exact value of cos15°. Solution You know exact values for 30°, 60° and 45°. Use the fact that 15° = 60° – 45° cos (A-B) = cos A cos B + sinA sinB cosine angle difference identity cos (60° – 45°)= cos 60° cos45° + sin60° sin45° substitute 60° for A and and 45° for B 144

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= 12 2 + 3 2

2 2

..................... replace with exact values

= 2 + 6 ................................... 4

simplify.

= 2 + 6 4

Therefore, cos 15º =

+ 6 .

Angle sum identities From A + B = A–(–B) and sin (–B) = – sin B, tan (–B) = – tan B, we have:

sin (A + B) cos (A + B)

= sin A cos B + cos A sin B = cos A cosB – sinA sinB

tan (A + B)

= 1 – tan A tan B

tan A + tan B

Example 23 Find the exact value of sin 105° Solution Use the fact that 105° = 60° + 45° sin (A + B) = sin A cos B + cos A sin B.................. single angle sum identity sin (A + B) = sin 60° cos 45° + cos 60° sin 45°..... substitute 60° for A and 45° for B

( )

= 3 2 + 12 2 ........................ replace with exact values simplify 2 2 2 = 6 + 2 ........................................... 4

Therefore, sin 105º

= 6 + 2 4

Double Angle identities You can use the angle sum identities to derive the double angle identities. Let Ɵ = A = B cos (A + B)

= cos A cos B – sin A sin B ......... cosine angle sum identity 145

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cos (Ɵ +Ɵ)

= cosƟ cosƟ – sin Ɵ sin Ɵ......... substitute Ɵ for A and B

cos 2Ɵ

= cos2Ɵ – sin2Ɵ ............................ simplify

You can use the pythagorean identity sin2Ɵ + cos2Ɵ = 1 in the form: sin2Ɵ = 1 – cos2Ɵ to derive another identity for cos 2Ɵ

cos2Ɵ = cos 2Ɵ – sin2Ɵ

= cos2Ɵ – (1–cos2Ɵ) substitute 1 – cos2Ɵ for sin2Ɵ = cos2Ɵ – 1 + cos2Ɵ

remove parentheses

=2cos Ɵ –1

simplify

You can use the pythagorean theorem in the form cos2Ɵ = 1 – sin2Ɵ to derive a third identity for cos2Ɵ

cos2Ɵ = cos2Ɵ - sin2Ɵ

= (1 – sin2Ɵ) – sin2Ɵ substitute 1 – sin2Ɵ for cos2Ɵ

= 1 – 2 sin2Ɵ

simplify

You can use the other angle sum identities to derive double angle identities for the sine and tangent. We have;

Cos2Ɵ = cos2Ɵ – sin2Ɵ

cos2Ɵ =2cos2Ɵ –1

cos2Ɵ = 1 – 2 sin2Ɵ

sin2Ɵ = 2sinƟ cosƟ 2tanƟ tan2Ɵ = 1 – tan2Ɵ

Example 24 Verify the identity cos 2Ɵ=

1 – tan2Ɵ 1 + tan2Ɵ

1 – tan2Ɵ ....................... pythagorean Identity = sec2Ɵ 1 + tan2Ɵ

1 – tan2Ɵ

1 sec2Ɵ

–

tan2Ɵ sec2Ɵ

........................................ write as two fractions

sin2Ɵ 2 = 1 – cos Ɵ .................... [express in terms of sin Ɵ and cosƟ] 1 1 sec2Ɵ cos2Ɵ 146

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= cos2Ɵ – sin2Ɵ ......................... simplify = cos2Ɵ......................................... double angle identity You can use double angle identities to derive half angle identities: let Ɵ = A 2

cos2Ɵ

= 2cos2Ɵ – 1

cosine double angle identity

cos2 ( A )

= 2 cos2 A – 1

substitute A for Ɵ

cosA + 1 2

= cos2 2

solve for cos2 A 2

± cosA + 1 =cos A

take the square root of each side

A similarly, sin 2 = ± cosA + 1 and tan A = ± 1–cosA 2

1+cosA

Half angle identities sin A = ± 1–cosA cos A =± 1+ cosA tan A = ± 1–cosA 2

1+cosA

Choose the positive or negative sign for each function depending on the quadrant in which A lies. 2

Example 25 Use the half angle identities to find the exact value of the following: (a) sin 15°

(b)

cos 150°

Solution (a) sin 15° = sin ( 30 )°....... rewrite 15° as 30° 2 2 = 1–cos30° ..... use the principal square root since sin 15° is positive 2

1–√ 3 2 2

......... substitute the exact value for cos 30°

2 – √3 4

...... simplify

2 – √3 2

...... simplify 147

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(b) cos 150° rewrite 150 as 300

cos 150° = cos( 300 )°

= – 1 + cos 300° use the negative square root, since cos 150° is negative 2

1+√ 1 2 2

=–

= – 3 4

replace with an exact value

simplify

= – 3

simplify

Example 26 –24

Given sin Ɵ = 25 and 180° < Ɵ < 270°. find sin Ɵ . 2 Solution First find cosƟ: cos2Ɵ + sin2Ɵ = 1.......... pythagorean identity cos2Ɵ + (– 24 )2 = 1....... substitute 25 cos2Ɵ =

252 – 7 cosƟ = 25 Now find sin Ɵ . 2

... solve for cos2Ɵ ... choose the negative square root since Ɵ is in quadrant III

Since 180° < Ɵ < 270°, 90° < Ɵ < 135° and Ɵ is in quadrant II 2

sin Ɵ = ± 1–cosƟ ..... half angle identity 2 2

1–(– 7 25

) ...... substitute. Choose the positive square root since Ɵ is in quadrant II. 2

= 4 ..................... simplify. 5

148

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Exercise 1.

Calculate, without using a calculator, cos 75°.

2. Show that:

1 – cos Ɵ = tan2 Ɵ 2 1 + cos Ɵ

(a)

(b) sin (x + y) – sin (x – y) = 2 cos x sin y The factor identities We have: sin(A + B) = sin A cos B + sin B cos A ............................. (1) sin (A – B) = sin A cos B – sin B cos A ............................ (2) Adding, side by side, (1) and (2): sin (A + B) + sin (A – B) = 2 sin A cos B Let A + B = P and A – B = Q Then A = P + Q and B = P – Q 2

sin (A + B) + sin (A – B) = 2 sin A cos B becomes sin P + sin θ = 2 sin P + Q cos P – Q 2

By subtracting (2) from (1) sin (A + B) – sin (A – B) = 2 cos A sin B Therefore, sin P – sin Q = 2 cos P + Q sin P – Q Also, 2

cos (A + B) = cos A cos B – sin A sin B .........................

(3)

and cos (A – B) = cos A cos B + sin A sin B .................

(4)

(3) + (4): cos (A + B) + cos (A – B) = 2 cos A cos B Therefore, cos P + cos Q = 2 cos P + Q cos P – Q . 2

(3) – (4): cos (A + B) – cos (A – B) = –2 sin A sin B. Therefore, cos P – cos Q = –2 sin P + Q sin P – Q 2

The factor formulas are: sin P + sin Q = 2 sin P + Q cos P – Q 2

sin P – sin Q = 2 cos P + Q sin P – Q

2 2 cos P + cos θ = 2 cos P + Q cos P – Q 2 2 P+Q P–Q sin cos P – cos Q = – 2 sin 2 2

149

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Example 27 sin A + 2 sin 3A + sin 5A sin 3A + 2 sin 5A + sin 7A

Simplify the expression Solution sin A + 2 sin 3A + sin 5A sin 3A + 2 sin 5A + sin 7A

A + 5A 2 3A + 7A 2 sin 2

2 sin A

= (sin A + sin 5A) + (2 sin 3A)

(sin 3A + sin 7A) + (2 sin 5A)

cos A – 5A + 2 sin 3A

2 3A – 7A cos + 2 sin 5A 2

2 sin3A cos 2A + 2 sin 3A 2 sin 5A cos 2A + 2 sin 5A

= (2 sin 3A)(cos 2A + 1)

(2 sin 5A) (cos 2A + 1)

= sin 3A . sin 5A

The product to sum identities We have: sin (A + B) = sin A cos B + sin B cos A ............

(1)

sin (A – B) = sin A cos B – sin B cos A ............. (2) Adding (1) and (2), side by side: sin (A + B) + sin (A – B) = 2 sin A cos B Therefore, sin A cos B = = 1 [ sin (A + B) + sin (A – B)] 2

Also, cos (A + B) = cos A cos B – sin A sin B .................. (3) cos (A – B) = cos A cos B + sin A sin B ..................... (4) (3) + (4) : cos (A + B) + cos (A – B) = 2 cos A cos B Therefore, cos A cos B = 1 [ cos (A + B) + cos (A – B)] 2

(3) – (4): cos (A + B) – cos (A – B) = –2 sin A sin B. Therefore, sin A sin B = – 1 [ cos (A + B) – cos (A – B)] 2

The identities: Sin A cos B = 1 [ sin (A + B) + sin (A – B)] 2

150

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cos A cos B = 1 [ cos (A + B) + cos (A – B)]

2 sin A sin B = – 1 2

[ cos (A + B) – cos (A – B)]

are called "product to sum identities" Example 28 Express as a sum: (a) sin 5x cos 7x

(b) cos x cos 3x

(c)

sin 4x sin 8x

Solution (a) sin 5x cos 7x = 1 [ sin (5x + 7x) + sin (5x – 7x)]

2 1 = (sin 12x – sin 2x) = 1 sin 12x – 1 sin 2x 2 2 2 1 (b) cos x cos 3 x = [cos (x + 3x) + cos (x – 3x)] 2 1 = (cos 4x + cos 2x) = 1 cos 4x + 1 cos 2x 2 2 2 1 (c) sin 4x sin 8x = – [ cos (4x + 8x) – cos (4x – 8x)] 2 1 =– (cos 12x – cos 4x) 2 = – 1 cos 12x + 1 cos 4x. 2 2

The t- identities Let t = tan x , where – π < x < π . 2

1 + t2 x 2

Fig. 7.32

From the figure above, we have: sin x =

opposite Hypotenuse

t 1 + t2

cos x =

Adjacent Hypotenuse

1 1 + t2

From double - angle formulas, 151

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(

t sin x = 2 sin x cos x = 2 1 + t2 2 2

) ( 1 1+ t ) = 1 2t+ t 2

cos x = cos2 x – sin2 x 2

( 1 1+ t ) – ( 1 +t t ) 2

1 1 + t2

tan x =

sin x cos x

–

t2 1 + t2

2 = 1 – t2

2t 1 + t2 1 – t2 1 + t2

1+t

= 1 2t . – t2

Example 29 cos x – sinx – 1 Use the change of variable tan x2 = t to simplify the expression sin x + cos x + 1

Solution cos x – sin x – 1 sin x + cosx + 1

1– t2 1 + t2 2t 1 + t2

– +

2t 1 + t2 1 – t2 1 + t2

–1 +1

1 – t2 – 2t – 1 – t2 1 + t2 2t + 1 – t2 + 1 + t2 1 + t2

= 2t + 2 = 2(t + 1)

= –t = – tan x2 .

–2t2 – 2t

–2t (t + 1)

Activity 7.6 1.

Given that sin x = 12 and π2 <x < π, 2

find cos x, sin (x + π) and tan (x + 2π).

Find, without using a calculator, the exact vlue of cos 75°.

(a)

cos θ = tan2 θ . Show that 11 +– cos θ 2

(b)

Factorise: sin (x + y) – sin (x – y).

152

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Without using a calculator, find the exact value of

π + sin π cos π . sin π9 cos 18 18 9

Find the value of each of the following:

(a)

cos 63° sin 27° + cos 27° sin 63°

(b)

cos 70° cos 10° + sin 70° sin 10°

(c)

cos 130° (cos 40°) + sin 130° (sin 40°)

without using a calculator, show that:

cos 11° + sin 11° = tan 56° (a) cos 11° – sin 11°

(c)

(b)

cos 15° – sin 15° cos 15° + sin 15°

= 3 3

tan ( π4 + θ) tan ( π4 – θ) = 1

7.3.7 Trigonometric equations and inequations Equations of the type cox x = cos ; sin x = sin

and tan x = tan .

The general solutions of the equations above are, respectively, x = ± α + 2kπ; (x = α +2kπ or x = π – α + 2kπ); x = α + kπ, where k is an integer. Example 30 Find the general solution of each of the following equations: (a) sin 2x – 1 = 0 (b) cos (2x + π3 ) = cos (5x – π4 ) (c) tan 3x = 1. Solution (a) The equation is equivalent to:

sin 2x = 1;

sin 2x = sin π2 ;

2x = π2 + 2kπ or 2x = π – π2 + 2kπ, where k∈ .

x = π4 + kπ 153

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(b) cos (2x + π3 ) = cos (5x – π4 );

2x + π3 = 5x – π4 + 2kπ or 2x + π3 = –5x + π4 + 2 kπ,

where k∈ .

+ 2kπ or x= –π + 2kπ x = 7π 36 3 84 7 (c) tan 3x = tan π4

3x = π4 + kπ

π kπ + 3 , where k∈ . x = 12

Quadratic trigonometric equations Example 31 Find the general solution of the equation: (a) sin2 x + 3 cos x – 3 = 0

(b)

cos2 x = 2 (sin x – 1)

(1 – cos2 x) + 3 cos x – 3 = 0;

– cos2 x + 3 cos x – 2 = 0

Let cos x = t. The equation becomes

–t2 + 3t –2 = 0;

t = 1 or t = 2 (rejected)

cos x = 1; x = 2kπ, where k∈ . (b) cos2 x = 2 (sin x – 1);

1 – sin2 x – 2 sin x + 2 = 0;

–sin2 x - 2 sin x + 3 = 0

Let sin x = t. The equation becomes

t2 + 2t – 3 = 0;

t = 1 or t = –3 (rejected)

sin x = 1; x = π2 + 2kπ, where k∈ . 154

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tan x = 1 + tan2x ; tan2x + tan x – 2 = 0.

Let tan x = t. The equation becomes

t2 + t – 2 = 0;

t=1

tan x = 1; x = π4 + kπ, where k∈ . Solving trigonometric equations by using trigonometric identities. Example 32 Find the general solution of the equation: 3 π (a) sin (x + 12 ) + sin (x + π4 ) = 2

(b) sin 5x – sin 3x – sinx = 0 Solution π ) + sin (x + π4 ) = 2 ; (a) sin (x + 12 3

[ (

2 sin

π 2x + 3 2

) cos π6 ] = 32 ;

sin 6x 6+ π = 3 ; 2

6x + π 6

= π3 + 2kπ or 6x 6+ π = 2π 2kπ, 3

where k∈

x = π6 + 2kπ or x = π2 + 2kπ;

S= { π6 + 2kπ; k∈ } { π2 + 2kπ; k∈ }

(b) sin 5x – sin 3x – sin x = 0;

2 sin x cos 4x – sin x = 0;

sin x (2 cos 4x – 1) = 0; 1

sin x = 0 or cos 4x = 2 ; 155

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x = 2kπ or x = π + 2kπ, where k∈

π π or x = 12 + kπ or x = – 12 + kπ . 2 12

Equations of the type a cos x + b sin x = c, where a ≠ 0, b ≠ 0 a cos x + b sin x = c; b

a (cos x + a sin x = c; Find θ such that tan θ = a a( cos x + tan θ sin x) = c; sin θ

a (cos x + cos θ sin x ) = c; a cos θ

(cos x cos θ + sin x sin θ)= C a

r (cos x cos θ + sin x sin θ) = c, let r = cos θ (r cos θ) cos x + (r sin θ) sin x = C. By identification to the original equation a = r cos θ; b = r sin θ and a2 + b2 = r2 c

Therefore, the equation a cos x + b sin x = c is equivalent to cos (x – θ) = r , where cos θ = 2 a 2 ; sin θ = 2 a 2 and r = a2 + b2 . a +b

a +b

Example 33 cos x +

Solve the equation:

(a)

(b)

sin x – 3 = 0

cos 3x + sin 3x = 2.

Solution (a) The equation is equivalent to: 1 cos (x – θ) = 3 , where cos θ = 2 and

5π 6

sin θ = π 6

1 2 o

3 2

. 1

It follows that θ = 3 x

cos (x – π2 ) = 3 ; 2

x – π3 = π6 + 2kπ or x – π3 = – π6 + 2πk,

k∈

Fig. 7.33 156

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x = π2 + 2kπ or x = π3 + 2kπ, where k∈ Solution set S = { π2 + 2kπ, k∈ } ∪ { π2 + 2kπ, k∈ } (b) The equation is eqivalent to:

1 cos (3x – θ) = 1, where cos θ = 3 and sin θ = 2 .

It follows that θ = π6

cos (3x – π6 ) = 1;

π 3x – 18 = 2kπ, where k∈

π x = 18 + 2kπ 3

π + 2kπ , k∈ } Solution set S = { 18 3

Note The equation a cos x + b sin x = c can also be solved by the change of variable tan 2x = t. Example 34 Solve, in the set of real numbers, the equation sin 2x – cos 2x – 1 = 0 Solution 2t

1 – t2

Let tan x = t. The equation becomes 1 + t2 – 1 + t2 – 1 = 0; 2t – 1 + t2 – 1 – t2 1 + t2

= 0;

t = 1.

tan x = 1; x = π4 + k, where k∈ . Trigonometric inequations Example 35 Solve in [0, 2π] the inequation. (a)

2 sin 3x – 1 ≥ 0

(b)

cos 2x + 2 sin 2x – 1 ≥ 0 157

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solution (a)

The inequation is equivalent to sin 3x ≥ 2 1

1 sin 3x = 2 and 0 ≤ x < 2π if and only if x = π2 or x = 5π 2

(sin 3x ≥ 2 ) if and only 1

+ 2kπ, (k∈ ) if π6 + 2kπ ≤ 3x ≤ 5π 2

π 18

The values of k for which x ∈ [0, 2π] are 0, 1 and 2

π ≤ x ≤ 5π For k =0, 18 18

≤x ≤ 17π For k = 1, 13π 18 18

≤ x ≤ 2918π For k = 2, 25π 18

The solution set is

, 25π ] ∪ [ 25π , 25π ] ∪ [ 25π , 25π ]. S = [ 25π 18 18 18 18 18 18

(b)

The inequation is equivalent to cos ( π3 – 2x) ≥ 2

+ 2kπ ≤ x ≤ 5π + 2kπ 3 18 2

– π3 + 2kπ ≤ π3 – 2x ≤ π3 + 2kπ; + 2kπ≤ –2x ≤ –2x≤ 2kπ; – 2π 3 y

–kπ≤ x≤ π3 – kπ

π 6

The values of k for which the solutions x are in the interval [0,2π] are k = 0 and k = 1 x

. For k = 0, 0 ≤ x ≤ π3 and for k = 1, π ≤ x ≤ 4π 3 ] Therefore, the solution set is S=[0, π3 ] ∪ [π, 4π 3

π Fig. 7.34 –

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Activity 7.7 1.

Solve the inequation 2 cos x + 1 < 0, 0≤x ≤ 2π

Solve the equation sin x + sin 5x – sin 3x = 0, x∈ .

Solve the euqation 2 cos2 x – sin x = 1, where – π ≤ x ≤π

Solve the equation cot x + tan x = 41 where – π ≤ x ≤π, for tan x.

7.4 Practice 7.4.1 Angle and its measurements 1.

Convert into degrees:

(a) 3π rad 8 2. Convert into radians:

π (a) 20 rad

(a)

(b)

Convert each angle to a decimal in degrees:

(a)

Convert each angle to degree, minute, second form:

(a)

15°

40° 10′ 25′′ 40.32°

(b) (b)

36° 9° 9′ 9′′ 18.255°

7.4.2 Unit circle and the six trigonometric ratios 1.

Complete the following table: 0°

30°

45°

60°

90°

sin cos tan sec cosec cot 2.

Show, on a unit circle, 135°, sin 135°, cos 135° and tan 135°.

Without using a calculator, determine the value of:

(a)

4 sin 30° cos 45°

(b)

sin2 270° + cos2 30° + tan 45° 159

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Given that the terminal side of angle θ in standard position passes through point (–1,1) find:

(a)

sin θ

(b)

cos θ

(c)

tan θ

7.4.3 Trigonometric identities 1. Simplify

1 (a) tan2θ cos2θ (b) – tan2θ cos2θ

In which quadrant is the terminal side of angle θ located?

(a)

State the sign of each of the following trigonometric ratio if the terminal side of θ lies in quadrant III

(a)

2 Given that cos θ = – 5 . where π2 < θ < π, find

(a)

sin θ

(b)

sec θ

(d)

tan θ

(e)

cosec θ

θ Show that cotcosec = cos θ θ + tan θ

225°

sec θ

(b)

85π 6

(b)

tan θ

rad.

(c)

cot θ

(c)

cot θ

(c)

cot θ

6. Simplify:

(a)

(1– sin2 θ) (1 + tan2 θ) – 1

(b)

Sin θ – 1 +sin2 θsin+ θcoscosθ θ

, find: Given that tan θ = 3, where π < θ < 3π 2

(a)

cos θ

(b)

sin θ

(d)

sec θ

(e)

coses θ

Show that:

tan x cos x (a) –sin = –1 x cos x 2

(b)

(– cot x) (–sin x) (– cox x) (– cos x)

= sec x

7.4.4 Calculations involving trigonometric ratios 1.

Determine, without using a calculator, the exact value of:

(a)

tan (–135°) –cos 180°

(b)

2 sin 120° + tan 150°

(c)

cos (–210°) – tan 135°

(d)

sin 120° cos 210° – sin 270° cos 315°

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Simplify each of the following without using a calculator: tanθ cosθ

cos x sin (180° – x)

(a) sin (90° – θ) sin θ (b) cos (180° – x) tan (360° – θ) cos (180° + θ

(a)

(c)

cos x 1 – sin2 x

= sec x

cos 260° cos 190° cos 170° sin 10° cos 350°

(d)

sin3 + sin x cos2 x cos x

(b)

= tan x

sin(180° + x) cos2 (90° – x) + sin (–x) sin2 (90° – x) = sin x

(

tan x + cos x (d)

)

4. Simplify:

3π

sin (π – a) cot ( 2 + a) cos (a – 2π) (a) π

3π

tan (π + a) tan ( 2 + a) cos ( 2 – a) cos(90° + a) cos (180° – a)

sin (180° – a) cos (180° + a)

+ sin (90° + a) cos (270° + a) (b) sin (90° – a) sin (a – 360°) π

(c)

tan ( π + a) tan( 2 + a) π

3π

tan (π + a) tan ( 2 + a) cos ( 2 – a)

cot ( 2 – a) cot (π – a) 3π

tan ( 2 + a)

sin ( π2 + a) cos (a – π) cot ( 2 – a) – sin ( 2 – a) cot ( a – 3π

(d)

3π

3π 2 )

7.4.5. Triangles 1.

In triangle ABC, right angled at A, BC = 6 and B = 30°

Find AC and AB

In triangle ABC, right angled at B, A = 30° BC = 3 and AC = 5.

Find sin A and tan A

In triangle ABC, right angled at C, B = 60° and BC = 3

Find AC and AB.

Solve the oblique triangle ABC on the figure: C

52°

48°

Fig. 7.35 161

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The sides of a triangle ABC are a= 6cm, b = 8 cmand c = 5 cm. Find the angles of the triangle. Find also the area of the triangle.

From a point A, an observer finds that the angle of elevation of the top of a tree is 21°. If he walks 10 metres towards the foot of the tree to a point B, he finds that the angle of elevation of the top of the tree is 34°. Calculate the height of the tree above the level of observation.

In triangle ABC, a = 10 cm, A = 30°, B = 40°, Find c.

In triangle ABC, a= 8cm, b = 10m and A = 30°. Find the size of angle C .

7.4.6 Angle identities 1.

Evaluate, without using calculator,

(a)

Given that sin A = 4 and cos B = 13 , sin B < 0, find cos (A – B), sin (A + B) and tan (A + B)

Given that sin (A + B) = 29 , sin A = 5 and cos A <0. Find sin B and cos B. Show that:

sec 105° (b)

tan 165°

(c)

π cosec 12

(a) cos4a – sin4 a = cos 2a = tan2 ( π4 – 2 ) (b) 1 + sin x 1 – sin x

5 –1

Given that cos x =

find the value of cos 2x and cos 4x

(a)

Given that sin θ = 13 , find cos 2 , sin 2 and tan 2 if cos θ < 0.

(b)

Given that cos θ = 7 , and sin θ < 0, find cos 2 , sin 2 and tan 2

Given that x + y = π4 and tan x = 5 , find tan y and y. 8. Factorise; 1

(a)

sin 5x + sin 3x

(b)

cos 3x – sin 7x

Show that cos2 a – cos2 b = – sin (a + b) sin (a – b)

10. Express as a sum (or difference);

(a)

cos 4x cos 2x

(b)

sin 4x cos 12 x

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7.4.7. Trigonometric equations and inequations 1.

Find the general solution of the equation 1 + tan x = 0

Solve for x the equation 2 cos2 x + 3 cos x – 2 = 0, where 0 ≤ x ≤2π.

Solve for x the equation 2 cos2 x – sin x – 1 = 0, where o ≤x ≤ 2π.

Solve the equation 3 cos x – sin x = 0, where 0 ≤ x ≤ 2π.

Solve the equation cos 2x – cos x = 0, where 0° ≤ x ≤ 360°.

Solve the equation 2 sin (x – 60°) = 1, where 0° ≤ x ≤ 360°.

Solve the equation tan 4x – 3 = 0, where 0° ≤ x ≤ 360°.

Solve the equation tan x + tan x = 4 for the value of tan x.

1 Solve the unequation cos (x – π3 ) ≥ – 2 , where 0 ≤ x ≤ 2π.

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Chapter 8 Linear algebra 8.1 Objectives In this chapter you will: • determine the sum of vectors and the product of a vector by a real number. •

verify that (R, V, +) and (R, R 2, +) are vector spaces.

•

verify if a given vector is a linear combination of other vectors.

•

differentiate linear dependent vectors from linear independent vectors.

•

determine a basic of a vector space and its dimension.

•

determine the image of vectors under linear transformation.

•

prove that (R, V, +) and (R, R 2, +) are isomorphic vector spaces.

•

determine the matrix of a linear transformation.

•

carry out operations on matrices.

•

calculate the determinants of matrices of order 2 and order 3.

•

solve problems related to a system of 2 or 3 linear equations by using matrices.

•

calculate the inverse of a matrix of order 2.

8.2 Key concepts Vector space

Linear combination

Linear independence Generating set

Basis Matrix

Dimension System

Components Determinant.

Linear transformation

8.3 Theory 8.3.1 Vector spaces

Definition The ordered pair (V, +) is said to be a vector space over the field (R, +, ) if and only if: (1) (V, +) is an abelian group (2) The scalar multiplication satisfies the following four properties P1: The scalar multiplication is distributive over the addition in V. .(u + v) = u + .v P2: The scalar multiplication is distributive over the addition in . ( + B).u = u + B.u 164

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P3: Associative property

( β)u = .(β.u)

P4: The identity property 1.u = u β

uV v w

( , V)→V ( , u)

The members of V are called vectors, the members of

are called scalars (operators).

In the set of ordered triples of real numbers

= {(a, b, c):a ∈ , b ∈ . c ∈ } Addition and scalar multiplication are defined as follows; if u = ( a, b, c) and v = (d, e, f) = (a, b, c) + (d, e, f) then u + v = (a + d, b + e, c + f) = . (a, b, c) and .u = ( a, b, c) It can be shown that ( 3, +) is a vector space over the field ( , +, .) of real numbers. More generally. 3

= {(x1, x2, x3, ....... xn), x1∈ ; x2 ∈ , ..... xn ∈ }

Then ( n, +) is a vector space over the field ( , + ,.)

Properties

1. 0.u = o for any vector u .o = o for any scalar u = o if and only if = o or u = o 2. 3. 4. 5.

( – β)u = u – βu for any vector u and any scalars and β (u – v) = u – v If u = v and ≠ o then u = v If u = βu and u ≠ o then = β

If u + v = w + v = then u = w If u + v = u + w then v = w

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Example 1 Vectors u , v and w are such that; –2v – 6u = o 3w – 18v – 9u = o

Express v and w in terms of u (u ≠ o ) Solution 2v – 6u = o – 2v = 6u 6 v = –2 u

–

v = –3u

3w – 18(–3u ) –9u = o 3w + 54u – 9u = o 3w + 45u = o w = – 45u 3

w = –15u

Linear combination Definition

w is said to be a linear combination of u and v if it can be expressed as w = xu + yv , where x and y are scalars. Example 2 u = (–1, 4); w = (2, –8) Check whether (a) w = (–7, 28); (b) t = (3, 1) are linear combinations of u and v . Solution w = x.u + y.v

(a)

(–7, 28)= x(–1, 4) + y(2, –8)

= (–x, 4x) + (2y, –8y)

(–7, 28)=(–x + 2y, 4x – 8y) –x + 2y= –7 –x + 2y = –7 4x – 8y = 28 ⇒ 4(x – 2y) = 4(7) x – 2y = 7

Where y is any real number. when y = 0 then x = 7; when y = 1 then x = 9 166

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when y = –1 then x = 5 w = 7.u + 0.v = 9u + 1. v

= 5u + 1.v Hence w is a linear combination of u and v t = xu + y.v

(3, 1) = x(–1, 4) + y(2, –8)

(3, 1) = (–x + 2y, 4x – 8y).

–x + 2y = 3 4x – 8y = 1

x –2y = –3 1 x – 2y = 4

No such values of x and y. Hence t is not a linear combination of u and v .

Basis Definition B = (e1 , e2 ) is said to be a basis of a vector space V if and only if: (i) Any vector in V is a linear combination of e1 and e2 , that is, for any u ∈v, there exists scalars x and y such that u = xe1 + y.e2 . We say that B generates V (ii)

o = x u + y.v implies that x = y = 0. We say that u and v are linearly independent.

Example 3

In the vector space let B = (e1 , e2 ) where e1 = (1, 0); e2 = (0, 1)

Show that B is a basis of

Solution (i) Let u = (a, b) be any vector in u = x.e1 + y.e1

(a, b) = x(1, 0) + y(0,1) = (x, 0) + (0, y) = (x, y)

x=a

y=b

(a, b) = a(1, 0) + b(0, 1)

∴u is a linear combination of e1 and e2 167

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(ii) u

= xe1 + y e2

(0, 0) = x(1, 0) + y(0, 1) = (x, y)

x = 0; y = 0

∴B is a basis

B = {(1, 0); (0, 1)} is said to be the standard basis of 2 .

The standard basis of

is B = {(1, 0, 0), (0, 1, 0); (0, 0, 1)}

Component of a vector in a basis If B = (e1 , e2 , e3 ,....,en ) is a basis of a vector space, V, then any vector u of V can be

expressed in only one way as u = x1e1 + x2e2 +....+xnen . The scalars x1, x2,....,xn are said to be the components (or coordinates) of vector u in the basis B. It is denoted u x1 or x2 • xn

u (x1, x2,.....,xn) Example 4

Show that B = (e1 , e2 ), where e1 = (–1, 1) and e2 = (2,3) is a basis of the vector space 2 over the field (R, +, .) and find the components of the vector u = (–13, –7) in the basis B. Solution

(i) Let u = (a, b) be any vector in

u = x.e1 + y.e2 (a, b) = x(–1, 1) + y(2, 3) = (–x + 2y, x + 3y) – x = a – 2y

x = –a + 2y; substituting for y = a +b gives 5 x = –a + 2(a+b) 5

x + 2y= a x + 3y = b 5y = a + b

–

y = a +b 5

x = –5a + 2a + 2b 5

x = –3a + 2b 5

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Such scalars x and y exist, therefore any vector u is a linear combination of e1 and e2 . If w = (a, b) then,

w = –3a + 2b e1 + a +b e2 5 5

(ii) let 0 = xe1 + ye2

= x(–1, 1) + y(2, 3)

i.e. (0, 0)

(0, 0) = (–x + 2y, x + 3y)

x + 2y=0 x + 3y = 0

–

5y = 0; y = 0 x=0 Since 0 = x.e1 + y.e1 implies x = y = 0, the vectors e1 and e1 are therefore linearly independent from (i) and (ii) it follows that B is a basis.

u = (–13, –7)

–13 + (–7) –3(–13) + 2(–7) e1 + e2 5 5

u = 5e – 4e2 . 1

Dimension of a vector space It can be shown that all the bases of the same vector space have the same number of members. The dimension of a vector space is the number of members in any basis of the vector space. •

The dimensions of vector spaces 2, 3, ......, h over the field ( , +, .) are 2, 3,..., n, respectively.

Example 5 •

In the vector space contained in 3.

, we consider E = {(a, b, c), a + b = 0}; E ia a vector space

Find (a) a basis of E

(b) the dimension of E.

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Solution (a)

= {(a, b, c); a + b = 0}

= {(a, b, c); b = –a}

= {(a, –a, c); a ∈ , c ∈ }

={(a, –a, o) + c(0, 0, 1}

Thus,

a basis of E is B = {(1, –1, 0); (0, 0,1)}

(b) Therefore, Dim E = 2.

Activity 8.1 1.

Given vectors u = (1,4), v = (–4, –2) and w = (–4, 6), find 2u – v + w

Solve for x the equation x + a = b if,

a = (3, –2) and b = (1,5). (a) a = (1, –2, 2) and b = (–2, 5, 4). (b) 3. Express w = (9, 2, 7) as linear combination of u = (1, 2, –1) and v = (6,4,2). 4.

B = ( i ,j ) is a basis of a vector space V.

B′= (e1 , e2 ), where e1 = 3 i – 2 j and e2 = 3 i + 2 j .

(a)

Show that B′ is a basis of V

x has components (3, –2) in basis B′. (b) Find the components of x in basis B.

Definition Let V and W be vector spaces over the same field ( , +, .) function f: V → W is said to be a linear mapping (a linear function) if and only if: u

f (u )

f ( u1 + u2 ) = f (u1 ) + f(u2 ) and f ( u ) = f ( u ) . 170

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for any vectors u1 , and u2 in V1 and any scalar , and vector u . If V = W, then the linear mapping is called linear operator. Example 6 Show that f: (x, y)

→

f (x, y) = (2x – y, x+y), is linear.

Solution let u1 = (x1, y1) and u2 = (x2, y2) Then u1 + u2 = (x1, y1) + (x2, y2)

f ( u1 + u2 )= f(x1+ x2, y1 + y2)

=(2x1 + x2)– (y1 + y2), (x1 + x2) + (y1 + y2)) =(2x1 – y2) + (2x2 – y1), (x1 + y1) + (x2 + y2)

= (2x1 – y1, x1 + y1) + (2x2 – y2, x2 + y2)

= f(x1, y1) + f(x2 + y2)

= f( u2 ) + f( u2 )

Let u = (x, y) ∈

and

∈ . Then

u = (x, y) = ( x, y)

f(α u ) = f ( x, y) = (2( x) – ( y), ( x) + ( y)) = (2x – y, x + y) = f (x,y) =

f( u )

Therefore f is linear.

Example 7 Check whether f is a linear in each of the following cases: (a) → x

f(x) = sin x

(b) → x

f(x) = 3x

Solution (a)

u1 = x; u2 = y 171

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Then ( u1 + u2 ) = x + y f( u1 + u2 ) = f (x + y) = sin (x + y) f(x) + f(y) = sin x + sin y since sin (x + y) ≠ sin x + sin y, f is not linear.

Alternatively: f( x) f(x) sin ( x) ≠

sin x

(b) f(x1 + x2) = 3(x1 + x2) = 3x1 + 3x2 = f(x1) + f(x2) f (∝x) = 3( x)

= (3x)

= f(x)

It follows that f is linear.

Note: The two properties are summarized as f( u + β u ) = f ( u ) + β f( u )

Operations 1.

Equality

Let f: V → W and g: V → W u →g( u ) be linear mappings f = g if and only if for all u ∈v, f( u ) = g( u )

Sum

(f +g) ( u ) = f( u ) + g( u ) for all u ∈ v

3. Multiplication by a scalar ( f)( u ) = [f( u )] for any u ∈ v 4.

Composite mapping f: V → W u f( u )

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g: W→ T g [f( u )] f( u ) The composite : (g o f) ( u ) = g[f( u )] 5.

Inverse mapping If f : V → V x

f ( u ) is one-to-one and onto

then, f –1: W→ V

f–1( u )

By definition f–1( u ) = v if and only if f ( v ) = u . Note: It can be shown that if f and g are linear mappings then f + g, f, g o f are also linear mappings. For example, for the composite, we have the following proof:

let f: V → W u → f (u )

and g: W → T v g o f (u ) composite g o f: V → T u gof(

(g o f) (u1 + u2 )

= g [ f (u1 + u2 )]: definition of the composite

= g [ f (u1 ) + f( u2 )]: f is linear

= g [ f(u1 )] + g [f(x2 )]: g is linear

= ( g o f) (x u1 ) + ( g o f) ( u2 ) definition of the composite

(2) ( g o f) ( u1 )

= g [f( u )]: definition of the composite

= g [ . f ( u )]: f is linear

= g[f( u )]: g is linear = (g o f) ( u ):

Therefore, g o f is linear. In the same way, for the sum of two linear mappings, we have:

(f + g)( u + v ) 173

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= f( u + v ) + g( u + v ): (definition of sum)

= f( u ) + f( v ) + g( u ) + g( v ): (f and g are linear)

= [f( u ) + g( u )] + [f( v ) + g( u )]: ( + is commutative; associative)

= (f + g)( u ) + (f + g)( v ):(definition of sum) (f + g)( u )

= f( u ) + g( u ):(definition of sum)

= f( u ) + g( u ) multiplication is distributive over addition = (f + g)( u ):(definition of sum). Therefore, f + g is linear.

Geometric linear transformations Identity mapping The identity mapping (Idv) of a vector space V is the lenear transformation that associates to a vector, the vector itself. Thus Idv ( u ) = u , for all u ∈V. If V = 2, then Idv (x,y) = (x,y). If V = 3, then Idv (x,y,z) = (x, y, z). Reflection in a line For the following mirror lines: the x-axis, the y-axis and the line y = x, We have: Reflection in the

Vector

Image (f(u)

x- axis

u = (x,y)

f(x,y) = (x, –y)

Illustration y x

y- axis

u = (x, y)

f(x,y) = (–x, y)

y x

line y = x

u (x.y)

f(x,y) = (y,x)

y x

Orthogonal projection We have the following particular cases: 174

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Orthogonal projection Vector u = (x, y)

x -axis

Image f(u) f(x,y) = (x, 0)

Illustration y x

u = (x,y)

y- axis

f(x,y) = (0, y)

y x

Enlargement 2 , An enlargement in the plane is the transformation f: 2 f(x,y) = (kx, ky), where k ≠ 0. (x,y) If |k| >1, the enlargement is said to be a dilation. If |k| <1, the enlargement is said to be a contraction. Rotation The rotation of the plane about the origin through angle θ is the transformation 2 , f: 2 f(x,y) = (x cos θ – y sin θ, x sin θ + y cos θ). (x,y) If θ = 180°, then f (x,y) = (–x, –y): the rotation is called the central symmetry about the origin. If θ = 90°, then f(x,y) = (–y, x). The null mapping V The linear operator f: V

f ( u )= o

is said to be the null mapping. Example 8 Show that the rotation about the origin through 90°, anticlockwise, is a linear operator and find the image of point A(–3, 5) under this rotation. Solution f(x, y) = (–y, x) Let u1 = (x, y) and u2 = (x′, y′) Then u1 + u2 = (x,y) + (x′, y′)

= (x + x′, y + y′) 175

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f ( u1 + u2 ) = f (x + x′, y + y′)

= (–y –y′, x + x′)

= (–y, x) + (–y′, x′)

= f (u1 ) + f (u2 )

If is a scalar and u = (x, y), then u = (x,y) = ( x, y) f ( u ) = f ( x, y) = (– y, x) = (–y, x) f (u )

It follows that f is linear. f(–3,5) = (–5, –3). Isomorphism of vector spaces Let V and W be two vector spaces over the same field ( , + ,.), and f a linear mapping from V to W. f is said to be an isomorphism if and only if f is one to one and onto. In this case, V and W are said to be isomorphic. Let π be the Cartesian plane ( the set of all points of π), V a two-dimensional vector space and 2 the set of all ordered pairs of real numbers. Consider the mappings f: π M

V f (M) = OM = x e1 + y e1 ,

where 0 is the origin and B = (e1 , e2 ) a basis of V. g:V u

g ( u ) = g OM = (x, y).

It is clear that f and g are one-to-one and onto, and linear. Therefore, π , V and 2 are isomorphic. It follows that any point M can be identified to its position vector OM , and xe1 + ye2 can be identified to ordered pair (x,y).

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Activity 8.2 1.

Determine whether f is linear or not in each of the following cases.

(a)

Linear transformation f is such that f (2 u – v ) = 2 u + 3 v and

f ( u + 3 v ) = 8 u – 2 v . Find f( u ) and f( v )

f(x,y) = (x + 1, 2y – 1) in

. (b) f(x,y) = (2x – y, x + y) in

3. Vectors e1 , e2 and e3 of a vector space V are such that f( e1 ) = 2e1 – e2 + e3 ; f(e2 ) = e1 + e2 – e3 and f (e3 ) = 3 e1 – 5 e2 + 4 e3 .

Given that u = – e1 + 2e2 – 3 e3 , find f ( u ).

Given that B = (e1 ,e2 ), where e1 = (–1, 1) and e1 = (1,1,),

(a)

Show that B is a basis of

(b) Given that f(e1 ) = e′1 and f (e2 ) = e′2 , where e′1 = (–3, 2) and e′2 = (2, 1), find an expression for (x,y) where u = (x,y). 8.3.3 Matrices

Definition Let f be a linear operator of a two-dimensional vector spce V and B = (e1 , e2 ) a basis of V. If f(e1 ) = a e1 + be2 and f(e2 ) = c e1 + de2 then, the table, in array form, a c M= is said to be the matrix of the linear operator f in basis B. b d If f is a linear operator of a three-dimensional vector space V, and B= (e1 , e2 ,e3 ) is a basis of V, and f (e1 ) = ae1 + be2 + ce3 , f (e2 ) = a′ e1 + b′e2 + c′e3 f (e3 ) = a′′e1 + b′′e2 + c′′e3 a then the matrix of f in basis B is M = b c

a b′ c′

a′′ b′′ . c′′

Example 9

B = ( i , j )is a basis of a vector space V and linear transformation f of V is such that f ( i ) = 2 i – j and f( y ) = 3 i + 5 y

Find the matrix of f in basis B. 177

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Solution 2 3 M = –1 5

Example 10 Linear transformation f is defined in vector space

by f(x, y) = (–x + y, 4x – y).

Determine the matrix of f in the standard basis B = (e1 , e2 ) of

Solution We have e1 = (1, 0) and e2 = (0,1) f (e1 ) = f(1, 0)

= (–1 + 0, 4(1) – 0)

= (–1, 4)

= –1 (1, 0) + 4(0, 1)

= –1 e1 + 4.e2

f(e2 ) = f (0,1)

= (–0 + 1, 4(0) – 1)

= (1, –1)

= 1.(1,0) – 1. (0,1)

= 1.e1 – 1. e2

Since f (e1 ) = –1.e1 + 4e2

and f(e2 ) = 1.e1 – 1.e2 ,

–1 1 the matrix of f is M = 4 –1 Note The matrix of a linear operator depends on the basis. More generally, a matrix is a table, in array, form, consisting of entries arranged in rows and columns. If matrix M has n rows and p columns, we say that matrix M is of order n × p. 178

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2 Thus 3 4

–1 5 is a 3 × 2 matrix. –9

Matrices of geometric transformations (in standard basis) Reflections Reflection in the x - axis

(i)

→

f(x,y) = (x, – y)

(x,y) y

u x

f(1,0) = (1,0) = 1.(1,0) + 0.(0,1) f(0,1) = (0, –1) = 0. (1,0) –1.(0,1)

f(u ) Fig. 8.1 1 0 The matrix is M = 0 –1 (ii)

Reflection in the y – axis

→

f(x,y) = (–x, y)

(x,y) y

f(u )

x o

f(1,0) = (–1,0) = –1.(1,0) + 0.(0.1) f(0,1) = (0, 1) = 0. (1,0) +1.(0,1)

Fig. 8.2 –1 0 The matrix is M = 0 1

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Orthogonal projection (i)

Projection on the x - axis

→

f(x,y) = (x, 0)

(x,y)

f(1,0) = (1,0) = 1.(1,0) + 0.(0.1)

f(0,1) = (0, 0) = 0. (1,0) +0.(0,1) x o 1 0 The matrix is M = 0 0 f(u ) Fig. 8.3 (ii)

Projection on the y - axis

→

f(x,y) = (0, y)

(x,y) y

f(u ) u x o

f(1,0) = (0,0) = 0.(1,0) + 0.(0,1) f(0,1) = (0, 1) = 0 (1,0) +1(0,1)

Fig. 8.4 0 0 The matrix is M = 0 1 Rotation y Let B = (e1 , e2 ) be a basis of vector space V, M′ S M where dim V = 2 Ɵ f(u ) u Ɵ1 u = x e1 + ye2 x o C′ C If the rotation about the origin through angle Ɵ,

is such that f ( u ) = u′ , where u′ = x′ e1 + y′ e2 .

Fig. 8.5 180

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Then: In the right triangle OCM, OC = x = OM cos Ɵ1 and CM = y = OM sin Ɵ1 Let OM = r. We have x = r cos Ɵ1 y = r sin Ɵ1. .......(1) In the right triangle OC′M′, x′ = r cos (Ɵ + Ɵ1) = r (cos Ɵ cos Ɵ1 – sin Ɵ sin Ɵ1) = (r cos Ɵ1) cos Ɵ – (r sin Ɵ1) sin Ɵ = x cos Ɵ – y sin Ɵ y′ = r sin (Ɵ + Ɵ1) = r (sin Ɵ cos Ɵ1 + sin Ɵ1 cos Ɵ) = (r cos Ɵ1) sin Ɵ + (r sin Ɵ1) cos Ɵ = x sin Ɵ + y cos Ɵ Therefore, f (x ,y) = (x cos Ɵ – y sin Ɵ, x sin Ɵ + y cos Ɵ) f (1, 0) = (cos Ɵ, sin Ɵ) = (cos Ɵ) (1, 0) + (sin Ɵ) (0, 1) f(0, 1) = (– sin Ɵ, cos Ɵ) = (– sin Ɵ) (1, 0) + (cos Ɵ) (0, 1) cos Ɵ – sin Ɵ The matrix is M = sin Ɵ cos Ɵ In particular, if the rotation is through 90°, since cos 90° = 0 and sin 90° = 1, the rotation is defined by f(x,y) = (–y, x); 0 –1 The matrix is M = 1 0 If the rotation is through 180°, since cos 180° = –1 and sin 180° = 0, the rotation is defined by f (x, y) = (–x, –y)

= – (x,y)

f ( u ) = – u : the rotation is called central symmetry about the origin. –1 0 The matrix of the central symmetry about the origin is M = 0 –1 Note: The linear transformation f such that, for all vector u , f( u ) = u is called the identity transformation. 181

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1 The matrix of the identity is I = 0

0 1 , called the identity matrix.

Example 11 Write down the matrix of the rotation about the origin, through 60°. Hence, find the –1 image of vector u = 3

Solution cos 60° – sin 60° M = sin 60° cos 60°

(

3 2

f (–1, 3 ) = – 1 – 3 , – 3 +

3 2

1 2

3 2

)

= (–2,0)

Note: The constant linear transformation f such that, for all vector u , f ( u ) = o is called the null transformation. 0 0 The matrix of the null transformation is M = 0 0 Case 2: V is a three- dimensional space, that is dim V = 3 Let B = (e1 , e2 , e3 ) be a basis of V and f a linear transformation of V.

If f (e1 ) = ae1 + be2 + ce3

f (e2 ) = a′ e1 , b′e2 , + c′e3

a a′ a′′ and f (e3 )= a′′e1 + b′′e2 + c′′e3 then, the table, in array form, M = b b′ b′′ c c′ c′′ the matrix of f in basis B.

Example 12 B = ( i , j , k ) is a basis of a vector space v. The linear operator f is defined by

f ( i ) = –2 i

f ( j ) = –2 j + 2 k

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f ( k ) = – 4j + 4 k Find the matrix of f.

Solution 2 0 0 – – M = 0 2 4 . 0 4 4 –

Example 13

A = {u1 , u2 , u3 } is a basis of

v1 = u2 +u3 ;

v2 = u3 ;

v3 = u1 + 2u2 + 5u3

The linear operator T is such that:

T(u1 ) = v1

T (u2 ) = v2

T (u3 ) = v3

Find the matrix M that represents T relative to basis A.

Solution T(u1 ) = v1 = u2 + u3 = 0.u1 + 1. u2 + 1.u3 T(u2 ) = v2 = u3 = 0.u1 + 0.u2 + 1.u3 T(u3 ) = v3 = u1 + 2u2 + 5u3 0 0 1 M = 1 0 2 . 1 1 5 183

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Example 14 f: 2→ (x, y)

f(x, y) = (x + 2y; –2y) is a linear operator. Find the matrix of f relative to the

standard basis 2. Solution B

= (e1 , e2 )

where e1 = (1, 0); e2 = (0, 1)

f (e1 ) = f(1, 0) = (1, 0) = e1

f (u1 ) f (0, 1) = (2, –2)

f(e1 ) = 1.e1 +0.e2

f(e2 ) = 2.e1 – 2.e2

The matrix is M = 1 2– . 2 0

Operations on matrices 1.

Equality of matrices

Let A = (aij) and B = (bij) be matrices of order nxp

A = B ⇔ the corresponding entries are equal

x 3 2x – 3 z–1 Thus, matrices A = 1 y–1 and B = t/2 5 are equal if and only if. x = 2x – 3 1 = t/2

3=z–1 y–1=5

Addition and substraction

A + B = (ai j + bi j)

We add the corresponding entries.

Note: If A is the matrix of a linear mapping f and B is the matrix of the linear mapping g; then A + B is the matrix of the linear mapping f + g. 184

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Example 15 5 If A = –2 1 –1 and B = 6 3 –4 0 Solution

–2 1 3 –4

A+B=

5 –1 + 6 0

–9 1

0 , find A + B. 7

–9 1

0 7

3 9

–8 –3

–1 7

Properties A + B = B + A (Addition is commutative)

(1)

(2) (A + B) + C = A + (B + C) (Addition is associative) 3.

Multiplication of a matrix by a scalar Let A = (a i j) and

a scalar

A = (a i j) = ( a i j)

Then

Each entry is multiplied by the scalar. Example 16 Given that A =

2 3 –2 find – 5A. –1 1 4

Solution –5A

2 3 –2 = –5 –1 1 4

–10 5

–15 10 . –5 –20

Properties For any matrices A and B of order n × p and for any scalars and β. 1.

(A + B) = A + B

( + β) A = A + βA

3. ( .β)A = (βA) 4.

IA = A 185

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Multiplication of matrices

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix.

A:n×p

B:p×r

AB=n×r

In this case, A and B are said to be conformable for multiplication.

Thus A = –1 8 –3 0 10 –4 A = –1 8 –3 0 10 –4

5 6 and B = –1 0 are conformable for multiplication. 0 3 and B =

2 3 4 2 are inconformable for multiplication.

Properties 1.

Multiplication of matrices is not commutative. A.B ≠ B.A

Example 17 0 1 If A = –1 3 and B = , find AB and BA and draw the conclusion. 7 –2 2 –4 Solution A: 2 × 2 2× 2 B: A.B 2×2 A . B = –1 3 0 3 = 21 –7 – 2 –4 7 –2 28 10 B.A=

0 –4 0 1 –1 3 = –11 29 7 –2 2 –4

A.B ≠ B.A Therefore, multiplication of matrices is not commutative.

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2. Multiplication of matrices is associative. (A B) . C = A (B C) for any matrices comformable for multiplication. 3.

For matrices the equality A C = B C and C ≠ 0 does not imply A = B.

4. Let f: V → w g: T. w Then gof:v T. If A is the matrix representing the linear mapping f and B is the matrix representing the linear mapping g, then the matrix representing g o f is B A. M (g o f) = M(g) . M(f) 5. The transpose of an n × p matrix M is a p × n matrix denoted MT whose rows are the columns of M and whose columns are the rows of M. Thus if M = 1 0 2 4 –3 5

then MT = 1 2 –3 . 0 4 5

Properties (1)

(A + B)T = AT + BT (transpose of a sum is the sum of transposes)

(2)

(AB)T = BT. AT (Transpose of a product is the product of transpose in reverse sequence)

5. Inverse of a Matrix

Definition Let M be a square matrix of order n that is, a matrix with n rows and n columns. M is said to be invertible if there exists a matrix M′, of order n, such that M.M′ = M′.M = I where I is the identity matrix of order n.

M′ is the inverse of M . It is denoted M–1

M.M–1 = M–1. M = I

A matrix that is not invertible is said to be a singular matrix.

Calculation of the inverse of an invertible matrix Example 18 Find the inverse of the matrix M = 3 2 1 1 187

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a c Let M–1 = b d be the inverse of M Solution 3 2 a c 1 0 Then M.M–1 = I 1 1 b d = 0 1 (i) ⇔ 3a + 2b = 1 a+b=0

(ii)

3c + 2d = 0 c+d=1

3c + 2d = 0 1 = c + d = 1 –2

3a + 2b = 1 –1 = a+b=0 2

3c + 2d = 0 c = –2; d = 3 –2c – 2d = –2

3a + 2b = 1 2a + 2b = 0

a = 1 ; b = –1

1 –2 M–1 = –1 3 .

Simultaneous linear equations If

ax + by = c aʹx + bʹy = cʹ

are two linear equations in two unknowns x and y.

a b The system can be written using matrices as a' b' If A is invertible, then :

–1 –1 A . (AX) = A B (A–1. A) X = A–1B I.X = A–1B X = A–1B.

x c = y c'

A . X = B A X = B.

Example 19 Solve , using matrices , the simultaneous linear equations 2x + 3y = 4 x + 2y = 3 188

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Solution

x 4 y = 3

2 3 1 2

x 4 2 3 ;X= y ; B= 3 1 2

X = A–1.B

The calculation of A–1 gives:

A–1 =

2 –3 ; – 12

– = 2 3 – 12

– 1 x = 2 y

X = A–1B 4 3

x = –1 y=2

Solution set S = { (–1, 2)}. Simultaneous linear equations in three unknowns a x + by + cz = d a'x + b'y + c'z = d' a''x + b''y + c''z = d''

These are three linear equations in three unknowns x, y, z

a b c x d Using matrices, the system can be expressed as: a' b' c' y d a'' b'' c'' z d'' a b c A = a' b' c' ; X = a'' b'' c''

x d y ; B = d' z d''

If A is invertible, then A–1 (AX) = A–1B

AX = B

⇔ (A–1A) X = A–1B

⇔ I. X = A–1B

⇔ X = A–1B.

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Activity 8.3 1.

A shop sells clothes for both children and adults and for both male and female. One Saturday, the shopkeeper sold 17 clothes for male children, 24 clothes fro female children, 4 clothes for male adults and 9 clothes for female adults. Represent the data in a table and write the corresponding matrix.

Write down, in standard basis of

(a)

A rotation about the origin through angle θ, anticlockwise.

(b)

Enlargement of scale factor k.

(c)

The reflection of the plane in the line y = x.

(d)

The orthogonal projection of the plane in the y-axis.

Find A + B in each of the following cases: (a) A = 5 –2 ; B = 0 0 2 3 –1 7

1 A = 1 5

0 4 2

4 0 1

, the matrix of;

0 3 5 ; B = 1 4 –3 2 –1 0

(b)

Find AB in each of the following cases. (a) A = 2 5 ; B = 3 1 3 1 0 4

(b)

(a)

(b)

1 –1 1 A = 2 1 0 3 –2 5

0 ; B = 1 2

2 0 4

4 5 3

Find the inverse of matrix A = 6 8 2 3 6x + 8y = 2 Use matrices to solve the simultaneous equations 2x + 3y = 1

Determinants Determinant of a 2 × 2 matrix

Definition Let M = a c bd

be a 2 × 2 matrix

The determinant of M is the real number denoted 190

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a c det M = a c and defined by = ad –bc b d b d Thus, M = 3 –1 = 3(1) – 4(–1) = 7; 4 1 –1 0 Given M = 6 10 = –1(10) –6(0) = –1;

Applications a.

Inverse of a Matrix M = a c b d

Matrix M is invertible if and only if det. M ≠ 0: i.e ad – bc ≠ 0 In this case, the inverse of M is M–1 =

1 d –c det. M –b a

If det M = 0 then matrix M is singular (not invertible). Example 20 – (a) Show that M = 1 2 is invertible and find M–1 3 1

(b) Given

–x + 2y = 4 , find the solution set using the inverse of a matrix method. 3x +y = 2

Solution (a) det.M =

–1 2 3 1

= (–1) (1) – (3)(2)

= –1–6 = –7

det. M = –7 ≠ 0, hence M is invertible. – M–1 = –1 1– –2 . 7 3 1

(b)

1 2 x = 4 ;B= 4 2 2 3 1 y

–

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0 – 12 1 1 –2 4 1 0 ; A–1 = –1 1– –2 X = A–1.B = – – = – –14 = 2 3 1 7 30 2 7 7 3 1

–

Hence (x , y) = (0,2)

Example 21 Find the value of x such that the matrix A = x – 1 3x

x + 1 is singular x

Solution A is singular if and only if det A = 0 x–1 x+1 i.e 3x x =0

⇔ (x – 1) x – 3x(x + 1) = 0 Either x = 0

⇔ x2 – x – 3x2 – 3x = 0

⇔ –2x2 – 4x = 0

⇔ –2x (x + 2) = 0.

or x = –2 A is singular if x = 0 or x = 2

Solving simultaneous linear equations by Cramer's method ax + by = c a'x + b'y = c'

a b c b Δ = a' b' ; Δx = ; Δy = a c c' b' a' c'

If Δ ≠ 0; then

ax + by = c is a Cramer's system and: a'x + b'y = c'

c c' Δx = x= a Δ a'

b b' ; y = Δy = b Δ b'

a a' a a'

c c' b b'

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Example 22 4x +3y = 2 by Cramer's method. x+y=1

Solve

Solution 4 3 Δ = 1 1 = 4(1) – 1(3) = 4 –3 = 1 ≠ 0 2 3 = 2(1) –1(3) = 2 –3 = –1 1 1

Δx= x= Δy = y=

Δx –1 = = –1 1 Δ 4 2 = 4(1) –1(2) = 4– 2 = 2 1 1

Δy 2 = =2 Δ 1

Solution set S = {(–1, 2)}. Linear dependence Let B = (e1 , e2 ) be a basis of a vector space V and u = ae1 + b e2 ; v = c.e1 + d. e2 u and v are linearly: (a)

a c dependent if and only if det ( u , v ) = b d = 0

(b) independent if det ( u , v ) ≠ 0 Example 23 Show that e1 = (–1, 1) and e2 = (2, 3) are linearly independent. Solution –1 2 (1) det (e , e ) = 1 3 1 2

= (–1) (3) – (1)(2) 193

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= –3–2 = –5 ≠ 0

Therefore e1 and e2 are linearly independent. Determinant of a 3 x 3 matrix

Definition a11 a12 a13 Let M = a21 a22 a23 be a 3 x 3 matrix a31 a32 a33 The determinant of M is the real number denoted and defined by : a11 a12 a13 det M = a21 a22 a23 a31 a32 a33 = (a11 a22 a33 + a12 a23 a31+ a13 a21 a32) – (a31 a22 a13 + a32 a23 a11 + a13 a21 a12)

Sarrus method Since it is not easy to master the value of the determinant, Sarrus method provides a technique. 1 2 –1

3 0 4

a11 a12

a21 a22 a23

a21 a22

a31 a32 a33

a31 a32

–3 1 = –19, in fact, from Sarrus method –2

From Sarrus' method, 1 3 –3 2 0 1 –1 4 –2

a11 a12 a13

1 2 1

3 0 4

The value of the determinant is (0 – 3 – 24) – (0 + 4 – 12) = – 19

Applications (a)

Inverse of a 3 × 3 matrix a11 a12 a13 Let M = a21 a22 a23 a31 a32 a33

be a 3 × 3 matrix

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M is invertible if and only if det M ≠ 0. To find the inverse of an invertible 3 × 3 matrix we proceed as follows: (1)

Find det M (det M ≠ 0)

(2)

Find the matrix of cofactors. A cofactor of an element is the signed determinant obtained by deleting the row and the column of the element.

(3)

Find the adjoint matrix, that is the transpose of the matrix of cofactors.

Adj (M).

(4)

The inverse is M–1 = 1 .Adj(M). det M

Example 24 1 Evaluate 2 –1

–1 0 1

2 1 2

Solution 1 2 –1

–1 0 1

1 Therefore, 2 –1

2 1 2 –1 0 1

1 2 –1

–1 0 1

2 1 = (0 + 1 + 4) – (0 + 1 – 4) = 8 2

Simultaneous linear equations

ax + by + cz = d a'x + b'y + c'z = d' a''x + b''y + c''z = d''

Method 1: Using matrices

a b c a' b' c' a'' b'' c''

A =

a b c d x a' b' c' ; X = y ; B = d' a'' b'' c'' d'' z

x y z

d = d' d''

A. X = B ⇔ X = A–1.B, where A is invertible. 195

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Example 25 Solve the simultaneous equations x+y+z=4 x + 2y + z = 6 using matrices x + y + 2z = 7 Solution 1 1 1 1 2 1 1 1 2

4 x 1 2 1 y = 6 A = 1 1 1 7 z 1 1 2

1 1 1 1 1 1 2 1 1 2 1 1 2 1 1

det A = (4 + 1+1)–(2 + 1+ 2) = 1.

Matrix of cofactors +

1 2 2 1 1 1 – + 1 1 1 2 1 2

1 1 – 1 1 + – 1 1 1 2 1 2 1 1

– 1 3 –1 1 1 0 = – 1 0 1 –

1 1 1 1 1 1 – 1 1 + 2 1 1 2 Adjoint of A = transpose of C +

– 1 3 –1 1 1 0 Adj (A) = – 1 0 1 –

A–1 =

1 . Adj. (A) det A

– 3 –1 1 1 1 0 – 1 0 1 –

– 3 –1 1 – –1 X = A .B = 1 1 0 – 1 0 1

– 4 1 6 = 2 7 3

x = –1 y=2 z=3

Solution set = {(–1, 2, 3)} 196

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Example 26 3 2 –1 Find the inverse of the matrix A = 4 3 –1 – 1 2 4 Solution 3 2 –1 (1) det A = 4 3 –1 = 1 ≠ 0 – 1 2 4

So the matrix is invertible.

(2) Matrix of cofactors:

3 2 –1 A = 4 3 –1 – 1 2 4

4 3 4 –1 M 11 = 3 –1 ; M12 = ; M13 = –1 2 –1 4 2 4 3 –1 3 2 M21 = 2 –1 ; M22 = –1 4 ; M23 = –1 2 2 4 – 3 –1 3 2 M31 = 2 1 ; M32 = ; M33 = 4 –1 4 3 3 –1

Matrix of cofactors:

3 –1 2 4

–

4 3 4 1 + –1 2 –1 4

2 –1 2 4

3 –1 – 3 2 –1 4 –1 2

2 –1 3 –1

3 –1 3 2 – 4 –1 + 4 3

14 = 10 1 –

–15 11 –1

11 –8 1

(3) Adjoint transpose of C: 14 Adj (A) = –15 11

–10 1 11 –1 –8 1 197

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1 (4) A–1 = det A . Adj (A) 14 –10 1 A–1 = 10 –11 –1 . 11 –8 1

Method 2: Cramer's rule a b c d b c a' b' c' Δ= ≠ 0 ; Δx = d' b' c' a'' b'' c'' d'' b'' c'' a d c Δy = a' d' c' a'' d'' c'' x=

a b d Δz = a' b' d' a'' b'' d''

Δx Δy Δz ; y = ; z = Δ Δ Δ

Example 27 Solve the simultaneous equations

x+y+z=4 x + 2y + z = 6 using cramer's method x + y + 2z = 7

Solution

1 1 1 Δ = 1 2 1 1 1 2

1 4 1 Δy = 1 6 1 1 7 2

Δx Δ

= 1;

Δx =

4 1 1 6 2 1 7 1 2

1 1 4 = 2; Δz = 1 2 6 1 1 7

= –1

–1 = 1 = –1; y = Δy = 2; z = Δz = 3 Δ Δ

Solution set S = {( –1, 2, 3)} 198

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Linear dependence Let u = ae1 + b e2 + ce3 and v = a' e1 + b′e2 + c′e3 be two vectors in a vector space V whose dimension is n = 3 u and v are linearly dependent if there exists a vector t such that v = t u . a a' b b' and v are linearly dependent if and only if Alternatively u c c' a a' a a' b b' b b' = c c' = c c' = 0

a a' a a' b b' If at least one of the partial determinants b b' , c c' , c c' is not zero, then u and 3 1 – v are linearly independent. u 2 and v –6 are linearly dependent since 3 1 v =–3u . –

3 –1 6 = –3 2 – 3 1 –

–1 3 –1 3 2 –6 2 –6 = 1 3 = 1 –3 = 0

–1 3 e1 = 1 ; e2 3 1 0

3 –1 ; 3 –1 ; 1 3 = 9 + 1 = 10 ≠ 0 0 1 0 1 1 3

e1 and e2 are linearly independent. Three vectors of a vector space of dimension n = 3 let u = a e1 + b e2 + ce3 ; v = a' e1 + b' e2 + c' e3 ; w = a'' e1 + b'' e2 + c'' e3 ; where (e1 , e2 ,e3 ) is a basis of V. a'' b'' w = c''

a b u= c

u , v and w are linearly dependent if and only if det( u , v , w )

a' v = b' c'

a a' a'' = b b' b'' = 0; c c' c''

Otherwise they are linearly independent.

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Properties of determinants The following properties are very useful: 1.

Interchanging the corresponding rows and columns of a determinant does not change its value, that is, det AT = det A.

Thus, 2.

3 –1 3 5 = . –1 2 5 2

If every element of a row or column of a determinant is zero, the value of the determinant is zero.

Thus

4 0 –1

–11 0 7

3 0 9

= 0.

If every element of a row or column of a determinant is multiplied by the same constant, the value of the determinant is multiplied by that constant.

Thus

If two rows or two columns of a determinant are interchanged, the sign of the determinant is changed, but its absolute value is unchanged.

Thus 5.

1 –2 5

5 3 –1

–1 4 –9

2 –6 1

1 –2 5

5 = – –1 3

3 1 –1

–1 –9 4

2 9 8

2 1 –6

If two rows ot two columns of a determinant are identical, the value of the determinant is zero.

9 Thus 9 1 6.

3 (7) 2 1(7) 9 –1(7) 8

–1 – 1 8

7 7 = 0. 11

The value of the determinant is not changed if each element multiplied by the same constant, is added to or subtracted from the corresponding element of any other row or column.

Thus,

1 5 –2

4 1 3

2 0 9

1 + 3(4) 5 + 3(1) –2 + 3(3)

4 1 3

2 0 . 9

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The determinant of the product of two matrices is equal to the product of determinants of the two matrices.

Thus det

The determinant of a triangular matrix is equal to the product of the entries on the main diagonal,

[ –13

4 1

5 2

–1 0

3 4 5 –1 . . –1 1 2 0

cos θ tan θ –1 Thus 0 sin θ 3 = 2 sin θ cos θ = sin 2θ 0 0 2 Example 28 Evaluate:

7 3 1

3 21 5

35 15 . 5

Solution C3 = 5C1, therefore

7 3 1

3 21 5

35 15 5

= 0.

Activity 8.4 1.

3–a 8 For what value of a is the matrix A = 1 –4–a singular?

2. Evaluate: 1 (a) 2 3

3 1 4

–2 4 . 2

2 (b) –3 4

5 2 1

0 3 0

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Given matrices A =

2 –2 3 and B = 1 4 5

4. Factorise: m (a) 1 1

1 m 1

1 1 . m

1 (b) 1 1

a b c

a2 b2 . c2

2 , find det [(AB)–1]. 4

8.4 Practice 8.4.1 Vector spaces 1. 2. 3.

Given that u = (2, 1, –5) and v = (1, 1, 3), find 2 u – 3v . Express w = (–2, –27) as linear combination of u = (1, –3) and v = (2,5). Determine whether S = { u , v } is a generating set of following cases:

or not, in each of the

u = (2, –1) and v = (4, 2) (a) u = (2, 1) and v = (2, –1) (b) 4.

Determine whether u and v are linearly dependent or not in each of the following cases:

u = (2, –1) and v = (–4, 2) (a) u = (2, 1) and v = (2, –1) (b) 5.

B = ( i , j ) is a basis of a vector space V. Vectors e1 and e2 are defined by e1 = e

– 2 j ; e2 = i + j . ) is a basis of V

(a)

Show that B′ = (e1 , e2

(b)

Find the components of vector u = 4 i , + 7 j in basis B′

(c)

Vector v has components (3, 1) in basis B′.

Find the components of v in basis B.

It can be shown that the subset E = {( x , y, z) ; 3x + y – z = 0} is a vector space contained in vector space 3 .

Find a basis and the dimension of E.

It can be shown that the subset E = {( x, y, z) ; x = y = z} of 3 is a vector space over thefield of real numbers. Find a basis and the dimension of E.

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8.4.2. Linear transformations. 1.

Determine whether function of defined in following cases.

is linear or not in each of the

(a) f(x,y) = (xy, x + y)

(b)

f (x,y) = (2x + 2y, –x –y)

Show that the reflection in the y-axis is linear.

B = ( e1 ,e2 ) is a basis of vector space V. Linear transformation f on V is such that

f(e1 ) = 2e1 – 3e2 and f (e2 ) = e1 + e2 4.

Find f( u ), where u = 5 e1 – e2 . B = ( e1 , e2 ) is a basis of vector space V,

Linear transformation f of V is such that

f( e1 + 7 e2 ) = 3 e1 + e2 and f (2e1 – 7 e2 ) = 6 e2 – 4 e2 . Find f ( e1 ) and f (e2 ). 8.4.3 Matrices 1.

Linear operator f is defined in 3 by f(x,y,z) = (–x + y + 3z, 2y + z, 2x – y + 5z). Find the matrix of f in the standard basis of 3.

Linear operator f is defined in

Find the matrix of f in the standard basis of

(a)

by f(x,y) = (–x, –x + y). 2

Find the matrix, in standard basis of the linear operator f defined in by f(x,y) = (0,–x + y)

(b) 4. 5.

Under a transformation f, (1,0) is mapped onto (0,–1) and (0,1) is mapped onto (1,0). Show that f is a rotation about the origin through –90°. 2 –1 , find M 2 , M3 amd M–1. Given the matrix M = 1 3 1 1 2 x = Solve for x and y in the matrix equation 4 –1 3 y

(a)

Given that A =

2 1 3 –1 and B = , find 2A – B. 3 –2 2 –3

(b)

Given that A =

3 1 3 –1 and B = , find (A + BT)–1. 3 –2 2 –3

Find matrix X if 3A + X = 2B where A =

Find AB in each of the following cases:

1 –2 2 –1 and B = . 4 5 7 8

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(a)

2 –1 –1 3 and B = . 3 2 4 0

(b)

A =

3 1 2 –1 and B = . 5 2 –5 3

Let A =

Given that matrix AB is singular, determine the values of x.

x –4 x 3 . and B = 1 2 1 x–2

1 10. Given that A = 1

1 2 1

–4 1 1 and B = 5 –2 2

11 –4 –1

16 –7 , –5

find the product AB.

8.4.4 Determinants 1. Evaluate: (a) 1 5 . 1 3 1 1 1 (b) –1 –2 –1 . 1 1 2 2.

Find the matrix of cofactors of the matrix.

1 A = 1 2

B = (e1 , e2 , e3 ),is a basis of vector space 3. Vectors u , v and w are defined

1 2 1

1 1 1

by u = e1 + 2e2 + 4e3 , v = 2e1 – e2 – 3e3 and w =4 e1 + 3e2 + 5e3 .

Determine whether u , v and w are linearly dependent or not.

Use Cramer's rule to solve the simultaneous equations:

– x –7y + 5z = –9 –3x – 3y – 3z = –9 –2x –8y + 4z = –12

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Chapter 9

Euclidean vector space 9.1 Objectivies In this chapter you will: • Calculate the scalar product of two vectors. • Determine the magnitude of a vector • Establish the parametric and cartesian equiatios of a straight line • Apply the scalar product to solve problems • Establish the parametric and cartesian equation of a circle

9.2 Key concepts Angle between two vectors Cartesian equation of a line Parametric equations of a line

Magnitude of a vector Scalar product Mid point.

Circle Distance

9.3 Theory 9.3.1 Scalar product

Definition Let V be a vector space over the field of real numbers. The dot product (scalar product) is the operation denoted and defined by: V ×V (u , ) u. To each ordered pair of vectors is associated the real number denoted u . and satisfying the following properties: 1. u . = . u (the scalar product is commutative) (u 1 + u 2) . = u 1 . = + u 2 2. u .(. 1 + . 2) = u . 1 + u . 2 The dot product is distributive over addition of vectors. ( u ). = (u . ) u( ) = ( u . ) where is a scalar. 205

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3. u . u ≥ 0 The dot product of a vector by itself is always a positive number 4. u . u = 0 if and only if u = Note: u . u is always denoted u 2; If u . v = 0, then u and v are said to be orthogonal (or perpendicular)

Properties 1.

.u = u . = 0

2. (u + )2 = u 2 + 2 u . +

(u + )2 = u 2 – 2u +

(u – ).(u + ) = u 2 –

Magnitude (or modulus) of a vector Definition The magnitude of a vector u is the positive number denoted and defined by || u || = u 2 If ||u || = 1, then u is said to be a “unit vector”.

Properties 1. ||u || = 0 if and only if u = 2. || u || > 0 if and only if u ≠ 3. || u || = | | || u || In particular, ||– u || = || u || | | : absolute value of the real number α 1 4. For any vector u ≠ , || || u is a unit vector u 1 5. u . = 2 (|| u +

||2 – || u ||2 –|| ||2)

6. | u . | ≤ ||u || || || (Cauchy - Schwarz inequality) 7. || u +

|| ≤ || u || + || || (Minkowski's inequality)

8. |u . | = ||u || || || if and only if u and are linearly dependent 206

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||u + || ≥ || u || – || ||

10. ||u + || = ||u || + || || if and only if u and

are linearly dependent

Example 1 Prove property number 5 and 6. Solution Property 5: ||u + ||2 = (u + )2 = u 2 + 2u . +

= ||u ||2 + 2u . + || ||2

2u . = ||u + ||2 – ||u ||2 – || ||2

∴ u . = ½ (||u + ||2 – ||u ||2 – || ||2) Property 6: (u + =

)2 = u 2 + 2 u . + 2

+ (2u . ) + u 2

But (u + 2

)2 ≥ 0

+ (2u . ) + u 2 ≥ 0

A quadratic inequality in . Discriminant ∆ = (2u . )2 – 4( 2)(u 2) ≤ 0 4(u . )2 ≤ 4u 2

(u . )2 ≤ ||u ||2 || ||2 (||u . ||)2 ≤ (||u || || ||)2 ∴ |u . | ≤ ||u || || || Other properties can be proved in the same way. Perpendicular vectors Vectors u and are said to be perpendicular or orthogonal if and only if u . = 0. We write u . 207

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Analytic expression of the dot product of vector (a)

Vectors in the plane

Consider a rectangular coordinate: let u = x e1 + ye2 and

= x′ e1 + y′ e2 where (e1 , e2 ) is a basis such that e1 ⊥ e2 and ||e1 || = ||e1 ||= 1

u . = (xe1 + ye2 ) (x′e1 + y′e2 ) = xx′e1 2 +xy′e1 .e2 + x′ye1 .e2 + yy′e2 2 x′ y′

x If u and y then u . = xx′ + yy′

⇔ xx′ + yy′ = 0

u ⊥

||u || =

u .u

= x2 + y2

||u || =

x2 + y2

Example 3 (a) ( , ) is a basis such that || || = || || = 1 and vectors and

Find ||–3 – 2

⊥ .

are defined by = 2 – 3 and

(b) Find a unit vector in the same direction as u = 3 – 4 . What is the unit vector in the opposite direction? (c) u = 2 –

and

Find ||u + ||

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Solution (a) 3 – 2

= –3(2 – 3 ) – 2( – 2 )

= –6 + 9 – 2 + 4

= –8

|| –3 + 2

+ 13 || = (–8)2 + (13)2

= (–8)2 + (13)2 =

233

(b) A unit vector in the direction of u is 1 1 = u 2 3 + (–4)2 u || u || 3 1 1 4 = u = (3 – 4 ) = – 5 5 5 5 –1 The unit vector in the opposite direction is || || u (c) u +

u =–

3 5

4 5

= 2 – + i + 5j

= 3 + 4 || u + || = (3)2 + (4)2

= 5.

Relation between dot product and magnitude (a)

Consider triangle ABC. Let θ denote the angle at vertex A. C

=u, = B A Fig. 9.1

|| and || BC||. But it is known The sides of the triangle have lengths || ||, || that in any triangle, the square of the length of one side is equal to the sum of the squares of the lengths of the two other sides, diminished by the product of the double of the product of the lengths of the two sides by the cosine of the angle between the two sides. 209

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Thus ||

||2 = ||

||2 – 2||

(Chasles Theorem)

–

|| cosӨ

= –u

= ||u ||2 + || ||2 – 2||u || || || cosӨ

But

||2 + ||

|| – u ||2 = ||u ||2 + || || || || cos Ө 2||u || || || cos Ө = ||u ||2 + || ||2 – ||u – ||2 = u 2 +

= u2 +

= 2u .

2 2

– (u – )2 – ( u2 – 2 u. +

)

Hence; u . = ||u || || || cos Ө

u. and cos Ө = || u || || ||

The dot product may be used to calculate the angle between two vectors.

(b)

Analytic expression of cos Ө

If u

x and y

cos Ө =

x′ then the angle Ө between u and y′

is such that

xx′ + yy′ 2 2 x +y . x′ + y′ 2

Activity 9.1 1.

Find, for the vectors u =

Find the dot product of u and in each of the following cases:

(a) u =3 + ;

=–

+ 2 and = 3 + , the acute angle θ between u and . –2

(b)

u =3 –4 ;

( , ) is the standard basis of 2 . 3. Vectors u and are such that || u || = 2, || || = 6 and the angle between u and is 30°. Find || u + || and || u – || θ 4. The angle between unit vectors u and is θ. Find 1 || u – || in terms of . 2 2 210

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9.3.2 Points Rectangular coordinates in a plane y

. P (x, y)

x o Fig. 9.2 Each point in a two dimensional space can be associated with an ordered pair of real numbers. If A (x1, y1) and B (x2, y2), then OA = x1 e1 + y1 e2 and OB = x2 e1 + y2 e2 .

AB = OB – OA = (x2 e1 + y2 e2 ) – (x1 e1 + y1 e2 ) = (x 2 – x1) e1 + (y2 – y1) e2 Therefore, AB

x2 – x1 , y2 – y1

The midpoint of line segment [AB] is point I such that AI = 1 AB ; 2 OI – OA = 1 (OB – OA ) 2 OI = 1 (OB + OA ) 2 = 1 [(x 2 e1 + y2 e2 ) + (x1 e1 + y1 e2 )] 2 = 1 [(x1 + x 2) e1 + (y1 + y2)e2 2 x +x y +y = 1 2 .e1 + 1 2 .e2 2 2 Therefore, I

(

y1 + y2 x1 + x2 , 2 2

]

By definition, the distance between points A and B is AB = ||AB || =

AB 2

If A (x1, y1) and B (x 2, y2) then the distance between A and B is AB = ||AB || = AB 2 = (x2 – X1)2 + (y2 – y1)2 211

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Note Points A, B and C are collinear if and only if AB and AC are linearly dependent. x1 The area of triangle ABC, where A(x1, y1), B(x 2, y2) and C(x 3, y3) is ± 1 x2 2 x 3 The sign is chosen such that the area is positive.

y1 y2 y3

1 1 2

Example 4 Given the points A(–3, 1) and B(1, 5), find: (a)

the components of vector

(b)

the coordinates of the midpoint of line segment [AB]

(c)

the distance between A and B

;

Solution 4 4

(a)

(b) Let I be the midpoint I

–3 + 1 , 1 + 5 2 2

I(–1, 3)

(1 + 3)2 + (5 – 1)2

16 + 16

32 = 4 2 units of length.

Circle In two-dimensional space, the collection of all points that are the same distance from some fixed point is called a circle. The constant distance is called the radius r and the fixed point is the centre of the circle. Any point M(x, y) on a circle of radius r and centre at point C obeys MC = r Therefore, the equation of a circle is

(x – x0)2 + (y – y0)2 = r

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•M

• C Fig. 9.3 Squaring both sides, (x – x0)2 + (y – y0)2 = r2

Expanding, x2 – 2x0x + x02 + y2 – 2y0y + y02 – r2 = 0

x2 + y2 – 2x0x – 2y0y + (x02 + y02 – r2) = 0; x2 + y2 +αx+ βy + γ = 0, where γ = –2x0; β = –2y0; γ = x02 + y02 – r2 β

x0 = – 2 ; y0 = 2 ; r =

x02 + y02 – γ

The circle with equation

x2 + y2 + x + β y + γ = 0

has centre at – , – β and radius 2 2

–2

+ –2 –r

Example 5 (a) Find the equation of the circle with centre at point (1, –2) and radius 3. (b) Find the centre and radius of the circle with equation 9x2 + 9y2 +6x – 35 = 0 Solutions (a) (x – 1)2 + (y + 2)2 = 32; x2 – 2x + 1 + y2 + 4y + 4 – 9 = 0

Hence the equation of the circle is, x 2 + y2 – 2x + 4y – 4 = 0.

(b) 9x2 + 6x + 9y2 = 35; 2

9(x2 + 3 x) + 9y2 = 35;

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9 x2 + 23 x +9 + 9(y + 0)2 = 35 + 1 9 x +13 2 + 9(y + 0)2 = 36 x + 13

+ (y + 0)2 = 22

Centre – 13 , 0 ; radius r = 2 The equation of the circle with A(x1, y1) and B(x2, y2) as end points of a diameter is (x – x1) (x – x2) + (y – y1) (y – y2) = 0 Parametric equations of a circle

(x – x0)2 + (y – y0)2 = r2 Dividing both sides by r2 x – x0 r

y – y0

But (cost)2 + (sint)2 = 1 r y – y0 r

= cost = sint

which is equivalent to

}

x = x0 + r cost y = y0 + r sint

}

By comparison,

x – x0

x = x0 + r cost y = y0 + r sint

, where

are the parametric equations of the circle with centre at point (x 0, y0) and radius r. Example 6 (a) Find the cartesian equation of the cirlce defined by parametric equations

}

x =–2 + 3 cost y = 3 sint

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Solution (x + 2)2 + (y – 0)2 = 32 x2 + y2 + 4x – 5 = 0 is the required cartesian equation (b) Find the parametric equations of the cirlce x2 + y2 – 6x + 8y = 0 Solution (x2 – 6x) + (y2 + 8y) = 0 (x2 – 6x + 9) + (y2 + 8y + 16) = 25 (x – 3)2 + (y + 4)2 = 52

}

x = 3 + 5 cost

Parametric equations

y = –4 + 5 sint

(c) Find the equation of the circle with C(1, –3) and D(2, 5) as end points of a diameter Solution (x – 1)(x – 2) + (y + 3)(y – 5) = 0 x2 – 3x + 2 + y2 – 2y – 15 = 0; x2 + y2 – 3x – 2y – 13 = 0 is the equation of the circle.

Activity 9.2 1.

Find the Cartesian and parametric equations of the circle with points A(1,1) and B(–2,5) as end points of a diameter.

Find the centre and the radius of the circle with equation x 2 + y2 + 4x – 2y – 4 = 0.

Circle (C) is centered at point (–1, 0) and passes through point (3,5). Find the Cartesian and parametric equations of the circle.

Given points A(2,3), B(4,7), C(7,3), find:

(a)

The length of AB

(b)

The midpoint of [AC]

(c)

The area of triangle ABC. 215

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9.3.3 Straight Lines When a vector has its initial point at the origin and terminal point at M(x, y), vector = is called the position vector at point M. Thus every point has a position vector associated with it. The vector equation of line (ℓ) through point A and parallel to vector is = where t

+t ,

Example 7 Find the vector equation of line (ℓ) through points A(1, 2) and B(4, 3) Solution 4–1

3–2

3 1

=3 +

The vector equation is = ( +2 ) + t(3 + ) = (1 + 3t) + (2 + t) , where t is a parameter. Parametric equations If

=x +y

= x0 + y0 = ⇔ x

, and = a + b , then

+t + y = (x0 + y0 ) + t(a + b ) = (x0 + at)

}

x = x0 + at

⇔

y = y0 + b t

+ (y0 + bt) ): parametric equations of line through point (x 0, y 0) and

parallel to vector

a b

Example 8 Find the parametric equations of the line through points A(1, 2) and B(4, 3)

}

Solution:

x = 1 + 3t y=2+t

, t

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Solving for t the equations x – x0

x – x0 a

y – y0 b

}

Cartesian equations

x = x0 + at y = y0 + bt

, a≠ 0, b≠0

: Cartesian equations.

Re-arranging; b(x – x0) – a(y – y0) = 0; b.x – a.y + (– b x0 + ay0) = 0 : α.x + βy + γ = 0 : another form of cartesian equation of the line. –β A vector parallel to line x + βy + γ = 0 is α The equation can be expressed as; y = mx + p. m = gradient of the line, (o, p) is the y - intercept of the line. If (ℓ1) : y = m1x + p1 and (ℓ2) : y = m2x + p2 are cartesian equations of lines (ℓ1) and (ℓ2), (ℓ1) and (ℓ2) are parallel if and only if m1 = m2 If (ℓ1) and (ℓ2) are expressed as (ℓ1): ax + by + c = 0 and (ℓ2): a′x + b′y + c′ = 0, then (ℓ1) a

and (ℓ2) are parallel if and only if a′ = b′ = c′ : the lines are coincident a

If a′ = b′ ≠ c′ : the lines are strictly parallel. Lines (ℓ1) and (ℓ2) are perpendicular if and only if m1m2 = –1.

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Distance (ℓ): ax + by + c = 0 P(x0, yo)

The distance from point P(x0, y0) to line (ℓ): ax + by + c = 0 is |ax0 + by0 + c| a2 + b2

Examples 9 Find the distance from point A(–1, 2) to line (ℓ): 3x – y + 1 = 0 Solution The distance is

|3(–1) – 2 + 1| (–1) + 3 2

= 10

4 10 10

10 = 2 . 5

Activity 9.3 1. Find the vector, parametric and Cartesian equations of the line through points A(–1,2) and B(3,1). 2.

Find the equation of the perpendicular bisector of line segment [AB], if A(1,7) and B(7,5).

Given that lines (2–m) x– (3m –1) y + 2 = 0 and 2x – y + 1 = 0 are perpendicular, find the value of m.

(a)

Find the angle between the lines y = 3 x + 2 and 3 y = x – 4.

(b)

Find the equations of the bisectors of the angles 5x + 12y + 4 = 0 and

3x – 4y + 2 = 0.

9.3.4 Scalar product and triangles Projection of vectors The projection of vector u onto vector v can be ilustrated as shown below: u

|| u || cos θ

Fig. 9.4

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Projection of u onto u = || u || cos θ = || u ||.

u .v ||u ||||v ||

u .v ||v ||

Therefore, the dot product of vector u and vector v equals the product of the projection of u on v by the magnitude of vector v . π If 0 < θ < 2 ( that is the angle between u and v is acute), then the projection of u onto v is positive. π If θ = 2 (that is u and v are perpendicular), then the projection of u onto v is zero. π If 2 < θ < π (that is the angle between u and v is obtuse) then the projection of u along v is negative. u v

Fig. 9.5

Relations in a triangle In triangle ABC: 1. A

C BC2 = AC2 + AB2 – 2 AC . AB

AB2 = CA2 + CB2 – 2 CA . CB

Fig. 9.6

AC2 = BA2 + BC2 – 2 BA . BC

These are the cosine rules in a triangle, where, for example, AC . AB

= || AB || || AC || cos A

= AB.AC. cos A

Common identities in a triangle: M

Fig. 9.7

For any points A, B, M, if O is the midpoint of line segment [AB], then: 219

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(a) MA .MB = ||OM ||2 – || OA ||2 (b) ||MA ||2 + ||MB ||2 = 2||OA ||2 + 2||OM ||2 (c) ||MA ||2 – ||MB ||2 = 2 AB . OM (d) ||MA ||2 – ||MB ||2 = 2 OH . AB , where H is the foot of the perpendicular to AB, through M. Example 10 Triangle ABC is such that H is the foot of the perpendicular to BC through A. Given that BC = 8cm, BH = 3cm, CH = 5 cm and AH = 4cm calculate. (a) BA .BC

(b)

HA .HC

(d) BH .BA

(e)

CA .CH

(c)

CA .CH

Solution (a) BA .BC = 3(8) = 24 A (b) HA .HC = 0 4

Fig. 9.8 (e) BC .AH = 0

Activity 9.4 1.

Isosceles triangle ABC is such that BC = 12, AB = AC = 10.

Calculate: (a) AB .BC . 2.

Given that AB =

(b) AB .AC . 2 1

, CD = and EF = , find 1 3

(a) AB . (3CD + 2EF ).

(b) (3AB – 5CD ). (AB + 2CD ).

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9.4 Practice 9.4.1 Dot product 1.

B = ( i , ) is the standard basis of

Given vectors u = 3 i – 2 , v = 2 i + 3 2 (a) (u .v ) w

and w = i – , calculate:

(b) (v . w) u 2.

B = ( i , ) is the standard basis of

Given vectors u = –2 i + a and v = 8 i + a j , find values of a such that u and v

are perpendicular. 3.

Given points A(–2, 1), B(1,3), C(–3, –2) and D(5,4), find the angle θ between vectors AB and CD .

Show that triangle ABC, where A(1,2), B(–2,1) and C(3,–4) is a right triangle.

9.4.2 Points 1.

Find the centre and radius of the circle with equation 4x2 + 4y2 – 4x + 12y + 9 = 0.

Find the equation of the circle with (0,0) and (2,2) as endpoints of a diameter.

Given the points P(–2, –3), Q(2,0) and R(8, –8).

(a)

Show that triangle PQR is right angled at Q.

(b)

Calculate the perpendicular distance from Q to PR.

The midpoint of line segment {AB}, where A(3,0) and B(5,6) is M.

Point C(t, 4t) is such that CM is perpendicular to AB.

(a)

Find the value of t

(b)

Find the area of triangle ABC.

9.4.3 Straight lines 1.

Find the equation of the line through the origin and concurrent with lines

2x – 5y – 3 = 0 and 3x – 4y + 2 = 0.

The midpoint of the line joining the points A(7,3) and B(–1, –5) is C. Find both values of k if the straight line joining C to pooint P(k2,k) is perpendicular to AB.

The perpendicular bisector of the line joining the points (1,2) and (5,4) meets the y-axis at point (o,k). Find the value of k. 221

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The equations of the sides of a triangle ABC are AB: x – 2y + 11 = 0, BC: y = 7; AC = 2x – y + 7 = 0.

Find: (a) The coordinates of the vertices of the triangle.

(b) The area of the triangle

9.4.4 Miscellineous exercises 3 –2 1. Given that a = and b = find a . b . –2 –4 –2

Given that BA = –5 perpendicular.

Find the angle between vectors BA =

and BC

–5

2 2 5

, show that BA

and BC = 5

Find unit vectors perpendicular to AB =

Find the angle between vector a =

Find the distance between points A(3,1) and B(6,–3). 4

1 2

–2

are

and BC

and b =

1 –5

Given that AB =

What do you conclude?

Find the Cartesian equation of the circle with centre at (3,–1) and radius 2.

Find the parametric equations of the circle with points A(–1, 5) and B(3, –1) as endpoints of a diameter.

and AC =

–1

. Find angle between AB and AC .

10. Find the centre and the radius of the circle x2 + y2 –6x + 8y = 0. 11. Find the value of m so that line 2y = mx + 4 passes through the intersection of lines y = 5x + 2 and 3y = 2x – 5s 222

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22. The sides of a triangle are carried by lines x – y + 3 = 0, 2y + 3x – 6 = 0 and 3y – x + 3 = 0. Find the vertices of the triangle, the equations of the lines through vertices and perpendicular to the sides opposite to the vertices. 13. Find the distance between the two parallel lines 4x – 3y + 9 = 0 and 4x – 3y – 1 = 0. 14. Find the distance from point P(3, –1) to line 4y – 3x – 2 = 0. 15. Calculate the perimeter of a triangle with vertices at points (–3, 4), (2, 6) and (0, –3)

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CHAPTER 10

Descriptive statistics 10.1. Objectives In this chapter you will: •

Revise measures of central tendency, more particularly the mode, median and mean of both ungrouped and groupd data, and revise the range.

•

Calculate and represent measures of dispersion around the mean by:

(a)

calculating quartiles and percentiles

(b) working out the range, mean deviation, variance, standard deviation and the coefficient of variation, and using the values of the measures of dispersion to reach conclusion about data sets. •

Draw cumulative frequency curves or ogives, and use these graphs to read off;

(a)

the median

(b)

the quartiles

10.2 Key words Mode Median Mean Data Quantiles Percentiles Modal class Range

Midpoint

Standard deviation

Deciles Interquartile range

10.3 Theory 10.3.1 Measures of central tendency There are several measures of an ‘average’ value for a given set of data. The mean, median and mode are three of the most commonly used. The mean Consider the set of n values x1, x2, x3, ---, xn. The mean of these values is denoted and defined by;

by x =

x1 + x2 + x3 + ---- +xn

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Finding the mean from row data. Example 1 Find the mean of the numbers 24, 26, 27, 29, 32, 42. Solution The mean is

= 1 ∑x = 1 (24 + 26 + 27 + 29 + 32 + 42) = 6 (180) = 30 6

Example 2 The mean numbe of match sticks in nine boxes is 51. How many matches must there be in the tenth box if the mean number of matches in all ten boxes is 52? Solution The total number of matches in the first nine boxes = 9 × 51 = 459. For the mean number in all ten boxes to be 52, there mut be a total of 10 × 52 = 520 match sticks in ten boxes. Therefore, the number of match sticks in the tenth box is 61. Finding the mean from a frequency distribution The mean value from a frequency distribution is given by

∑fx ∑f

Example 3 Find the mean from the following frequency distribution: x 10 11 12 13 14 Frequency 5 10 11 8 6 Solution x 10 11 12 13 14

f 5 10 11 8 6 ∑ f=40

Therefore the mean is

fx 50 110 132 104 84 ∑ fx = 480 = ∑fx = 480 = 12 ∑f

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If the data has been grouped into intervals, we cannot know the values of the variable for all instances within the interval and so we must estimate the mean of these values. To achieve this, we assume that the interval frequency is spread ‘evenly’ throughout the interval by using the mid-interval value for all values of the variable. Example 4 Estimate the mean of the following frequency distribution. Mass (Kg)

32–36

Frequency (f) 6

37–41 42–46

47–51

52–56

57–61

Solution The mid-interval values are 34[= 1 (31.5 + 36.5)], 39, 44, 49, 54 and 59 2

Mass (Kg) 32–36 37–41 42–46 47–51 52–56 57–61

Mid-interval (x) 34 39 44 49 54 59

f 6 14 24 15 12 9 ∑ f = 80

fx 204 546 1056 735 648 531 ∑ fx = 3720

3720 Therefore the mean = ∑fx = 80 = 46.5 ∑f

The Median If a set of numbers is arranged in ascending order, the middle number or the mean of the two middle numbers is called the median. Thus there are as many numbers less than the median as there are greater than the median. Consider the number x1, x2, x3 ----, xn written in ascending order. 1

If n is odd, the median is xi where i = 2 (n + 1) 1

If n is even, the median is 2 (xi + xi + 1 ) where i = 2 n.

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Example 5 Find the median of the numbers (a) 12, 9, 17, 16, 10, 13; (b) 56, 46, 61, 57, 48, 50, 47,, 44. Solution (a) In ascending order the numbers are: 9, 10, 10, 12, 13, 16, 17. There are 7 numbers 1

and so the median is in position 2 (7 + 1) = 4. Therefore the median is 12. (b) In ascending order the numbers are: 44, 46, 47, 48, 50, 56, 57, 61. There are 8 1

numbers and so the two middle numbers are in 2 (18)= 4th and 5th positions. 1

Therefore the median is 2 (48 + 50) = 49 For grouped data, the median like the mean, can only be estimated. Linear interpolation is generally used. Example 6 Estimate the median from the following grouped data. Class 0–4 5–9 10–14 15–19 2-0–24

Frequency 4 5 5 4 2 ∑ f = 20

Solution There are 20 values and so the median is the mean of the 10th and 11th values in ascending order, i.e. the ‘10.5th’ value. There are 9 values in the first two classes and so the median must lie in the next class. This class has a lower class boundary of 9.5, a width of 5 and contains 5 values. The difference between 10.5 and 9 is 1.5. 1.5

Therefore, the median is approximately 9.5 + 5 × 5 = 11.

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The mode The mode is the value with the largest frequency. For grouped data, the modal class is the class with the greatest frequency. There may of course be more than one mode or modal class.

Cumulative frequency, Quartiles and Percentiles To find the cumulative frequency corresponding to a given class we sum the frequencies up to the upper class boundary. Example 7 The frequencies of the scores of 80 students in a test are given in the following table. Complete the corresponding cumulative frequency table. Test score 0–9 10–19 20–29 30–39 40–49

Frequency 8 18 24 20 10

Solution A suitable table is as follows: Test score

Cumulative frequency

≤ –0.5 ≤ 9.5 ≤ 19.5 ≤29.5 ≤ 39.5 ≤49.5

0 8 26= (8 + 18) 50 = (26 + 24) 70= (50 + 20) 80= (70 + 10)

The information provided by a cumulative frequency table can be displayed in graphics form by plotting the cumulative frequencies given in the table against the upper class boundaries and joining these points with a smooth curve.s The cumulative frequency curve corresponding to the data in the previous example is as follows:

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Cumulative frequency

• •

60 • 40 • 20 • 0

Test score

Fig. 10.1

Note: If we join the dots with straight lines, we form a cumulative frequency polygon. Example 8 The results obtained by 200 students in a mathematics test (maximum mark = 50) are given in the following table. Mark 1–10 11–20 21–30 31–40 41–50

Frequency 22 35 72 60 11

Draw a cumulative frequency curve and use it to estimate (a) the median mark; (b) the number of students who scored less than 22 marks; (c) the pass mark if 120 students passed the test; (d) the minimum mark required to obtain an A grade if 10% of the students received an A grade. Solution The required cumulative frequency curve is as follows: 229

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Cumulative frequency

200

• •

•

100

•

• • 0.5

10.5

Test score 20.5

30.5

22 26 28

40.5 38

50.5

Fig. 10.2

From the graph: (a) the median mark was 26; (b) approximately 69 students scored less than 22 marks; (c) the pass mark was 28; (d) the minimum mark required to obtain an A grade was 38.

Quartiles and Percentiles Twenty five percent of the observaions have values which are less than the lower quartile, Q1, and twenty five percent of the observations have values which are greater then the upper quartile; Q3. Thus the lower quartile, median and upper quartile divide the distribution into four equal parts. When data is grouped, we can only estimate the values of the quartiles and we use a cumulative frequency curve for this purpose. A useful measure of the dispersion of the observations is the interquartile range Q3 – Q1. This is not affected by any extreme values at either end of the range of values. The middle 50% of the population have values between Q1 and Q3 and so the interquartile range provides a measure of the spread of the middle half of the population. 1

A similarly useful measure of dispersion is the semi-interquartile range, 2 (Q3 – Q1). The nth percentile, Pn, has n% of the observations with values less than Pn. Thus 20% of the observations have values less than P20 and 20% observations have values greater than P80. As with quartiles, we estimate the percentiles from grouped data by reading information from the corresponding cumulative frequency curve. 230

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Examples 9 The heights of 500 students in a school were measured and the results were as follows: Height (cm) 140–144 145–149 150–154 155–159 160–164

Frequency 8 19 35 68 81

Height (cm) Frequency 165–169 92 170–174 88 175–179 53 180–184 42 185–189 14

Draw a cumulative frequency curve and use it to: (a) estimate the median height of the students; (b) determine the interquartile range (c) determine P90 – P10 (the 10th to 90th percentile range) Solution The cumulative frequencies are given in the following table: Height (cm) Cumulative Frequency Height (cm) Cumulative Frequency <139.5 0 <169.5 303 <144.5 8 <174.5 391 <149.5 27 <179.5 444 <154.5 62 <184.5 486 <159.5 130 <189.5 500 <164.5 211 The cumulative frequency curve is as follows: 500

•

400

•

300 •

200 •

100 • 140

150

160

170

P10 Q1 Mediam Q 3

180

P90

190

Fig. 10.3 231

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(a) Median height = 162 cm (b) Interquartile range = (168 – 154) cm = 14 cm (c)

P90 – P10 =(176 – 148) cm = 28 cm

Activity 10.1 1.

Find the mean, median and mode of the data listed below:

1; 0; 2; 4; 1; 2; 1; 1; 2; 5; 5; 0; 1; 2; 3.

The frequency chart at the left shows the scores out of 10 achieved by a class in a maths test. 7 6

frequency

5 4 3 2 1 0

3 4 test score

10 Fig. 10.4

Calculate the mean, median and mode for this data.

The history test scores for a group of 40 students are shown in the frequency table below. Score Frequency Mid-interval value Frequency × mid-interval value 0 ≤ s < 20 2 9.5 19 20 ≤ s < 40 4 29.5 118 40 ≤ s < 60 14 49.5 693 60 ≤ s < 80 16 69.5 1112 80 ≤ s < 100 4 89.5 358

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(a)

Calculate an etimate for the mean test result.

(b)

What is the modal class?

10.3.2 Measures of Dispersion for Discrete Datas The simplest measure of the dispersion (spread) of data is the range. The range is equal to the difference between the largest value and the smallest value and is completely determined by these extreme values. Consider the following data sets: (a)

8, 9, 10, 11, 12;

(b)

–45, –8, 2, 42, 59

Each has a mean of 10, but the second set is more spread out than the first. In fact, the range of the first is 12 – 8 = 4 but the range of the second is 59 – (– 45) = 104. The mean Deviation from the mean The mean deviation (M.D) can be measured from the arithmetic mean. For n values x1, x2, ..., xn, if x is the arithmetic mean, then the mean deviation from the mean is |x – x | + | x2 – x | + ... + | xn – x | M.D = 1 |xi – x | = 1 n n For a frequency distribution x1

...

xp fp

the mean deviation from the mean is: M.D = 1 n

fi |xi – x | =

f1 |x – x | + f 2| x2 – x | + ... + f p| xp – x | where n = f. n

Example 10 Calculate the mean deviation from the mean of the distribution. x fi

12.5 2

17.5 12

22.5 27

27.5 41

32.5 30

37.5 7

Solution The mean is x = 3207.5 = 26.95. 119

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12.5 17.5 22.5 27.5 32.5 37.5

2 12 27 41 30 7

|xi – x | 14.45 9.45 4.45 0.55 5.55 10.55

f|xi – x | 28.90 113.40 120.15 22.55 166.50 73.85

f = 119; f | xi – x | = 525.35 Mean deviation = 525.35 = 4.41. 119 We have already met at least one other measure of spread – the interquartile range. Its disadvantage is that it is quite ‘insensitive’ to changes in the lower and upper quartiles of the data. Much more useful measures of the spread of data are the variance and its square root, the standard deviation. Variance: Let the observed values be; x1, x2, x3, ----, xn. Then the variance is denoted and defined by

s2 = 1 (x – )2 n i=1 i

where

x is the mean of the data = 1 n i=1 i

Standard Deviation: The standard deviation of the observed values is denoted and defined by 1 (x – )2 n i=1 i

which is clearly the square root of the variance.

The variance is thus the mean of the squared deviation from the mean and the standard deviation is simply the square root of this. The following results follow directly from the definitions of mean and standard deviation. 1.

When all data values are mutliplied by a constant a, the new mean and new standard deviation are equal to a times the original mean and standard deviation.

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The mean of ax1, ax2, ax3 ---., axn is a , and the standard deviation is as.

When a constant value, b, is added to all the data values, then the mean is also increased by b. However, the standard deviation does not change. The mean of x1 + b, x2 + b, ---, xn+ b, is + b and the standard deviation is s.

Example 11 The six runners in a 200 metre race clocked times (in seconds) of 24.2, 23.7, 25.0, 23.7, 24.0 and 24.6 (a) Find the (i) mean and (ii) standard deviation of these times; (b) These readings were found to be 10% too low due to fautly time keeping. Write down the (i) new mean and (ii) standard deviation. Solution = 24.2 + 23.7 + 25.0 +6 23.7 + 24.0 + 24.6 = 24.2 seconds

(a) (i)

2 2 2 s = (24.2 – 24.2) + (23.7 – 24.2) + ... + (24.6 – 24.2)

(ii)

(b) (i) (ii)

0 + 0.25 + 0.64 + 0.25 +0.04 + 0.16

= 0.473 seconds We must divide each time by 0.9 to find the correct time. The new mean = 24.2/0.9=26.9 seconds, the new standard deviation = 0.4726/0.9 = 0.025 seconds.

The method which uses the formula for the standard deviation is not necessarily the most efficient. Consider the following: 2 Variance = ∑(x – ) n

2 2 = ∑(x – 2x + ( ) ) n

= n – 2 n + ( )2 n (since is a costant)

= n –2

∑x2

∑x

∑x2

∑1

+ ( )2

∑x2

2 n –( )

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Example 12 The heights (in metres) of six children are 1.42, 1.35, 1.37, 1.50, 1.38 and 1.30. Calculate the (i) mean height (ii) standard deviation of the heights. Solution (i) Mean = 16 (1.42 + 1.35 + 1.37 + 1.50 + 1.38 + 1.30) M = 1.39 (2 d.p) (ii) Variance = 16 (1.422 + 1.352 + 1.372 + 1.502 + 1.382 + 1.302) – 1.3872

= 0.003862M2

Standard deviation = 0.00386 = 0.0621m.

The Mean and Variance for Grouped Data The mean for grouped data where the mid-interval value for the ith group is xi which occurs with frequency fi (so that ∑fi = n) is

∑fi xi ∑fi

∑f x

= ni i

the variance is =

∑fi (xi – )2 ∑fi

2 ∑f x 2 = ∑fi (xi – ) = i n i –

and the standard deviation, , as alwalys is the square root of the variance. In the first example we will use the above formulae, but a GDC should be used in practice. Example 13 The number of customers served lunch in a restaurant over a period of 60 days is as follows: Number of customers served lunch 20–29 30–39 40–49 50–59 60–69 70–79

Number of days in the 60–day period 6 12 16 14 8 4

Find the mean and standard deviation of the number of customers served lunch using this grouped data. 236

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Solution In the 40–49 group, we do not know exactly how many of the 16 days, 44 customers say, were served lunch. So we assume that on each of these days, 44.5 customers were served. We choose the mid-interval value ( 21 [40 + 49]) as representative of the group as a whole and use this to estimate the mean and standard deviation. Group

Mid–inerval value (xi)

Frequency (fi) fixi

fixi2

20–29 30–39 40–49 50–59 60–69 70–79

24.5 34.5 44.5 54.5 64.5 74.5

6 12 16 14 8 4 ∑fi = 60

3601.5 14283.0 31684.0 41583.5 33282.0 22201.0 ∑fixi2 = 146635

The mean is =

∑fi xi ∑fi

147 414 712 763 516 298 ∑fixi = 2850

= 2850 = 47.5 60

2 The standard deviation is = ∑fixi –

∑fi

146635 – 47.52 60

= 13.7.

Example 14 The age distribution of the population of a small country town on January 1, 2001 is given in the following table: Age group (years)

Mid–interval (xi years)

Frequency (fi)

0–14 15–29 30–44 45–59 60–64 75–89 90–104

7.5 22.5 37.5 52.5 67.5 82.5 97.5

856 1120 1054 792 651 288 96

Calculate the (i) mean (ii) standard deviation of the population age on January 1, 2001. 237

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It should be noted that people from age 15 to almost 30 lie in the group 15–29 and so it is correct to take 12 (15 + 30) = 22.5 as the mid-interval value. From the GDC, (i)

= 39

(ii) s = 23.6 years.

Combining sets of numbers Consider the two distributions Distribution I: x1, x2, ..., xn: Number of items: n Mean:

= n1

Variance: S2x = ( n1

xi – x 2

2 Standard deviation Sx = Sx

Distribution II: y1, y2, ..., ym: Number of items: m 1 y Mean: y = m i 1 Variance: S2y = ( m

xi – y2

2 Standard deviation Sy = Sy

If the two distributions are combined in one: x1, x2, ..., xn, y1, y2, ..., ym, that is z1, z2, ... , zn, zn + 1, zn + 2, ..., zn + m, where x1 = z1, x2 = z2, ..., xn = zn, y1 = zn + 1, y2 = zn + 2, ..., ym = zn + m Then: the number of items is: n + m. The mean for the combined set: z= =

z1 + z2 + ... + zn + zn + 1 + ... + zn+ m n+m

(x1 + x2 + ... + xn) + (y1 + y2 + ... + ym) n+m

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x+ y

=n+m

1 1 But x = n x and y = m y

imply x = nx and y = my n x + my Therefore, the mean for the combined set is z = n + m .

Variance for the combined set: 1

S2 = ( n + m Σ z2)– z 2 z

(x21 + x22 + ... + x2n ) + (y21 + y22 + ... + y2m ) n+m

x2 + y2

= n+m

– z2

But S2 = ( n x2) –x 2 and S2 = ( m y2) – y 2 x y imply x2 = n (S2 + x 2) and y2 = m(S2 + y 2) x

Therefore, the variance for the combined set 2

2 n(Sx is S2 =

+ x 2) + m (S2 + y 2) n+m

y – (n xn ++ m m )

Example 15 A set of 12 numbers has mean 4 and standard deviation 2. A second set of 20 numbers has a mean of 5 and standard deviation 3. Find the mean and standard deviation of the combined set of 32 numbers. Solution For the first set of numbers, x, n = 12, x = 4 and S2 = 4 y

For the second set of numbers, y, m = 20, y = 5 and S2 = 9 y

y The mean of the combined set is z = n xn ++ m m =

12(4) + 20(5) 12 + 20

= 4.625

= 7.359.

Therefore, the standard deviation of the combined set is Sz = 7.359 = 2.71. 239

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Coefficient of Variation For a distribution x1, x2 , ... , xn, the coefficient of variation is the ratio S

C.V = xx where Sx is the standard deviation and x is the mean. It may be expressed as a percent. Example 16 Given the data 5, 6, 7, 8, 9, find the coefficient of variation. Solution The mean is x =

5+6+7+8+9

The variance is S2 =

52 + 62 + 72 + 82 + 92

â&#x20AC;&#x201C; 72 = 2

The standard deviation is Sx = 2

Therefore, the coefficient of variation is C.V = xx = 7

Activity 10.2 1.

The following marks were obtained by 12 candidates in a mathematics examination: 90; 69; 50; 40; 35; 80; 45; 73; 65; 84; 38; 75. Find: (a) The range (b) The quartiles and the interquartile range of the results. 2. The table below represents the number of people who read books in library ALPHA. Number of books read Number of readers

7 19

8 18

9 7

10 6

(a) (b)

Find the average number of books read per person. Calculate the standard deviation.

The table below shows the results in Mathematics of six combinations of senior 4. Class Mean Standard deviation Number of students in the class

4CEM 72 10.1 35

4MCB 67 7 25

4MEG 68 8.6 30

4MPC 72 12.5 23

4MPG 4PCM 58 60 10.1 8.6 20 27

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(a) (b) (c)

In which class do students have almost the same level? In which class do students have quite different levels? What is the mean mark for the combined classes?

10.4 Practice 10.4.1 Measures of central tendency 1.

Find the mean, median and mode of each of the following sets of numbers. (a)

6, 10, 4, 13, 11, 9, 1, 6, 12;

(b)

193, 195, 202, 190, 189, 195;

(c)

0.77, 0.73, 0.61, 0.73, 0.65,, 0.83, 0.81, 0.65, 0.73, 0.69, 0.74, 0.76

Find the mean and median of each of the following frequency distributions: (a) x f

1 32

x f

10 23

2 27

3 26

4 35

5 33

6 27

(b) 11 19

12 16

13 24

14 18

2 28

x f

3.2 4

3 32

5 7 11 35 17 6

13 2

(d) 3.6 9

4.0 26

4.4 35

4.8 42

5.2 34

3. Estimate the mean and median of the following grouped frequency distributions. (a) x f

0–2 2–4 4–6 6–8 20 19 25 16

x f

10–14 15–19 61 50

(b) 20–24 25–29 30–34 37 27 16

35–39 9

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0–9 10–19 20–29 30–39 3 24 46 38

40–49 28

50–60 11

The frequency table giving the ages in completed years of the population of a small country town is as shown population. Estimate the mean age and the median age of the. Age (x years) Frequency (f)

0–19 20–29 30–39 40–49 258 761 906 8332

Age (x years) Frequency (f)

60–69 70–79 484 305

50–59 756

80–89 148

90–99 44

100–109 6

10.4.2 Measures of dispersion 1.

Find the mean and standard deviation of 25.2, 22.8, 22.1, 25.3, 24.6, 25.0, 24.3 and 22.7.

(a)

Using the mean and standard deviation of the set of numbers {3, 5, 6 , 8, 10}, find the mean and standard deviation of each of the following sets of numbers.

i) {6, 8, 89, 11, 13}

ii) {9, 15, 18, 24, 30}

iii) {2.7, 4.5, 5.4, 7.2, 9.0}

(b) Find the mean and standard deviation of the set containing numbers which are 5% higher than those in the origin set.

The mean height of a group of 5 people is h = 155cm and the standard deviation of their heights is 5 cm. (a)

Calculate ∑h and ∑h2 for this data

(b) If an extra person of height 165m is added to the group, calculate the (i) new mean

(ii) standard deviation of the heights.

The mean and variance of x1, x2, x3, ---, xn are and mean and variance of 2 – 3x1, 2 – 3x3, ---, 2 – 3xn.

Twenty values of a random variable have a mean of 15 and a variance of 1.5. Another thirty values of the same random variable have a mean of 14 and a variance of 1.4. Find the mean and variance of the combined fifty values.

respectively. State the

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The mean daily maximum temperature at a fixed location for the month of January was 220C and the standard deviation of these daily maxima was 20C. For the following February which was not a leap year, the mean was 24 0C and the standard deviation was 4 0C. Calculate the mean and standard deviaton of the daily maximum temperatures for the two months combined.

A sample of 165 values of a random variable have a mean of 23.4 and a standard deviaton of 1.6. When combined with another 219 values, the mean of all 384 values was 24.8 and the standard deviation was 2.2. Find the mean and standard deviation of the 219 values which were added to the original sample.

Twenty values of a random variable have a mean value of 112.5 and a variance of 1.35. If two more values are added to the original 20, the mean remains at 12.5 but the variance is increased by 0.082. Fidn the two values added.

Calculate the mean and standard deviation from the following data. (b) (a) Test score 1–10 11–20 21–30 31–40 41–50

Frequency 6 17 29 23 10

Test mark

Frequency

1–20 21–40 41–60 61–80 81–100

4 25 71 38 12

10. The ages of the people living in a country town are grouped as follows: Age group 0-19 10–19 20–29 30–39 40–49 50–59

Frequency 1205 1528 2006 1857 1691 1483

Age group 60–69 70–79 80–89 90–99 100–109

Frequency 954 532 187 56 4

Calculate the (i) mean (ii) standard deviation of the ages of the people in the town.

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11. The heights in centimeters of a group oe people are given in the following table. Calculate the mean and standard deviation of these heights. Height (cm)

Frequency

Height (cm) Frequency

140–144 145–149 150–154 155–159 160–164 165–169

5 18 27 62 87 115

170–174 175–179 180–184 185–189 190–194 195–199

173 130 72 35 12 2

12. The life-times (in hours) of 200 light bulbs were determined and the results are given in the following table. Life–time 800–849 850–899 900–999 950–999

Frequency 4 18 23 28

Life–time 1000–1049 1050–1099 1100–1149 1150–1199

Frequency 31 48 38 10

Calculate the (i) mean (ii) standard deviation of the life-times of the bulbs.

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Model Exam Papers Model Exam Paper 1 Section A (55) Marks by f(x, y) = (19x + 4y, x + 6y).

Linear operator f is defined in

Find the matrix M representing f in the standard basis of

Show that cos 2x – sin 2x = sec 4x + tan 4x. (4 marks)

The roots of the equation x2 – 2x – 10 = 0 are

(3 marks)

cos 2x + sin 2x

and β. Determine a quadratic 1

equation in x whose roots are (2 + )2 and (2 + β )2 .

(4 marks)

The parabola y = 4 x2 – 1 and the line 2y = x + 10 intersect at points A and B.. Calculate the coordinates of A and B. (3 marks)

Write, in interval form, the solution of the inequality |3x + 2 | ≤ 4 . (3 marks)

Given that A =

. ( 4 marks) A.X.B = C and find X 2000

Given that vectors u = (cos 3t, sin 3t), v = (cos 2t, sin 2t) and w = (cos t, sin t), where t is a real number,

(a)

Show that u + w and v are linearly dependent.

(b)

Find the values of t such that u + w = v .

(4 marks)

Find the value of m for which the equation.

( 4 marks)

(2 – m) x2 + my2 – 2m (2 – m) x – 2y = m3 – 3m2 + 2m – m represents a circle.

Find the centre and the radius of the circle.

Let H =

Determine whether, or not, H is closed under:

(a)

2 1 2 1 –3 0 ,B=a and C= , find matrix X such that 3 3 1 2 –3 3

{ a2b b0

(4 marks)

}

; a∈ , b∈ .

Addition

(b) Multiplication (4 marks) 10.

Solve for x in the equation 4 x . 16 x + 2 = 64 x + 1 . (4 marks) 245

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11.

In a certain test, twenty 4PCM students have a mean of 15 and a variance of 1.5, while thirty MPG students have a mean of 14 and a variance of 1.4. Calculate the mean and variance of the combined students. (4 marks)

12.

The binary operator is defined in the set of integers by a b = a + b + ab. Solve for x the equation 2 x 3 = 23. (Assume is associative). (3 marks)

13.

Consider the statement:

(a)

Write down the converse of the statement.

(b)

Write down the negation of the converse.

14.

Rationalize the denominator of each of the expressions:

If x is natural number, then x is an integer»

(a) 2

(3 marks)

2 6 – 3– 5 2

(b) 1 + 3 ( 4 marks) 15.

Solve for x the equation

log (x + 1) + log (x – 1) = 2 log (x – 2).

( 4 marks)

Section B (45 marks) 16. The table below shows the quantity of powdered milk consumed by babies aged two months per day in grams. (15 marks) 60; 41; 53; 66; 42; 70; 85; 65; 70; 81; 55; 47; 57; 69; 68; 43; 61; 80; 88; 53; 49; 50; 75; 80; 60; 44; 81; 70; 76; 66; 64; 70; 72; 52; 56; 68; 54; 72; 81; 48; 52; 65; 51; 61; 68; 73; 48; 50; 49; 66. (a)

Group the data into 10 classes of equal width, starting with 40 ‑ 44 and ending with 85-89.

Construct a detailed frequency table.

(b) Find the modal class. (c)

Calculate the average quantity of powdered milk consumed, using 67.5 as assumed mean.

(d) Find the variance, standard deviation and the coefficient of variation. (15 marks) 246

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17.

A path, 3 meters wide, turns at right angle and its width is now 1 meter only (as shown on the figure). A straight line passes through point O, makes an angle x with one of the walls and cuts the other two walls at points A and B.

0 B 1

(a)

Express, in terms of x, the lengths OA; OB and AB.

(b)

If AB = f(x), show that f(x) =

18.

(a) 7x + 3 – 53x = 2 (7x + 3 + 53x – 1).

(b)

(16 – x2)2 – (x + 4)2 x4 – 4

AB = 4 OA = OB

4 cos (x – π6 ) Sin 2x

(15 marks)

≤ 0.

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Model Exam Paper 2 Section A (55 marks) 1.

Use the change of variable tan 2x = t to solve for x in the equation

5 cos x – 2 sin x = 2. (4 marks)

A metal worker wants to make an open box from a sheet of metal width 12 cm and length 16 cm, by cutting equal squares of side x cm from each corner.

(a)

Write an expression for the length, width and height of the box.

(b)

Express the volume of the box as a polynomial in x.

3. Express

x 3x – 9x + 6 2

–

2x + 1 3x2 + x – 6

(3 marks)

as a single simplified fraction, stating the

restrictions on x. (3 marks)

(a)

If tan θ = – 32 and sin θ > 0, find the value of cos θ sin θ.

(b)

8 , find sin Y. ( 4 marks) If triangle XYZ is right angled at Z and tan X = 15

Solve the simultaneous equations; 1

2 x . 2 y = 32 6 2x. 2y = 3

(4 marks)

Solve for x in the equation x4 – x3 – 10x2 – x + 1 = 0.

The resultant of forces;

(4 marks)

F1 = 3 i + (a – c) j , F2 = (2a + 3c) i + 5 j and F3 = 4 i + 6 j

acting on a particle is 10 i + 12 j .

(a)

Find the values of a and c.

(b)

Find the magnitude of force F2 . (4 marks)

Consider the proposition:

(a)

What is the truth value of the proposition?

(b)

Write down the negation of the proposition.

Use a calculator to evaluate:

If x (x + 1) = 4, then x = 4 or x + 1 = 4».

log 0.25 – log2 12

(a)

log 4.3

tan 28.3° tan4 12°

(b)

1 – cos 333°

(3 marks)

, round to three decimal places. . (4 marks)

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10.

(a) Given that tan a and tan b are the roots of the equation x 2 + px + q = 0, express tan (a + b) in terms of p and q.

n (b) Given that x = m2 m , where m > 0 and n>0, express logm x in terms of m and n.

11.

Gicven the polynomial;

p(x) = 2x5 + 4x 4 – 5x3 – 10x2 + 3x + 6,

(a)

Find P(–2).

(b)

Solve the equation P(x) = 0.

12.

Line PQ , where P(7,8) and Q (2t + 3, t – 9) is perpendicular to line x – 2y = 6. Find the value of t. (4 marks)

13.

Find Matrix X such that

2X + 3A = B, if A =

14.

Binary operation T is defined in the set of integers by xTy = xy – 2.

(a)

Calculate (–5) T (6 T8) and [(–5) T6]T8.

(b)

Determine whether T is:

–3 1 3

–1 0 –4

0 2 4

and B =

3 . 0 –1

(4 marks)

(3 marks)

(i) Commutative

(ii) Associativce

15.

The numbers 4; 6; 12; 4; 10; 12; 3; x, y have a mean of 7 and a mode of 4, and

x < y.

(a)

(b) When two additional numbers 7 + n and 7 – n are included, the standard deviation of all eleven numbers is found to be 4.

Write down the mean of these eleven numbers and calculate the value of n. (4 marks)

(4 marks)

Find the two numbers and the median of the set of the nine numbers.

Section B (45 marks) xy = 16

16.

(a)

Solve for (x, y) the simultaneous equations log2 x + log2 y = 5 log24 2

(b)

The sides of a triangle are carried by the lines x + 3y – 1 = 0,

5x + 3y + 4 = 0 and x + 2y + 1 = 0.

Find: (i) the coordinates of the vertices of the triangle.

(ii) the area of the triangle.

(15 marks) 249

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17.

The weight of 60 people are measured and grouped into four classes of equal width a;

[U0, U1 [; [U1, U2[; [U2, U3[; [U3, U4[.

Let n∈ {1, 2, 3, 4}.

(a)

Express Un in terms of U0, a and n.

(b)

For U0 = 38 kg, and U4 = 70 kg, find the value of a.

(c)

Hence, find the different classes.

(d)

Given that the frequencies of the classes are 5; 15; 30 and 10,

respectively, find the mean and the standard deviation

of the distribution. (15 marks) 18.

(a)

(b)

Given that sin x + cos x = 21 , find the value of sin4 x + cos4x

Solve the inequality cos 2x + sin x ≥ 0, where 0 ≤ x ≤ 2π. (15 marks)

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Model Exam Paper 3 Section A (55 marks) 1.

(a)

Given that sint = x, where t is an acute angle, x

(i) Express

(ii) Find the value of

1 – x2

in terms of t. x 1 – x2

for t = π .

(4 marks)

(b) If x = sin t and y = cos t, where t is an acute angle, simplify

1 – y2

1 – x2

(2 marks) 2.

Express f(x) = x –|x + 2| + |1 – x | without "absolute value sign" and

graph f(x) on interval [–3, 2].

A teacher records the following scores for a 20-point quiz: 20; 19; 18;

17; 16; 15; 15; 12; 10; 6; 6. Find the mode, median and coefficient

of variation.

(3 marks)

(4 marks)

Given that x + 1 = 3, expand (x + 1 ) and find the value of x3 + 13 . (3 marks)

Given that triangle XYZ, where X (2,2) Y (a,7) and Z(15,1) is right

angled at Y, find possible values of a.

The number x of errors, on each of 200 pages of a book was recorded.

The results showed that x = 920, x2 = 5032.

(a)

(3 marks)

Calculate the mean and the standard deviation of the number of errors perpage.

(b) A further 50 pages were considered and it was found that the mean was 4.4 errors and the standard deviation was 2.2 errors.

Find the mean and the standard deviation of the number of errors per page for 250 pages. (4 marks)

From a ship A, two other ships B and C are on bearings of 340°

and 040° respectively. If B is 5 km from A and C is 3 km from A,

find the distance between B and C.

(3 marks)

x2 – x + 1 x–1

Find the values of k for which the equation = k has repeated roots. What are the repeated roots? (4 marks)

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4 2 2 1 and B = , find matrix C such that 2A–1 + C = B. 3 1 –2 3 (4 marks) 9.

Given that A =

10.

Given that f(x) = 3x, find a simplified expression for

f(x) + f( 1 ) + 1 + f–1(x). (4 marks) x

f(x)

11.

2 Solve for x in the equation 12x –260x + 75 = 3 .

12.

Find the truth value of the proposition [(P Q) ⇒ Q ] ⇔ (P Q)

when P and Q are both true.

13.

(a)

8x – 50

(4 marks) (3 marks)

Given that the polynomials

P(x) = 5x – 8 and Q(x) = a(x –2) + b (x – 1) are equal,

find the values of a and b.

(b) Simplify

|4 – x2 | x–2

for –2 < x < 2.

(4 marks)

1 cos x + cos y = 2

14.

Given that cos 2x + cos 2y = –1 , (4 marks) 2

Find cos x and cos y.

15.

(a)

Find the value of x and the value of y if x2y + 1 = x3y – 1 = 243.

(b)

Express as a single logarithm 3 log x – 12 log y + 1.

(4 marks)

Section B (45 marks) 16.

Discuss, with respect to the values of parameter m, the number and sign of the solutions of the equation in x.

x2 2

17.

(a) Consider the right angled triangle on the figure below;

– (m – 2)x + m 2 = 0. (15 marks)

C x+2

A (i)

3–x

Express the area of the triangle as a function of x.

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(ii) Find the value of x for which the area is maximum and find the maximum area.

(b) Express E = sin2 t – 2 (1 – cos t) in terms of sin t2 , then discuss with respect to m, the real solutions of the equation 2x2 (1 – cos t) –2 x sin t + 1 = 0 18.

(15 marks)

(a) B = ( i , j ) is a basis of a vector space V. Vectors u and v

are defined by u = 2 i – 2 j and v = – i – 3 j . (i)

Show that B′ = (u , v ) is a basis of V.

(ii) Vector w has components (x, y) in basis B, and (x′ y′) in basis B′. Express x′ and y′ in terms of x and y.

(b) Discuss, with respect tothe valuesof parameter m, the solutions of the –2x + m2 y = m (15 marks) simultaneous equation x – my = 1 – m .

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Model Exam Paper 4 Section A (55 marks) 1.

Show that the set of all multiples of 5 is a subgroup of (

(a)

Rationalize

(b)

Simplify x6 – a6 .

Function f(x) = a sin x + b, where 0 ≤ x ≤ 2 π, is such that

f ( π2 ) = 2 and f ( 3π 2 ) = – 8.

Find the values of a and b, and then x, if f (x) = 0.

(a) Determine the set of values of x for which the expression

1 4

2 –1

, +).

(4 marks)

x12 – a12

(4 marks)

(3 marks)

logx (–x2 – x + 2) is valid. Solve for x the equation log3x – 7 (x2 – 2x – 3) = 1.

(b)

Find, without using a calculator, the exact value of

(4 marks)

tan 480° sin 300° cos 14° sin (–135°) . (4 marks) sin 104° cos 225°

Find the equation of the circle that passes through points

A(–1, 4), B(2,5) and C(0,1).

Given that u = x i + 2 j , v = i + j and || u + v|| = 5,

find possible values of x. (3 marks)

1 , where A and B are acute angles, If tan A = 51 and tan B = 239

find 4A – B.

(a)

(4 marks)

4x x–5 x x–5 is the inverse of , 1 – 3x x 3x –1 4x find the values of x.

(b)

Given the matrices A =

find 4A + BC.

Given that

0 1 1 3 2 3 ,B and = , –1 2 –2 2 3 –1

(4 marks)

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10.

The table below shows the distribution of heights of people in a group. Heights (cm) 150–154 155–159 160–164 165–169 Frequency 30 25 23 22

Find (a) The mean height.

(b) The standard deviation and the coefficient of variation. (4 marks)

11.

The binary operation T is defined in the set of real numbers by x

Ty = x + y + 21 .

(a)

Find (i) The identity element

(ii) The inverse of –3 under operation T.

Solve for x the equation (x T 1) Tx = 5.

(b)

(3 marks)

12.

Vectors = u = (m – 1) i + (2 – m) j and v = (–2m) i + (2m – 3) j , where B = ( i , j ) is a basis of 2, are linearly dependent.

(a)

Find the value of m

(b)

Hence, express v in terms of u

13.

(a)

Given that x = 4 (3–y), make y the subject of the formula

(4 marks)

4 and find the value of y when x = 27 .

(b)

Given that cos A cos B = 43 and sin A sin B = 21 , find cos (A,B) and

cos (A – B). Hence, find the values of A and B given that

A + B < π2 . (4 marks)

14.

Given that u =

(a) Express w as a linear combination of u and v .

(b)

15.

(a)

4 , = v –3

2 and = 22 , w –11 4

Find the cosine of the angle between u and v . Expand and simplify

(4 marks)

(cos x + sin x)2 + (cosx – sinx)2.

(b)

If cos x + sinx =

2 2

, find cos x – sinx,

then find cos x and sin x. (4 marks)

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Section B (45 marks) 16.

Given the curves (C1): y = 25 – 5x – 2x2 and (C2): y = 15 – 2x – x2,

(a)

For each curve, find the vertex, its nature, the axis, the intercepts.

(b)

Find the intersection of the two curves.

(c)

Graph (C1) and (C2) on the same figure and shade the region

enclosed between the two curves.

17.

Find the values of m so that the equation x – 3m x + m – 1 = 0 Sas:

(a)

(i) 4 roots

Solve the simultaneous equations

(b)

(ii)

3 roots

(15 marks)

(iii)

2 roots

log (–x + 2y) + 1 = log2 (3y – 2x) 35x + y =

81 3–x – 7y

(15 marks) 18.

The mean of the set of numbers 3, 1,7, 2, 1, 1, 7, x, y, where x and y are single digit positive integers is 4.

(a)

Find the mode if:

(i) x = y (ii) x ≠ y

(b)

If the standard deviation is

76 , find x and y, if x ≤ y. 3

(15 marks)

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Model Exam Paper 5 Section A (55 marks) 1.

Point p(a,b) is such that the product of the square of the distances to lines

4x – 3y + 11 = 0 and 4x + 3y – 5 = 0 is 144 25 . Find an equation connecting a and b. (3 marks) 1 2. Calculate 0 5

–2 1 6

3 4 1 1

8 3

7 . 3

Find the remainder of the division of kx2 + x – 1 by x + 2k.

Find the smallest positive integer which satisfies the congruence

3x + 4 x + 7 (mod 7).

(3 marks)

(3 marks) (3 marks)

The following marks are obtained by students in an English examination:

96; 81; 23; 62; 44; 18; 62; 70; 72; 40; 81; 70;

30; 28; 23; 02; 00; 60; 20; 48; 50; 19; 33; 32;

58; 71; 62; 19; 12; 83; 53; 81; 73; 75; 52; 25;

71; 61; 46; 64; 35; 59; 82; 82; 42; 63; 43; 17; 35;

72; 50; 37; 54; 47; 76; 18; 44; 65; 45; 70; 38;

63; 38; 63; 27; 23; 03; 15; 63; 25; 52; 53; 38;

57; 53; 71; 70; 63; 89; 31; 37; 37; 93; 58; 58; 90

(a)

(b) Find the mean mark, the median mark and the standard deviation. (8 marks)

In a certain shop, 5 ball point pens and 2 exercise books cost 700 FRW and 1 ball point pen and 3 books cost 660 FRW. Use matrices to find the cost of 1 ball point pen and the price of 1 exercise book. (3 marks)

Solve for x in the equation 2 x – 1 – x + 4 = 1.

(3 marks)

Show that (P Q) ( P Q ) is a contradiction.

(3 marks)

Let G = {I, R, S, T}, where I =

Using class intervals of 10 marks, and starting with the class 0 – 9, construct a frequency distribution table.

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–1 0 0 –1 0 1 ,S= and T = 0 –1 1 0 –1 0

Show that (G, .) is a group.

10.

(a)

Simplify cosec ( π2 + θ) sin ( 3π 2 – θ).

(b)

Solve the equation tan 2θ tan 4θ = 1, for 0° ≤ θ ≤ 90°.

11.

The roots of the quadratic equation x 2 + bx + 2 = 0 are and β.

Given that =

(b)

Find the value of

(c)

Form a quadratic equation whose roots are

12.

Let A(4, 0), B(0,3) and P(x,y). Given that line (AP) is

perpendicular to line (BP), Find an equation connecting x and y.

13.

Given that u = –2 i + 4 j , v = –5 i + 10 j and

(4 marks)

5 + 3 , find b. 2

β + β2 . 2

+β and

– β2 . (4 marks) (4 marks)

w = 3 i + 4 j , find:

(a) || u + 2 v – 3 w ||

(b)

The angle between u + v and w

(3 marks)

14. (a) If loga (2 + a) = 2, find a –1

Simplify 1 + log ( x44 ) 2 –2 log x.

(b)

15.

Solve for x in the inequation

4x – 3 2x + 1

≤ 2.

(3 marks) (4 marks)

Section B (45 marks) 16.

(a)

Show that 3 + 2 2 = 1 + 2 .

(b)

Solve the equation 2y2 + (1 – 2 )y –

(c)

Solve the inequality 2y2 + ( 1 –

(d)

Solve the equation

2 cos2x + (1 – 2 ) cos x –

2 2

2 )y –

2 2 2 2

= 0. ≤ 0.

≤ 0, where 0 ≤ x ≤ 2π.

17.

(a)

Solve the equation cos4 x + sin4x = 87 , where – π< x < π.

(b)

Show that sin 3θ = 3sinθ – 4 sin3 θ.

(15 marks)

258

Rwanda Sr. 4 Chap 1 new.indd 258

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Solve the equation 2 sin 3θ = 1, 0 ≤ θ ≤ 2π.

Hence, solve the equation 8x3 – 6x + 1 = 0.

18.

A body initially at point (2, 1) has a velocity vector

(a)

(15 marks)

3 i – 4 j . If the distance is in metre and the time is in second, find:

(i) The position vector of the body after 8 seconds.

(ii) The speed of the body.

(b)

If a body moving in the direction of the vector 7 i + 24 j , where the unit of distance is km and the speed of the body is 200 m/s. Find the body's velocity vector.

(c)

Find the set of values of x, 0≤ x ≤ 2π, for which the expression

f(x) =

1 + cot x 1 – cot x

is valid.

(15 marks)

259

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