essentials maths methods 12

8/21/2019 Essentials Maths Methods 12

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C H A P T E R

1Functions and relationsObjectives

To understand and use the notation of sets, including the symbols,,,,and\.To use the notation forsets of numbers.

To understand the concept of relation.

To understand the terms domainand range.

To understand the concept of function.

To understand the termone-to-one.

To understand the terms implied domain,restriction of a function,hybrid

function, andodd and even functions.

To understand themodulus function.

To understand and use sumsand products of functions.

To definecomposite functions.

To understand and findinverse functions.

To apply a knowledge of functions to solving problems.

In this chapter, notation that will be used throughout the book will be introduced. The

language introduced in this chapter is necessary for expressing important mathematical ideas

precisely. If you are working with a CAS calculator it is appropriate to work through sections

B1 to B4 of the Computer Algebra System Appendix.

1.1 Set notationSet notationis used widely in mathematics and in this book it is employed where appropriate.This section summarises much of the set notation you will need.

Asetis a collection of objects. The objects that are in the set are known as the elementsor

members of the set. Ifxis an element of a setAwe writex A .This can also be read as xis amember of the setA or xbelongs toA or xis inA.

The notationx /Ameansxisnotan element ofA.For example: 2 / set of odd numbers.A setBis called asubsetof a setAif and only ifxB impliesx A .

1


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2 Essential Mathematical Methods 3 & 4 CAS

To indicate thatBis a subset ofA, we writeBA .This expression can also be read as Biscontained inA or AcontainsB.

The set of elements common to two setsAandBis called the intersectionofAandBand is

denoted byA B.ThusxA Bif and only ifx AandxB .If the setsAandBhave no elements in common, we sayAandBaredisjoint, and write

A B= .The set is called theemptyset ornullset.Theunionof setsAandB, writtenA B , is the set of elements that are either inAor inB.

This does not exclude objects that are elements of bothAandB.

Example 1

A= {1, 2, 3, 7};B= {3, 4, 5, 6, 7}Find:

a A B b A BSolution

a A B= {3, 7} b A B= {1, 2, 3, 4, 5, 6, 7}Note:In this example, 3 Aand 5 /Aand{2, 3} A.

Finally, theset differenceof two setsAandBis denotedA\B, where:

A\B= {x :x A ,x /B}

e.g., forAandBin Example 1,A\B = {1, 2}andB\A = {4, 5, 6}There will be a further discussion of set notation in Chapter 14, which will provide the

additional notation necessary for the study of probability.

Sets of numbersThe elements of the set {1, 2, 3, 4, . . .} are called the natural numbers. The set of naturalnumbers will be denoted byN.

The elements of{. . . ,2,1, 0, 1, 2, . . .} are calledintegers. The set of integers will bedenoted byZ.

The numbers of the form p

qwithpandqintegers,q = 0, are calledrational numbers. The

rational numbers may be characterised by the property that each rational number may be

written as a terminating or recurring decimal. The set of rational numbers will be denoted byQ.The real numbers that are not rational numbers are calledirrational(e.g., and

2).

The set of real numbers will be denoted byR.

It is clear thatN Z QRand this may be represented by the diagram:

N Z Q R


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Chapter 1 Functions and relations 3

Note: {x : 0


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Exercise 1A

1 ForX= {2, 3, 5, 7, 9, 11},Y= {7, 9, 15, 19, 23}andZ= {2, 7, 9, 15, 19}, find:a X

Y b X

Y

Z c X

Y d X\Y

e Z\Y f XZ g [2, 8] X h (3, 8] Yi (2,) Y j (3,) Y

2 ForX= {a,b,c,d,e}andY= {a,e,i,o,u), find:a X Y b X Y c X\Y d Y\X

3 ForA = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},B = {2, 4, 6, 8, 10}andC= {1, 3, 6, 9}, find:a B C b B\C c A\B d (A\B) (A\C)e A\(B C) f (A\B) (A\C) g A\(B C) h A B C

4 Use the appropriate interval notation, i.e. [a,b], (a,b) etc., to describe each of the

following sets:

a {x : 3x


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1.2 Identifying and describing relationsand functionsAnordered pair, denoted (a,b), is a pair of elementsaandbin whichais considered to be

the first element andbthe second. In this section, only ordered pairs of real numbers are

considered.

Two ordered pairs (a,b) and (c,d) are equal ifa = candb = d.Arelationis a set of ordered pairs. The following are examples of relations:

S= {(1, 1), (1, 2), (3, 4), (5, 6)}T= {(3, 5), (4, 12), (5, 12), (7,6)}

Every relation determines two sets defined as follows:

Thedomainof a relationSis the set of all first elements of the ordered pairs inS.

Therangeof a relationSis the set of all second elements of the ordered pairs inS.

In the above examples:domain ofS= {1, 3, 5}; range ofS= {1, 2, 4, 6}domain ofT= {3, 4, 5, 7}; range ofT= {5, 12, 6}

A relation may be defined by a rule which pairs the elements in its domain and range. Thus

the set

{(x,y):y=x+ 1,x {1, 2, 3, 4}}

is the relation

{(1, 2), (2, 3), (3, 4), (4, 5)}

When the domain of a relation is not explicitly stated, it is understood to consist of all real

numbers for which the defining rule has meaning. For example:

S= {(x,y):y=x 2}

is assumed to have domainRand

T= {(x,y):y= x}

is assumed to have domain [0, ).

Example 3

Sketch the graph of each of the following relations and state the domain and range of each.

a {(x,y):y=x 2} b {(x,y):yx+ 1}c {(2,1), (1,1), (1, 1), (0, 1), (1,1)} d {(x,y):x2 +y2 =1}e {(x,y): 2x+ 3y=6,x0} f {(x,y):y=2x 1,x[1, 2]}
http://../Activities/Circles_3_4.xls


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Solution

0x

ya b

x1

1

0

y

Domain =R; range= R+ {0} Domain =R; range= R

x02 1

1

2

1 2

1

2

yc d

x

1

101

1

y

Domain ={2,1, 0, 1};range

={1, 1

}

Domain ={x : 1x1};range

={y:

1

y

1

}e f

x

2

1

10

(0, 2)

2 3

y

x

1

1

2

y

(2, 3)

(1, 3)

Domain = [0,);range = (, 2]

Domain = [1, 2];range = [3, 3]

Sometimes the set notation is not used in the specification of a relation.

For the above example:

a is written asy=x2b is written asyx+ 1e is written as 2x+ 3y=6,x0


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Afunctionis a relation such that no two ordered pairs of the relation have the same first

element. For instance, in Example 3,a,eandfare functions butb,canddare not.

LetSbe a relation with domainD. A simple geometric test to determine ifSis a function is

as follows.

Consider the graph ofS. If all vertical lines with equationsx=

a, a

D , cut the graph ofS

only once, thenSis a function.

For example,

x0

y

x0

y

x2 +y2 =1 is not a function y=x 2 is a functionFunctions are usually denoted by lower case letters such asf,g,h.

The definition of a function tells us that for eachxin the domain offthere is a unique

element,y, in the range such that (x,y) f. The elementyis called theimageofxunderforthevalueoffatxand is denoted byf(x) (read fofx).

If (x,y) f, thenxis called apre-imageofy.This gives an alternative way of writing functions.

1 For the function {(x,y):y=x 2}, write:

f:RR ,f(x)=x 2

2 For the function{(x,y):y = 2x 1,x [0, 4]}write:

f: [0, 4]R ,f(x)=2x 1

3 For the function

(x,y):y= 1

x

,write:

f:R\{0} R ,f(x)=1

x

If the domain isRwe often just write the rule, for example in 1f(x) =x2.Note that in using the notationf:X Y,Xis the domain butYis not necessarily the range.

It is a set that contains the range and is called the codomain. With this notation for functions

the domain offis written as domfand range offas ranf.


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Using a CAS calculatorFunction notation can be used with a CAS calculator.

Press F4 and select1:Define. Enter f(x)

=4x

3

in the entry line and press ENTER.

In the entry line enter f(3) and press ENTER.In the entry line enter f({1, 2, 3}) to obtain the

values of f(1),f(2)andf(3).

Example 4

If f(x)=2x2 +x,findf(3),f(2) and f(x 1).

Solution

f(3)=2(3)2 + 3=21f(2)=2(2)2 2=6

f(x 1)=2(x 1)2 +x 1=2(x2 2x+ 1) + (x 1)=2x2 3x+ 1

Example 5

If f(x)=2x+ 1,findf(2) and f

1

a, a=0.

Solution

f(2)=2(2) + 1= 3

f

1

a

=2

1

a

+ 1= 2

a+ 1

Using a CAS calculatorPress F4 and select1:Define.

Enter Define f(x)=2x+ 1

Press ENTER.Now complete f(2) and f

1

a

.

Example 6

Consider the function defined by f(x)=2x 4 for all xR .a Find the value of f(2),f(1)andf(t). b For what values oftisf(t)=t?c For what values ofxisf(x)x? d Find the pre-image of 6.


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Solution

a f(2)=2(2) 4=0

f(1)=2(1) 4

= 6f(t)=2t 4

b f(t)=t2t 4=tt 4=0t=4

c f(x)x2x 4xx 40x4

d f(x)=62x 4=6

x=55 is the pre-image of 6.

Using a CAS calculator

Press F4 and select 1:Define.Enter Define f(x)=2x 4FromF2select1:solve(.

Now entersolve(f(t)=t, t) and press ENTER.Entersolve(f(x)>=x ,x).

Restriction of a functionConsider the following functions:

f

x0

f(x)

g

x01 1

g(x)

h

x0

h(x

f(x)=x2,xR g(x)=x 2,1x1 h(x)=x 2,xR+ {0}The different letters,f,gandh, used to name the functions, emphasise the fact that there are

three different functions even though they each have the same rule. They are different because

they are defined for different domains. We callgandhrestrictions offsince their domains are

subsets of the domain off.


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Example 7

For each of the following, sketch the graph and state the range:

a f: [

2, 4]

R ,f(x)

=2x

4 b g: (

1, 2]

R ,g(x)

=x 2

Solution

a b

x

4

0 2

(4, 4)

(2, 8)

y

x0

(1, 1)

(2, 4)

y

Range=[8, 4] Range=[0, 4]

Exercise 1B

1 State the domain and range for the relations represented by each of the following graphs:

y

x

x1

0

1

2

1

2

2

ya b c

d e f

x

0

1 2

1

2

y

(2, 4)

(3, 9)

x

(1, 2)

(3, 6)

y

x2 1

1

2

0

y

x

1

2

0

y


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2 Sketch the graph of each of the following relations and state the domain and range of

each:

a {(x,y):y=x 2 + 1} b {(x,y):x2 +y2 =9}c {(x,y): 3x+ 12y=24,x0) d y=

2x

e {(x,y):y=5 x,x[0, 5]} f y=x2

+ 2,x[0, 4]g y=3x 2,x[1, 2] h y=4 x2

3 Which of the following relations are functions? State the domain and range for each.

a {(1, 1), (1, 2), (1, 2), (3, 4), (2, 3)}b {(2, 0), (1,1), (0, 3), (1, 5), (2,4)}c {(1, 1), (1, 2), (2,2), (2, 4), (4, 6)}d {(1, 4), (0, 4), (1, 4), (2, 4), (3, 4)} e {(x, 4):xR}f {(2,y):yZ} g y= 2x+ 4

h y3x+ 2 i {x,y):x2 +y2 =16}

4 Consider the functiong(x)=3x2 2.a Findg(2),g(4). b State the range ofg.

5 Let f(x)=2x2 + 4xandg(x)=2x3 + 2x 6.a Evaluate f(1),f(2) and f(3). b Evaluateg(1),g(2) andg(3).c Express the following in terms ofx:

i f(2x) ii f(x 2) iii g(2x) iv g(x+ 2) v g(x2)

6 Consider the function f(x)=2x 3.Find:

a the image of 3 b the pre-image of 11 c {x :f(x)=4x}7 Consider the functionsg(x)=6x+ 7 andh(x)=3x 2.Find:

a {x :g(x)=h(x)} b {x :g(x) >h(x)} c {x : h(x)=0}

8 Rewrite each of the following using the f:XYnotation:a {(x,y):y=2x+ 3} b {(x,y): 3y+ 4x=12}c {(x,y):y=2x 3,x0} d y=x 2 9,xRe y=5x 3, 0x2

9 Sketch the graphs of each of the following and state the range of each:

a y=x+ 1,x[2,) b y= x+ 1,x[2,)c y=2x+ 1,x[4,) d y=3x+ 2,x(, 3)e y=x+ 1,x(, 3] f y=3x 1,x[2, 6]g y= 3x 1,x[5,1] h y=5x 1,x(2, 4)

10 For f(x)=2x2 6x+ 1 andg(x)=3 2x :a Evaluate f(2),f(3),f(2). b Evaluateg(2),g(1) andg(3).


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c Express the following in terms ofa:

i f(a) ii f(a+ 2) iii g(a) iv g(2a)v f(5 a) vi f(2a) vii g(a) + f(a) viii g(a) f(a)

11 For f(x)

=3x2

+x

2, find:

a {x :f(x)=0} b {x :f(x)=x } c {x :f(x)= 2}d {x :f(x) >0} e {x :f(x) >x } f {x :f(x) 2}

12 For f(x)=x 2 +x , find:a f(2) b f(2) c f(a) in terms ofa d f(a) + f(a) in terms ofae f(a) f(a) in terms ofa f f(a2) in terms ofa

13 Forg(x)=3x 2, find:a {x :g(x)=4} b {x :g(x) >4} c {x :g(x)=a}

d {x :g(x)=6} e {x :g(2x)=4} f x : 1g(x)=6 ,g(x)=014 Find the value ofkfor each of the following if f(3)=3, where:

a f(x)=kx 1 b f(x)=x 2 k c f(x)=x 2 + kx+ 1d f(x)= k

xe f(x)=kx2 f f(x)=1 kx2

15 Find the values ofxfor which the given functions have the given value:

a f(x)=5x 4,f(x)=2 b f(x)= 1x,f(x)=5

c f(x)= 1x2

,f(x)=9 d f(x)=x+ 1x,f(x)=2

e f(x)=(x+ 1)(x 2),f(x)=0

1.3 Types of functions and maximal domains

One-to-one and many-to-one functionsA functionfis said to be one-to-oneif fora, b domf,a=b, then f(a)= f(b). In otherwordsfis called one-to-one if every image underfhas a unique pre-image.

The function f(x)=2x+ 1 is a one-to-one function. The function f(x)=x 2 is not aone-to-one function as, for example, f(3)=9 and f(3)=9; i.e., 9 does not have a unique

pre-image.The function f(x)=5 is not a one-to-one function as there are infinitely many pre-images

of 5.

The function f(x)=x 3 is a one-to-one function.A geometric test for a function to be one-to-one is as follows.

If for anya ranfthe horizontal line,y = a, crosses the graph offat only one point, thefunction is one-to-one.


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y =x2

x0

y

y = 2x + 1

x0

y

f(x)= 5

0

x

y

not one-to-one one-to-one not one-to-one

y =x3

0 x

y

3 30

3

x

y

one-to-one not one-to-one

A function that is not one-to-one is many-to-one.

Implied domains (maximal domains)If the domain is unspecified, then the domain is the largest subset ofRfor which the rule is

defined. When the domain is not explicitly stated, it is implied by the rule.

Thus for the function, f(x)= xthe implied domain (maximal domain) is [0, ). Wewrite:

f: [0,)R ,f(x)= x

Example 8

Find the implied domain of the functions with the following rules:

a f(x)=2

2x 3 b g(x)= 5 xc h(x)= x 5 + 8 x d f(x)=

x2 7x+ 12

Solution

a f(x) is not defined when 2x 3=0, i.e. whenx= 32.

Thus the implied domain isR\

3

2

.


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b g(x) is defined when 5 x0, i.e. whenx5.Thus the implied domain is (, 5].

c h(x) is defined whenx 50 and 8 x0, i.e. whenx5 andx8.Thus the implied domain is [5, 8].

d f(x) is defined whenx2

7x

+12

0.

x2 7x+ 120is equivalent to (x 3)(x 4)0.Therefore,x4 orx3.Thus the implied domain is

(, 3] [4,). x0 1 2 3 4 5

y =x2 7x+ 12

y

Hybrid functions

Example 9

Sketch the graph of the functionfgiven by:

f(x)=

x 1 for x


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Anevenfunction has the property that f(x)= f(x).For example, f(x)=x 2 1 is an even functionsince f(x)=(x)2 1

=x 2 1= f(x)

x

y =x21

1

0

y

The graphs of even functions are symmetrical about they-axis.

The properties of odd and even functions often facilitate the sketching of graphs.

Exercise 1C

1 State which of the following functions are one-to-one:

a {(2, 3), (3, 4), (5, 4), (4, 6)} b {(1, 2), (2, 3), (3, 4), (4, 6)}c {(x,y):y=x 2 + 2} d {(x,y):y=2x+ 4}e f(x)=2 x2 f y=x 2,x1

2 The following are graphs of relations.

a State which are the graphs of a function.

b State which are the graphs of a one-to-one function.

x0

1

2

yi ii iii

x

2

2

0

y

x210

1

2

y

iv v vi

x44

0

y

x0

y

x0

y


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vii viii

x0

y

x0

y

3 The graph of the relation {(x,y):y2 =x+ 2,x 2}is shown. From this relation, form two functions and

specify the range of each.

x012

y

4 a Draw the graph ofg:RR ,g(x)=x 2 + 2.b By restricting the domain ofg, form two one-to-one functions that have the same rule

asg.

5 State the largest possible domain and range for the functions defined by the rule:

a y=4 x b y= x c y=x 2 2 d y=

16 x2e y

=1

xf y

=4

3x2 g y

=

x

3

6 Each of the following is the rule of a function. In each case write down the implied

domain and the range.

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

9 x2 d g(x)= 1x 1

7 Find the implied domain for each of the following rules:

a f(x)= 1x 3 b f(x)=

x2 3 c g(x)=

x2 + 3

d h(x)= x 4 + 11 x e f(x)= x2 1

x+ 1

f h(x)=

x2

x 2 g f(x)=1

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

1

x+ 2i f(x)=

x 3x2 j h(x)=

25 x2 k f(x)= x 3 + 12 x

8 Which of the following functions are odd, even or neither?

a f(x)=x 4 b f(x)=x 5 c f(x)=x 4 3xd f(x)=x 4 3x2 e f(x)=x 5 2x3 f f(x)=x 4 2x5


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9 a Sketch the graph of the function:

f(x)=

2x 2, x


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find:

a f(4) b f(0) c f(4) d f(a+ 3) in terms ofae f(2a) in terms ofa f f(a 3) in terms ofa

15 Given that

f(x)=

x 1, x14, x


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Example 10

Evaluate each of the following:

a i

|3

2

| ii

|3

| |2

|b i42

ii |4||2|c i|6 + 2| ii|6| + |2|

Solution

a i |3 2| = |6| =6 ii |3| |2| =3 2=6Note: |3 2| = |3| |2|b i

42 = |2| =2 ii |4||2| = 42=2

Note: 42 =

|4||2|

c i |6 + 2| = | 4| =4 ii |6| + |2| =6 + 2=8Note: |6 + 2| = | 6| + |2|

Consider two pointsAandBon a number line:

O A B

ba

On a number line the distance between pointsAandBis |a b| = |b a|.Thus|x 2| 3 can be read as on the number line, the distance ofxfrom 2 is less than or equal to

3, and|x | 3 can be read as on the number line, the distance ofxfrom the origin is less thanor equal to 3. Note that |x | 3 is equivalent to 3x3 orx[3, 3].

The graph of the function f:RR ,f(x)= |x| is as shown here.

x

(1, 1)

0

(1, 1)

y

Note that |x | = |x |, i.e. |x | is an even function.

Example 11

Illustrate each of the following sets on a number line and represent the sets using interval

notation.

a {x : |x |


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Solution

a (4, 4)

101234 2 3 4

b (,4] [4,)

024 2 4

c [3, 5]

10123 2 3 4 5

Example 12

Sketch the graphs of each of the following functions and state the range of each of the

functions:

a f(x) = |x 3| + 1 b f(x) = |x 3| + 1

Solution

First, note that |a b| = a b ifa band |a b| = b aifb a.

a f(x) = |x 3| + 1 =

x 3 + 1 if x 3

3 x + 1 if x


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Press ENTERand now in theY= menu

complete asY1= f(x).Note that the expression could be directly

entered in theY=window but this gives you greater

flexibility to do other things with the function ifrequired.

Exercise 1D

1 Evaluate each of the following:

a |5| + 3 b |5|+ |3| c |5| |3|d |5| |3| 4 e |5| |3| |4| f |5|+ |3| |4|

2 On a number line, illustrate each of the following sets and represent the sets using interval

notation:

a {x : |x |


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Solution

a domf domg=[2, 4](f+g)(x)= f(x) +g(x)

=

x 2 +

4 x

dom(f+g)=[2, 4]

b (f+g)(3)=

3 2 +

4 3=2

c (f g)(x)= f(x)g(x)=

(x 2)(4 x)dom(f g)=[2, 4]

d (f g)(3)=

(3 2)(4 3)=1

Exercise 1E

1 For each of the following, find (f+g)(x) and (f g)(x) and state the domain for both f+gandfg:

a f(x)=3x,g(x)=x+ 2b f(x)=1 x2 for allx[2, 2] andg(x)=x 2 for allxR+c f(x)= xandg(x)= 1

xforx[1,)

d f(x)=x 2,x0 andg(x)= 4 x, 0x4

2 Functions f,g, h, andkare defined by:

i f(x)=x2 + 1,xR ii g(x)=x ,xRiii h(x)

=1

x2,x

=0 iv k(x)

=1

x

,x

=0

a State which of the above functions are odd and which are even.

b Form the functions of f+ h,f h,g+ k,gk,f+g,f g, stating which are odd andwhich are even.

1.6 Composite functionsA function may be considered to be similar to a machine for which the input (domain) is

processed to produce an output (range).

For example, the following diagram represents an

f-machine where f(x)

=3x

+2

INPUT

11

3

OUTPUT

f(3) = 3 3 + 2 = 11

f-machine

An alternative diagram is: Domain,R

3 11

Range,Rf


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With many processes, more than one machine operation is required to produce an output.

Suppose an output is the result of one function being applied after another,

e.g., f(x)=3x+ 2followed by g(x)=x 2

This is illustrated diagrammatically on the right. INPUT

OUTPUT

11

3

121

(3) = 3 3 + 2 = 11

g(11) = 112= 121

f-machine

g-machine

A new functionhis formed.

The rule forhish(x)=(3x+ 2)2The diagram shows f(3)=11 and

theng(11)=121.This may be written:

h(3)=g(f(3))=g(11)=121Similarly,h(2)=g(f(2))=g(4)=16

his said to be the compositionofgwithf.This is writtenh =gf(read composition offfollowed byg) and the rule forhis defined

byh(x)=g(f(x)). In the example we have considered:h(x)=g(f(x))

=g(3x+ 2)=(3x+ 2)2

The domain of the functionh=g f= domain off.In general for the composition ofgwithfto be defined, range off domain ofg.When this composition (or composite function) ofgwithfis defined it is denotedgf.

For functionsfandgwith domainsXandYrespectively and such that the range off Y, wedefine the composite function ofgwithf:

g f:XR , whereg f(x)=g(f(x))

x

X = domain off range off

Y = domain ofg

f g

g(f(x))

g(f(x))

f(x)

Example 14

Find both f

gandg

f, stating the domain and range of each where:

f:RR ,f(x)=2x 1 and g:RR ,g(x)=3x2

Solution

To determine the existence of a composite

function, it is useful to form a table of

domains and ranges.

Domain Range

g R R+ {0}

f R R


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fgis defined since rang domf, andgfis defined since ranf domg.

fg(x)= f(g(x))= f(3x2)

=2(3x2)

1

=6x2 1and dom fg= domg=R and ran fg=[1,)

g f(x)=g(f(x))=g(2x 1)=3(2x 1)2=12x2 12x+ 3

domg f= dom f=R

rang f=[0,)

It can be seen from this example that in general fg=g f.

Using a CAS calculatorDefine f(x)=2x 1 andg(x)=3x2. The rulesfor f gandg fcan now be found using f(g(x))andg(f(x)).

Example 15

For the functionsg(x)=2x 1,xRand f(x)= x,x0:a State which of fgandg fis defined.b For the composite function that is defined, state the domain and rule.

Solution

a Range of f domain ofgbut range ofg domain of f.g fis defined but fgis not defined.

Domain Range

g R R

f R+ {0} R+ {0}b g f(x)=g(f(x))=g(x)=2x 1

domg f= dom f=R+ {0}


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Example 16

For the functions f(x)=x 2 1,xR , andg(x)= x,x0:a State whyg

fis not defined.

b Define a restriction foffsuch thatg fis defined and findg f.

Solution

a Range of f domain ofg.g fis not defined. Domain Range

f R [1,)

g R+ {0} R+ {0}b Forg fto be defined,

range of f domain ofg, i.e. range off

R+

{0

}. For range offto be a

subset ofR+ {0}, the domain offmust be restricted to a subset of:

{x :x 1} {x :x1}, orR\(1, 1).

So we define fby:

y =f(x)

x01

1

1

f:R\(1, 1)R ,f(x)=x 2 1g f(x)=g(f(x))

=g(x2 1)=

x2 1

domg f= dom f=R\(1, 1)The composite functiong fis:

g f:R\(1, 1)R ,g f(x)=

x2 1

Compositions involving the modulus functionFunctions with rules of the formy= |f(x)| andy= f(|x |) are considered in this section.

Functions of the form y==|||f(x)||||f| is the compositiong fwhereg(x)= |x|. The functionfis applied first and then themodulus function. The following observation enables the graph of functions with rule of theformy= |f(x)| to be sketched if the graph ofy= f(x) is known:

|f(x)| = f(x) if f(x)0 and |f(x)| = f(x) if f(x)


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Solution

a The graph ofy=x 2 4 is drawnand the negative part reflected in

thex-axis.

x02 2

y

b The graph ofy = 2x 1 isdrawn and the negative part

reflected in thex-axis.

x

y = 1

y = 1

0

y

Functions of the formy==f(|||x|||)The graphs of functions with rules of the form y= f(|x |) wherexRare sketched byreflecting the graph ofy= f(x), forx0, in they-axis. The function with rule f(|x |) is theresult of the composition fgwhereg(x)= |x |.

Example 18

Sketch the graphs of each of the following:

a y= |x |2 2|x | (This is the rule for the function fgwhere f(x)=x 2 2xandg(x)= |x |.)

b y=2|x |(This is the rule for the function fgwhere f(x)=2x andg(x)= |x |.)Solution

a

x2 20

y

The graph ofy=x 2 2x,x0, isreflected in they-axis.

b

x

1

0

y

The graph ofy=2x,x0, isreflected in they-axis.


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Exercise 1F

1 For each of the following, find f(g(x)) andg(f(x)):

a f(x)=

2x

1,g(x)=

2x b f(x)=

4x+

1,g(x)=

2x+

1

c f(x)=2x 1,g(x)=2x 3 d f(x)=2x 1,g(x)=x 2e f(x)=2x2 + 1,g(x)=x 5 f f(x)=2x+ 1,g(x)= |x |

2 For the functions f(x)=2x 1 andh(x)=3x+ 2, find:a f h(x) b h(f(x)) c f h(2) d h f(2)e f(h(3)) f h(f(1)) g f h(0)

3 For the functions f(x)=x 2 + 2xandh(x)=3x+ 1, find:a f h(x) b h f(x) c f h(3)d h

f(3) e f

h(0) f h

f(0)

4 For the functionsh:R\{0} R , h(x)= 1x2

andg:R+R ,g(x)=3x+ 2, find:a hg(state rule and domain) b g h(state rule and domain)c hg(1) d g h(1)

5 fandgare the functions given by f:RR ,f(x)=x 2 4 andg:R+ {0} R ,g(x)= x .a State the ranges offandg. b Find fg, stating its range.c Explain whyg fdoes not exist.

6 Letfandgbe functions given by:

f:R\{0} R ,f(x)= 12

1

x+ 1

g:R\

1

2

R ,g(x)= 1

2x 1Find:

a fg b g f, and state the range in each case

7 The functionsfandgare defined by f:RR , f(x)=x2 2 and

g: {x :x0} R , whereg(x)= x .a Explain whyg fdoes not exist. b Find fgand sketch its graph.

8 a For f(x)=4 xandg(x)= |x |, find fgandg fand sketch the graphs of eachof these functions.

b For f(x)=9 x2 andg(x)= |x |, find fgandg fand sketch the graphs of eachof these functions.

c For f:R\{0) R ,f(x)= 1x

andg:R\{0)R ,g(x)= |x |, find fgandg fand sketch the graphs of each of these functions.


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9 f: {x:x3} R , f(x)=3 xandg:RR ,g(x)=x2 1a Show that fgis not defined.b Define a restrictiong* ofgsuch that fg* is defined and find fg*.

10 f:R+R , f(x)=x 1

2

andg:RR ,g(x)=3 xa Show that fgis not defined.b By suitably restricting the domain ofg, obtain a functiong1such that fg1is defined.

11 Let f:RR , f(x)=x 2 and letg: {x :x3} R ,g(x)= 3 x .State with reasonswhether:

a fgexists b g fexists

12 Let f:SR , f(x)=

4 x2 andSbe the set of all real values ofxfor which f(x) isdefined. Letg:RR , whereg(x)=x2 + 1.

a FindS. b Find the range offand the range ofg.c State whether or not fgandg fare defined and give a reason for each assertion.

13 Letabe a positive number, let f: [2,)R , f(x)=axand letg: (, 1]R ,g(x)=x 2 + a. Find all values ofafor which fgandg fboth exist.

1.7 Inverse functionsIffis a one-to-one function, then for each numberyin the range offthere is exactly one

number,x, in the domain offsuch that f(x)=y .Thus iffis a one-to-one function, a new function f1, called the inverse off, may be

defined by:

f1(x)=yif f(y)=x ,forxran f,y dom f

It is not difficult to see what the relation betweenfand

f1 means geometrically. The point (x,y) is on thegraph of f1 if the point (y,x) is on the graph off.Therefore, to get the graph of f1 from the graphoff, the graph offis to be reflected in the liney=x .

(y,x)

(x,y)

y =x

f1

f

x0

y

From this the following is evident:

dom f1 =ran fran f1 =dom f

A function has an inverse function if and only if it is one-to-one.

We note f f1(x)=x , for allxdom f1f1 f(x)=x , for allxdom f
http://../Activities/inverse_relation.xls


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Example 19

Find the inverse function f1 of the function f(x)=2x 3.

SolutionMethod 1

The graph offhas equationy=2x 3 and the graph of f1 has equationx=2y 3, i.e.xandyare interchanged. Solve fory.

x+ 3=2yand y= 1

2(x+ 3)

f1(x)= 12

(x+ 3)

and dom f1

=ran f

=RMethod 2

We require f1 such that:

f(f1(x))=x2f1(x) 3=x

f1(x)= 12

(x+ 3)

and dom f1 =ran f

=R

Example 20

fis the function defined by f(x)= 1x2

,xR\{0}.Define a suitable restriction for f,f, suchthat f1 exists.

Solution

fis not a one-to-one function. Therefore the inverse functionf1 is not defined. Thefollowing restricted functions offare one-to-one.

f1: (0,)R ,f1(x)=1

x2 Range f1=(0,)

f2: (, 0)R ,f2(x)=1

x2 Range f2=(0,)

Let fbe f1and determine f11. x0

y


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Method 1

Interchangingxandy:

0 1

1

x

f1(x)=1

x2

y = x

f11(x)=

1

x

y

x= 1y2

y2 =1

x

y= 1x

But range f11 =domain f1=(0,)

f11 =1x, ran f11 =(0,) and dom f11 = ran f1=(0,)

f11 : (0,)R ,f11 (x)=1x

Method 2

We require f11 such that:

f1f11 (x)

=x

1f11 (x)

2= x f11 (x)=

1x

But range f11 =domain f1=(0,)

f11 = 1x , ran f11 =(0,) and dom f11 = ran f1=(0,)

f11 : (0,)R ,f11 (x)=1x

Exercise 1G

1 For each of the following, find the rule for the inverse:

a f:R

R ,f(x)

=x

4 b f:R

R ,f(x)

=2x

c f:RR ,f(x)= 3x4

d f:RR ,f(x)= 3x 24

2 Find the inverse of each of the following functions, stating the domain and range for

each:

a f: [2, 6]R ,f(x)=2x 4 b g(x)= 19 x ,x >9

c h(x)=x 2 + 2,x0 d f: [3, 6]R ,f(x)=5x 2e g: (1,)R ,g(x)=x 2 1 f h:R+R , h(x)= x


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3 Consider the functiong: [1,)R ,g(x)=x 2 + 2x .a Findg1, stating the domain and range. b Sketch the graph ofg1.

4 Let f:SR , whereS= {x : 0x3} and f(x)=3 2x . Find f1(2) and the domainof f1.

5 Find the inverse of each of the following functions, stating the domain and range of

each:

a f: [1, 3]R ,f(x)=2x b f: [0,)R ,f(x)=2x2 4c {(1, 6), (2, 4), (3, 8), (5, 11)} d h:RR , h(x)= xe f:RR ,f(x)=x 3 + 1 f g: (1, 3)R ,g(x)=(x+ 1)2g g: [1,)R ,g(x)= x 1 h h: [0, 2]R , h(x)=

4 x2

6 For each of the following functions, sketch the graph of the function and on the same set of

axes sketch the graph of the inverse function. For each of the functions state the rule,

domain and range of the inverse. It is advisable to draw in the line with equationy =xforeach set of axes.

a y=2x+ 4 b f(x)= 3 x2

c f: [2,)R ,f(x)=(x 2)2 d f: [1,)R ,f(x)=(x 1)2

e f: (, 2]R ,f(x)=(x 2)2 f f:R+R ,f(x)= 1x

g f:R+R ,f(x)= 1x2

h h(x)= 12

(x 4)

7 Copy each of the following graphs and on the same set of axes draw the inverse of each of

the corresponding functions:

x

(3, 3)

(0, 0)

ya b

x(2, 1)

(3, 4)

0 1

y c

x

2

0 3

y

d

x0 1

4

y e f

x3

3

0

y

x

y

(1, 1)

(0, 0)

(1, 1)


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g h

x0

y

x02

y

8 Match each of the graphs ofa,b,canddwith its inverse.

a b

x0

y

x0

y

c d

x01

yx=1

x

0

y= 1

y

A B

x

x= 1

0

y

x0

y

C D

x0 y = 1

x0

y


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9 a Let f:AR ,f(x)= 3 x .IfAis the set of all real values ofxfor which f(x) isdefined, findA.

b Letg: [b, 2]Rwhereg(x)=1 x2. Ifbis the smallest real value such thatghas aninverse function, findbandg1(x).

1.8 Applications

Example 21

The cost of a taxi trip in a particular city is $1.75 up to and including 1 km. After 1 km the

passenger pays an additional 75 cents per kilometre. Find the functionfwhich describes this

method of payment and sketch the graph ofy = f(x).

Solution

Letxdenote the length of the trip.

Thenf(x)=

1.75 for 0x11.75 + 0.75(x 1) forx >1

x

4

3

(2, 2.50)

y =f(x)

2

1

0 1 2 3 4

y

Example 22

A rectangular piece of cardboard has dimensions

18 cm by 24 cm. Four squares eachxcm byxcm

are cut from the corners. An open box is formed

by folding up the flaps.

Find a function forV, which gives the volume of

the box in terms ofx, and state the domain of the

function.

18 cm

x

x

24 cm

Solution

The dimensions of the box will be:

24 2x, 18 2xandx

The volume of the box is determined by the function:

V(x)=(24 2x)(18 2x)x


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For the box to be formed:

24 2x >0 and 18 2x >0 andx >0

Therefore,x0 and 12 12x9 >0

x>0 andx


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Exercise 1H

1 The cost of a taxi trip in a particular city is $4.00 up to and including 2 km. After 2 km the

passenger pays an additional $2.00 per kilometre. Find the functionfwhich describes this

method of payment and sketch the graph ofy= f(x), wherexis the number of kilometrestravelled. (Use a continuous model.)

2 A rectangular piece of cardboard has dimensions 20 cm by 36 cm. Four squares eachxcm

byxcm are cut from the corners. An open box is formed by folding up the flaps. Find a

function forV, which gives the volume of the box in terms ofx, and state the domain for

the function.

3 The dimensions of an enclosure are shown. The

perimeter of the enclosure is 160 m.

x m

20m

12 m

y m

a Find a rule for the area,Am

2

, of the enclosurein terms ofx.

b State a suitable domain of the functionA(x).

c Sketch the graph ofAagainstx.

d Find the maximum possible area of the enclosure

and state the corresponding values ofxandy.

4 A cuboid tank is open at the top and the internal

dimensions of its base arexm and 2xm.

The height ishm.

The volume of the tank isVcubic metres and the

volume is fixed. LetSm2 denote the internal

surface area of the tank.

x m2xm

hm

a FindSin terms of:

i xandh ii Vandx

b State the maximal domain for the function defined by the rule ina ii.

c If 2


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view


Chapter summary

Arelationis a set of ordered pairs.

Thedomainof a relationSis the set of all first elements of the ordered pairs inS. The

domain offis denoted bydom f.

Therangeof a relationSis the set of all the second elements of the ordered pairs inS. The

range offis denoted by ran f.

Afunctionis a relation such that no two ordered pairs of the relation have the same first

element.

For eachxin the domain of a functionfthere is a unique element,y, in the range such that

(x,y) f. The elementyis called the imageofxunderfor thevalueoffatxand is denotedby f(x).

A functionfis said to be one-to-oneif fora,b domf,a=b, then f(a)= f(b).A function which is not one-to-one is many-to-one.

If the domain of a function is unspecified, then the domain is the largest subset ofRfor

which the rule is defined. This set is called theimplied domainormaximal domainof the

function rule.

A functionfisoddif f(x)= f(x) for allxin the domain off.A functionfisevenif f(x)= f(x) for allxin the domain off.Letfandgbe functions such that dom f domg= . Thesum f+g, and the product,

fg, as functions on dom f domgare defined by:

(f+g)(x)= f(x) +g(x) and (f g)(x)= f(x) g(x)

Iffis a one-to-one function, a new function, f1, called theinverseoff, may be defined by:

f1(x)=yif f(y)=x , forxranf,ydomf

For a functionfand its inverse f1:

dom f1 = ran fran f1 = dom f

The inverse of any relation may be defined. The inverse relation is not a function unless the

initial relation is a one-to-one function.

For a relationS

={(a,b)}the inverse relation is{(b,a)}.

The modulus or absolute value of a real numberxis denoted by |x| and is defined by:

|x | =x ifx0x ifx


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iew


The function |x | has the following properties: |ab| = |a||b|

a

b = |a|

|b

| |a+ b| |a| + |b|. Ifaandbare both non-negative or both non-positive, then equalityholds.

Ifa0, |x | ais equivalent to axa Ifa0, |x k| ais equivalent tok axk+ a The composition of functionsfandgis denoted byfg. The rule is given byf(g(x)).

The domain offgis the domain ofg. The compositionfgis defined if therange ofg the domain off.

Multiple-choice questions

1 For the function with rule f(x)= 6 2x , which of the following is the maximal domain?A (, 6] B [3,) C (, 6] D (3,) E (, 3]

2 For f: [1, 3)R ,f(x)= x2, the range is:A R B (9, 0] C (, 0] D (9,1] E [9, 0]

3 For f(x)=3x2 + 2x,f(2a)=A 20a2 + 4a B 6a2 + 2a C 6a2 + 4a D 36a2 + 4a E 12a2 + 4a

4 For f(x)=2x 3,f1(x)=A 2x+ 3 B 1

2x+ 3 C 1

2x+ 3

2 D

1

2x

3 E

1

2x 3

5 For f: (a, b]R ,f(x)=10 xwherea


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9 For which one of the following functions does an inverse function not exist?

A f:RR ,f(x)=2x 4 B g: [4, 4]R ,g(x)=

16 x2C h: [0,)R , h(x)= 1

5x2 D p:R+R ,p(x)= 1

x2

E q:RR , q(x)=2x3

510 The graph of the functionfwith ruley= f(x) is

shown on the right.

x

2

20

2

2

y

Which one of the following is most likely to be the graph of the inverse function off?

A

x

2

0 22

2

y B

x

2

202

2

y C

x2

2

0

2

2

y

D

x2

2

0

2

2

y E

2

0

2

2

x

y

2

Short-answer questions (technology-free)

1 Sketch the graph of each of the following relations and state the implied domain and

range:

a f(x)=x 2 + 1 b f(x)=2x 6 c {(x,y):x2 +y2 =25}d {(x,y):y2x+ 1} e {(x,y):y


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iew


2 For the functiong: [0, 5]R ,g(x)= x+ 32

:

a Sketch the graph ofy=g(x). b State the range ofg.c Findg1,stating the domain and range ofg1.

d Find{x:g(x)=4}. e Find{x :g1

(x)=4}.3 Forg(x)=5x+ 1,find:

a {x :g(x)=2} b {x :g1(x)=2} cx :

1

g(x)=2

4 Sketch the graph of the functionf, for which

f(x)=

x+ 1 for x >2x2 1 for 0x2x2 forx


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4 The base of a 3 m ladder leaning against a wall isxmetres from the wall.

a Express the distance,d, from the top of the ladder to the ground as a function ofxand

sketch the graph of the function.

b State the domain and range of the function.

5 A car travels half the distance of a journey at an average speed of 80 km/h and half at an

average speed ofxkm/h.

Define a function,S, which gives the average speed for the total journey as a function ofx.

6 A cylinder is inscribed in a sphere with a radius of length 6 cm.

For the cylinder:

a Define a function,V1, which gives the

volume of the cylinder as a function of

the height (h).

(State the rule and domain.)

b Define a function,V2, which gives the

volume of the cylinder as a function of

the radius of the cylinder (r).

(State the rule and domain.)

h

r

7 Let f:RRandg:RR , where f(x) =x + 1 andg(x)=2 +x3.a State whyg fexists and findgf(x).b State whyg fis a function and find (g f)1(10).

8 A functionfis defined as follows:

f(x)=x2 4, forx(, 2)x, forx[2,)

a Sketch the graph off.

b Find the value of:

i f(1) ii f(3)c Giveng:SRwhereg(x)= f(x), find the largest setSso that the inverse ofgexists

and1S.d Ifh(x)=2x , find f(h(x)) andh(f(x)).

9 Find the rule for the area,A(t), enclosed by the graph of the function:

f(x)=

3x, 0x13, x >1

thex-axis, they-axis and the vertical linex=t(t0). State the domain and range of thefunction.


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iew


10 Let f:R\d

c

R ,f(x)= ax+ b

cx+ da Find the inverse function f1.b Find the inverse function when:

i a=3, b=2, c=3, d=1 ii a=3, b=2, c=2, d= 3iii a=1, b= 1, c= 1, d= 1 iv a= 1, b=1, c=1, d=1

c Determine the possible values ofa,b,canddif f= f1.11 The radius of the incircle of the right-angled triangle

ABCisrcm. Find:

X

A Z B

Y

x cm

r cm

r cm

r cm

3cm

C

a i YBin terms ofr

ii ZBin terms ofr

iii AZin terms ofrandx

iv CY

b Use the geometric resultsCY=C XandA X= A Zto find an expression forrin termsofx.

c i Findrwhenx=4. ii Findxwhenr=0.5.d Use a CAS calculator to investigate the possible valuesrcan take.

12 Let f(x)= px+ qx+ r wherexR\{r, r}.

a If f(x)=

f(

x) for allx, show that f(x)=

p forx

R\{

r, r}.

b If f(x)= f(x) forx=0, find the rule for f(x) in terms ofq.c Ifp=3, q=8 andr= 3:

i Find the inverse function off.

ii Find the values ofxfor which f(x)=x .13 a Let f(x)= x+ 1

x 1.i Find f(2),f(f(2)),f(f(f(2))). ii Find f(f(x)).

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

Find f(f(x)),f(f(f(x))).


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C H A P T E R

2Revising linear functionsand matrices

ObjectivesTorevise:

methods forsolving linear equations

methods for solvingsimultaneous linear equations

calculating the gradientof a straight line

interpreting and using the general equationof a straight line: y= mx+ c a method for determining the gradient of a line perpendicularto a given line

finding thedistance between two points

finding themidpointof a straight line

calculating theangle between two intersecting straight lines matrix arithmetic

To apply a knowledge of linear functions to solving problems.

It is assumed that the material in this chapter has been covered by students inEssential

Mathematical Methods 1 & 2. The chapter provides a framework for revision with worked

examples and practice exercises.

2.1 Linear equations

Exercise 2A

1 Solve the following linear equations:

a 3x 4 = 2x+ 6 b 8x 4 = 3x+ 1 c 3(2 x) 4(3 2x) = 14

d3x

4 4 = 17 e 6 3y= 5y 62 f 2

3x 1=3

7

42
http://../Activities/simultaneous_Equations.xls


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Chapter 2 Revising linear functions and matrices 43

g2x 1

3 = x+ 1

4h

2(x 1)3

x+ 42

= 56

i 4y 3y + 42

+ 13= 5(4 y)

3j

x+ 12x 1=

3

4

2 Solve each of the following pairs of simultaneous linear equations:

a x 4=y4y 2x=8

b 9x+ 4y=132x+y=2

c 7x=18 + 3y2x+ 5y=11

d 5x+ 3y=137x+ 2y=16

e 19x+ 17y=02xy=53

fx

5+ y

2=5

xy=4

3 An aircraft, used for fire spotting, flies from its base to locate a fire at an unknown

distance,xkm away. It travels straight to the fire and back, averaging 240 km/h for the

outward trip and 320 km/h for the return trip. If the plane was away for 35 minutes, find

the distance,xkm.

4 A group of hikers is to travelxkm by bus at an average speed of 48 km/h to an unknown

destination. They then plan to walk back along the same route at an average speed of

4.8 km/h and to arrive back 24 hours after setting out in the bus. If they allow 2 h for

lunch and rest, how far must the bus take them?

5 The length of a rectangle is 4 cm more than the width. If the length were to be decreased

by 5 cm and the width decreased by 2 cm, the perimeter would be 18 cm. Calculate the

dimensions of the rectangle.

6 In a basketball game a field goal scores two points and a free throw scores one point. John

scored 11 points in the game and David 19 points. David scored the same number of free

throws as John but twice as many field goals. How many field goals did each score?

7 The weekly wage, $w, of a salesman consists of a fixed amount of $800 and then $20 for

each unit he sells.

a If he sellsnunits a week, find a rule for his weekly wage,w, in terms of the number of

units sold.

b Find his wage if he sells 30 units.

c How many units does he sell if his weekly wage is $1620?

8 Water flows into a tank at a rate of 15 litres per minute. At the beginning the tankcontained 250 litres.

a Write an expression for the volume,Vlitres, of water in the tank at timetminutes.

b How many litres of water are there in the tank after an hour?

c The tank has a capacity of 5000 litres. How long does it take to fill?

9 A tank contains 10 000 litres of water. Water flows out at a rate of 10 litres per minute.

a Write an expression for the volume,Vlitres, of water in the tank at timetminutes.

b How many litres of water are there in the tank after an hour?

c How long does it take for the tank to empty?


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10 The cost, $C, of hiring diving equipment is $100 plus $25 per hour.

a Write a rule which gives the total charge, $C, of hiring the equipment forthours

(assume that parts of hours are paid for proportionately).

b Find the cost of hiring the equipment for:

i 2 hours ii 2 hours 30 minutesc For how many hours can the equipment be hired if the following amounts are

available?

i $375 ii $400

2.2 Linear literal equations and simultaneouslinear literal equationsA literal equation inxis an equation whose solution will be expressed in terms of pronumerals

rather than numbers.

2x+ 5 = 7 is an equation whose solution is the number 1.In the literal equationax+ b= c, the solution isx= c b

aLiteral equations are solved in the same way as solving numerical equations or transposing

formulas. Essentially, the literal equation is transposed to makexthe subject.

Example 1

Solve the following forx.

a px q= r b ax+ b= cx+ d c ax

= b2x

+ c

Solution

a px q= rpx= r+ q

x= r+ qp

b ax+ b=cx+ dax cx=d b

x(a c)=d bx= d b

a cc

a

x= b

2x+ c

Multiply both sides of the equation by 2x .

2a=b + 2xc2a b=2xc2a b

2c=x

Simultaneous literal equationsSimultaneous literal equations are solved by the methods of solution of simultaneous

equations, i.e. substitution and elimination.


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Example 2

Solve each of the following pairs of simultaneous equations forxandy.

a y

=ax

+c

y=bx+ db ax

y

=c

x+ by=d

Solution

a ax+ c=bx+ d ax bx=d c

x(a b)=d cx= d c

a band therefore

y=a d ca b+ c= ad bc

a b

b axy= c . . . (1)x+ by= d. . . (2)

Multiply (1) byb:

abx by= cb . . . (1)Add (1) and (2):

abx+x= cb + dx(ab + 1) = cb + d

x= cb + dab + 1

Substitute in (1) :

a

cb + dab + 1

y= c

y= a

cb + dab + 1

c

=

ad cab

+1

Exercise 2B

1 Solve each of the following forx:

a ax+ n= m b ax+ b= bxc

ax

b+ c= 0 d px= qx+ 5

e mx+ n= nx m f 1x+ a =

b

x

gb

x

a= 2b

x+

ah

x

m+ n= x

n+ m

i b(ax+ b) = a(bx a) j p2(1 x) 2pq x= q2(1 +x)

kx

a 1 = x

b+ 2 l x

a b +2x

a + b =1

a2 b2

mp qx

t+p= qx t

pn

1

x+ a +1

x+ 2a =2

x+ 3a

2 For the simultaneous equationsax+ by=p andbx ay= q , show thatx= ap + bqa2 + b2

andy= bp aqa2 + b2


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3 For the simultaneous equationsx

a+ y

b= 1 andx

b+ y

a= 1,show thatx=y= ab

a + b4 Solve each of the following pairs of simultaneous equations forxandy.

a ax+y=cx+ by=d

b ax by=a2

bx ay=b2c ax+ by=t

ax by=sd ax+ by=a2 + 2ab b2

bx+ ay=a2 + b2e (a + b)x+ cy=bc

(b + c)y + ax= abf 3(x a) 2(y + a)=5 4a

2(x+ a) + 3(y a)=4a 1

5 Writesin terms ofaonly in the following pairs of equations:

a s=ahh=2a + 1

b s=ahh=a(2 + h)

c as=

a+

h

h + ah= 1d as

=s

+h

ah=a + he s=h2 + ah

h=3a2f as= a + 2h

h= a s

g s= 2 + ah + h2

h= a 1a

h 3s ah=a2as+ 2h=3a

2.3 Linear coordinate geometryThe following is a summary of the material that is assumed to have been covered inEssential

Mathematical Methods 1 & 2.

Gradient of a straight line joining two points

x

B(x2,y2)

A(x1,y1)

0

yGradient,m= y2 y1

x2 x1

The general equation of a straight liney= mx+ c

wheremis the gradient andcis the value of the intercept on they-axis.
http://../Activities/Gradient.pps


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Equation of a straight line passing through agiven point (x1, y1) and having a gradient, m

x

P(x,y)

A(x1,y1)

0

yEquation of line is:

y y1= m(xx1)

Equation of a line passing through two givenpoints (x1, y1) and (x2, y2)

(x,y)

x

B(x2,y2)

A(x1,y1)

0

yEquation of line is:

y y1= m(xx1) wherem= y2 y1

x2 x1

The intercept form of the equation of astraight line

x0

(0, b)

(a, 0)

yFor a line passing through the points (a, 0) and (0,b),

the equation is:

x

a +y

b = 1

Tangent of the angle of slope ()

tan = y2 y1x2

x1

where is the angle the line makes with the positive direction of thex-axis.

Product of gradients of two perpendicularstraight linesIf two straight lines are perpendicular to each other, the product of their gradients is 1.

m1m2= 1


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Distance between two points

x0

B(x2,y2)

A(x1,y1)

y

AB =

(x2 x1)2 + (y2 y1)2

MidpointThe midpoint of a straight line joining (x1,y1) and (x2,y2) is the point

x1 +x2

2 ,

y1 +y22

.

The angle between intersecting straight lines

x

1

0

2

y

= 2 1

Example 3

A fruit and vegetable wholesaler sells 30 kg of hydroponic tomatoes for $148.50 and 55 kg of

hydroponic tomatoes for $247.50. Find a linear model for the cost,Cdollars, to buyxkg of

hydroponic tomatoes. How much would 20 kg of tomatoes cost?

Solution

Let (x1, C1) = (30, 148.5) and (x2, C2) = (55,247.5).The equation of the straight line is given by:

C C1= m(xx1) where m=C2 C1x2 x1

Nowm= 247.5 148.555 30 = 3.96 so C 148.5 = 3.96(x 30)

Therefore the straight line is given by the equationC=

3.96x+

29.7

Substitutex= 20: C= 3.96 20 + 29.7 = 108.9It would cost $108.90 to buy 20 kg of tomatoes.

Exercise 2C

1 Find the coordinates ofM, the midpoint ofAB, whereAandBhave the following

coordinates:

a A(1, 4),B(5, 11) b A(6, 4),B(1, 8) c A(1, 6),B(4, 7)


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2 Usey= mx+ cto sketch the graph of each of the following:a y= 3x 3 b y= 3x+ 4 c 3y + 2x= 12d 4x+ 6y= 12 e 3y 6x= 18 f 8x 4y= 16

3 Find the equations of the following straight lines:

a gradient + 2, passing through (4, 2) b gradient 3, passing through (3, 4)c passing through the points (1, 3) and (4, 7)

d passing through the points (2, 3) and (2, 5)

4 Use the intercept method to find the equation of the straight lines passing through:

a (3, 0) and (0, 2) b (4, 0) and (0, 6)

c (4, 0) and (0, 3) d (0, 2) and (6, 0)

5 Write each of the following in intercept form and hence draw their graphs:

a 3x+

6y=

12 b 4y

3x=

12

c 4y 2x= 8 d 32x 3y= 9

6 A printing firm charges $35 for printing 600 sheets of headed notepaper and $46 for

printing 800 sheets. Find a linear model for the charge,Cdollars, for printingnsheets.

How much would they charge for printing 1000 sheets?

7 An electronic bank teller registered $775 after it had counted 120 notes and $975 after it

had counted 160 notes.

a Find a formula for the sum registered, ($C), in terms of the number of notes (n)

counted.b Was there a sum already on the register when counting began?

c If so, how much?

8 Find the distance between each of the following pairs of points:

a (2, 6), (3, 4) b (5, 1), (6, 2) c (1, 3), (4, 5)d (1, 7), (1, 11) e (2, 6), (2, 8) f (0, 4), (3, 0)

9 a Find the equation of the straight line which passes through the point (1, 6) and is:

i parallel to the line with equationy= 2x+ 3ii perpendicular to the line with equationy

=2x

+3

b Find the equation of the straight line which passes through the point (2, 3) and is:

i parallel to the line with equation 4x+ 2y= 10ii perpendicular to the line with equation 4x+ 2y= 10

10 Find the equation of the line which passes through the point of intersection of the lines

y=xandx+y= 6 and which is perpendicular to the line with equation 3x+ 6y= 12.

11 The length of the line joiningA(2, 1) andB(5,y) is 5 units. Findy.


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12 Find the equation of the line passing through the point (1, 3) which is:a parallel to the lines with b perpendicular to the lines with equations below.

equations below

i 2x+ 5y 10 = 0 ii 4x+ 5y + 3 = 0

13 Find the angle that the lines joining the given points make with the positive direction of

thex-axis:

a (4, 1), (4, 6) b (2, 3), (4, 6) c (5, 1), (1, 8) d (4, 2), (2, 8)

14 Find the acute angle between the linesy= 2x+ 4 andy= 3x+ 6

15 Given the pointsA(a, 3),B (2, 1) andC(3, 2), find the possible value ofaif the lengthofABis twice the length ofBC.

16 Three points have coordinatesA(1, 7),B(7, 5) andC(0, 2). Find:

a the equation of the perpendicular bisector ofABb the point of intersection of this perpendicular bisector andBC.

17 The point (h, k) lies on the liney =x + 1 and is 5 units from the point (0, 2). Write downtwo equations connectinghandkand hence find the possible values ofhandk.

18 PandQare the points of intersection of the liney

2+ x

3= 1 with thexandyaxes

respectively. The gradient ofQRis1

2, whereRis the point withx-coordinate 2a,a >0.

a Find they-coordinate ofRin terms ofa.

b Find the value ofaif the gradient ofPRis 2.

19 The figure shows a triangleABCwith

A(1, 1), andB(1, 4).The gradients ofAB,ACandBCare

3m, 3mandmrespectively.

x

C

A(1, 1)

B(1, 4)

0

y

a Find the value ofm.

b Find the coordinates ofC.

c Show thatAC= 2AB.

20 In the rectangleABCD,AandBare the points (4, 2)

and (2, 8) respectively. Given that the equation of

ACisy=x 2, find:

x0

B

A

D

C

y

a the equation ofBC

b the coordinates ofC

c the coordinates ofD

d the area of rectangleABCD.


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21 ABCDis a parallelogram, lettered anticlockwise, such thatAandCare the points (1, 5)and (5, 1) respectively.

a Find the coordinates of the midpoint ofAC.

b Given thatBDis parallel to the line whose equation isy + 5x = 2, find the equation of

BD.c Given thatBCis perpendicular toAC, find:

i the equation ofBC ii the coordinates ofB iii the coordinates ofD.

2.4 Applications of linear functions

Example 4

There are two possible methods for paying gas bills:

method A: a fixed charge of $25 per quarter+ 50c per unit of gas usedmethod B: a fixed charge of $50 per quarter+ 25c per unit of gas used.

Determine the number of units which must be used before method B becomes cheaper than

method A.

Solution

Let C1=charge in $ using method AC2=charge in $ using method Bx=number of units of gas used.

NowC1=25 + 0.5xC2=50 + 0.25x

It can be seen from the graph that if the

number of units exceeds 100, then

method B is cheaper.

C2= 0.25x+ 50

C1= 0.5x+ 25

0

50

100

C ($)

25 50 75 100 125 150 x(units)

The solution could also be obtained by solving simultaneous linear equations:

C1=C225 + 0.5x=50 + 0.25x

0.25x=25x=100

Exercise 2D

1 A car journey of 300 km lasts 4 h. Part of this journey is on a freeway at an average speed

of 90 km/h. The rest is on country roads at an average speed of 70 km/h.

LetTbe the time (in hours) spent on the freeway.

a In terms ofT, state the number of hours travelling on country roads.


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b State:

i the distance travelled on the freeway in terms ofT

ii the distance travelled on country roads in terms ofT.

c Find:

i T ii the distance travelled on each type of road.

2 A farmer measured the quantity of water in a storage tank 20 days after it was filled and

found it contained 3000 litres. After a further 15 days it was measured again and found to

have 1200 litres in it. Assume that the amount of water in the tank decreases at a constant

rate.

a Find the relation betweenL, the number of litres of water in the tank, andt, the number

of days after the tank was filled.

b How much water does the tank hold when it is full?

c Sketch the graph ofLagainsttfor a suitable domain.

d State this domain.e How long does it take for the tank to empty?

f At what rate does the water leave the tank?

3 On a small island two rival taxi firms have the following fare structures:

firm A: fixed charge of $1 plus 40 cents per kilometre

firm B: 60 cents per kilometre, no fixed charge.

a Find expressions forCA, the charge of firm A in terms ofn, the number of

kilometres travelled, andCB, the charge of firm B in terms of the number of kilometres

travelled.

b Sketch the graph of the charges of each of the firms against the number of kilometrestravelled on the one set of axes.

c Find the distance for which both firms charge the same amount.

d On a new set of axes sketch the graph ofD= CA CBagainstnand explain what thisgraph represents.

4 A boat leaves fromOto sail to two islands. The boat arrives at a pointAon Happy Island

with coordinates (10, 22.5) (units are in kilometres).

x

Happy Island

Sun Island

B

A

O

ya Find the equation of the line through points

OandA.

b Find the distanceOAto the nearest metre.

The boat arrives at Sun Island at pointB .

The coordinates of pointBare (23, 9).

c Find the equation of lineAB.

d A third island lies on the perpendicular bisector

of line segmentAB. Its port is denoted byC. It is

known that thex-coordinate ofCis 52.

Find they-coordinate of the pointC.


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5 ABCDis a parallelogram with

coordinatesA(2, 2),B(1.5, 4) andC(6, 6).

xO

D

C

B

A

y

a Find the gradient of:

i lineAB

ii lineADb Find the equation of:

i lineBC

ii lineCD

c Find the equation of the diagonalsACandBD.

d Find the coordinates of the point of intersection

of the diagonals.

2.5 Review of matrix arithmeticA matrix is a rectangular array of numbers.

Two matrices AandBare equal when: each has the same number of rows and the same number of columns they have the same number or element at corresponding positions.

3 6 56 10 612 1 0

=

a b cd e fg h i

impliesa= 3, b= 6, c= 5, etc.

The size ordimensionof a matrix is described by specifying the number of rows (m) and

the number of columns (n). The dimension is writtenm n. The dimensions of thefollowing matrices in order are:

3 2, 1 4, 3 3, 1 11 23 4

5 6

2 1 5 6

2 3

0 0 12 0

[5]

Addition will be defined for two matrices only when they have the same dimension. The

sum is found by adding corresponding elements.a b

c d

+

e f

g h

=

a + e b + fc +g d+ h

Subtraction is defined in a similar way.

IfAis anm nmatrix andkis a real number,kAis defined to be anm nmatrix whoseelements arektimes the corresponding element ofA.

k

a b

c d

=

ka kb

kc kd


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IfAis anm n matrix andBis ann rmatrix, then the productABis them rmatrix

whose entries are determined as follows.

To find the entry in rowiand columnjofAB, single out rowi in matrixA and columnj

in matrixB. Multiply the corresponding entries from the row and column and then add up

the resulting products.

The productAB is defined only if the number of columns ofA is the same as the

number of rows ofB.

ForA =

2 3 4

5 6 7

andB =

4 01 2

0 3

Ais a 2 3 matrix andB is a 3 2 matrix. Therefore AB is a 2 2 matrix.

AB =

2 3 4

5 6 7

4 0

1 2

0 3

=

2 4+ 3 1+ 4 0 2 0+ 3 2+ 4 3

5 4+ 6 1+ 7 0 5 0+ 6 2+ 7 3

=

11 18

26 33

The identity matrix for the family ofn n matrices is the matrix with ones in the

top left to bottom right diagonal and zeros in all other positions. This is denoted by I.

IfA andB are square matrices of the same dimension andAB = BA= IthenA is said to

be the inverse ofBandBis said to be the inverse ofA. The inverse ofAis denoted byA1.

If A = a bc d

thenA1 = d

ad bc

b

ad bc

cad bc

aad bc

det (A) = ad bc is thedeterminantof matrixA.

A square matrix is said to beregularif its inverse exists. Those square matrices which do

not have an inverse are calledsingularmatrices. The determinant of a singular matrix

is 0.

Using a CAS calculatorThere are two ways of entering matrices.

Matrix Editor

Press the APPS key and select Data/Matrix Editor

and then3:New.

Complete the screen as shown opposite.

From the Type menu select2:Matrix.

Call this first matrixa and define it as a 2 2

matrix.


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Press ENTER to obtain the edit screen.

Note that the status in the top left of the screen

is nowMat2 2.The entries are made in the usual way. This is

defined as matrixa.

The matrixb =

3 6

5 6.5

is defined in a similar way.

Return to the home screen.

The two matrices can be viewed by enteringaand

thenbin the entry line.

Entering matrices in the home screen

This can be done row by row. For the matrix

3 66 7

enter [[3, 6][6, 7]] and press ENTER.

The matrix can be named by using STO.

In the entry line [[3, 6][6, 7]] a

Use1:Definefrom the F4 menu to definecas shown.

Addition, subtraction and multiplying

by a scalar

Onceaandbare defined as abovea + b, a bandkacan be determined in the home screen.

Multiplication

Multiplication ofA =

3 6

6 7

andB =

3 6

5 6.5

.

ABandBAare shown.

Inverse and determinant

The operation of matrix inverse is obtained by entering

a(1) in the entry line. The operation determinant isobtained through theMATHmenu,

which is obtainable by pressing 2nd 5 and selecting

4:Matrixand then the appropriate operation.


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Example 5

For the matrixA =

3 2

1 6

find:

a det(A) b A1 c XifAX = 5 67 2

d YifYA = 5 6

7 2

Solution

a det(A) = 3 6 2 = 16 b A1 = 116

6 21 3

c AX =

5 6

7 2

Multiply both sides (from the left) by A1.

A1

AX = A1 5 6

7 2

IX = X = 116

6 21 3

5 6

7 2

= 1

16

16 32

16 0

=

1 2

1 0

d YA =

5 6

7 2

Multiply both sides (from the right) by A1.

YAA1=

1

165 6

7 2 6 21 3

YI = Y = 116

24 8

40 8

Y =

3

2

1

25

2

12

Exercise 2E

1 For the matricesA =

2 1

3 2

andB =

2 2

3 2

find:

a A + B b AB c BA d A B e kA f 2A + 3Bg A 2B h det(A) i A1 j det(B) k B1 l det(AB)


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2 Find the inverse of the following regular matrices (kis any non-zero real number):

a

3 14 1

b

3 1

2 4

c

1 0

0 k

3 IfA,Bare the regular matrices A = 2 10 1,B = 1 03 1, findA1, B1.Also findABand hence find, if possible, (AB)1.

Also find fromA1, B1, the productsA1B1 andB1A1. What do you notice?

4 Consider the matrixA =

4 3

2 1

.

a FindA1. b IfAX =

3 4

1 6

, findX. c IfYA =

3 4

1 6

, findY.

5 IfA = 3 21 6

,B = 4 1

2 2

andC = 3 4

2 6

, find:

a Xsuch thatAX + B = C b Ysuch thatYA + B = C6 IfAis a 2 2 matrix,a12= a21= 0, a11= 0, a22= 0 then show that Ais regular and

findA1.

7 ForA =

3 6 56 10 6

12 1 0

andB =

2 1 20 1 6

2 1 0

find, using a calculator:

a A

+B b AB c BA d 3A e B1

f A B g A 2B h A + 3B i (AB)1 j B1A1

8 ForA =

3 6 5 610 6 12 1

0 4 1 6

0 0 1 1

andB =

0 1 1 1

2 1 2 0

1 6 2 1

0 7 8 0

find, using a calculator:

a AB b BA c 3A d B1 e A B f A 2B9 Solve each of the following matrix equations forX:

a AX = C b AX + D = C

whereA =2 1 20 1 6

2 1 0

,C =20

2

andD =10

3

10 Solve each of the following matrix equations forX:

a AX = C b AX + D = C

whereA =

0 1 1 1

2 1 2 0

1 6 2 1

0 7 8 0

,C =

1

0

2

3

andD =

10

3

1


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2.6 Solving systems of linear simultaneousequations in two variablesInverse matrices can be used to solve certain sets of simultaneous linear equations. Consider

the equations:

3x 2y= 55x 3y= 9

This can be written as:3 25 3

x

y

=

5

9

IfA =

3 25 3

the determinant ofAis 3(3) 5(2) = 1 which is not zero and so A1

exists.

A1 =3 25 3

Multiplying the matrix equation

3 25 3

x

y

=

5

9

on both sides on the left-hand side

byA1 and using the fact that A1A = Iyields the following:

A1

A

x

y

= A1

5

9

I

x

y

= A1

5

9

x

y

=

3

2

sinceA1

5

9

=

3

2

This is the solution to the simultaneous equations.

Check by substitutingx= 3,y= 2 in the equations.When dealing with simultaneous linear equations in two variables which represent straight

lines that are parallel, then a singular matrix results.

For example the system

x+ 2y=32x 4y=6


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has associated matrix equation 1 2

2 4

x

y

=

3

6

Note that the determinant of 1 22 4 = 1 4 (2 2) = 0

There is no unique solution to the system of equations.

Example 6

Solve the simultaneous equations

3x 2y=67x+ 4y=7

Solution

The matrix equation is

3 27 4

xy

=

67

Let A =

3 27 4

Then A1 = 126

4 2

7 3

and

x

y

= 1

26

4 2

7 3

6

7

= 1

26

38

21

Using a CAS calculator

Enter

3 27 4

(1)

6

7

Remember that a 2 2 matrix is said to be singular if its determinant is equal to 0. The matrixbeing singular can correspond to one of two situations:

1 There are infinitely many solutions.

2 There is no solution.


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Example 7

Explain why the simultaneous equations 2x+ 3y= 6 and 4x+ 6y= 24 have no solution.

SolutionThe equations have no solution as they

correspond to parallel lines and they

are different lines. There is no

point of intersection.

Each of the lines has

gradient 23

. x

y

4

2

4x+6y= 24

2x+3y= 6

3 60

The matrix of the coefficients ofxandyis

2 3

4 6

and the determinant of this

matrix is 0. That is, the matrix is singular.

If, for a system of two linear equations with two variablesxandy, the 2 2 matrix of thecoefficients ofxandyis singular, then the system has either no solutions as discussed above or

infinitely many solutions. This second case arises when the two equations represent the same

line.

Example 8

The simultaneous equations 2x+ 3y= 6 and 4x+ 6y= 12 have infinitely many solutions.Describe these solutions through the use of a parameter.

Solution

The parameter is a third variable. Note that the two equations represent the same

straight line. They both have gradient 23

andy-axis intercept 2.

The1:Solve(command from the Algebra menu will be used. Rememberandis

obtained fromCATALOG.

Let be this third variable. In this case lety= . Thenx= 3( 2)2

and the

line can be described by3 ( 2)2

,

: R

This may seem to make the situation unnecessarily

complicated but it is the solution given by the calculator

as shown opposite. The symbol 1 takes the place of.

The 1 indicates that it is the first parameter used

in this session with the calculator.


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Example 9

Consider the simultaneous linear equations

(m 2)x + y = 2 andm x + 2y = k

Find the values ofm andksuch that the system of equations has:

a a unique solution b no solution c infinitely many solutions

Solution

Using a CAS calculator to find the solution

x =4 k

m 4and y =

k(m 2) 2m

m 4

a The solution is unique ifm = 4 andkcan be any real number.

b Ifm = 4, the equations become

2x + y = 2 and 4x + 2y = k

There is no solution ifm = 4 andk= 4.

c Ifm = 4 andk= 4 there are infinitely many solutions as the equations are the

same.

This method of expressing a solution generalises to the more complicated situation in three

dimensions. This is also discussed in the next section.

Again it is noted that for a system of linear equations in two unknowns, the matrix of the

coefficients ofx andy being singular corresponds to either no solutions (parallel lines) or

infinitely many solutions (the same line).

Exercise 2F

1 Solve each of the pairs of simultaneous linear equations using matrix methods:

a 3x + 2y = 6 andx y = 7 b 2x + 6y = 0 andy x = 2

c 4x 2y = 7 and 5x + 7y = 1 d 2x y = 6 and 4x 7y = 5

2 Explain why the simultaneous equations 2x + 3y = 6 and 4x + 6y = 10 have no solution.

3 The simultaneous equationsx y = 6 and 2x 2y = 12 have infinitely many solutions.

Describe these solutions through the use of a parameter.

4 Find the value ofm for which the simultaneous equations

(m + 3)x + my = 12

(m 1)x + (m 3)y = 7

have no solution.

5 Find the value ofm for which the simultaneous equations

3x + my = 5 and (m + 2)x + 5y = m have:

a an infinite number of solutions

b no solutions


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6 Consider the simultaneous equations

mx+ 2y= 84x (2 m)y= 2m

a Find the values ofmfor which there are:

i no solutions ii infinitely many solutionsb Solve the equations in terms ofm, for suitable values ofm.

7 a Solve the simultaneous equations 2x 3y= 4 andx+ ky= 2, wherekis a constant.b Find the value ofkfor which there is not a unique solution.

8 Find the values ofbandcfor which the equationsx+ 5y= 4 and 2x+ by= chave:a a unique solution b an infinite set of solutions c no solution

2.7 Simultaneous linear equations with

more than two variablesConsider the general linear system of three equations in three unknowns:

ax+ by + cz= dex+ f y +gz=hkx+ my + nz=p

It can be written as a matrix equation:

a b c

e f g

k m n

x

y

z

=

d

h

p

LetA =

a b ce f g

k m n

, X =

xy

z

andB =

dh

p

The equation isAX = B

We recall that for 3 3 matricesI =

1 0 0

0 1 0

0 0 1

andDI = D = IDfor all 3 3 matricesD.

If the inverseA1 exists (this is not always the case) the equation can be solved bymultiplyingAX, andB, on the left by A1, andA1(AX) = A1BandA1(AX) = (A1A)X = IX = X, whereIis the identity matrix for 3 3 matrices.

HenceX = A1B, which is a formula for the solution of the system. Of course it dependson the inverse A1 existing, but onceA1 is found then equations of the formAX = Bcan besolved for all possible 3 1 matricesB.

In this course you are not required to find the inverse of a 3 3 matrix by hand but anunderstanding of matrix arithmetic is necessary.


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Example 10

Consider the system of three equations in three unknowns:

2x

+y

+z

= 1

3y + 4z= 76x+z= 8

Using matrix methods to solve the system of equations.

Solution

Enter 3 3 matrixAand 3 1 matrixBinto the calculator.

A =

2 1 1

0 3 4

6 0 1

, X =

x

y

z

andB =

17

8

The equations can be written as the matrix equation:

AX = BMultiply both sides byA1.

A1AX=A1BIX=A1B

X=A1B

X=

1

52

Hencex= 1,y= 5 andz= 2

It should be noted that, just as for two equations in two unknowns, there is a geometric

interpretation for three equations in three unknowns. There is only a unique solution if the

equations represent three planes intersecting at a point.

A CAS calculator can be used to solve systems of three equations in the same way that was

used for two simultaneous equations. The solution of equations without CAS is considered in

Exercise 2G.

Example 11

Solve the linear simultaneous equations

xy +z= 6, 2x+z= 4, 3x+ 2y z= 6 forx,yandz.

Solution

Use the1:solve(command from the Algebra menu.

In the entry line enter1:solve(xy +z= 6and2x+z= 4and3x+ 2y z= 6, {x,y,z}).

The solution isx= 143

, y= 203

andz= 163


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As a linear equation in two variables defines a line, a linear equation in three variables

defines a plane. The coordinate axes in three dimensions are drawn as shown. The point

P(2, 2, 4) is as marked.

x

2

4P(2, 2, 4)

z

y

2

An equation of the formax+ by + cy= e

defines a plane. For example the equation

xy +z= 6 corresponds to the graph shown here.

Press MODE and from the Graphsubmenu

choose5:3D. Enterz1 = 6 x+y. This ischosen by makingzthe subject from the equation.

The arrow keys may be used to look at the graph

from different perspectives.

The solution of simultaneous linear equations in three variables can correspond to

a point a line a planeThere also may be no solution.

The situations are as shown:

Diagram 1 Diagram 2 Diagram 3

Intersection at a point Intersection, a line No intersection

Diagram 4 Diagram 5

No common intersection No common intersection


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Example 12

The simultaneous equationsx + 2y + 3z= 13,x 3y + 2z= 2 andx 4y + 7z= 17

have infinitely many solutions. Describe these solutions through the use of a parameter.

Solution

The equations have no unique solution. For example, the point (9, 5, 4) satisfies all

three equations but it is certainly not the only one. We use the CAS calculator to find

the solution in terms of a fourth variable,.

Letz= , then y = 5( 3) andx = 43 13

If = 4,x = 9,y = 5 andz= 4.

Note that as zincreases by 1 theny increases by 5

andx decreases by 13. All of the points which

satisfy the equations lie on a straight line. The

situation is similar to that shown in diagram 2 on the previous page. The calculator

uses the parameter 2 for the parameter.

Exercise 2G

1 Solve each of the following sets of simultaneous equations using matrix methods:

a 2x + 3y z= 12

2y + z= 7

2y z= 5

b x + 2y + 3z= 13

x y + 2z= 2

x + 3y + 4z= 26

c x + y = 5

y + z= 7

z+ x = 12

d x y z= 0, 5x + 20z= 50, 10y 20z= 30

2 Consider the simultaneous equationsx + 2y 3z= 4 andx + y + z= 6

a Subtract the second equation from the first to findy in terms ofz.

b Letz= . Solve the equations to give the solution in terms of.

3 Consider the simultaneous equations

x + 2y + 3z= 13 (1)

x 3y + 2z= 2 (2)

x 4y + 7z= 17 (3)

a Add equation (2) to equation (1) and subtract equation (2) from equation (3).

b Comment on the equations obtained ina.

c Letz= and findy in terms of.

d Substitute forzandy in terms of in equation (1) to findx in terms of.


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4 Solve each of the following pairs of simultaneous equations, giving your answer in terms of

a parameter. Use the technique introduced in Question 2.

a xy +z= 4x+y +z= 6

b 2xy +z= 6xz= 3

c 4x 2y +z= 6x+y +z= 4

5 The system of equationsx+y +z+ w= 4,x+ 3y + 3z= 2,x+y + 2z w= 6 hasinfinitely many solutions. Describe this family of solutions and give the unique solution

whenw= 6.

6 Find all solutions of the following sets of equations:

a 3xy +z= 4 andx+ 2y z= 2 andx+y z= 2b xy z= 0 and 3y + 3z= 5c 2xy +z= 0 andy + 2z= 2

7 a Find the inverse of the matrix2 a 13 4

(a

+1)

10 8 a 4 in terms ofa.

b For what values ofadoes the inverse not exist?

c Find the value ofafor which there are infinitely many solutions to the equations

2x+ ay z= 03x+ 4y (a + 1)z= 1310x+ 8y + (a 4)z= 26

d For the value ofafound inc, solve the equations.

8 Find a value ofpfor which the system of equations

3x+ 2y z= 1 andx+y +z= 2 andpx+ 2y z= 1

has more than one solution.

(Hint:Find the inverse of the matrix of coefficients in terms ofp.)

Solve the system of equations for this value ofp.


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Rev

iew


Chapter summary

Gradient of a straight line joining two points (x1,y1) and (x2,y2) ism=y2 y1x2 x1

.

Different forms for the equation of a straight line:

y= mx+ c wheremis the gradient andcis they-axis intercepty y1= m(xx1) wheremis the gradient and (x1,y1) is a point on the linex

a+ y

b= 1 where (a, 0) and (0,b) are points on the line intersecting the axes.

Tangent of the angle of slope (), tan = y2 y1x2 x1

, where is the angle the line makes with

the positive direction of thex-axis.

If two straight lines are perpendicular to each other, the product of their gradients is 1,i.e.m1m2= 1.Distance between two points (x1,y1) and (x2,y2)

= (x2 x1)2

+(y2

y1)

2.

Midpoint of a straight line joining (x1,y1) and (x2,y2) is the point

x1 +x2

2 ,

y1 +y22

.

The angle between intersecting straight lines is as shown:

x

1

0

2

y

= 2 1

A matrix is a rectangular array of numbers.

Two matrices AandBare equal when: each has the same number of rows and the same number of columns they have the same number or element at corresponding posi

essentials maths methods 12

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