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Sets, Vectors & Functions. IGCSE Chapter 8. Note 1 : Sets. A ∩ B is green. A. B. ∩ - intersection U – Union - ‘is a subset of’. A. B. ∩. B. A. ∩. A B. Note 1 : Sets. X. a. C - ‘is a member of’ ‘belongs to’ b є X ξ – ‘universal set’ ξ = { a,b,c,d,e }

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note 1 sets
Note 1: Sets

A ∩ B is green

A

B

∩ - intersection

U – Union

- ‘is a subset of’

A

B

B

A

A B

note 1 sets1
Note 1: Sets

X

a

C - ‘is a member of’

‘belongs to’

bєX

ξ – ‘universal set’

ξ = {a,b,c,d,e}

- ‘complement of’

b

c

ξ

d

a

c

b

e

ξ

B

A ‘

A

A U A’ = ξ

note 1 sets2
Note 1: Sets

A

a

n(A) - ’the number of elements

in set A’

A = {x : x is an integer, 2 ≤ x ≤ 9}

Reads: A is the set of elements x such that x is an integer and 2 ≤ x ≤ 9

b

n(A) = 3

c

The set A is {2,3,4,5,6,7,8,9}

Ø or { } - ‘empty set’

Ø A for any set A

note 1 sets3
Note 1: Sets

ξ

T

R

e.g. In the venn diagram

8

5

11

ξ = {students in year 10}

9

13

15

R = {Yr 10 students who play Rugby}

12

17

V = {Yr 10 students who play Volleyball}

V

T = {Yr 10 students who play Tennis}

a.) How many students play Rugby?

b.) How many students do not play Volleyball?

c.) How many students play Rugby and Tennis?

d.) How many students in total?

e.) How many play only 1 of these 3 sports?

40

41

14

90

31

IGCSE

Ex 1 Pg 242-243

slide6

e.g.

If X = {1,2,3,..10}

ξ

Y = {1,3,5,…19}

Y

X

13

1

4

Z = {x : x is an integer, 5≤x≤11}

15

3

17

Find:

2

5

7

19

10

{1,3,5,7,9}

9

a.) X ∩ Y

b.) Y ∩ Z

c.) X ∩ Z

d.) n(X U Y)

e.) n(Z)

f.) X’ U Y’

11

6

8

{5,7,9,11}

{5,6,7,8,9,10}

Z

15

7

Ø or { }

State whether true or false:

False

a.) 7 εX ∩ Y

b.) {5,7,9,11} Z

c.) Z X U Y

True

IGCSE

Ex 2 Pg 244

True

slide7

e.g.

Draw and shade this diagram to show the following sets:

a.) X∩ Y

b.) (X U Y)’

c.) X’ ∩ Y

ξ

X

Y

IGCSE

Ex 3 Pg 245-246

slide8

e.g.

Logical Problems

If A = { sheep }

B = { sheep dogs }

C = { ‘intelligent animals’ }

D = { good pet }

Express the following in set language:

a.) No sheep are ‘intelligent’ animals

b.) All sheep dogs make good pets

c.) Some sheep make good pets

A ∩ C = Ø

B D

A ∩ D = Ø

Interpret the following statements

a.) B C

b.) B U C = D

c.) AεD

All sheep dogs are intelligent animals

All sheep dogs and all intelligent animals make good pets

Sheep do not make good pets

slide9

e.g.

Logical Problems

Of 27 students in the class, 18 play chess, 15 play piano and 7 do both. How many do neither?

C

P

27 students = 11 + 7 + 8 + X

7

11

8

27 – 11 – 7 – 8 = X

X = 1

1

There is only 1 student who does not play either piano or chess

slide10

e.g.

Logical Problems

The math results from the Hockey Team show that all 16 players passed at least 2 subjects, 8 passed at least 3 subjects and 5 students passed 4 subjects or more

ξ

How many passed exactly 2 subjects?

What fraction passed exactly 3 subjects?

3

2

3

8

8

5

4

3/16

IGCSE

Ex 4 Pg 247-249

note 2 vectors
Note 2: Vectors

A vector is a quantity that has both magnitude and direction

Vectors can be added using scale drawings.

Head to Tail method:

a

a

b

a + b

Therefore:

a + b = b + a

b

* Notice that this is the same result as b + a

note 2 vectors1
Note 2: Vectors

A scalar is a quantity that has magnitude but no direction.

e.g. ordinary numbers, & quantities like temperature, mass & volume.

We can multiply a vector by a scalar.

e.g.x multiplied by 2 gives 2x

x

x

x

x

x

x

-3x

2x

e.g.x multiplied by -3 gives -3x

* The negative sign reverses the direction of the vector

note 2 vectors2
Note 2: Vectors

The result of a – b is the same as a + -b

e.g.Find 3a – b

a

a

a

a

a

a

a

a

b

b

b

b

3a – b

e.g.Find -4a + 2b

-4a + 2b

note 2 vectors3
Note 2: Vectors

d

Starting from Eeach time, find vectors for the following:

c

e.g. Find:

2c

3c + d

-c + d

-d

c – 4d

-2c – 4d

EG

EA

EC

EF

EL

EK

C

A

B

E

F

D

G

J

H

I

M

L

K

note 2 vectors4
Note 2: Vectors

d

The result of a – b is the same as a + -b

c

e.g. Find:

DE

DG

HJ

GB

BE

GC

CG

= 2c

C

A

B

= c – 2d

F

E

D

= c– 2d

= 2c + 3d

= -c – d

G

H

= 4c + 3d

= -4c – 3d

J

I

note 2 vectors5
Note 2: Vectors

Write each vector in terms of a, b and c

e.g.

  • AB

DE

DC

HD

FE

BE

EA

DG

A

C

2a

a

= -2a

a

B

D

= -a-c

b

c

c

= -4a

F

= 2a+c

b

= b +a

a

a

G

H

E

= -2b

( c )

= -2b + 2a

= 2b – 2a

note 2 vectors6
Note 2: Vectors

Write each vector in terms of a, bonly

e.g.

DE

HD

A

C

2a

a

a

B

D

= -3a+ 2b

b

c

c

= 4a– 2b

F

b

a

a

G

H

E

IGCSE

Ex 5 Pg 251-252

starter
Starter

OA = a and OB = b

e.g. Using the figure, express each of the following vectors in terms of a and/or b

AP = OA

OB = BQ

NP = QN

= a

a.) AP

b.) AB

c.) OQ

d.) PO

e.) PQ

f.) PN

g.) ON

h.) AN

i.) BP

j.) QA

= -a + b

= 2b

= -2a

= -2a + 2b

= ½PQ

= -a + b

Q

= 2a + PN

= a + b

b

= a + PN

= b

B

N

b

= -b + 2a

A

P

O

= -2b + a

a

a

note 2 vectors7
Note 2: Vectors

OA = a and OB = b

e.g. Using the figure, express each of the following vectors in terms of a and/or b

AP = 3OA

OB = 1/2BQ

NP = QN

= 3a

a.) AP

b.) AB

c.) OQ

d.) PO

e.) PQ

f.) PN

g.) ON

h.) AN

i.) BP

j.) QA

= -a + b

= 3b

= -4a

= -4a + 3b

= ½PQ

= -2a + 3/2b

Q

= 4a + PN

= 2a + 3/2b

b

= 3a + PN

= a + 3/2b

N

b

B

= -b + 4a

b

A

P

O

= -3b + a

a

a

a

a

note 2 vectors8
Note 2: Vectors

ABCDEF is a regular hexagon with AB representing the vector s and AF representing the vector t. Find the vector representing AC

t

s

BC

= ?

= s + t

C

B

s

= AB + BC

AC

A

D

t

= s + s + t

E

IGCSE

Ex 6 Pg 253-255

F

= 2s + t

note 3 column vectors
Note 3: Column Vectors

The vector AB can be written as a column vector

( )

2

Horizontal Component (Movement in x direction)

AB =

3

Vertical Component (Movement in y direction)

F

C

B

( )

( )

( )

-2

5

0

-3

3

0

E

A

H

G

D

note 3 column vectors1
Note 3: Column Vectors

We can easily add column vectors.

AB + CD =

AB =

CD =

B

= ( )

( )

( )

( )

( ) +

3

7

3

7

10

C

-3

-3

2

2

-1

A

AB + CD

D

* A vector is described by its length and direction, NOT its position

note 3 column vectors2
Note 3: Column Vectors

Similarly, subtracting column vectors.

AB − CD =

AB =

CD =

C

= ( )

( )

( )

( )

( ) −

6

5

6

5

1

-2

-2

1

1

3

AB – CD

B

D

A

* A vector is described by its length and direction, NOT its position

note 3 column vectors3
Note 3: Column Vectors

Multiplying by a scalar

2AB =

Each component is multiplied by 2

AB =

= ( )

2( )

( )

6

12

6

1

2

1

B

2AB

A

* A vector is described by its length and direction, NOT its position

note 3 column vectors4
Note 3: Column Vectors

Vectors are parallel if they have the same direction

i.e. Both components must be in the same ratio

e.g. Is parallel to

IGCSE

Ex 7 Pg 257-258

e.g. Is parallel to

( )

( )

k( )

( )

( )

( )

-10

12

5

a

6

a

4

2

1

b

-2

b

In general, a vector is parallel if it is a scalar multiple

Is parallel to

note 3 column vectors5
Note 3: Column Vectors

y

If A has the coordinate (1,2) and B has the coordinate (6,4), find the column vector for AB

B

AB =

A

( )

( )

5

-5

2

-2

Find the column vector for BA

x

BA =

note 3 column vectors6
Note 3: Column Vectors

y

Given the following diagram, what would be the coordinate of point A such that ABCD is a parallelogram.

C

D

A = (1,0)

( )

( )

1

4

1

3

B

Notice the column vector for

AB = DC

x

A

AD = BC

note 3 column vectors7
Note 3: Column Vectors

y

y = - x

y = x

Find the image of the vector after reflection in the following lines

a.) y = 0

b.) x = 0

c.) y = x

d.) y = -x

x

( )

( )

( )

( )

( )

-5

3

5

-5

5

5

-3

-3

3

3

IGCSE

Ex 8 Pg 258-259

note 4 modulus of a vector
Note 4: Modulus of a Vector

The modulus of a vector a is written a .

This represents the magnitude (or length) of the vector

As shown in the diagram:

B

a =

( )

5

3

3

We can find the length using Pythagoras’ Theorem

A

5

|a|= √(52+32)

|a|= √34 units

note 4 modulus of a vector1
Note 4: Modulus of a Vector

e.g. If AB = & BC = , find AC

AB + BC =

C

B

=

( )

( )

( ) +

( )

( )

1

-4

5

-4

5

3

3

1

1

4

We can find the length using Pythagoras’ Theorem

A

IGCSE

Ex 9 Pg 260-261

|AC|= √(12+42)

|AC|= √17units

note 5 vector geometry
Note 5: Vector Geometry

Position vectors give the position of a point relative to the origin, O.

In the diagram we see that

OD = 2OA, OE = 4OB, OA = a and OB = b

b.) Express BA in terms of a and b

a.) Express OD and OE in term of a and b respectively

d.) Given that BC = 3BA, express OC in terms of a and b

c.) Express ED in terms of a and b

C

OD = 2a

OE = 4b

BA = -b + a

ED = -4b + 2a

OC = OB + BC

D

A

a

b

O

E

B

OC = b + -3b +3a

OC = -2b +3a

note 5 vector geometry1
Note 5: Vector Geometry

Position vectors give the position of a point relative to the origin, O.

In the diagram we see that

OD = 2OA, OE = 4OB, OA = a and OB = b

IGCSE

Ex 10 Pg 262-264

e.) Express EC in terms of a and b

C

EC = -6b + 3a

EC = -4b – 2b + 3a

EC = EO + OC

EC = -4b + OC

ED = -4b + 2a

From part (d)

D

A

a

e.) Show that E, D and C lie on a straight line

b

O

E

B

From part (c)

OC = -2b +3a

EC and ED are parallel vector that pass through the same point, E

slide34

Note 6: Functions

Function Notation

f(x) = x2+ 3 or f : xx2 +3

We can read this as “the function f, such that, x is mapped on x2 +3

If f(x) = 7x – 5 and g(x) = 2x2 +5, find:

a.) f(2) b.) f(-4) c.) g(3) d.) g(0)

f(2) = 7(2) - 5

f(-4) = 7(-4) - 5

g(3) = 2(3)2 + 5

g(0) = 0 + 5

f(-4) = -33

g(2) = 23

g(2) = 5

f(2) = 9

slide35

Note 6: Functions

If f : x and g : x √[5(x+2)], find:

a.) x if f(x) = 40 b.) x if g(x) = 10

5x2

2

√[5(x+2)] = 10

= 40

[5(x+2)] = 102

5x2 = 2 x 40

5(x+2) = 100

5x2 = 80

(x+2) = 20

x2 = 16

x = 18

x = ± 4

slide36

Note 6: Functions (flow diagrams)

We can illustrate functions using flow diagrams.

e.g.The function (7x + 8)2 consists of 3 simpler functions

7x

x

7x+8

(7x+8)2

multiply by 7

add 8

square

slide37

Note 6: Functions

e.g. The functions h and k are defined as follows:

h: x x2 + 1, k:x ax + b, whereaandbare constants.

If h(0) = k(0) and k(2) = 15, find values of aandb

h(0) = 02 + 1

= 1

k(2) = 15

a(2) + 1 = 15

IGCSE

Ex 11 Pg 265-267

k(0) = 1

a(0) + b = 1

2a + 1= 15

2a = 14

a = 7

b = 1

slide38

Note 7: Composite Functions

Consider f(x) = x4and g(x) is 2x + 3

fg means that g converts x to 2x + 3, ….. and then……

f converts 2x + 3 to (2x + 3)4

slide39

Note 7: Composite Functions

Consider f(x) = x4and g(x) is 2x + 3

fg means that g converts x to 2x + 3, ….. and then……

f converts 2x + 3 to (2x + 3)4

Notice how this is different from gf

gf means that f converts x to x4 , ….. and then……

g converts x4to 2(x4) + 3

You must get used to reading and applying the composite function from right to left

In general, fg=gf

slide40

Note 7: Composite Functions

Other common notations show fg as f(g(x)) and gf as g(f(x))

e.g. Given f(x) = 2x + 1 and g(x) = 3 – 4x, find in simplest form:

a.) fg b.) gf

= f(g(x))

= g(f(x))

=2(3 – 4x) + 1

=3 – 4(2x + 1)

=6 – 8x +1

=3 – 8x – 4

=7 – 8x

=– 8x – 1

slide41

Note 7: Composite Functions

e.g. Given f:x x3

g:x 4 + 5x

h:x 2x

Find:

a.) fg b.) ghc.)fgh

= g(h(x))

=fg(h(x))

= f(g(x))

=4 + 5(2x)

=(4 + 5x)3

=f(4 + 5(2x))

=4 + 10x

=f(4 + 10x)

d.) gf(2)

d.) ff(-1)

=(4 + 10x)3

= g(f(x))

=(x3)3

=x9

=4 + 5x3

=(-1)9

=4 + 5(2)3

=44

=-1

slide42

Note 7: Composite Functions

e.g. Given f:x 2x – 8

g:x x + 4

h:x 7x2

Find:

a.) x if fg= 0 b.) ggg(5)c.) x if f(x) = g(x)

2x – 8 = x + 4

g ((x+4) + 4))

f(g(x)) = 0

= g (x+ 8)

x = 4 + 8

2(x + 4) – 8 = 0

x = 12

= (x+4) + 8

2x + 8 – 8 = 0

= x + 12

2x = 0

= 5 + 12

x = 0

IGCSE

Ex 12 Pg 268-269

#1-8

= 17

slide43

Note 8: InverseFunctions

Reversing (undoing) a function

The operations: + and –

× and ÷

x2and √x

are all inverse operations because one undoes the other

If a function f maps a number n onto m then its inverse function f-1 will map m onto n.

f-1

recall column vectors
Recall: Column Vectors

y

y = x

Find the image of the vector after reflection in the following lines

a.) y = 0

b.) x = 0

c.) y = x

d.) y = -x

x

( )

( )

( )

( )

( )

5

-5

5

-5

3

3

5

-3

3

-3

slide45

Note 8: InverseFunctions

We can find the inverse of a function using a flow diagram

f(x) =

4x+5

x

x

4x

3x

3x-5

Multiply by 3

Subtract 5

Divide by 4

Now, start on the right hand side and replace each operation with its inverse

Add 5

Multiply by 4

Divide by 3

Thus, f-1(x) =

you try
You try!

Find the inverse of f(x) = 4x-2

*4

-2

x

4x-2

(x+2)/4

/4

+2

x

So f -1 (x)=(x+2)/4

IGCSE

Ex 12 Pg 268-269

#9-22

a neat little trick
A neat little trick…
  • As always in maths, there is a trick to this…
  • Write function as a rule in terms of y and x.
  • Swap ‘x’ and ‘y’
  • Rearrange to get in terms of y.

f(x) = 5x + 7

y = 5x + 7

x = 5y + 7

x -7 = 5y

y = (x-7)/5

f-1(x) =

slide49
Inverse functions only exist for one-one functions.

i.e. functions where each input (x) can

only lead to one possible output ( f(x) )

ad