Success Additional Mathematics SPM Free Chapte

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Year

Paper

Number of questions 3

1

0

2

3

1

0

2

2

1

1

2

3

1

0

2

3

1

0

2

3

1

0

2

3

1

0

2

3

1

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2004 2005 2006 2007 2008 2009 2010 2011

SPM Topical Analysis

RELATIONS

HUBUNGAN

FUNCTIONS

FUNGSI

Types of Relations Jenis Hubunganbull One-to-onebull Many-to-one

bull One-to-manybull Many-to-many

Absolute Value FunctionsFungsi Nilai Mutlak

f x rarr|g( x )|

Composite FunctionsFungsi GubahanThe function of f

followed by g is gf

Graphs of Absolute Value FunctionsGraf Fungsi Nilai Mutlak The graph of a linear absolutevalue function has a V shape

Problems that involve CompositeFunctions and Inverse FunctionsMasalah melibatkan Fungsi Gubahandan Fungsi Songsang

Inverse FunctionsFungsi SongsangGiven that y = f ( x )

then f ndash1( y ) = x

DomainDomain

CodomainKodomain

Objects

Objek

ImagesImej

Range Julat

1 Functions

CHAPTER FORM 4

ONCEPT MAP

Learning ObjectivesCOMPANION W EBSITE



Functions 2

Facts1 A relation from set A to set B is the linking (or pairing) of theelements of set A to the elements of set B

2 A relation between two sets can be represented by

(a) an arrow diagram (b) ordered pairs (c) a graph

11 Relations

1

A relation from set A = 12 14 23 25 43 to set B = 3 5 7 is defined by

lsquosum of digits of rsquo Represent the relation by

(a) an arrow diagram(b) ordered pairs

(c) a graph

Solution

(a) Arrow diagram

3

5

7

12

14

23

25

43

AB

lsquosum of digits ofrsquo

(b) Ordered pairs

(12 3) (14 5) (23 5) (25 7) (43 7)

(c) Graph7

5

3

12 14 23 25 43Set A

Set B

Try Question 1 Self Assess 11

In a relation between set ( A) and another set (B)bull the first set ( A) is known as domainbull the second set (B) is known as codomainbull the elements in the domain are known as objectsbull the elements in the codomain that are linked to the objects are known

as imagesbull the set of images is known as range

A set is a well-defined collectionof objects

For exampleUniversal set = x 10 x 30

x is an integerSet A = Factors of 36Set B = Prime numbersSet C = Numbers where the sum

of the digits is 3The list of elements of each of thesets A B and C is as follows = 10 11 12 13 14 15 16 17

18 19 20 21 22 23 2425 26 27 28 29 30

A = 12 18B = 11 13 17 19 23 29C = 12 21 30

11a Representation of a Relation

11b Domain Codomain Objects Images and Range



3 Functions

2


A relation from set P = 16 36 49 64 to set

Q = 4 6 7 8 11 is defined by lsquo factor(s) of rsquo

(a) Represent this relation using an arrow diagram

(b) State

(i) the domain

(ii) the codomain

(iii) the images of 36 (iv) the images of 64

(v) the object of 7

(vi) the objects of 4

(vii) the range

of this relation

Solution

(a) The arrow diagram that represents the given

relation is as shown in the next column

16

36

49

64

4

6

7

8

11

P Qlsquofactor(s) ofrsquo

Range

Non-imageDomain Codomain

(b) (i) The domain is 16 36 49 64

(ii) The codomain is 4 6 7 8 11

(iii) The images of 36 are 4 and 6

(iv) The images of 64 are 4 and 8

(v) The object of 7 is 49

(vi) The objects of 4 are 16 36 and 64

(vii) The range is 4 6 7 8

11c Types of Relations 3

State the type of relation shown by each of the

following arrow diagrams

(a) (b)

A Blsquoreciprocal ofrsquo

3

4

5

1

3

1

4

1

5

10

21

2

3

5

7

lsquofactors ofrsquo P Q

Solution(a) One-to-one relation

(b) One-to-many relation


4

Draw arrow diagrams to represent the relationsbecause it is easier to interpret arrow diagramscompared to ordered pairs or graphs

State the type of relation shown by

(a) the ordered pairs

(4 8) (4 10) (4 12) (6 10) (7 12)

(b) the graph

Set X

Set Y

M

P

5 7 9 12 15Solution

Types of Relations

1 One-to-one relation

Each object in the

domain has only one

image in the

codomain

2 Many-to-one relation

There are more than

one object in the

domain that have the

same image in the

codomain

3 One-to-many relation

There is at least one

object in the domain

that has more than

one image in thecodomain

4 Many-to-many relation There is at least one


that has more than

one image in the

codomain and there is

at least one element in

the codomain that is

linked to more than

one object in the

domain

Domain Codomain

Domain Codomain

Domain Codomain

Domain Codomain



Functions 4

11

1 Represent each of the following relations by

(a) an arrow diagram

(b) ordered pairs

(c) a graph

(i) The relation lsquoremainder when divided

by 5rsquo from set A = 6 12 18 24 to set

B = 1 2 3 4

(ii) The relation lsquo factors of rsquo from set P to

set Q where P = Q = 2 3 4 8

(iii) The relation lsquodifference of digits of rsquofrom set X = 14 21 23 34 48 to set

Y = 1 3 4

2 A relation from set P = 26 34 45 62 to set

Q = 12 20 24 48 is defined by lsquo product of

digits of rsquo

(a) Represent the relation by an arrow diagram

(b) State

(i) the domain

(ii) the codomain

(iii) the image of 45

(iv) the objects of 12

(v) the range

of the relation

3 State the type of relation shown by each of the


(a) M N lsquocapital ofrsquo

Sarawak

Perak

Kedah

Kuching

Ipoh

Alor Setar

(b) P Qlsquoelements ofrsquo

methane

water

carbondioxide

carbon

hydrogen

oxygen

4 State the type of relation of


(4 0) (4 1) (5 2) (6 7) (6 8)


(a)4

6

7

8

10

12

there4 The above arrow diagram shows

many-to-many relation

(b) X Y

5

7

9

12

15

P

M


many-to-one relation

1

The following ordered pairs represent a relation

from set A = 3 5 7 to set B = 9 15 21

(3 9) (5 15) (7 21)

(a) State the type of the above relation

(b) State the range

(c) Using function notation write down a relationbetween set A and set B

Solution

3

5

7

9

15

21

A B

Based on the arrow diagram

(a) the relation is a one-to-one relation(b) the range is 9 15 21

(c) the function notation is

f ( x) = 3 x


SPMClone

rsquo07

SPMClone

rsquo06

Each image is three timesits objectsIt is better to draw an arrow diagram to represent

the above relation because it is easier to interpretan arrow diagram compared to ordered pairs



5 Functions

(b) the graph

3 4 6 7 9

Set B

Set A

11

10

8

5 The following ordered pairs represent a relation

from set P = 2 6 10 to set Q = 1 3 5

(2 1) (6 3) (10 5)


(b) State the range

(c) Using function notation write down a

relation between set P and set Q

1 A function is a special type of relation where each and every object inthe domain has only one image

2 The types of relations that are considered as functions are(a) one-to-one relation (b) many-to-one relation

x

y

z

a

b

c

a

b

c

d

w

x

y

z

3 The main properties for a relation to be considered a function are(a) Each and every object must have one and only one image [If there are objects that have two or more images the relation is not a

function](b) Two or more objects are allowed to have the same image [ Many-to-one relation is considered a function](c) It is not necessary that all elements in the codomain are images [ Non-image elements are allowed in a function]

4 If a function links x to x2 ndash 3 x + 5 by using function notation it is written as f x rarr x2 ndash 3 x + 5 or f ( x) = x2 ndash 3 x + 5 It is read as lsquofunction f

that maps x onto x2 ndash 3 x + 5rsquo

12 Functions

Facts

Function is also known asmapping

Facts

The types of relations that are not

considered as functions are(a) one-to-many relation

a

b

x

y

z

[Not a function becauseelement lsquoarsquo has more thanone image]

(b) many-to-many relation

a

b

c

x

y

z

[Not a function becauseelement lsquobrsquo has more thanone image]

(c) relation where there areelements in the domainthat do not have an image

x

y

z

a

b

c

d

[Not a function becauseelement lsquodrsquo does not havean image]

5

Explain whether each of the following relations is a function

(a) (b) (c)

A Blsquohasrsquo

January

June

August

28 days

30 days

31 days

P Qlsquoconsists of rsquo

Bronze

Brass

cuprum

stanum

zinc

1

3

57

3

9

15

M N lsquothree times ofrsquo

Solution

(a) It is a function because

bull each and every object has only one image

bull many-to-one relation is allowed

bull non-image element ie lsquo28 daysrsquo is allowed

(b) It is not a function because there are objects that have more than one image

(c) It is not a function because there is an element in the domain that does not

have an image ie lsquo7rsquo

12a To Recognise Functions




Functions 6

6

Try Questions 2ndash3 Self Assess 12

Given the function f x rarr x 2 ndash 4 x ndash 1 find

(a) the image of 2

(b) the objects that have the image 4

Solution(a) f x rarr x 2 ndash 4 x ndash 1

f ( x ) = x 2 ndash 4 x ndash 1

When x = 2

f (2) = 22 ndash 4(2) ndash 1

= 4 ndash 8 ndash 1

= ndash5

Hence the image of 2 is ndash5

(b)

When f ( x ) = 4

x 2 ndash 4 x ndash 1 = 4


( x + 1)( x ndash 5) = 0

x = ndash1 or 5

Hence the objects that have the image 4 are

ndash1 or 5

Object

12b To Solve Problems involving Functions

7

The arrow diagram below shows the function

15 f x rarr mdashmdashmdashmdash ax + b

ndash3

x f

ax +b

ndash4

ndash5

ndash3

15

(a) Find the value of a and of b

(b) State the value of x such that the function f is

undefined

Solution

15(a) f x rarr mdashmdashmdashmdash ax + b

15 f ( x ) = mdashmdashmdashmdash ax + b

f (ndash3) = ndash5

15 ndashndashndashndashndashndashndash = ndash5 ndash3a + b

15a ndash 5b = 15

3a ndash b = 3 1

f (ndash4) = ndash3

15 ndashndashndashndashndashndashndashndash = ndash3 ndash4a + b

12a ndash 3b = 15

4a ndash b = 5 2

1 ndash 2 ndasha = ndash2

a = 2

From 1 3(2) ndash b = 3

b = 3

(b)

15 f ( x ) = mdashmdashmdashmdashmdash 2 x + 3

When the denominator 0

2 x + 3 0

3 x ndash ndashndash

2

1 x ndash1ndashndash

2

Hence the value of x such that the

1 function f is undefined is ndash1ndashndash

2

Substitute x = ndash4



When the function f has an image 4 it meansthat f ( x) = 4

A function f is undefined when its denominatoris zero



7 Functions

9


(a) Given the function f ( x ) = 3 sin x ndash tan x find

the image of 30deg

(b) Given the function g( x ) = 2 cos x find the

value of x such that the function g has the

image 16 for the domain 0deg x 90deg

Solution

(a) f ( x ) = 3 sin x ndash tan x

f (30deg) = 3 sin 30deg ndash tan 30deg

= 3(05) ndash 05774

= 09226

Hence the image of 30deg is 09226

(b) g( x ) = 16

2 cos x = 16

16cos x = ndashndashndash

2

cos x = 08

x = 3687deg

The value of x such that the function g has the

image 16 is 3687deg

Press SHIFT cos

08 =

Answer display 3686989765

10


12c Domain Range Objects and Imagesof a Function

The arrow diagram represents the function

f x rarr 2 x 2 ndash 5 State

(a) the domain

(b) the range

(c) the image of 0

(d) the objects of (i) 3

(ii) ndash3

Solution

(a) Domain = ndash2 ndash1 0 1 2

(b) Range = ndash5 ndash3 3

(c) The image of 0 is ndash5

(d) (i) The objects of 3 are 2 and ndash2

(ii) The objects of ndash3 are 1 and ndash1

8


3 x + k A function f is defined by f x rarr ndashndashndashndashndashndash for all 2 x ndash 4

values of x except x = p and k is a constant

(a) State the value of p

(b) Given that the value 5 is mapped onto itself

under f find

(i) the value of k (ii) another value of x that is mapped onto

itself

Solution

3 x + k (a) For f x rarr ndashndashndashndashndashndashndash 2 x ndash 4

Denominator ne 0

2 x ndash 4 ne 0

x ne 2

It is given that x ne p there4 p = 2

(b)

(i) 5 is mapped onto itself

thus f (5) = 5

3(5) + k mdashmdashmdashmdashmdash = 5 2(5) ndash 4

15 + k mdashmdashmdashmdash = 5 6

15 + k = 30

k = 15

(ii) For self-mapping

f ( x ) = x

3 x + 15 mdashndashmdashmdashmdash = x 2 x ndash 4

3 x + 15 = 2 x 2 ndash 4 x

2 x 2 ndash 7 x ndash 15 = 0 (2 x + 3)( x ndash 5) = 0

3 x = ndash ndashndash or 5 2

Hence another value of x that is mapped

1 onto itself apart from 5 is ndash1ndashndash 2

x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5

Self-mapping is given by f ( x) = x where both theobject and the image have the same value



Functions 8

12

1 State whether each of the following relations is a

function

(a) A Blsquotype of numberrsquo

6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u

P Qlsquoconsist of lettersrsquo

(c)

21

32

5364

2

6

15

M N lsquoproduct of digits ofrsquo

(d) X Y lsquohasrsquo

PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels

2 Given the function f x rarr 18 9mdash x ne ndashndash find

2 x ndash 9 2

(a) the image of (i) 0 (ii) 12 (b) the object that has the image (i) 2 (ii) 6

a 3 Given that f x rarr ndashndashndashndashndashndash f (3) = ndash5 and

x ndash b

f (ndash5) = ndash1 find

(a) the value of a and of b

(b) the value of x such that the function f is

undefined

4 The arrow diagram

shows the function

b f x rarr ax + ndashndash x Find


(b) the value of x such that

the function f is undefined

(c) the object that has the image 7 apart from

x = 2

2 x 5 The function g is defined by g x rarr mdashmdashmdashmdash

x + m

If g(5) = 3g(2) find the value of m Hence find

(a) the image of 10

1(b) the object that has the image ndash ndashndash 2

(c) the value of x such that the function f is

undefined

6 Given that f x rarr

f (3) = 4 find

(a) the value of p and of q

4(b) the values of x such that f ( x ) = ndashndash x

3

7 Given that g x rarr a + bx g(1) = ndash3 and

g(ndash2) = 3 find


(b) the values of n if g(n2 + 1) = 5n ndash 6

8 A function f is defined by f x rarr 5 x ndash 2 Find

(a) the object that has the image 5

(b) the object that is mapped onto itself

a 9 A function f is defined by f x rarr mdashmdashmdash The

b ndash x

values 3 and 5 are mapped onto themselves under

f



undefined

10 The arrow diagram

shows the function

f x rarr px + qx 2

Find

(a) the value of p and

of q

(b) the values of x that

are mapped onto themselves

11 (a) Given the function f x rarr sin x + cos x find

the image of (i) 45deg (ii) 60deg

(b) Given the function g x rarr tan x find the

value of x such that the image is 1527 if x is

an acute angle


represents the function

f x rarr x 2 + 2 State

(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions

13 Absolute Value Functions

1 The absolute value of x is written as 983135 x983135 and it is read aslsquothe modulus of x rsquo

2 The definition of 983135 x983135 is

x if x 0 983135 x 983135 =

ndash x if x 0

3 The absolute value function is defined by

f ( x) if f ( x) 0 983135 f ( x)983135 =

ndash f ( x) if f ( x) 0

Facts

983135 x 983135 is the numerical value of x iepositive numbers are still positivebut negative numbers will bechanged to positive For example

9831355983135 = 5 and 983135ndash5983135 = 5

Facts

The graph of a linear absolutevalue function has a V shape

2

Given the function f x rarr

9831354 x ndash 5983135 find(a) the image of

(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135

f ( x ) = 9831354 x ndash 5983135 (i) f (ndash1) = 9831354(ndash1) ndash 5983135 = 983135ndash9983135 = 9

Hence the image of ndash1 is 9

(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11

Hence the image of 4 is 11

(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3

4 x ndash 5 = plusmn 3

The first equation is

4 x ndash 5 = 3

4 x = 8

x = 2

The second equation is

4 x ndash 5 = ndash3

4 x = 2

x =

1

ndashndash2 Hence the objects that have the image 3 are

2 or 1ndashndash2


SPMClone

rsquo07

3

Sketch the graph of each of the following absolute

value functions

(a) f ( x ) = 983135 x + 3983135 for the domain ndash 4 x 1

(b) f ( x ) = 9831354 x ndash 7983135 for the domain 0 x 4

State the corresponding range of values of f ( x )

Solution

(a) Prepare a table as shown below

The sketch of the graph of f ( x ) = 983135 x + 3983135 is as shown

below

ndash4 ndash3 ndash 2 ndash1 O 1

y

x

4

3

2

1

Range

Domain

SPMClone

rsquo08

x

ndash4 ndash3 ndash2 ndash1 0 1

f ( x ) 1 0 1 2 3 4

For any 983135 f ( x)983135 = k there are two equations that canbe formed ie f ( x) = k or f ( x) = ndashk



Functions 10

1 If f is a function which maps set A onto set B and g is a function whichmaps set B onto set C then gf is a composite function of f followedby g which maps set A onto set C

x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g

14 Composite Functions

14a To Find Composite Functions

Facts

bull fg ( x) means f [ g ( x)]bull In general fg ne gf bull f 2 = ff f 3 = fff or ff 2 and so on

13

1 Given the function f x rarr 983135 x 2 ndash 4 x ndash 3983135 find the

image of

(a) ndash3 (b) 0 (c) 2

2 Given the function f ( x ) = 983135 2 ndash 5 x 983135 find

(a) the image of

(i) 2 (ii) ndash2


3 Sketch the graph of each of the following

absolute value functions

(a) f ( x ) = 983135 x + 2983135 for the domain ndash3 x 3

(b) f ( x ) = 983135 2 x ndash 5983135 for the domain 0 x 7

(c) f ( x ) = 983135 3 ndash 2 x 983135 for the domain ndash3 x 4 (d) f ( x ) = 983135 3 x ndash 5983135 for the domain ndash2 x 4



there4 The corresponding range of values of

f ( x ) is 0 f ( x) 4

(b)

At the x -axis y = 0

Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7

7 3 x = ndashndash = 1ndashndash

4 4

Hence the graph touches the x -axis at 13ndashndash4

0Prepare a table as shown below

The sketch of graph of f ( x ) = 9831354 x ndash 7983135 is as shown

below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9

The corresponding range of values of f ( x) meansthe range from the smallest value of y to the largest

value of y based on the given domain

First of all determine the point where the graphtouches the x-axis



11 Functions

11

The functions f and g are defined by f x rarr 2 x + 1

and g x rarr x 2 ndash 2 respectively Find

(a) the value of fg(3) and of gf (ndash2)

(b) the composite functions

(i) fg (iii) f 2

(ii) gf (iv) g2

(c) the values of x if gf ( x ) = 23Solution

(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15

gf (ndash2) = g[2(ndash2) + 1] = g(ndash3) = (ndash3)2 ndash 2

= 7

(b) f x rarr 2 x + 1 g x rarr x 2 ndash 2

f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3

there4 fg x rarr 2 x2 ndash 3

(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1

there4 gf x rarr 4 x2 + 4 x ndash 1

(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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

there4 f 2 x rarr 4 x + 3

(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)

= ( x 2 ndash 2)2 ndash 2

= x 4 ndash 4 x 2 + 4 ndash 2

= x 4 ndash 4 x 2 + 2

there4 g2 x rarr x4 ndash 4 x2 + 2

(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3


Substitute the x inf ( x ) = 2 x + 1

darrwith ( x 2 ndash 2)

Substitute the x in g( x ) = x 2 ndash 2

darrwith (2 x + 1)


darrwith (2 x + 1)



12


expression for each of the following functions

(a) f 2 (b) f 8 (c) f 9

Solution

x + 1(a) f x rarr mdashmdashmdash x ndash 1

x + 1 f ( x ) = mdashmdashmdash x ndash 1

f 2( x ) = ff ( x )

x + 1 = f mdashmdashmdash x ndash 1

x + 1

= mdashmdashmdash + 1 x ndash 1 = mdashmdashmdashmdashmdashmdash x + 1 = mdashmdashmdash ndash 1 x ndash 1

x + 1 + x ndash 1 mdashmdashmdashmdashmdashmdashndashmdash x ndash 1 = mdashmdashmdashmdashmdashmdashndashmdash x + 1 ndash ( x ndash 1) mdashmdashmdashmdashmdashmdashndashmdash x ndash 1

2 x = ndashndash 2

= x



Functions 12

4

(a) The function f is defined by f x rarr 2 x + 4

Another function g is such that

fg x rarr 3 x ndash 8 Find the function g

(b) The function f is defined by f x rarr x ndash 2


gf x rarr x ne 11 ndash 3 x 3

1 11 Find the function g

Solution

(a) Case where the function g that has to

be determined is situated lsquoinsidersquo

It is given that f x rarr 2 x + 4 and

fg x rarr 3 x ndash 8

fg( x ) = 3 x ndash 8

f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12

3 x ndash 12 g( x ) = mdashmdashmdashndashndashndashmdash 2

3 x ndash 12 there4 g x rarr mdashmdashmdashndashndashndashmdash 2

(b)Case where the function g that has to

be determined is situated lsquooutsidersquo

It is given that f x rarr x ndash 2 and

1 gf x rarr mdashndashmdashmdashmdash 11 ndash 3 x

1 gf ( x ) = mdashndashmdashmdashmdash 11 ndash 3 x

1 g[ f ( x )] = mdashndashmdashmdashmdash 11 ndash 3 x

1 g( x ndash 2) = mdashndashmdashmdashmdash 11 ndash 3 x

Let x ndash 2 = u then x = u + 2

1 there4 g(u) = mdashndashmdashmdashmdashmdashmdashmdash 11 ndash 3(u + 2)

1 = mdashndashmdashmdashmdashmdashmdashmdash 11 ndash 3u ndash 6

1

= mdashmdashmdashmdashmdash 5 ndash 3u there4 g x rarr mdash x ne ndashndash

5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside

Change each x tothe terms in u

14b To Find the Related Function Given theComposite Function and One of the Functions


(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )

x + 1 = mdashmdashndashndashndashndashndashndashndashndashndash x ne 1 x ndash 1

f 2( x ) = x

f 8( x ) = x

In the SPM marking schemeit is not compulsory for

students to write down x ne 1in the answer



13 Functions

14c Further Examples on Composite Functions

5

Given the functions f x rarr hx + k

g x rarr ( x + 1)2 + 2 and fg x rarr 2( x + 1)2 + 1 find

(a) the value of g2(2)

(b) the value of h and of k

Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)

It is given that f x rarr hx + k and

g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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

But it is given that fg( x ) = 2( x + 1)2 + 1

Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14

1 Given the functions f x rarr 983135 4 ndash 5 x 983135 and

g x rarr 983152 983151983151983151983151983151983151983151

x ndash 2 x 2 find the value of

(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)

2 Find the composite functions fg and gf for each

of the following pairs of functions f and g

(a) f x rarr x 2 ndash 1 (c) f x rarr x ndash 3

g x rarr 3 x + 1 g x rarr983135 x 983135(b) f x rarr ( x + 1)2 (d) f x rarr 2 ndash x

g x rarr 1 ndash 3 x 1 g x rarr mdashmdashmdashmdash

x 2 + 2

3 The functions f and g are defined by

f x rarr 4 x ndash 3 and g x rarr x + 1 respectively

13

Try Question 8 SPM Exam Practice 1 ndash Paper 2

Given the functions f ( x ) = 3 x + 7 and fg( x ) = 22 ndash 3 x

find gf ( x )

Solution

Find g( x ) first

fg( x ) = 22 ndash 3 x

f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x

3g( x ) = 22 ndash 3 x ndash 7

3g( x ) = 15 ndash 3 x

15 ndash 3 x g( x ) = mdashmdashmdashmdashmdash 3

g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)

= 5 ndash 3 x ndash 7

= ndash3 x ndash 2

Substitute the x in f ( x ) = 3 x + 7

darr with g( x )

Substitute the x in g( x ) = 5 ndash x

darr with (3 x ndash 7)

The following function problem can be solved bymaking a comparison



Functions 14

1 If f x rarr y is a function that maps x onto y its inverse function isdenoted by f ndash1 Inverse function is a function that maps y back to x

x y

f

f ndash1

2 The main step to find the inverse function islsquoLet y = f ( x) then x = f ndash1 (y )rsquo or lsquoLet f ndash1( x) = y then f (y ) = xrsquo

Find the expressions for f 2 and g2 Hence find

the value of x such that

(a) f = g (b) f 2 = g2


f x rarr mx + 2 and g x rarr kx ndash 3 respectively If

fg = gf find the relation between m and k If

m = 5 find the value of x that satisfies each of the

following equations

(a) f 2 = f (b) g2 = g


f x rarr ( x + 1)2 and g x rarr x ndash 2 respectively

Find

(a) the composite functions fg and gf

(b) the values of x if

(i) fg = 9

(ii) gf = 14

(c) the value of x if fg = gf

6 Given the functions f x rarr 5 x ndash 7 and

g x rarr 2 x

x ne 0 find the expression for

each of the following functions

(a) fg (c) g6

(b) g2 (d) g7

7 If f is defined by f x rarr x ndash 1

x + 1 x ne ndash1 find the


(a) f 2 (c) f 16

(b) f 4 (d) f 17

8 The function f and the composite function fg aredefined as follows Find the function g

(a) f x rarr x + 2 fg x rarr 3 x ndash 2

(b) f x rarr 3 x + 2

fg x rarr 2 x + 5

x ndash 2 x ne 2

(c) f x rarr x 2 ndash 1 fg x rarr x 2 + 4 x + 3

9 The function f and the composite function gf are

defined as follows Find the function g

(a) f x rarr

3 x + 1 gf x rarr x ne 2

x ndash 2

(b) f x rarr ndashndash rarr x ne ndashndashndash x

11 5

10 x ndash 1gf x 10

1

(c) f x rarr 3 x + 2 gf x rarr 9 x 2 + 9 x + 2

10 Find the expressions for the functions g and h in

each of the following

(a) f x rarr 2 x fg x rarr 4 x ndash 12

2 x + 1 hf x rarr mdashmdashmdashmdash 2

2 x ndash 1(b) f x rarr 2 x ndash 2 gf x rarr mdashmdashmdashmdash 3

fh x rarr 2 x 2

(c) f x rarr x 2

ndash 2 fg x rarr x 2

+ 6 x + 7 hf x rarr 2 x 2 ndash 7


f x rarr 1 ndash x and g x rarr px 2 + q respectively If

the composite function gf is given by

gf x rarr 3 x 2 ndash 6 x + 5 find


(b) the value of g2(0)

12 The functions f and f 2 are such that

f x rarr hx + k and f 2 x rarr 9 x + 16

(a) Find the values of h and the corresponding

values of k

(b) Considering h 0 find the values of x such

that f ( x 2) = 8 x

15 Inverse Functions

15a To Find Inverse Functions

Facts

bull (fg)ndash1 = gndash1f ndash1 and ( gf )ndash1 = f ndash1 g ndash1

bull (f2

)ndash1

= (fndash1

)2

bull ff ndash1( x ) = f ndash1f ( x ) = x

Facts

The conditions for the existenceof inverse functions are(a) the function must be one-to-

one(b) that every element in the

codomain must be linked toan object



15 Functions

1 + 2 y there4 f ndash1( y) = mdashmdashndashmdash 2 ndash y

1 + 2 x f ndash1( x ) = mdashmdashndashmdash 2 ndash x

there4 f ndash1

Method 2

Let f ndash1( x ) = y

Thus f ( y) = x

2 y ndash 1 mdashmdashndashmdash = x y + 2

2 y ndash 1 = x ( y + 2)

2 y ndash 1 = xy + 2 x

2 y ndash xy = 2 x + 1

y(2 ndash x ) = 2 x + 1

2 x + 1 y = mdashmdashndashmdash 2 ndash x

2 x + 1 f ndash1( x ) = mdashmdashndashmdash 2 ndash x

there4 f ndash1 x rarr

In the SPM marking schemeit is not compulsory forstudents to write x ne 2

Rearrange the formulamaking y the subject

14


Find the inverse function of f x rarr x ne ndash2

2 x ndash 1

x + 2

Solution

Method 1

2 x ndash 1 f x rarr mdashmdashndashmdash x + 2

2 x ndash 1 Let y = mdashmdashndashmdash x + 2

y( x + 2) = 2 x ndash 1

yx + 2 y = 2 x ndash 1

yx ndash 2 x = ndash1 ndash 2 y

x ( y ndash 2) = ndash1 ndash 2 y

ndash1 ndash 2 y x = mdashndashndashndashmdashndashmdash y ndash 2

ndash(1 + 2 y) x = mdashndashndashndashmdashndashmdashndashndashmdash ndash(2 ndash y)

1 + 2 y x = mdashmdashndashmdash 2 ndash y

Rearrange making x the subject

Cross multiplication

Expand the left-handside

Group the terms in xtogether

Factorise the

left-hand side

Making x the subject

15

Given the functions f ( x ) = 6 ndash 2 x and g( x ) =1ndashndash

x

x ne 0 find f ndash1gndash1

Solution

First of all find the inverse functions f ndash1 and gndash1

f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y

6 ndash x y = mdashndashndashndashmdash 2

6 ndash x there4

f

ndash1

( x ) = mdashndashndashndashmdash 2

g( x ) =1ndashndash x

Let gndash1( x ) = y

Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x

there4 gndash1( x ) =1ndashndash x


6 SPM

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Given the function f ndash1 x rarr

find the value of f ndash1(3)

Solution

Let y = f ndash1

(3) f ( y) = 3

y + 8 mdashmdashmdash = 3 y ndash 6

y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13

there4 f ndash1(3) = 13



Functions 16

7


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Given the function f x rarr mdashmdashmdashmdash and its

x hx ndash 5

x ne 2 x ndash 2

inverse function f ndash1 x rarr

mdashmdashmdashmdash2 x + k

x ne 3 find x ndash 3

(a) the value of h and of k

(b) the values of t such that f (t ) =4ndashndash3

t

Solution

(a) Let f ndash1( x ) = y Thus f ( y) = x

hy ndash 5 mdashmdashmdashmdash = x y ndash 2

hy ndash 5 = x ( y ndash 2)

hy ndash 5 = xy ndash 2 x

2 x ndash 5 = xy ndash hy

2 x ndash 5 = y( x ndash h)

2 x ndash 5 y = mdashmdashmdashmdash x ndash h

2 x ndash 5 there4 f ndash1( x ) = mdashmdashmdashmdash

x ndash h

2 x + k But it is given that f ndash1( x ) = mdashmdashmdashmdash x ndash 3

Hence by comparison k = ndash5 h = 3

(b) f(t ) =4ndashndash3

t

3t ndash 5 mdashmdashmdashmdash =

4ndashndash3

t t ndash 2

3(3t ndash 5) = 4t (t ndash 2)

9t ndash 15 = 4t 2 ndash 8t

0 = 4t 2 ndash 17t + 15

0 = (4t ndash 5)(t ndash 3)

t =5ndashndash4 or 3


Hence f ndash1gndash1 ( x )

= f ndash11ndashndash x 6 ndash 1ndashndash x = mdashndashmdashmdashndashndashmdash 2

= mdashndashmdashmdashndashndashmdash x ne 02 x

6 x ndash 1

Apply the compositefunction of gndash1 followed by f ndash1

8 SPM

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13ndash2

10

x

f

y z

g

The above diagram shows the representation of the

mapping of y onto x by the function

f y rarr py + q and the mapping of y onto z by

the function

5 qg yrarrmdashmdashndashmdash ndashndash Find

3 y ne

3 y ndash q


(b) the function that maps x onto y

(c) the function that maps x onto z

Solution

5(a) f y rarr py + q g y rarr mdashndashmdashndashmdash 3 y ndash q

5 f ( y) = py + q g( y) = mdashndashmdashndashmdash 3 y ndash q

f 3ndashndash2 = 1 g3

ndashndash2 = 10

5

3ndashndash2

p + q = 1 mdashndashmdashmdashmdashmdashmdash = 10

33ndashndash2 ndash q

3 p + 2q = 2 1 5mdashndashmdashmdash = 10




17 Functions


The inverse of an inverse function f ndash1( x) will give

us back the original function f ( x)

15b To Find a Function Given its InverseFunction

15c Further Examples on InverseFunctions

16

The function f is defined by f ( x ) =

2 x + 1 x ne 1

x ndash 1

If f ndash1(k ) =4ndashndash3

k find the values of k

Solution

f ndash1(k ) =4ndashndash3

k

k = f 4ndashndash3

k

24k ndashndash3 + 1 k = mdashmdashmdashmdashmdashmdash

4k ndashndash3

ndash 1

8k + 3 mdashmdashmdashmdash 3 k = mdashmdashmdashmdashmdashmdash 4k ndash 3 mdashmdashmdashmdash 3

8k + 3 3 k = mdashmdashmdashmdashmdashmdashmdashmdash 3 4k ndash 3

8k + 3 k = mdashmdashmdashmdash 4k ndash 3

If f ndash1 ( x ) = y then x = f ( y )

5 mdashndashmdashmdashndashmdash = 10 9 ndash 2q mdashndashmdashmdashmdash 2

10 mdashndashmdashmdashmdash = 10 9 ndash 2q

1 mdashndashmdashmdashmdashmdash = 1 9 ndash 2q

1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2

(b) The function that maps x onto y is f ndash1( x )

It has been found that f ( y) = ndash2 y + 4

Let f ndash1( x ) = w

Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x

w = mdashndashndashmdash 2 4 ndash x there4 f ndash1( x) = mdashmdashmdashmdash 2

(c) Based on the diagram the function that maps x

onto z is gf ndash1( x )

x y z

gf ndash1

gf ndash1

gf ndash1( x )

4 ndash x = gmdashndashndashmdash 2

5 = mdashndashmdashmdashmdashmdashmdashndashmdash 4 ndash x 3mdashndashndashmdash ndash 4 2

5 = mdashndashmdashmdashmdashndashmdashmdash 12 ndash 3 x ndash 8 mdashmdashmdashmdashndashmdashmdashmdash 2

10 = mdashndashmdashmdash 4 ndash 3 x

there4 gf ndash1 x rarr 10

4 ndash 3 x x ne

4

3

9


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Given that gndash1( x ) =2 x + 1

2 x ndash 1 x ne

1

2 find g( x )

Solution

2 x + 1 Let y = mdashmdashmdashmdash 2 x ndash 1

y(2 x ndash 1) = 2 x + 1

2 xy ndash y = 2 x + 1

2 xy ndash 2 x = y + 1

x (2 y ndash 2) = y + 1

y + 1 x = mdashmdashmdashmdash 2 y ndash 2

y + 1 g( y) = mdashmdashmdashmdash 2 y ndash 2

there4 g( x) = x + 1

2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3

4k 2 ndash 11k ndash 3 = 0

(4k + 1)(k ndash 3) = 0

k = ndash1ndashndash4 or 3

Given that f ndash1( x ) = 1

x x ne ndashm and x + m

g( x ) = 3 ndash x find

1 ndash 3 x (a) the value of m if f ( x ) = mdashndashmdashndashndashmdash x

(b) the values of k if ff ndash1(k 2 ndash 7) = g[(k + 2)2]

Solution

(a) Let f ndash1( x ) = y

f ( y) = x

1 ndash 3 y

mdashndashmdashndashndashndashmdash = x

y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)

1 y = mdashndashmdashmdash x + 3

1 there4 f ndash1( x ) = mdashndashmdashmdash x + 3

1 But it is given that f ndash1( x ) = mdashndashmdashndashndashndashndashndashndashndashmdash x + m

Hence by comparison m = 3


18

Given the function f x rarr 5 x ndash 2 find ( f 2)ndash1 in the

same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)

= 5(5 x ndash 2) ndash 2 = 25 x ndash 10 ndash 2

= 25 x ndash 12

Next find the inverse of f 2( x )

Let ( f 2)ndash1 ( x ) = y

Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12

x + 12 y = mdashmdashndashndashndashmdash 25

x + 12 ( f

2

)ndash1

( x ) = mdashmdashndashndashndashmdash 25

x + 12there4 ( f 2)ndash1 xrarr mdashmdashndashndashndashmdash 25

If the question requiresus to state the answerin the lsquosame formrsquo wehave to express theanswer in the form x + 12(f

2

)ndash1

x rarr mdashmdashmdashmdashmdash 25and not

x + 12(f 2)ndash1 ( x ) = mdashmdashmdashmdashmdash 25


17

The inverse of g ndash1 will give us back g

2 ndash kx Given that gndash1( x ) = mdashmdashmdashmdash and f ( x ) = 3 x find 4

(a) g( x ) in terms of k

(b) the values of k such that fg(1) = ndash3ndashndash2

k

Solution

2 ndash kx (a) gndash1( x ) = mdashmdashmdashmdash 4

2 ndash kx Let y = mdashmdashmdashmdash 4

4 y = 2 ndash kx

kx = 2 ndash 4 y 2 ndash 4 y x = mdashmdashmdashmdash k

2 ndash 4 y g( y) = mdashmdashmdashmdash k

2 ndash 4 x g( x) = mdashmdashmdashmdash k

(b) fg(1) = ndash3ndashndash2

k

2 ndash 4(1) f mdashmdashmdashmdashndashndashmdash = ndash

3ndashndash2

k k

f ndash2mdashmdash

k = ndash3ndashndash2

k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15

1 Evaluate f ndash1(4) for each of the following

functions

(a) f x rarr 5 ndash 4 x

5(b) f xrarr 6 ndash mdash x ne 0

x

(c) f xrarr x ne ndash3

2

3 x + 2

2 x + 3

2 Find the inverse function of each of the

following functions

(a) f x rarr 7 x ndash 4

3 x ndash 4(b) g x rarr mdashmdashndashmdashmdash 2

(c)3

h xrarr 9 ndashmdash x ne 0 x

(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151

2 ndash x ndash x x 2

3 Given the functions

4mdash x

f xrarr x ne 0 and

g x rarr 2 x + 3 find each of the followingfunctions

(a) fg (d) g2 (g) f ndash1gndash1

(b) gf (e) f ndash1 (h) gndash1 f ndash1

(c) f 2 (f) gndash1

3 x ndash 1 4 The function f is defined by f x rarr mdashmdashndashmdashmdash

x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1

5 Given the functions f x rarr 2 ndash 4 x and

3g xrarr x ne 1 find

x ndash 1

(a) f ndash1 (c) f ndash1gndash1 (e) (gf )ndash1

(b) gndash1 (d) gf

Is (gf )ndash1 = f ndash1gndash1


g xrarrmdashndashmdashmdash x ndash 2

x + 2 x ne 2 find

(a) f ndash1 (d) fg

(b) gndash1 (e) ( fg)ndash1

(c) gndash1 f ndash1

Is ( fg)ndash1 = gndash1 f ndash1

7 The function f is defined by f x rarr 2 x ndash 1

(a) Find the expressions for f 2 dan f ndash1

(b) Show that ( f ndash1)2 = ( f 2)ndash1

8 Given the function f x rarr 4 x + h and its inverse

kx + 5 function f ndash1 x rarr mdashndashndashmdashmdashmdash find 4


(b) the expression for f ndash1 f

9 Given the function f xrarr x

x + p x ne 5 and

x ndash 5

qx + 6 its inverse function f ndash1 x rarr mdashndashndashndashmdashmdash

x ndash 1

x ne 1 find

(a) the value of p

and ofq

(b) the values of x for which f ndash1( x ) = 8 x

10

ndash2

2

ndash1

x y z

The diagram shows the representation of the


f y rarr my + n and the mapping of y onto z by

the function g yrarr m mdash y nemdash Findn ndash 2 y 2

n

(a) the value of m and of n



3 x + 10 11 (a) Given that f ndash1( x ) = mdashmdashmdashmdashmdash find f ( x ) 4

2 x + 3 (b) Given that gndash1( x ) = mdashmdashmdashmdash find g( x ) x + 4

(b) ff ndash1(k 2 ndash 7) = g[(k + 2)2]

k 2 ndash 7 = 3 ndash (k + 2)2

k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)

k 2 ndash 7 = 3 ndash k 2 ndash 4k ndash 4

2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1

ff ndash1( x ) = x is always true and it hasto be memorised



Functions 20

(a) State the value of k (b) Using the function notation

express f ( x ) in terms of x

[2 marks]

8 The diagram shows the relation

between set X and set Y in the form

of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State

(a) the relation in the form of

ordered pairs

(b) the type of the relation

(c) the domain of the relation

[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1

The arrow diagram represents

the function f x rarr a

bx ndash 2 x ne k

where a b and k are constants

Find


(b) the value of k

SPMClone

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SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9

The arrow diagram shows the


Based on the arrow diagram state

(a) the domain(b) the range

(c) the codomain

2 The arrow diagram shows the

relation between set A and set B

1

5

6

8

A B

a

e

(a) Represent the given relation

using ordered pairs

(b) State the type of the given

relation

3 Set Q

11

9

7

5

Set P 2 6 8

The above graph represents therelation from set P = 2 6 8 to set

Q = 5 7 9 11 State

(a) the images of 6

(b) the objects of 7

(c) the range

4 P = 2 6 8

Q = 1 3 5 7 9SPMClone

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Paper Short Questions1

Based on the above informationthe relation from P to Q is defined

by the ordered pairs (2 1) (2 3)

(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]

5 The diagram below shows the


x

y

z

10

2030

40

Set P Set Q

State

(a) the range of the relation


[2 marks]

6 In the diagram below set Q shows

the images of the elements of set P

3

2

ndash2

ndash3

81

16

Set P Set Q

(a) State the type of relation

between set P and set Q

(b) Using function notation

write down a relation between

set P and set Q

[2 marks]

7 The following arrow diagram

represents a linear function f

x f ( x )

2

5

7

k

1

4

6

8

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21 Functions

10 (a) Sketch the graph of the

absolute value function

f ( x ) = 3 ndash x for the domain

0 x 4

(b) Hence state the corresponding

range of values of f ( x )

11 Given the function g( x ) = 2 x ndash 9find the possible values of

x if g( x ) = 4

12 The diagram shows the function

m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h

Find the value of h [2 marks]

13 Given the function f x rarr 983135 x ndash 4983135find the values of x such that

f ( x ) = 9 [2 marks]

14 The given diagram shows the graph

of the function f ( x ) = |2 x ndash 3| for thedomain 0 le x le 4

y

x

3

4O k

State(a) the value of k

(b) the range of values of f ( x )

corresponding to the given

domain [3 marks]



f x rarr 3 x ndash 2 and g x rarr x + 2

respectively Calculate fg(5)

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16 Given the functions g x rarr 4 + x 2

and h x rarr 2 x ndash 4 find gh

17 Given the function f x rarr 4 ndash 5 x

find f 2

18 Given that f x rarr 2 x + m and

f 2 x rarr px + 6 find the value of m

and of p

19 Given the functions f x rarr 3 x ndash 7

and fg x rarr x 2 + 1 find the

function g

20 The function f is defined by

f x rarr x ndash 2 Another function g is

such that gf x rarr x 2 + 1 Find the

function g

21 Given the functions f ( x ) = 2 x ndash 4

and fg( x ) = 8 x + 2 find gf ( x )

22 Given the functions h( x ) = 7 x ndash 1

and the composite function

hg( x ) = 35 x + 13 find

(a) g( x )

(b) the value of x when

gh( x ) = 4

[4 marks]

23 Given the function g x rarr px + q

and its composite function

g2 x rarr 49 x ndash 32 find the value of p

and of q such that p gt 0

[3 marks]

24 Given the function f ( x) = x + 4 and

g( x ) = tx ndash 6 find

(a) f (6)

(b) the value of t such that

gf (6) = 24

[3 marks]

25 Given the function g x rarr 4 x ndash 1

and h x rarr 8 x find

(a) hg ( x )

(b) the value of x if hg( x ) = 2g( x )

[4 marks]


26 Given the function f ( x ) =4 x + 3

x + 5

x ne ndash5 calculate the value of

f ndash1(2)

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27 Given the function g x rarr 4 x + 1

x + 4

x ne ndash4 find gndash1

28 Given the functions f ( x ) =2

x ndash3

x ne 3 and g( x ) = 4 x ndash 1 find

f ndash1g

1ndashndash

2

29 Given the function f x rarr 3 x + h

and its inverse function

f ndash1 x rarr kx ndash2ndashndash3

find the value of h

and of k

30 Given the function f ( x ) =24

px + q

f (1) = 8 find


(b) the values of k if f ndash1 (k ) = k

31 Given the inverse function

f ndash1( x ) = 2 x ndash 5 find

(a) f ( x )

(b) the values of k if

f ndash1 f (k ) = k 2 ndash 12

32

x fndash14 x + p

ndash2

ndash6


f ndash1( x ) = 4 x + p where

p is a constant Find

(a) the value of p(b) f ( x )


3 x + 1

(a) gndash1 f ( x )

(b) the values of x which are

mapped onto themselves

under the function f



Functions 22

34 Given that g x rarr 4 x ndash 1 and

h x rarr x 2 ndash 3 x + 5 find

(a) gndash1(7)

(b) hg( x )

[4 marks]

35 Given the functions h x rarr mmdashndash x

ndash 3

x ne 0 and h ndash1 x rarr

10mdashmdashmdash x + k

x ne ndashk

where m and k are constants find

the value of m and of k

[3 marks]

36 In the following arrow diagram the

function f maps x onto y and the

function g maps y onto z

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Paper Long Questions2


1 The following diagram shows the

functions f and g that are defined

by f xrarr ax + b and g x rarr10

c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find

(a) the value of a of b and of c(b) the value of m

(c) the expression for gf

2 The functions f and g are

defined by f x rarr hx ndash 5 and

g x rarr x 2 + 3 x + 5 respectively

If the composite function fg is

defined by fg x rarr 2 x 2 + kx + 5

find the value of h and of k

3 The functions f and g are definedax

by f xrarr mdashmdashmdash x ne 1 and1 ndash x

g x rarr bx ndash 1 respectively If

f (4) = ndash4 and g(2) = 3 find


(b) the value of x for which

fg =gf

4 Given the functions f x rarr 3 ndash x

1and g xrarr mdashmdashmdash x ne 1 find the

1 ndash x

expression for each of the

following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19


f x rarr 2 x + 3 Another function g is

such that gf x rarr 4 x 2 + 12 x + 15

Find

(a) the function g

(b) the values of c if g(c) = 7c

(c) the values of x if fg = gf

6

5

89

x y

f

g

z

The above arrow diagram shows

the representation of the mapping

of x onto y by the function f x rarr

4 x ndash a and the mapping of y onto z

b by the function g y rarr mdashmdashndashndashmdash

12 ndash y

y ne 12 Find


(b) the expression for the function

that maps x onto z

(c) the element x that does not

change when it is mapped

onto z


defined by f x rarr 2 x + 3 and

g x rarr x 2 + bx + c respectively


given by fg x rarr 2 x 2 + 4 x ndash 3

find

(a) the value of b and of c


x y f g

z

4

1

ndash2

Write down the value of (a) gndash1(4)

(b) gf (1)

[2 marks]

37 The functionsh and w are defined

by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)

(b) wndash1( x ) [4 marks]

38 Given the functions m x rarr 4 x ndash 6

and n xrarr 3

mdash x

x ne 0 find nm ndash1

[3 marks]

39 Given the function g x rarr 6 ndash 2 x

find(a) g(ndash4)

(b) the value of p such that

gndash1( p) = 8

[3 marks]

40 Given the functions g x rarr 2 x + 3

and h x rarr 5 x ndash 8 find

(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions


and fg x rarr 4 ndash 15 x find

(a) the expression for gf

(b) the values of n if

gf (n2 ndash 1) = 6n + 6

9 The functions f and f 2 are such

that f x rarr ax + b and

f 2

x rarr 9 x + 8(a) Find the values of a and

the corresponding values

of b

(b) Considering a 0 find the

values of k for which

f (2k + 1) = k (k + 2)

10 The function g is defined by

g x rarr x + 2 Another function f

is such that fg x rarr x 2 + 5 x + 7

Find(a) the function f ( x )

(b) the values of c if f (2c) = 7c



f x rarr mx + k where m and k

are constants The function g is

defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1

(a) Find the expression for gndash1

(b) Find the expression for fg in

terms of m and k

(c) If f (3) = gndash1(3) and

fg(ndash2) = ndash2 find the value of m

and of k

12 Given the functions f x rarr x + 1

x ndash 2

x ne 2 and g xrarrmdashmdashmdashmdash

x

kx + 3 x ne 0

find the value of k if gf ndash1(0) = 3

13 Given the function f x rarr x + p

x + q

x ne ndashq and its inverse function

f ndash1 x rarr2 ndash 3 x

1 ndash x x ne 1 find


(b) the values of k if f 2(k ) =1mdash3 k

14

7

5

3

ndash1

A B

f

gndash1

C

The diagram shows the mapping of

the functions f and gndash1 such that

f x rarr

2 x + a andg x rarr bx + c Given that g maps 1

onto itself find

(a) the value of a

(b) the value of b and of c



f ndash1 x rarr kx +2mdash5 find


(b) (i) f (3) (ii) f ndash1 f (3)

mx ndash n 16 Given the function f x rarr mdashmdashmdashndashndashndashmdash x ndash 2

x ne 2 and its inverse function

f ndash1 x rarrndash

mdashmdashmdashmdash x ne 2 find5 ndash 2 x

2 ndash x


(b) the value of k for which

f (k ) = k + 2

17 The following diagram represents

the mapping of y onto x by the

function f y rarr hy + k and

the mapping of y onto z by the

function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find


(b) the function that maps x

onto y

(c) the function that maps x

onto z


3 x and g( x ) = mdashmdashmdash x ne ndash3 find

x + 3

(a) f ndash1g( x )


gf (ndash x ) = f 2(2)

19 Given that f x rarr p ndash qx find

(a) f ndash1( x ) in terms of p and q

(b) the value of p and of q if

f ndash1(8) = ndash1 and f (1) = ndash2

3 ndash kx 20 Given that gndash1( x ) = mdashmdashmdashmdash and

2

f ( x ) = 2 x 2 ndash 3 find



g( x 2) = 2 f (ndash x )

21 Given the functions f x rarr hx + k

h 0 and f 2 x rarr 25 x ndash 18

find


(b) ( f ndash1)2 in the same form

22 Given that f ndash1( x ) =

1 x ne k

k ndash x

and g( x ) = 2 + x find

(a) f ( x ) in terms of k

(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]

23 Given that f x rarr 2 x ndash 3 and

g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]

(b) the function h such that

hg x rarr 2 x + 4

[3 marks]

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1 The function f and its inverse function f ndash1 are

defined by f ( x ) =hx

x ndash 3 x ne 3 f ndash1( x ) =

kx

x ndash 2 x ne 2

respectively where h and k are constants Another

function g is defined by g( x ) =1

x x ne 0

(a) Find the value of h and of k [4 marks]

(b) If gf ndash1( x ) = ndash5 x find the values of x [3 marks]

2 Given the functions f xrarr x

2 ndash 2 and g x rarr 3 x + k

where k is a constant find

(a) f ndash1(3) [2 marks]

(b) the value of k if f ndash1g x rarr 6 x ndash 4 [2 marks]

(c) h( x ) such that hf x rarr 9 x ndash 3 [2 marks]

Long Questions

Paper 2

1 The relation between set

X = 6 12 15 21 and set

Y = 3 5 7 is lsquo factor of rsquo

(a) Find the image of 12

(b) Express the relation in the

form of ordered pairs

[3 marks]

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2 Given the functions g( x ) = 5 x ndash 11

and h( x ) = 3 x find the value of

gh(2)

[2 marks]

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3 The inverse function hndash1 is defined

by hndash1 x rarr 3

4 ndash x x ne 4 Find

(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1

1 The graph below shows the relation

between set A and B

40

30

20

10

S e t B

Set A

4321

State

(a) the object of 40


[2 marks]

2 The relation between two variables

is represented by the following set

of ordered pairs

(ndash4 16) (ndash3 9) (ndash2 4) (2 4)

(3 9) (4 16)

(a) State the type of the above

relation

(b) Represent the above relation

using a function notation

[2 marks]

3 Given the function f ( x ) = px + q and

f 2 ( x ) = 4 x + 9 where p and q are

constants find the value of p and of

q if p lt 0 [3 marks]

4 Given the function f x rarr x + 4


gf x rarr x 2 + 6 x + 2 find

(a) g( x )

(b) fg(4) [4 marks]

5 Given the function f ( x ) =

2 x + p

5 and

its inverse function f ndash1( x ) =

5 x + 3

q

find the value of p and of q

[4 marks]

Short Questions

Paper 1



Year

Paper

Number of questions 3

1

0

2

3

1

0

2

2

1

1

2

3

1

0

2

3

1

0

2

3

1

0

2

3

1

0

2

3

1

0

2

2004 2005 2006 2007 2008 2009 2010 2011

SPM Topical Analysis

RELATIONS

HUBUNGAN

FUNCTIONS

FUNGSI

Types of Relations Jenis Hubunganbull One-to-onebull Many-to-one

bull One-to-manybull Many-to-many

Absolute Value FunctionsFungsi Nilai Mutlak

f x rarr|g( x )|

Composite FunctionsFungsi GubahanThe function of f

followed by g is gf

Graphs of Absolute Value FunctionsGraf Fungsi Nilai Mutlak The graph of a linear absolutevalue function has a V shape

Problems that involve CompositeFunctions and Inverse FunctionsMasalah melibatkan Fungsi Gubahandan Fungsi Songsang

Inverse FunctionsFungsi SongsangGiven that y = f ( x )

then f ndash1( y ) = x

DomainDomain

CodomainKodomain

Objects

Objek

ImagesImej

Range Julat

1 Functions

CHAPTER FORM 4

ONCEPT MAP

Learning ObjectivesCOMPANION W EBSITE



Functions 2




11 Relations

1




(c) a graph

Solution

(a) Arrow diagram

3

5

7

12

14

23

25

43

AB


(b) Ordered pairs

(12 3) (14 5) (23 5) (25 7) (43 7)

(c) Graph7

5

3

12 14 23 25 43Set A

Set B








18 19 20 21 22 23 2425 26 27 28 29 30

A = 12 18B = 11 13 17 19 23 29C = 12 21 30





3 Functions

2





(b) State

(i) the domain

(ii) the codomain


(v) the object of 7


(vii) the range

of this relation

Solution



16

36

49

64

4

6

7

8

11


Range


(b) (i) The domain is 16 36 49 64










(a) (b)


3

4

5

1

3

1

4

1

5

10

21

2

3

5

7





4




(4 8) (4 10) (4 12) (6 10) (7 12)

(b) the graph

Set X

Set Y

M

P

5 7 9 12 15Solution

Types of Relations


Each object in the

domain has only one

image in the

codomain


There are more than

one object in the


same image in the

codomain




that has more than




that has more than

one image in the




linked to more than

one object in the

domain

Domain Codomain

Domain Codomain

Domain Codomain

Domain Codomain



Functions 4

11



(b) ordered pairs

(c) a graph



B = 1 2 3 4




Y = 1 3 4



digits of rsquo


(b) State

(i) the domain

(ii) the codomain



(v) the range

of the relation




Sarawak

Perak

Kedah

Kuching

Ipoh

Alor Setar


methane

water

carbondioxide

carbon

hydrogen

oxygen



(4 0) (4 1) (5 2) (6 7) (6 8)


(a)4

6

7

8

10

12



(b) X Y

5

7

9

12

15

P

M



1


from set A = 3 5 7 to set B = 9 15 21

(3 9) (5 15) (7 21)


(b) State the range


Solution

3

5

7

9

15

21

A B




f ( x) = 3 x


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5 Functions

(b) the graph

3 4 6 7 9

Set B

Set A

11

10

8


from set P = 2 6 10 to set Q = 1 3 5

(2 1) (6 3) (10 5)


(b) State the range





x

y

z

a

b

c

a

b

c

d

w

x

y

z





12 Functions

Facts


Facts



a

b

x

y

z



a

b

c

x

y

z



x

y

z

a

b

c

d


5


(a) (b) (c)

A Blsquohasrsquo

January

June

August

28 days

30 days

31 days


Bronze

Brass

cuprum

stanum

zinc

1

3

57

3

9

15


Solution












Functions 6

6



(a) the image of 2




When x = 2

f (2) = 22 ndash 4(2) ndash 1

= 4 ndash 8 ndash 1

= ndash5


(b)

When f ( x ) = 4



( x + 1)( x ndash 5) = 0

x = ndash1 or 5


ndash1 or 5

Object


7



ndash3

x f

ax +b

ndash4

ndash5

ndash3

15



undefined

Solution



f (ndash3) = ndash5


15a ndash 5b = 15

3a ndash b = 3 1

f (ndash4) = ndash3


12a ndash 3b = 15

4a ndash b = 5 2


a = 2


b = 3

(b)



2 x + 3 0


2


2



2








7 Functions

9



the image of 30deg




Solution



= 3(05) ndash 05774

= 09226


(b) g( x ) = 16

2 cos x = 16


2

cos x = 08

x = 3687deg


image 16 is 3687deg

Press SHIFT cos

08 =


10





(a) the domain

(b) the range

(c) the image of 0


(ii) ndash3

Solution






8






under f find


itself

Solution


Denominator ne 0

2 x ndash 4 ne 0

x ne 2


(b)


thus f (5) = 5



15 + k = 30

k = 15


f ( x ) = x


3 x + 15 = 2 x 2 ndash 4 x





x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5




Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


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3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 2




11 Relations

1




(c) a graph

Solution

(a) Arrow diagram

3

5

7

12

14

23

25

43

AB


(b) Ordered pairs

(12 3) (14 5) (23 5) (25 7) (43 7)

(c) Graph7

5

3

12 14 23 25 43Set A

Set B








18 19 20 21 22 23 2425 26 27 28 29 30

A = 12 18B = 11 13 17 19 23 29C = 12 21 30





3 Functions

2





(b) State

(i) the domain

(ii) the codomain


(v) the object of 7


(vii) the range

of this relation

Solution



16

36

49

64

4

6

7

8

11


Range


(b) (i) The domain is 16 36 49 64










(a) (b)


3

4

5

1

3

1

4

1

5

10

21

2

3

5

7





4




(4 8) (4 10) (4 12) (6 10) (7 12)

(b) the graph

Set X

Set Y

M

P

5 7 9 12 15Solution

Types of Relations


Each object in the

domain has only one

image in the

codomain


There are more than

one object in the


same image in the

codomain




that has more than




that has more than

one image in the




linked to more than

one object in the

domain

Domain Codomain

Domain Codomain

Domain Codomain

Domain Codomain



Functions 4

11



(b) ordered pairs

(c) a graph



B = 1 2 3 4




Y = 1 3 4



digits of rsquo


(b) State

(i) the domain

(ii) the codomain



(v) the range

of the relation




Sarawak

Perak

Kedah

Kuching

Ipoh

Alor Setar


methane

water

carbondioxide

carbon

hydrogen

oxygen



(4 0) (4 1) (5 2) (6 7) (6 8)


(a)4

6

7

8

10

12



(b) X Y

5

7

9

12

15

P

M



1


from set A = 3 5 7 to set B = 9 15 21

(3 9) (5 15) (7 21)


(b) State the range


Solution

3

5

7

9

15

21

A B




f ( x) = 3 x


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5 Functions

(b) the graph

3 4 6 7 9

Set B

Set A

11

10

8


from set P = 2 6 10 to set Q = 1 3 5

(2 1) (6 3) (10 5)


(b) State the range





x

y

z

a

b

c

a

b

c

d

w

x

y

z





12 Functions

Facts


Facts



a

b

x

y

z



a

b

c

x

y

z



x

y

z

a

b

c

d


5


(a) (b) (c)

A Blsquohasrsquo

January

June

August

28 days

30 days

31 days


Bronze

Brass

cuprum

stanum

zinc

1

3

57

3

9

15


Solution












Functions 6

6



(a) the image of 2




When x = 2

f (2) = 22 ndash 4(2) ndash 1

= 4 ndash 8 ndash 1

= ndash5


(b)

When f ( x ) = 4



( x + 1)( x ndash 5) = 0

x = ndash1 or 5


ndash1 or 5

Object


7



ndash3

x f

ax +b

ndash4

ndash5

ndash3

15



undefined

Solution



f (ndash3) = ndash5


15a ndash 5b = 15

3a ndash b = 3 1

f (ndash4) = ndash3


12a ndash 3b = 15

4a ndash b = 5 2


a = 2


b = 3

(b)



2 x + 3 0


2


2



2








7 Functions

9



the image of 30deg




Solution



= 3(05) ndash 05774

= 09226


(b) g( x ) = 16

2 cos x = 16


2

cos x = 08

x = 3687deg


image 16 is 3687deg

Press SHIFT cos

08 =


10





(a) the domain

(b) the range

(c) the image of 0


(ii) ndash3

Solution






8






under f find


itself

Solution


Denominator ne 0

2 x ndash 4 ne 0

x ne 2


(b)


thus f (5) = 5



15 + k = 30

k = 15


f ( x ) = x


3 x + 15 = 2 x 2 ndash 4 x





x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5




Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


SPMClone

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3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


SPMClone

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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



3 Functions

2





(b) State

(i) the domain

(ii) the codomain


(v) the object of 7


(vii) the range

of this relation

Solution



16

36

49

64

4

6

7

8

11


Range


(b) (i) The domain is 16 36 49 64










(a) (b)


3

4

5

1

3

1

4

1

5

10

21

2

3

5

7





4




(4 8) (4 10) (4 12) (6 10) (7 12)

(b) the graph

Set X

Set Y

M

P

5 7 9 12 15Solution

Types of Relations


Each object in the

domain has only one

image in the

codomain


There are more than

one object in the


same image in the

codomain




that has more than




that has more than

one image in the




linked to more than

one object in the

domain

Domain Codomain

Domain Codomain

Domain Codomain

Domain Codomain



Functions 4

11



(b) ordered pairs

(c) a graph



B = 1 2 3 4




Y = 1 3 4



digits of rsquo


(b) State

(i) the domain

(ii) the codomain



(v) the range

of the relation




Sarawak

Perak

Kedah

Kuching

Ipoh

Alor Setar


methane

water

carbondioxide

carbon

hydrogen

oxygen



(4 0) (4 1) (5 2) (6 7) (6 8)


(a)4

6

7

8

10

12



(b) X Y

5

7

9

12

15

P

M



1


from set A = 3 5 7 to set B = 9 15 21

(3 9) (5 15) (7 21)


(b) State the range


Solution

3

5

7

9

15

21

A B




f ( x) = 3 x


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5 Functions

(b) the graph

3 4 6 7 9

Set B

Set A

11

10

8


from set P = 2 6 10 to set Q = 1 3 5

(2 1) (6 3) (10 5)


(b) State the range





x

y

z

a

b

c

a

b

c

d

w

x

y

z





12 Functions

Facts


Facts



a

b

x

y

z



a

b

c

x

y

z



x

y

z

a

b

c

d


5


(a) (b) (c)

A Blsquohasrsquo

January

June

August

28 days

30 days

31 days


Bronze

Brass

cuprum

stanum

zinc

1

3

57

3

9

15


Solution












Functions 6

6



(a) the image of 2




When x = 2

f (2) = 22 ndash 4(2) ndash 1

= 4 ndash 8 ndash 1

= ndash5


(b)

When f ( x ) = 4



( x + 1)( x ndash 5) = 0

x = ndash1 or 5


ndash1 or 5

Object


7



ndash3

x f

ax +b

ndash4

ndash5

ndash3

15



undefined

Solution



f (ndash3) = ndash5


15a ndash 5b = 15

3a ndash b = 3 1

f (ndash4) = ndash3


12a ndash 3b = 15

4a ndash b = 5 2


a = 2


b = 3

(b)



2 x + 3 0


2


2



2








7 Functions

9



the image of 30deg




Solution



= 3(05) ndash 05774

= 09226


(b) g( x ) = 16

2 cos x = 16


2

cos x = 08

x = 3687deg


image 16 is 3687deg

Press SHIFT cos

08 =


10





(a) the domain

(b) the range

(c) the image of 0


(ii) ndash3

Solution






8






under f find


itself

Solution


Denominator ne 0

2 x ndash 4 ne 0

x ne 2


(b)


thus f (5) = 5



15 + k = 30

k = 15


f ( x ) = x


3 x + 15 = 2 x 2 ndash 4 x





x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5




Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


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3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

rsquo06






kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 4

11



(b) ordered pairs

(c) a graph



B = 1 2 3 4




Y = 1 3 4



digits of rsquo


(b) State

(i) the domain

(ii) the codomain



(v) the range

of the relation




Sarawak

Perak

Kedah

Kuching

Ipoh

Alor Setar


methane

water

carbondioxide

carbon

hydrogen

oxygen



(4 0) (4 1) (5 2) (6 7) (6 8)


(a)4

6

7

8

10

12



(b) X Y

5

7

9

12

15

P

M



1


from set A = 3 5 7 to set B = 9 15 21

(3 9) (5 15) (7 21)


(b) State the range


Solution

3

5

7

9

15

21

A B




f ( x) = 3 x


SPMClone

rsquo07

SPMClone

rsquo06





5 Functions

(b) the graph

3 4 6 7 9

Set B

Set A

11

10

8


from set P = 2 6 10 to set Q = 1 3 5

(2 1) (6 3) (10 5)


(b) State the range





x

y

z

a

b

c

a

b

c

d

w

x

y

z





12 Functions

Facts


Facts



a

b

x

y

z



a

b

c

x

y

z



x

y

z

a

b

c

d


5


(a) (b) (c)

A Blsquohasrsquo

January

June

August

28 days

30 days

31 days


Bronze

Brass

cuprum

stanum

zinc

1

3

57

3

9

15


Solution












Functions 6

6



(a) the image of 2




When x = 2

f (2) = 22 ndash 4(2) ndash 1

= 4 ndash 8 ndash 1

= ndash5


(b)

When f ( x ) = 4



( x + 1)( x ndash 5) = 0

x = ndash1 or 5


ndash1 or 5

Object


7



ndash3

x f

ax +b

ndash4

ndash5

ndash3

15



undefined

Solution



f (ndash3) = ndash5


15a ndash 5b = 15

3a ndash b = 3 1

f (ndash4) = ndash3


12a ndash 3b = 15

4a ndash b = 5 2


a = 2


b = 3

(b)



2 x + 3 0


2


2



2








7 Functions

9



the image of 30deg




Solution



= 3(05) ndash 05774

= 09226


(b) g( x ) = 16

2 cos x = 16


2

cos x = 08

x = 3687deg


image 16 is 3687deg

Press SHIFT cos

08 =


10





(a) the domain

(b) the range

(c) the image of 0


(ii) ndash3

Solution






8






under f find


itself

Solution


Denominator ne 0

2 x ndash 4 ne 0

x ne 2


(b)


thus f (5) = 5



15 + k = 30

k = 15


f ( x ) = x


3 x + 15 = 2 x 2 ndash 4 x





x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5




Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


SPMClone

rsquo07

3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

SPMClone

rsquo08

x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


SPMClone

rsquo06

g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


SPMClone

rsquo07

14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

Clone

rsquo03

SPMClone

rsquo05

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SPMClone

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


SPMClone

rsquo04


x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

Clone

rsquo05

13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


SPMClone

rsquo11



2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

rsquo09

SPMClone

rsquo10

1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

SPMClone

rsquo04

SPMClone

rsquo06

SPMClone

rsquo07



21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

rsquo06

SPMClone

rsquo07

SPMClone

rsquo08




find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

SPMClone

rsquo04

SPMClone

rsquo07

SPMClone

rsquo08

SPMClone

rsquo10

SPMClone

rsquo09


x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

rsquo03

SPMClone

rsquo08

SPMClone

rsquo04

SPMClone

rsquo05






c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

rsquo05

(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

SPMClone

rsquo05

SPMClone

rsquo09

SPMClone

rsquo10



23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

rsquo06






kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



5 Functions

(b) the graph

3 4 6 7 9

Set B

Set A

11

10

8


from set P = 2 6 10 to set Q = 1 3 5

(2 1) (6 3) (10 5)


(b) State the range





x

y

z

a

b

c

a

b

c

d

w

x

y

z





12 Functions

Facts


Facts



a

b

x

y

z



a

b

c

x

y

z



x

y

z

a

b

c

d


5


(a) (b) (c)

A Blsquohasrsquo

January

June

August

28 days

30 days

31 days


Bronze

Brass

cuprum

stanum

zinc

1

3

57

3

9

15


Solution












Functions 6

6



(a) the image of 2




When x = 2

f (2) = 22 ndash 4(2) ndash 1

= 4 ndash 8 ndash 1

= ndash5


(b)

When f ( x ) = 4



( x + 1)( x ndash 5) = 0

x = ndash1 or 5


ndash1 or 5

Object


7



ndash3

x f

ax +b

ndash4

ndash5

ndash3

15



undefined

Solution



f (ndash3) = ndash5


15a ndash 5b = 15

3a ndash b = 3 1

f (ndash4) = ndash3


12a ndash 3b = 15

4a ndash b = 5 2


a = 2


b = 3

(b)



2 x + 3 0


2


2



2








7 Functions

9



the image of 30deg




Solution



= 3(05) ndash 05774

= 09226


(b) g( x ) = 16

2 cos x = 16


2

cos x = 08

x = 3687deg


image 16 is 3687deg

Press SHIFT cos

08 =


10





(a) the domain

(b) the range

(c) the image of 0


(ii) ndash3

Solution






8






under f find


itself

Solution


Denominator ne 0

2 x ndash 4 ne 0

x ne 2


(b)


thus f (5) = 5



15 + k = 30

k = 15


f ( x ) = x


3 x + 15 = 2 x 2 ndash 4 x





x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5




Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


SPMClone

rsquo07

3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

SPMClone

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


SPMClone

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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


SPMClone

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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

Clone

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


SPMClone

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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

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SPMClone

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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 6

6



(a) the image of 2




When x = 2

f (2) = 22 ndash 4(2) ndash 1

= 4 ndash 8 ndash 1

= ndash5


(b)

When f ( x ) = 4



( x + 1)( x ndash 5) = 0

x = ndash1 or 5


ndash1 or 5

Object


7



ndash3

x f

ax +b

ndash4

ndash5

ndash3

15



undefined

Solution



f (ndash3) = ndash5


15a ndash 5b = 15

3a ndash b = 3 1

f (ndash4) = ndash3


12a ndash 3b = 15

4a ndash b = 5 2


a = 2


b = 3

(b)



2 x + 3 0


2


2



2








7 Functions

9



the image of 30deg




Solution



= 3(05) ndash 05774

= 09226


(b) g( x ) = 16

2 cos x = 16


2

cos x = 08

x = 3687deg


image 16 is 3687deg

Press SHIFT cos

08 =


10





(a) the domain

(b) the range

(c) the image of 0


(ii) ndash3

Solution






8






under f find


itself

Solution


Denominator ne 0

2 x ndash 4 ne 0

x ne 2


(b)


thus f (5) = 5



15 + k = 30

k = 15


f ( x ) = x


3 x + 15 = 2 x 2 ndash 4 x





x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5




Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


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3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

SPMClone

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


SPMClone

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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


SPMClone

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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


SPMClone

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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



7 Functions

9



the image of 30deg




Solution



= 3(05) ndash 05774

= 09226


(b) g( x ) = 16

2 cos x = 16


2

cos x = 08

x = 3687deg


image 16 is 3687deg

Press SHIFT cos

08 =


10





(a) the domain

(b) the range

(c) the image of 0


(ii) ndash3

Solution






8






under f find


itself

Solution


Denominator ne 0

2 x ndash 4 ne 0

x ne 2


(b)


thus f (5) = 5



15 + k = 30

k = 15


f ( x ) = x


3 x + 15 = 2 x 2 ndash 4 x





x f

23

2 x 2 ndash 5

1

0

ndash1

ndash2

ndash3

ndash5




Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


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3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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SPMClone

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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 8

12


function


6

14

15

24

multiple of 3

multiple of 7

(b)

Consonants

Vowels

i

v

w

u


(c)

21

32

5364

2

6

15



PERAK

KEDAH

SELANGOR

2 vowels

3 vowels

4 vowels


2 x ndash 9 2



x ndash b




undefined

4 The arrow diagram

shows the function






x = 2


x + m


(a) the image of 10



undefined


f (3) = 4 find



3


g(ndash2) = 3 find







b ndash x


f



undefined


shows the function

f x rarr px + qx 2

Find


of q







an acute angle




(a) the domain

(b) the range

(c) the image of 0

(d) the objects of

(i) 3

(ii) 6

ndash5

7

2

ndash1

x f b ndash

x ax +

ndash5

ndash16

ndash1

ndash2

x f

px + qx 2

6

x f x 2 + 2

32

1

0

ndash1

ndash2

2



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


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3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



9 Functions




x if x 0 983135 x 983135 =

ndash x if x 0


f ( x) if f ( x) 0 983135 f ( x)983135 =


Facts


9831355983135 = 5 and 983135ndash5983135 = 5

Facts


2



(i) ndash1 (ii) 4


Solution

(a) f x rarr 9831354 x ndash 5983135



(ii) f (4) = 9831354(4) ndash 5983135 = 98313511983135 = 11


(b)

f ( x ) = 3

9831354 x ndash 5983135 = 3



4 x ndash 5 = 3

4 x = 8

x = 2



4 x = 2

x =

1


2 or 1ndashndash2


SPMClone

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3


value functions




Solution



below


y

x

4

3

2

1

Range

Domain

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x


f ( x ) 1 0 1 2 3 4




Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 10


x f ( x ) g[f ( x )] = gf ( x )

gf

A B C

f g



Facts


13


image of

(a) ndash3 (b) 0 (c) 2


(a) the image of

(i) 2 (ii) ndash2










f ( x ) is 0 f ( x) 4

(b)


Thus 9831354 x ndash 7983135 = 0

4 x ndash 7 = 0

4 x = 7


4 4




below

O 1 2 3 4 x

y

10

8

6

4

2

3ndash4

1

Range

Domain


f ( x ) is 0

f ( x)

9

x 0 1 2 3 4

f ( x ) 7 3 1 5 9






11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



11 Functions

11





(i) fg (iii) f 2

(ii) gf (iv) g2


(a) fg(3) = f [g(3)] = f (32 ndash 2) = f (7) = 2(7) + 1

= 15


= 7


f ( x ) = 2 x + 1 g( x ) = x 2 ndash 2

(i) fg( x )

= f [g( x )]

= f ( x 2 ndash 2)

= 2( x 2 ndash 2) + 1

= 2 x 2 ndash 3


(ii) gf ( x )

= g[ f ( x )]

= g(2 x + 1)

= (2 x + 1)2 ndash 2

= 4 x 2 + 4 x + 1 ndash 2

= 4 x 2 + 4 x ndash 1


(iii) f 2( x )

= ff ( x )

= f (2 x + 1)

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


(iv) g2( x )

= gg( x )

= g( x 2 ndash 2)



= x 4 ndash 4 x 2 + 2


(c) gf ( x ) = 23

4 x 2 + 4 x ndash 1 = 23

4 x 2 + 4 x ndash24 = 0

x 2 + x ndash 6 = 0

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

x = 2 or ndash3





darrwith (2 x + 1)


darrwith (2 x + 1)



12



(a) f 2 (b) f 8 (c) f 9

Solution



f 2( x ) = ff ( x )


x + 1



2 x = ndashndash 2

= x



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 12

4








Solution






f [g( x )] = 3 x ndash 8

2g( x ) + 4 = 3 x ndash 8

2g( x ) = 3 x ndash 12













1


5 ndash 3 x 3

51


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g is inside


darrwith g( x )

g is outside




(b) f 8( x ) = f 2 f 2 f 2 f 2( x )

= f 2 f 2 f 2( x )

= f 2 f 2( x )

= f 2( x )

= x

(c) f 9( x ) = ff 8( x )

= f ( x )


f 2( x ) = x

f 8( x ) = x





13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

SPMClone

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



13 Functions


5





Solution

(a) g2(2) = gg(2)

= g[(2 + 1)2 + 2]

= g(11)

= (11 + 1)2 + 2

= 146

(b)


g x rarr ( x + 1)2 + 2

Thus fg( x ) = f [( x + 1)2 + 2]

= h[( x + 1)2 + 2] + k

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


Hence by comparison

h = 2 and 2h + k = 1

2(2) + k = 1

k = ndash3


SPMClone

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14


g x rarr 983152 983151983151983151983151983151983151983151


(a) fg(6) (c) f 2(0)

(b) gf (2) (d) g2(27)






x 2 + 2



13



find gf ( x )

Solution

Find g( x ) first


f (g( x )) = 22 ndash 3 x

3g( x ) + 7 = 22 ndash 3 x


3g( x ) = 15 ndash 3 x


g( x ) = 5 ndash x

Hence gf ( x ) = g(3 x + 7)

= 5 ndash (3 x + 7)


= ndash3 x ndash 2


darr with g( x )






Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


SPMClone

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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

SPMClone

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

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SPMClone

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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

SPMClone

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SPMClone

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

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gh(2)

[2 marks]

SPMClone

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 14


x y

f

f ndash1




(a) f = g (b) f 2 = g2





following equations

(a) f 2 = f (b) g2 = g



Find



(i) fg = 9

(ii) gf = 14



g x rarr 2 x



(a) fg (c) g6

(b) g2 (d) g7




(a) f 2 (c) f 16

(b) f 4 (d) f 17




fg x rarr 2 x + 5

x ndash 2 x ne 2




(a) f x rarr


x ndash 2


11 5

10 x ndash 1gf x 10

1







fh x rarr 2 x 2

(c) f x rarr x 2












values of k


that f ( x 2) = 8 x



Facts


bull (f2

)ndash1

= (fndash1

)2


Facts






15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


SPMClone

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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

Clone

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


SPMClone

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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

SPMClone

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SPMClone

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

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SPMClone

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SPMClone

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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

SPMClone

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SPMClone

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SPMClone

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

SPMClone

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

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gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



15 Functions



there4 f ndash1

Method 2


Thus f ( y) = x


2 y ndash 1 = x ( y + 2)



y(2 ndash x ) = 2 x + 1






14



2 x ndash 1

x + 2

Solution

Method 1



y( x + 2) = 2 x ndash 1











Factorise the

left-hand side


15


x


Solution


f ( x ) = 6 ndash 2 x


Thus f ( y) = x

6 ndash 2 y = x

6 ndash x = 2 y


6 ndash x there4

f

ndash1




Thus g( y) = x

1ndashndash y

= x

y =1ndashndash x



6 SPM

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Solution

Let y = f ndash1

(3) f ( y) = 3


y + 8 = 3( y ndash 6)

y + 8 = 3 y ndash 18

2 y = 26

y = 13




Functions 16

7


SPMClone

rsquo04


x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

Clone

rsquo05

13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


SPMClone

rsquo11



2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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SPMClone

rsquo10

1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 16

7


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x hx ndash 5

x ne 2 x ndash 2






t

Solution









x ndash h




t


4ndashndash3

t t ndash 2



0 = 4t 2 ndash 17t + 15







6 x ndash 1


8 SPM

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13ndash2

10

x

f

y z

g




the function


3 y ne

3 y ndash q




Solution




ndashndash2 = 10

5

3ndashndash2







17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



17 Functions






16


2 x + 1 x ne 1

x ndash 1



Solution


k

k = f 4ndashndash3

k


4k ndashndash3

ndash 1








1 = 9 ndash 2q

2q = 8 q = 4

From 1 When q = 4 3 p + 2(4) = 2

p = ndash2




Thus f (w) = x

ndash2w + 4 = x

4 ndash x = 2w

4 ndash x




x y z

gf ndash1

gf ndash1

gf ndash1( x )






4 ndash 3 x x ne

4

3

9


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2 x ndash 1 x ne

1

2 find g( x )

Solution


y(2 x ndash 1) = 2 x + 1



x (2 y ndash 2) = y + 1




2 x ndash 2 x ne 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

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gh(2)

[2 marks]

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

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Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 18

19


4k 2 ndash 3k = 8k + 3


(4k + 1)(k ndash 3) = 0







Solution


f ( y) = x

1 ndash 3 y


y 1 ndash 3 y = xy

1 = xy + 3 y

1 = y( x + 3)






18


same form

Solution

Find f 2( x ) first

f 2( x ) = ff ( x )

= f (5 x ndash 2)


= 25 x ndash 12



Thus f 2( y) = x

25 y ndash 12 = x

25 y = x + 12


x + 12 ( f

2

)ndash1




2

)ndash1




17





k

Solution



4 y = 2 ndash kx





k


3ndashndash2

k k

f ndash2mdashmdash


k

3ndash2mdashmdash


k

ndash6

k =

ndash3k

2

ndash3k 2 = ndash12

k 2 = 4

k = plusmn 2




19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

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SPMClone

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SPMClone

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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

SPMClone

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SPMClone

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

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by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



19 Functions


15


functions



x


2

3 x + 2

2 x + 3


following functions



(c)3


(d) x ne3

55 x ndash 3m xrarr

2 x + 2

(e) n x rarr 983152 983151983151983151983151983151983151983151



4mdash x

f xrarr x ne 0 and




(c) f 2 (f) gndash1


x ndash 2

x ne k find

(a) the value of k

(b) f 2

(c) f ndash1



x ndash 1


(b) gndash1 (d) gf




x + 2 x ne 2 find

(a) f ndash1 (d) fg












x + p x ne 5 and

x ndash 5


x ndash 1

x ne 1 find

(a) the value of p

and ofq


10

ndash2

2

ndash1

x y z





n








k 2 ndash 7 = 3 ndash (k 2 + 4k + 4)


2k 2 + 4k ndash 6 = 0

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0

k = ndash3 or 1




Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

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SPMClone

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SPMClone

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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

SPMClone

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

SPMClone

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

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kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

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gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 20



[2 marks]



of a graph

n

m

k

h

2 4 6 8

Set Y

Set X

State


ordered pairs



[3 marks]

12 Functions


9 x f a

10bx ndash 2

4

1 1



bx ndash 2 x ne k


Find


(b) the value of k

SPMClone

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SPMClone

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1

11 Relations

1

15

24

35

45

P Q

6

7

8

9





(c) the codomain



1

5

6

8

A B

a

e


using ordered pairs


relation

3 Set Q

11

9

7

5

Set P 2 6 8


Q = 5 7 9 11 State

(a) the images of 6


(c) the range

4 P = 2 6 8


rsquo03




(6 5) (6 7) State

(a) the images of 2

(b) the object of 7

[2 marks]



x

y

z

10

2030

40

Set P Set Q

State



[2 marks]



3

2

ndash2

ndash3

81

16

Set P Set Q





set P and set Q

[2 marks]



x f ( x )

2

5

7

k

1

4

6

8

SPMClone

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SPMClone

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21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

rsquo06

SPMClone

rsquo07

SPMClone

rsquo08




find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

SPMClone

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SPMClone

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SPMClone

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SPMClone

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SPMClone

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

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SPMClone

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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

SPMClone

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

rsquo06






kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



21 Functions




0 x 4




x if g( x ) = 4


m x rarr x ndash h

h where h is

a constant

7

x x ndash hm

2

h



f ( x ) = 9 [2 marks]



y

x

3

4O k




domain [3 marks]





SPMClone

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SPMClone

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SPMClone

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find f 2



and of p



function g




function g





hg( x ) = 35 x + 13 find

(a) g( x )


gh( x ) = 4

[4 marks]





[3 marks]



(a) f (6)


gf (6) = 24

[3 marks]



(a) hg ( x )


[4 marks]



x + 5


f ndash1(2)

SPMClone

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SPMClone

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x + 4



x ndash3


f ndash1g

1ndashndash

2




find the value of h

and of k


px + q

f (1) = 8 find





(a) f ( x )



32

x fndash14 x + p

ndash2

ndash6






3 x + 1

(a) gndash1 f ( x )






Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

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SPMClone

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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

SPMClone

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SPMClone

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SPMClone

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

rsquo06






kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



Functions 22



(a) gndash1(7)

(b) hg( x )

[4 marks]


ndash 3



x ne ndashk



[3 marks]




SPMClone

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SPMClone

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SPMClone

rsquo04

SPMClone

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c ndash x

x ne c respectively

ndash2

ndash 4

4

17

ndash3

m

g

g

f

f

Find















fg =gf



1 ndash x


following

(a) fg

(b) f 2

(c) f 13

(d) g

2

(e) g3

(f) g19




Find

(a) the function g



6

5

89

x y

f

g

z






12 ndash y

y ne 12 Find



that maps x onto z



onto z






find



x y f g

z

4

1

ndash2


(b) gf (1)

[2 marks]


by h( x ) = 3 x + 5 and w( x ) =2

1 ndash 4 x

SPMClone

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(a) hndash1(6)



and n xrarr 3

mdash x


[3 marks]


find(a) g(ndash4)


gndash1( p) = 8

[3 marks]



(a) gndash1( x )

(b) hgndash1(11)

[3 marks]

SPMClone

rsquo05

SPMClone

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SPMClone

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23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

rsquo06






kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1



23 Functions





gf (n2 ndash 1) = 6n + 6



f 2



of b



f (2k + 1) = k (k + 2)










defined by

g xrarr12

mdashmdashmdash

x + 1

x ne ndash1



terms of m and k



and of k


x ndash 2


x

kx + 3 x ne 0



x + q






14

7

5

3

ndash1

A B

f

gndash1

C



f x rarr


onto itself find

(a) the value of a






(b) (i) f (3) (ii) f ndash1 f (3)





2 ndash x



f (k ) = k + 2





function g y rarr6

2 y ndash k y ne

k

2

3 2

ndash2

x y z

Find



onto y


onto z



x + 3

(a) f ndash1g( x )








2




g( x 2) = 2 f (ndash x )



find




1 x ne k

k ndash x



(b) the value k if

ff ndash1(k 2 + 2) = g[(5 + k )2]


g x rarr x

2 + 2 find

(a) f ndash1

g [3 marks]


hg x rarr 2 x + 4

[3 marks]

SPMClone

rsquo06






kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1






kx

x ndash 2 x ne 2



x x ne 0









Long Questions

Paper 2


X = 6 12 15 21 and set





[3 marks]

SPMClone

rsquo11



gh(2)

[2 marks]

SPMClone

rsquo11


by hndash1 x rarr 3


(a) h( x )


h( x ) = ndash14

[4 marks]

SPMClone

rsquo11

Short Questions

Paper 1


between set A and B

40

30

20

10

S e t B

Set A

4321

State



[2 marks]



of ordered pairs


(3 9) (4 16)


relation



[2 marks]








(a) g( x )

(b) fg(4) [4 marks]


2 x + p

5 and


5 x + 3

q


[4 marks]

Short Questions

Paper 1

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Success Additional Mathematics SPM Free Chapte