LABUAN ADDITIONAL MATHEMATICS PROJECT WORK 2015
SUGGESTED ANWSER BY www.addmathsprojectwork2015.blogspot.com
PART 1
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a) Sets of Son = {C,D}
Sets of Daughter = {A,B}
Sets of Grand-son = {G,H,J,K}
Sets of Grand-daughter = {E,F,L}
Sets of Great Grand-daughter = {M,P}
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Daughter A
Daughter B
Son C
Son D
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



Zulkifli
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b) 1. Father of
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(i) Domain
= { Zulkefli }
Codomain = { Daughter A, Daughter B, Son C, Son D }
Objects
= Zulkefli
Images
= Daughter A, Daughter B, Son C, Son D
Range
= { Daughter A, Daughter B, Son C, Son D }
(ii) Type of Relation = One – to – many Relation
(iii) { (Daughter A, Zulkefli) , (Daughter B, Zulkefli) , (Son C, Zulkefli) , (Son D, Zulkefli) }
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Son C
Son D
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2. Son of
 Zulkefli
 Aminah
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(i) Domain
= { Son C, Son D }
Codomain = { Zulkefli, Aminah }
Objects
= Son C, Son D
Images
= Zulkifli, Aminah
Range
= { Zulkifli, Aminah}
(ii) Type of Relation = Many – to – many Relation
(iii) { (Son C, Zulkefli) , (Son C, Aninah), (Son D, Zulkefli) , (Son D, Aninah) }
3. Grand-daughter of
Grand-daughter E
Grand-daughter F
Grand-daughter L
 Zulkefli
 Aminah
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(i) Domain
= { Grand-daughter E, Grand-daughter F, Grand-daughter L }
Codomain = { Zulkefli, Aminah }
Objects
= Grand-daughter E, Grand-daughter F, Grand-daughter L
Images
= Zulkifli, Aminah
Range
= { Zulkifli, Aminah}
(ii) Type of Relation = Many – to – many Relation
(iii) { (Grand-daughter E, Zulkefli) , (Grand-daughter E, Aninah), (Grand-daughter F, Zulkefli) ,
(Grand-daughter F, Aninah), (Grand-daughter L, Zulkefli) , (Grand-daughter L, Aninah) }
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c) A function is a relation in which NO two ordered pairs have the same first coordinate. So, from the
three arrow diagrams stated above, only the first relation (Relation 1 – Father of) arrow diagram is
considered a function. This is because, when observing the first and second arrow diagram :
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1- { (Daughter A, Zulkefli) , (Daughter B, Zulkefli) , (Son C, Zulkefli) , (Son D, Zulkefli) }
The first coordinates does not have any repetition.
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2- { (Son C, Zulkefli) , (Son C, Aninah), (Son D, Zulkefli) , (Son D, Aninah) }
The first coordinates have repetition for both Son C and Son D.
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This proof that, there is only one relation (Relation 1) which is a function.
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PART 2 – DO IT ON YOUR OWN
PART 3
a) Function of f
f(x) = x + 3
Function of g
g(x) = x2
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b) Inverse function of f
f ( x)  x  3
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1
( x)  x  3
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Let f 1 ( x)  y
f ( y)  x
y 3 x
y x3
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Inverse function of g
g ( x)  x 2
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Let g 1 ( x)  y
or
g ( y)  x
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y2  x
g 1 ( x)  x
fg ( x)  f [ g ( x)]
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fg ( x)  x 2  3
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c) Composite function of fg
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y x
Composite function of gf
gf ( x)  g[ f ( x)]
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gf ( x)  ( x  3) 2
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gf ( x)  x 2  6 x  9
Composite function of f2
Composite function of g2
ff ( x)  f [ f ( x)]
ff ( x)  ( x  3)  3
gf ( x)  g[ f ( x)]
ff ( x)  x  6
gf ( x)  x 4
gf ( x)  ( x 2 ) 2
d) Composite function of g-1f-1
g 1 f
1
( x)  g 1 [ f
1
g 1 f
1
( x)  x  3
( x)]
Inverse function / (fg)-1
( fg ) 1 ( x)  x 2  3
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Let ( fg ) 1 ( x)  y
fg ( y )  x
y2  3  x
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y x3
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( fg ) 1 ( x)  x  3
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Yes g-1f-1(x) = (fg)-1(x)
e) Composite function of ff-1
ff
( x)  f [ f
1
( x)  ( x  3)  3
1
( x)  x
( x)]
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1
k.
ff
1
or
ff
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Composite function of gg-1
gg 1 ( x)  g[ g 1 ( x)]
gg 1 ( x)  x
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gg 1 ( x  1)  x  1
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gg 1 ( x)  ( x ) 2
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Therefore, ff-1 is not equal to gg-1
Deduce ff-1 (k2 + 2)
( x)  x
1
(k 2  2)  k 2  2
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1
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FURTHER EXPLORATION
x 1
, x  1
x 1
x 1
1
g 2 : x  gg : x  x  1
x 1
1
x 1
g:x

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i)
x  1  1( x  1)
x 1
x
x 1
x  1  1( x  1)
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a)
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1
1

x
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g4 : x  g2g2 : x 
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 x 1
 

 x 1 
k.
1
x 1
x 1
or
g3 : x  g2g : x  
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  2  x  1 



 x  1  2 x 
1

x
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x
x 1
x 1
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g5 : x  g4g : x 
x 1
1
6
5
x

1
g : x g g : x 
x 1
1
x 1

x  1  1( x  1)
x 1
x
x 1
x  1  1( x  1)
  2  x  1 



 x  1  2 x 
1

x
By finding all the functions of g2,g3,g4,g5, and g6 we can observe the trend of the function
Power of
2,6,10,14,18,22,26,30
Composite Function
1
x
 x  1


 x  1
x
x 1
x 1

3,7,11,15,19,23,27
g
4,8,12,16,20,24,28
5.9,13,17,21,25,29
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1
or g 2 n : x  x
x
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g 2n : x  
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or
x 1
 x  1
2 n 1
g 2 n 1 : x  
:x
 or g
x 1
 x  1
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ii)
1
x
k.
Thus, g 30 : x  
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b)
ax  b
cx  d
1
f ( x)  y
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ay  b
x
cy  d
ay  b  cyx  bx
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f :x
f ( y)  x
ay  cyx  bx  b
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y (a  cx )  bx  b
bx  b
a  cx
bx  b
a
f 1 ( x) 
,x
a  cx
c
y
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function
i) Inverse of f : x 
2x  1
x3
a
,
c
3x  1
2x
a 2
where 
c 1
3x  1
thus f 1 ( x) 
,x2
2x
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or
( x) 
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1
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f
k.
check
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based on the inverse function above, x ≠
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f 1 ( x)  y
f ( y)  x
2y 1
x
y3
2 y  1  xy  3 x
2 y  xy  3 x  1
y (2  x)  3 x !
3x  1
y
2x
3x  1
f 1 ( x) 
2x
ii) Inverse of f : x 
3x
1  4x
x
3  4x
a 3
where 
c 4
3x  1
3
thus f 1 ( x) 
,x
2x
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( x) 
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1
or
check
f
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a
,
c
k.
based on the inverse function above, x ≠
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f 1 ( x)  y
f ( y)  x
3y
x
1 4y
3 y  x  4 xy
3 y  4 xy  x
y (3  4 x)  x !
x
y
3  4x
x
f 1 ( x) 
3  4x
iii) Inverse of f : x 
2x
5x  2
a
,
c
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based on the inverse function above, x ≠
2x
2  5x
a 2
where 
c 5
2x
2
thus f 1 ( x) 
,x
2  5x
5
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or
( x)  
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1
k.
check
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REFLECTION
WRITE YOUR OWN REFLECTION
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f 1 ( x)  y
f ( y)  x
2y
x
5y  2
2 y  5 xy  2 x
2 y  5 xy  2 x
y (2  5 x)  2 x!
2x
y
2  5x
2x
f 1 ( x)  
2  5x