VCE Mathematical Methods Unit 3 Workshop
Mathematical Methods Unit 3 Workshop
Solutions
Extended Response Question
a)
b)
c)
f)
V  lwh
1
V  l  l  (l  2), as w  12 l and h  l  2
2
1
V  l3  l2
2
A  lw  2lh  2wh
1
1
A  l  l  2l  (l  2)  2  l  (l  2)
2
2
1
A  l 2  2l 2  4l  l 2  2l
2
7
A  l 2  6l
2
A
e)
1 6  2 149 3 6  2 149 2
V (
) (
)
2
7
7
V  22.13 cm3
A  lw
6  2 149 1 6  2 149
)( 
)
7
2
7
A  9.44 cm 2
A(
Yes there is enough black plastic
Page 5
0  3x 2  x3
f '( x)  6 x  3 x 2
0  x 2 (3  x)
x  0, x  3
(0, 0) and (3,0)
0  6 x  3x 2
0  3 x(2  x)
x  0, x  2
(0, 0) and (2, 4)
7 2
l  6l
2
dA
 7l  6
dl
0  7l  6
6
l  cm
7
BUT
h l 2h 0l20
l2
l  2
7 2
d) A  l  6l
2
7
40  l 2  6l
2
6  2 149
l
7
6  2 149
l
, as l  2
7
l  (2,
1|P a g e
6  2 149
]
7
© Conquest Education
VCE Mathematical Methods Unit 3 Workshop
Page 7
Page 11
B
x  [0,  ]  2 x  [0, 2 ]
3 tan(2 x)  1
1
tan(2 x) 
3
 7
2x  ,
6 6
 7
x ,
12 12
kx  3  x 2  8 x
x 2  (8  k ) x  3  0
0
(8  k ) 2  12  0
k 2  16k  64  12  0
k 2  16k  52  0
k  8  2 3, k  8  2 3
Page 14
Page 8
B
f ( x)  ( x  (a ))( x  (b))( x  (c))( x  ( d ))
f ( x)  ( x  a )( x  b)( x  c)( x  d )
f ( x)  ( x  a )((b  x))((c  x))( x  d )
f ( x)  ( x  a )(b  x)(c  x )( x  d )
Page 9
Using long division
Asym : 2 x 
y  1
3
2x 1
Vertical: x 
1
2
Horizontal: y  1
2|P a g e



3 2
5
Asym : 2 x 
6
5
Asym : x 
12
Remaining asymptotes spaced one period apart
Asymptotes at:
x
7

5
11
, x , x
, x
12
12
12
12
© Conquest Education
VCE Mathematical Methods Unit 3 Workshop
  x    
7
 5
 2x  
3
3
3

0  2 tan(2 x  )
3

tan(2 x  )  0
3

(k  1)
x
4
3
(k  1)
y
x
(k  2)
(k  2)
y  2
m1  m2
 2 ,  , 0, 
3
5   2
x   , , ,
6
3 6 3
2x 
Page 18
(k  1)
3

4
( k  2)
(k  1)( k  2)  12

k 2  3k  2  12  0

f (0)  2 tan(0  )
3
f (0)  2 3
k 2  3k  10  0
(k  5)(k  2)  0
Page 16
k  5, k  2
c1  c2
log 2 ( x 2  x  2)  2 log 2 2
2
(k  1)
( k  2)
2k  4  k  1
k 5
x2  x  2  4
x2  x  6  0
( x  3)( x  2)  0
x  3, x  2
x  3 as x  2
 k  2
Page 19
f ( x)  sin( x)  x
2 f ( x)  2sin( x)  2 x
2 f ( x)  3  2sin( x)  2 x  3
Page 17
C
Reflected about the x-axis and translated to the
right.
_________________________________________
D
Sketch an imaginary graph with a local max with a
y-value of -3 and a local minimum with a y-value of
-8. Function cannot be shifted up between 3-8
units.
Therefore
3|P a g e
c  3 or c  8
© Conquest Education
VCE Mathematical Methods Unit 3 Workshop
Page 20
Domain of f ( x )
 x '   2 0    x  1  
T 
    
 y '  0 2    y  0  
 x '   2 0   x  1
T 


 y '  0 2   y 
 x ' 2 x  2
T 

 y '  2 y 
x  0
x0
x '  2x  2
y '  2 y
x ' 2
x
2
y'
y
2
y'
3( x ' 2)
  cos(
)
2
2
3
y  2 cos( ( x  2))
2
Page 22
Implied domain of f [ g ( x)]
3
 log e ( x  )  0
2
3
log e ( x  )  0
2
3
x  1
2
1
x
2
Domain of g ( x )
3
0
2
3
x
2
x
Therefore domain of f [ g ( x)] :
3 1
x  ( ,  ]
2 2
_________________________________________
Implied domain of f [ g ( x)]
3
 log e ( x  )  0
2
3
log e ( x  )  0
2
3
x  1
2
1
x
2
3
0
2
3
x
2
3
1
 x
2
2
x
Domain of f ( x )
1  x  0
3
0
2
3
x
2
3
1
 x
2
2
x
4|P a g e
Domain of g ( x )
1  x  
Therefore domain of f [ g ( x)] :
1
x  (1,  ]
2
© Conquest Education
VCE Mathematical Methods Unit 3 Workshop
Page 23
dy
 f (2 x)  g ''(h( x 2 ))  h '( x 2 )  2 x 
dx
2 f '(2 x)  g '(h( x 2 ))
dy
 f (2 x) g ''(h( x 2 ))h '( x 2 )2 x  2 f '(2 x) g '(h( x 2 ))
dx
A
f '( x)  3x 2  6 x
0  3x 2  6 x
0  x(3x  6)
x  0, x  2
Page 28
Therefore the first turning point from the left hand
side is at
x  0 , so a  0
f '( x)  x  2 x log e ( x)
0  x(1  2 log e ( x))
x  0 or 1  2 log e ( x)  0
Page 24
x  0 or x  e
E
xe
x 1
1
x

1
f ( f ( x)) 
x 1
1
x 1
x 1  x 1
f ( f ( x)) 
x  1  ( x  1)
2x
f ( f ( x)) 
2
f ( f ( x))  x


1
2

1
2
1
2

1
2
as Arg  0
1

1
2
f (e )  e log e (e )
f (e )  
(
1
2e
1
1
, )
e 2e
_________________________________________
g '( x)  x 2  2 cos(2 x)  sin(2 x)  2 x
g '( x)  2 x 2 cos(2 x)  2 x sin(2 x)
Page 25

Page 26
E
y  (1  f ( x))



1
2
1

dy 1
 (1  f ( x)) 2   f '( x)
dx 2
dy
 f '( x)

dx 2 1  f ( x)
Page 30
B
The derivative function has 2 x-ints so the original
function must have 2 turning points.
_________________________________________
5|P a g e

g '( )  2( ) 2 cos(2( ))  2( ) sin(2( ))
6
6
6
6
6
2



3
g '( ) 
 
6
36
3 2
2


 3
g '( ) 

6
36
6
2 x cos( x 2  2)
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