Transcript THE MAV REVISION LECTURE 2010

THE MAV REVISION LECTURE 2011
Mathematical Methods CAS Units 3 and 4
Allason McNamara
Lecture Notes
 MAV Revision Lecture Series 2011.doc
EXAMINATION 1
Structure of the paper
 Short answer questions – 40 marks
 Time limit: 60 minutes writing time
15 minutes reading time
Added Content to Exam 1
 General solutions to trigonometric equations
 Average value of a function
 Functional Equations
 Matrices – transformations, transition matrices and steady
state (Further Mathematics), solving simultaneous equations
(up to 5 unknowns)
 (Compound and Double Angle formulae)
Sample VCAA Questions and Frequently
Asked Questions
 Solutions to VCAA 2010 Sample Questions.doc
 mmcasfaqs2011.pdf
EXAMINATION 2
Structure of the paper
Part I
 22 multiple choice questions – 22 marks
Part II
 Extended answer questions – 58 marks
 Time limit:
120 minutes writing time
15 minutes reading time
General Advice
 No calculator syntax
 Always give exact answers unless the question asks for an
approximate answer.
 Draw graphs properly: scale axes, give coordinates if asked
for……
 For questions worth more than one mark show appropriate
working. On Exam 2, this is the rule you are using and the
answer.
 For “show that” questions show all the steps.
Algebra
Don’t forget to put the multiplication
sign between a and x and the b and x
on the calculator.
ax  bx
3
 x ( ax  b )
2
 x( a x 
E
b )( a x 
b)
f ( x )  2 x  3 x  7 x  11
4
3
x 1 0
x  1
f (  1)  2 (  1)  3 (  1)  7 (  1)  11
4
 2  3  7  11
9  0
No f is not exactly divisible by x + 1.
3
There are 4 types of relations (Horizontal line: Vertical Line Test)
many:many
circle
1:many
inverse is a function
y x
many:1
parabola
function
1:1
line
function
inverse is a function
If they ask for f give the domain and rule.
If they ask for f(x) give the rule.
Be careful if the square root is in the denominator.
1
2x  3
,2 x  3  0
Solving Equations 2010 Exam 2
dT
dx
8000 cos( x ) sin( x )( 3 cos ( x )  1)
2

3
1  2 cos
2
( x)

0
 If the calculator does not give a solution, just solve the
numerator equal to zero.
8000 cos( x ) sin( x )( 3 cos ( x )  1)  0
2
Using transformations
A
B
New to Exam 1 2010
x  4x  1
x 
A
x ' 1
4
y   2 y  3
y ' 3
 x ' 1 


2
 4 
y 
y ' 3
2
3
3
 ( x ' 1)
 x ' 1 
y '   2
3
 3
4
32


3
psandqs[1].doc
n: even over odd >1
n: odd over odd >1
psandqs[1].doc
n: odd over even less than 1
n: odd over odd less than one
a
b
D
c
e
5
x  2 5x
5 x  10
10
 Don’t flick your graphs back – show asymptotic behaviour
 Domain R\{b}
 Range R\{c}
 Give the equations of the asymptotes
 Scale the axes properly
 Domain R\{b}
 Range y > c
D
-1
y 
1
2 x
1 
1
x2
 x2 x3
1
x2
1
E
Check the a value.
A
New to Exam 1 2010
Note both curves go through the origin.
Solve the determinant equal to zero.
By-hand
a ( a  1)  3  2  0
a a60
2
( a  3 )( a  2 )  0
B
B
D
By-hand
Solve 2 k  1  k  1
 2k  1  k  1
y
2
|x – 3 x |
-1
1
2
3
4
x
2
x – 3x
x  3x  4  0
2
( x  4 )( x  1)  0
{ x : x  4}  { x : x   1}
 x  3x  4  0
2
No solution
g  f ( x ) 

x4
 x 1

2
3
y
g [ f( x ) ] = x – 1
6
5
4
3
(4, 3)
2
1
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
x
-1
-2
-3
-4
-5
-6
Domain is the same as the domain of f(x).
B
Which one is a many:1 function?
Asymptotes of f

x  2
y  1
Asymptotes of inverse
y  2
E
x  1
Swap x and y and solve for y on the calculator.
C
B
x  3e
f
1
2y
x 
D om f
1
 x 1
log e 

2
 3 
1
1
:   1,  
C
e
ax

2
5
2
ax  log e  
5
x 
C
2
log e  
a
5
1
 x2 
1
  log e  
log e 

2
 x 3
x
2
x3
2x
2

1
2x  x  3  0
2
( 2 x  3 )( x  1)  0
2
 x3
CHECK ANSWERS
x
3
2
or
x  1
D
Period 

b


1
4
D
 4
The amplitude is 2.
The graph has been translated 1 unit up.
The period is about 0.4. 2 
5
 0 .4
A
y 1   sin( 4 x )  1
y 2   sin( x )  1
Reflection in the x-axis
Dilation from the y-axis (reciprocal)
Change the domain.



0

4
,

4
 0 , 2  


2


E
2 x
 
3
2 x

, 2 
3

3
x  2,
4
,
3
5
2
....

3
5
3
...
...
Period 
6
2
3
x  2 or x  2 . 5
 The first maximum, where x > 0, of f occurs at
x 
P
4

3
4
The graph of f has been shifted one unit to the right to get the graph of g.
x 
3
4
1 
7
4
cos( x )(cos( x )  2 )  0
cos( x )  0 as cos( x )   2
A
New to Exam 1 2010
A
 

tan    
4
3




 
 
tan    tan  
3
4
 
 
1  tan   tan  
3
4
3 1
1
3 1
3 1
3 1
3 1
3 1

42 3
2
3 1
3 1
 2
3
New to Exam 1 2010
f ( u ) f ( v )  (1  e
 1 e
u
v
)(1  e
e
u
v
)
u
e e
v
u
 f (u )  f ( v )  1  e e
 f ( u )  f ( v )  (1  e
v
 (u  v )
 f (u )  f ( v )  f (u  v )
)
f ( x  y )  ( x  y )  x  2 xy  y
2
2
f ( x)  f ( y)  x  y
2
D
2
2
2x 
f '( x) 
1
x
 2  log e ( 2 x )
4x

1  log e ( 2 x )
2x
E
2
2
g ( x )  2 ( x  a ) g ( x )  ( x  a ) g ( x )
2
 ( x  a )( 2 g ( x )  ( x  a ) g ( x ))
E
C
 Can get directly from some calculators.
dy
 8 x  12 x
3
2
y  2x  4x
4
3
dx
m  64  48  16
 32  32  0
(2, 0)
D
y  y1  m ( x  x1 )
y  16 ( x  2 )
at x = 2
f (3)  f ( 0 )
30

C
25  1
3

26
3
 Find the equation of the normal to the line.
y   2 x  10
Solve
x
  2 x  10 for x .
2
( 0 , 0 ),
y 
1
2
mN 
x
1
2
x  4 and y  2 (4, 2)
d 
4 2
2
2
 2 5
y
-5
-4
-3
-2
-1
1
2
3
4
5
x
 b=4
 Solve
 4  8a (2  4)
a 
1
4
f ( x) 
x
3
( x  4) 
x
4
f (4) 
256
4
x
3
4
 64  0 (4, 0)
4
3
2
f ( x )  x  3 x
m  f ( 4 )  64  48  16
y t  16 ( x  4 )  16 x  64
f ( 2  0 . 2 )  f ( 2 )  0 . 2 f ( 2 )
D
Do 2010 Exam 1 last question.
Using the y coordinate of the tangent line to f(x)
to estimate the new f(x).
 Write down the rule.
dx

dt
dx

dv

dx
dv
dv
dt
dx
8
dv

8
6 x
8

6 4
 Put your units in.

2
3
cm/s
1
 6x 2


3
3

 ( 2 x  1) 2





dx


 3( 2 x  1)
3  2
1 2

3
2
dx
( 2 x  1)
  3 ( 2 x  1)


1
2
c
1
2
c
B
 sin(
 
1
2
D
2 x )  24 x
3
dx
cos( 2 x )  6 x  c
4
What is wrong with my
calculator?
Put your calculator in
mode radians
3
 2  f ( x ) dx 
1
3
 3 dx
1
 2  5  3 x 1
3
 10  ( 9  3 )  4
A
 cos( 3 x )  3 x sin(
  3 x sin(
  3 x sin(
3 x ) dx  x cos( 3 x )  c
3 x ) dx  x cos( 3 x ) 
3 x ) dx  x cos( 3 x ) 
 cos( 3 x ) dx  c
sin( 3 x )
3
 c1
1
sin( 3 x ) 
  x sin( 3 x ) dx   3  x cos( 3 x )  3 
A
D
dy
dx
 log e ( 2 x ) 
2x
2x
 1  log e ( 2 x )
e
e
2
 log
e
( 2 x ) dx   x log e ( 2 x )  x  12
1
2
2
e 1
1
e
  log e ( e )     log e (1)  
2 2
2
2

1
2

1

2
D
0
2
 cos( x ) dx
0
f (0)  2
f ( x ) 
1
2
f ( 0 ) 
1
2
m N  2
x
e2
y  2 x  2
 Top curve minus bottom curve
0  2 x  2
x 1
 x

 e 2  1  (  2 x  2 ) dx


0 

1
 x

   e 2  2 x  1) dx



0 
 1

0
2

 2e  1  1  2e  0  0





1
1
 x

2
2
 2e  x  x 

0
1
 2e 2  2

f (1)  f ( 2 )  f ( 3 )
 2  9  28  39
D
D
Pr( X  2 ) 
7
30
D

10
30

4
30

21
30

7
10
E(X ) 
 x . Pr( X
 x)
3
6
 0
 1
30

66
30
B

30
11
5
 2
7
30
 3
10
30
 4
4
30
 0 .3

 0 .7
2
0 . 6   1   0 . 51 
    

0 . 4   0   049 
Answer 0.49
OR
1  0 .3  0 .7  1  0 .7  0 .4
 0 .9

 0 .1
4
0 . 6   350   597 . 975 
 
  

0 . 4   350  102 . 025 
598 Tea
102 coffee
New to Exam 1 2010
 0 .9

 0 .1
0 .6  
x

0 . 4   700 
600 Tea
100 coffee
OR
0 .1
0 .1  0 .6
 700  100
x
 
  
x   700 


x
7
10
D

6
9

5
8

7
24
1
X ~ Bi (10 , )
2
C
Pr( X  8 )  0 . 0547
E ( X )  np  1 . 2
Var ( X )  npq  0 . 72
1 . 2 q  0 . 72
E
q  0 .6
p  0 .4
n 3

1
1

sin(
x
)
dx



 2
4

a
Solve
a = 2.09
E
for a.
3
 x 
  12 dx
1
3
x2 
 

24

1
9

24

1
3

1
24
5
 x
  12
a
5

dx 
8

5
x 
5



8
 24  a
2
25

24
a
a

24
2

24
a
2
2
5
8
5
12
 10
a 
10
D
2
X ~ N ( 4 . 7 , 1.2 )
Z ~ N ( 0 , 1)
3.5 4.7
Pr( X  3 . 5 )  Pr( Z   1)  Pr( Z  1)
B
2
X ~ N ( 72 , 8 )
Pr( X  80 )  Pr( Z  1)  0 . 16
72 80
Pr( 64  X  72 )  Pr(  1  Z  0 )  Pr( 0  Z  1)  0 . 84  0 . 5  0 . 34
64 72
Pr( X  64 \ X  72 ) 

Pr( X  64 )  Pr( X  72 )
Pr( X  72 )
0 . 16
0 .5

8
25
64 72
X ~ N ( 200 , 100)
Pr( X  208 )  0 . 7881
D
200 208
2
X ~ N (130 , 2.7 )
Pr( X  a )  0 . 35
a  129
D
130