Transcript No Slide Title

REVISION LECTURE
MATHEMATICAL METHODS
(CAS)
UNITS 3 AND 4
Presenter: MICHAEL SWANBOROUGH
Flinders Christian Community College
EXAMINATION 1
Short-answer questions (40 marks)
•
Questions are to be answered without the use
of technology and without the use of notes
Time Limit:
• 15 minutes reading time
• 60 minutes writing time
EXAMINATION 2
Part I: Multiple-choice questions
•
22 questions (22 marks)
Part II: Extended response questions:
•
58 marks
Time limit:
• 15 minutes reading time
• 120 minutes writing time
Examination Advice
General Advice
• Answer questions to the required degree of
accuracy.
• It is assumed that students will provide
exact answers to questions unless specified
otherwise.
• When an exact answer is required,
appropriate working must be shown.
Examination Advice
General Advice
• When an instruction to use calculus is
stated for a question, an appropriate
derivative or antiderivative must be shown.
• Label graphs carefully – coordinates for
intercepts and stationary points; equations
for asymptotes.
• Pay attention to detail when sketching
graphs.
Examination Advice
General Advice
• Marks will not be awarded for questions
worth more than one mark if appropriate
working is not shown.
• Correct mathematical notation is expected
and should always be used. Calculator
syntax should not be used.
Examination Advice
Notes Pages
• Well-prepared and organised into topic
areas.
• Prepare general notes for each topic.
• Prepare specific notes for each section of
Examination 2.
• Include process steps as well as specific
examples of questions.
Examination Advice
Notes Pages
• Include key steps for using your CAS
calculator for specific purposes.
• Be sure that you know the syntax to use
with your calculator.
Examination Advice
Strategy - Examination 1
• Use the reading time to carefully plan an
approach for the paper.
• Momentum can be built early in the exam
by completing the questions for which you
feel the most confident.
• Read each question carefully and look for
key words and constraints.
Examination Advice
Strategy - Examination 2
• Use the reading time to plan an approach
for the paper.
• Make sure that you answer each question in
the Multiple Choice section. There is no
penalty for an incorrect answer.
• It may be sensible to obtain the “working
marks” in the extended answer section
before tackling the multiple choice
questions.
Examination Advice
Strategy - Examination 2
• Some questions require you to work
through every multiple-choice option –
when this happens don’t panic!!
• Eliminate responses that you think are
incorrect and focus on the remaining ones.
• Multiple Choice questions generally require
only one or two steps – however, you
should still expect to do some calculations.
Examination Advice
Strategy - Examination 2
• If you find you are spending too much time
on a question, leave it and move on to the
next.
• When a question says to “show” that a
certain result is true, you can use this
information to progress through to the next
stage of the question.
Revision Quiz
1
2
3
4
5
Question 1
The range of the function
f  x   5  x is
a)
 2,7
b)  2,7 
c)
 2, 
d)
e)
R
 2,7 
f : 2,7  R,
A
Question 2
2
f  x 
1
x3
The equations of the asymptotes of the graph of the
inverse function are
a) x  3, y  1
b) x  1, y  3
c) x  1, y  3
d) x  3, y  1
e) x  1, y  3
C
Bonus Prize!!
Question 4
The smallest positive value of x for which
tan  2x   1 is
a) 0

c)
4
e) 

b)
8

d)
2
B
Question 5
What does V.C.A.A. stand for?
a) Vice-Chancellors Assessment Authority
b) Victorian Certificate of Academic Aptitude
c) Victorian Combined Academic Authority
d) Victorian Curriculum and Assessment Authority
e) None of the above
D
Question 1
The complete set of linear factors of
the polynomial ax3  bx
ax  bx
3
 x  ax  b 
2
x

ax  b
ANSWER: E

ax  b
VCAA 2004: 49%

Question 2
Use the remainder theorem
2 x  3 x  7 x  11
4
3
P  1  2  1  3  1  7  1  11
4
3
 2  3  7  11
9
2 x  3x  7 x  11
divisible by x  1
Therefore
4
3
is not exactly
Simultaneous Linear Equations
Unique solution
Infinite solutions
Different gradients
Same gradients
No solution
Same gradients
Same y-intercept Different y-intercept
Question 4
Infinitely many solutions for the
simultaneous linear equations
ax  3 y  0
2 x   a  1 y  0
ax
y
3
2 x
y
a 1
1
 2
a
2

3 a 1
2
a a6  0
 a  2  a  3  0
a  2,  3
ANSWER: B
VCAA 2008: 45%
Functions and Their Graphs
Vertical line test - to determine whether a
relation is a function
f : A  B, where f ( x)  rule
A represents the DOMAIN
B represents the CODOMAIN (not the range!)
Interval Notation
 a, b   x : a  x  b 
 a, b   x : a  x  b 
 a, b   x : a  x  b 
 a, b   x : a  x  b 
Square brackets [ ] – included
Round brackets ( ) – excluded
Maximal (or implied) Domain
The largest possible domain for which the
function is defined
A function is undefined when:
a) The denominator is equal to zero
b) The square root of a negative number is
present.
c) The expression in a logarithm results in a
negative number.
Consider the function
f ( x)  2x  3
2x  3  0
So the maximal domain is:

x : x 

3  3

 or  ,  
2  2

Using Transformations
When identifying the type of transformation that
has been applied to a function it is essential to
state each of the following:
NATURE – Reflection, Dilation, Translation
MAGNITUDE (or size)
DIRECTION
1.Translations
a) Parallel to the x-axis – horizontal translation.
b) Parallel to the y-axis – vertical translation.
To avoid mistakes, let the bracket containing x
equal zero and then solve for x.
If the solution for x is positive – move the graph x
units in the positive x-direction (ie. to the right).
If the solution for x is negative – move the graph x
units in the negative x-direction (ie. to the left).
2. Dilations
a) From the x-axis – the dilation factor is the number
outside the brackets.
b) From the y-axis – the dilation factor is the
reciprocal of the coefficient of x.
Note: A dilation of a from the x-axis is the same as a
1
dilation of
from the y-axis.
a
3. Reflections
a) Reflection about the x-axis
y   f (x)
b) Reflection about the y-axis
y  f ( x)
c) Reflection about both axes
y   f (  x)
d) Reflection about the line y  x
Question 5
y
y  f (x)
3
2
1
1
1
2
3
1
2
3
Determine the graph of
y  1  f (2 x)
x
y
3
2
1
1
-
1
2
3
x
1
2
3
-
y   f  2x 
4
-
Reflection about the x-axis
Dilation by a factor of
1
2
from the y-axis
y
3
2
1
1
1
2
3
x
1
2
3
4
y  1  f  2x 
Translation of 1 unit in the positive y-direction
VCAA 2009: 66%
ANSWER: E
Question 6
Graph of f  x   3x
5
2
Reflected in the x-axis, Translated 3 units to the
right, Translated 4 units down
Reflection:
y  3x
5
2
Horizontal Translation:
Vertical Translation:
VCAA 2007: 95%
y  3  x  3
5
2
5
2
y  3  x  3  4
ANSWER: B
Matrix Transformations
Question 7
T : R  R maps y  loge  x  onto
2
2
y  loge  2x  4  3
y  loge  2  x  2  3
1
Dilation by a factor of
from the y-axis
2
Translation of 2 units in the positive x-direction
Translation of 3 units in the positive y-direction
Dilation:
  x   0.5 0  x 
T    



  y    0 1  y 
Translations:
  x   0.5 0  x   2
T    
 



  y    0 1   y   3
VCAA 2006: 33%
ANSWER: A
Graphs of Power Functions
yx
n
y
y = x
6
y = x
4
y
4
y = x
3
-2
4
2
2
1
1
1
-1
2
x
-2
-1
1
-1
-2
-2
-3
-3
-4
-4
5
y = x
3
2
-1
y = x
2
3
x
Square Root Functions
 x  b  c
ya
y
4
y =
The graph is:
x – 2 + 1
• translated 2 units in the
positive x direction
3
2
1
-2
(2 , 1)
1
-1
-1
-2
2
3
4
5
x
• translated 1 unit in the
positive y direction
Question 9
The rule of the graph shown could be
1
a) y  3
x
y
c) y  x
O
x
2
3
e) y  x
b) y  x
d) y  x
1
3
ANSWER: D
3
2
1
3
Graphs of Rational Functions
Question 10
The equations of the horizontal and vertical
asymptotes of the graph with equation
x3
1
f  x 
 f  x 
1
2 x
x2
Vertical:
x20
x2
VCAA 2008: 86%
Horizontal:
y  1
ANSWER: D
Absolute Value Functions
Question 12
2k 1  k 1
2k  1  k  1
2k  1  0
k 0
  2k  1  k  1
2k  1  0
3k  2
2
k 
3
ANSWER: D
VCAA 2006: 75%
Question 13
Part of the graph of y  x  3x is shown below.
2
y
-1
1
2
3
4
a) Sketch the graph of y  x  3 x
2
x
b) Find the set of values of x for which
x  3x  4
2
From the graph, solve x  3x  4
2
y
x  3x  4  0
2
 x  4  x  1  0
 x : x  4  x : x  1
-1
1
2
3
4
x
Composite Functions
For the composite function f  g  x   to be defined
Range g  Domain f
When the composite function f  g  x   is defined
Dom f  g  x    Dom g
Investigating Composite Functions
Step 1: Complete a Function, Domain, Range
(FDR) table.
Step 2: Check that the range of g is contained in
the domain of f .
Step 3: Substitute the function g(x) into the
function f (x).
Step 4: Remember that: Dom f  g  x    Dom g
Question 14
f :  4,    R, f  x   x  4
g : R  R, g  x   x  3
2
F
f
g
D
R
 4,   0,  
R
3,  
g  f  x    g  x  4 


x4

2
3
y
 x 1
6
5
dom g  f  x    dom f
4
3
(4, 3)
2
 x   4,  
1
-1
1
-1
2
3
4
5
6
x
Inverse Functions
Key features:
The original function must be one-to-one
Reflection about the line y = x
Domain and range are interchanged
Intersections between the graph of the function
and its inverse occur on the line y = x
ran f  dom f
1
ran f
1
 dom f
To find the equation of an inverse function
Step 1: Complete a Function, Domain, Range
(FDR) table.
Step 2: Interchange x and y in the given
equation.
Step 3: Transpose this equation to make y the
subject.
Step 4: Express the answer clearly stating the
rule and the domain.
Question 16
1
f : R  R, where f ( x) 
3
x

F
D
f
f
R
1

  3,  
R
  3,  
R

1
f ( x) 
3
x
Inverse
1
x
3
y
f
1
 x 
1
 x  3
2
1
x3
y
1
y
x3
ANSWER: D
VCAA 2008: 89%
Question 17
Graph of the inverse function
y
y = 1
x
O
ANSWER: C
x = 1
VCAA 2005: 85%
Question 18
f : B  R, f ( x)  4x  3x  1
3
2
The function, f, will have an inverse function for:
a)
c)
e)
BR
1

B   , 
2

 1 
B   ,  
 2 
b)
d)
1 
B   ,
2 
1

B   , 
2

f : B  R, f ( x)  4x  3x  1
3
2
1 

The function will have an inverse for B   ,  
2 
as the function is a one-to-one function on this
domain.
ANSWER: B
VCAA 2008: 50%
Algebra of Functions
Sum and Difference of Functions
Key features:
• Points of intersection of the two functions.
• Points where either graph crosses the x-axis
• The dominant function in different parts of the
domain.
y
d
c
f( x )
f( x ) + g ( x )
b
a
a
x
g(x)
e
Logarithmic and Exponential Functions
Question 21
y  b  3a
y  3a  b
2x
y b
a 
3
 y b 
2 x  log a 

 3 
2x
2x
1
 y b 
x  log a 

2
 3 
ANSWER: B
VCAA 2006: 84%
Change of Base Rule for Logarithms
logb x
log a x 
logb a
Question 23
log e 6
log 5  6  
log e 5
 1.113
ANSWER: D
Circular (Trigonometric) Functions
f ( x)  a sin(b( x  c))  d
f ( x)  a cos(b( x  c))  d
Amplitude: a
Period: 2
b
Horizontal translation:
c units in the negative
x-direction
Vertical translation: d
units in the positive
y-direction
Question 24
y
4
3
2
1
-1
- 0.5
0.5
-1
-2
f t   a cos bt   c
1
t
Amplitude: a  2
Period: 5 complete cycles in 2 seconds
2
Therefore the period is second
5
2 2

b 5
b  5
Translation: 1 unit in positive y, c  1
ANSWER: A
f t   2cos 5 t  1
Question 25
Find the range of the function
  
f :  ,   R, f  x   2sin  2 x 
8 3 
x

8
 
y  2sin  
4
y 2
x

3
 2 
y  2sin 

 3 
y 3
x
Hence, the range is:

4
 
y  2sin  
2
y2
 2, 2


ANSWER: C
VCAA 2008: 45%
y
Question 26
y = f (x)
2
f  x   sin  4x  1
1

2

3
2
2
x
ANSWER: E
VCAA 2008: 36%
-1
y = g(x)
-2
Reflection in the x-axis
g  x    sin  x   1
x   0, 2 
Dilation by a factor of 4 from the y-axis
Solving Trigonometric Equations
• Put the expression in the form sin(ax) = B
• Check the domain – modify as necessary.
• Use the CAST diagram to mark the relevant
quadrants.
• Solve the angle as a first quadrant angle.
• Use symmetry properties to find all solutions
in the required domain.
• Simplify to get x by itself.
Question 27
3
 2 x 
sin 

2
 3 
a)
0 x3
3
2 x
 2 x 
sin 
0
 2

2
3
 3 
2 x


   , 2 
3
3
3
2 x 4 5

,
3
3 3
5
x  2,
2
S
A
T
C
b) g  x   3 f  x  1  2
  2  x  1  
 3 sin 
  2
3
 
 

The maximum value occurs when the angle is
2
2  x  1 

3
2
3
x 1 
4

x
4
Question 28
cos  x   2cos  x   0
2
cos  x  cos  x   2   0
cos  x   0
S
A
T
C
cos  x   2
but cos  x   2
  x : cos  x   0
ANSWER: A
VCAA 2007: 27%
General solutions
Question 29
2x  
x


4
2
sin  2x   1
 2n , n  Z
 n , n  Z
ANSWER: A
VCAA 2009: 55%
S
A
T
C
Functional Equations
Question 30
f  f  x   x
 x 1

 1
x 1 

f  f  x  
 x 1 

 1
 x 1 
x 1 x 1

x  1   x  1
2x

2
x
ANSWER: E
VCAA 2007: 47%
Revision Quiz
1
2
3
4
5
Question 1
The derivative of
sin x
a) (cos x)e
b)
sin x
e
sin x
e
is equal to
c)
e
cos x
d) (cos x)ecos x e) (cos x)e x
A
Bonus Prize!!
Question 3
Angie notes that 2 out of 10 peaches on her peach
tree are spoilt by birds pecking at them. If she
randomly picks 30 peaches the probability that
exactly 10 of them are spoilt is equal to
10
a) 0 .2
b) (0.2) (0.8)
10
c) (0.2) (0.8)
e)
30
20
20
d)
10
C10 (0.2) (0.8)
30
20
10
C10 (0.2) (0.8)
D
20
Question 4
The total area of the shaded region shown is given
y
by
2
1
a)  f ( x) dx
b)
2
 f ( x) dx
y = f(x)
2
1
0
1
2
0
c)  f ( x) dx   f ( x) dx
x
0
1
2
0
d)  f ( x) dx   f ( x) dx
e)
1
0
0
2
 f ( x) dx   f ( x) dx
-2
1
D
-1
Question 5
Which one of the following
sets of statements is true?
a) 1   2 and  1   2
b) 1 >  2 and  1   2
c) 1   2 and  1 >  2
X ~ N (  ,2 )
1
1 1
d) 1 >  2 and  1 >  2
e) 1 >  2 and  1   2
X ~ N (  ,2 )
2
2 2
A
DIFFERENTIAL CALCULUS
Chain Rule:
dy dy du


dx du dx
Product Rule:
d
du
dv
(uv )  v
u
dx
dx
dx
Quotient Rule:
du
dv
v
u
d u
dx
dx
 
2
dx  v 
v
Further Rules of Differentiation
Square Root Functions
y  f ( x)
dy
f ( x)

dx 2 f ( x)
Further Rules of Differentiation
Logarithmic Functions
y  loge  f ( x)
dy f ( x)

dx
f ( x)
y  loge (5x  7)
dy
5

dx 5 x  7
Further Rules of Differentiation
Exponential Functions
ye
ye
f ( x)
( x 5 x 3)
2
dy
f ( x)
 f ( x)  e
dx
dy
( x 2  5 x  3)
 (2 x  5)e
dx
Further Rules of Differentiation
Trigonometric Functions
y  sin f ( x)
dy
 f ( x) cos f ( x)
dx
y  cos f ( x)
dy
  f ( x) sin f ( x)
dx
y  tan f ( x)
dy
2

 f ( x) sec f ( x)
dx
Question 31
u  sin  4x 
y  f u 
d
 f  sin  4 x   
dx
ANSWER: E
dy dy du
VCAA 2006: 50%


dx du dx
dy
 f   u   4 cos  4 x 
dx
dy
 4 cos  4 x  f   sin  4 x  
dx
Question 33
y  4x  27 x  30x  10
3
2
The gradient of the graph is negative on the
interval
ANSWER: D
VCAA 2007: 68%
Question 35
f  x   x  2x
3
f  5   5  10
3
 135
f 1  1  2
3
3
f  5   f 1 135  3

5 1
4
 33
ANSWER: E
VCAA 2010: 69%
Question 36
2 x  y  10  0
 y  10  2 x
The gradient of this line is 2
The shortest distance between the origin and the line
will be the perpendicular distance.
1
This line will therefore have a gradient: m 
2
x
y
2
y  10  2 x
1
 0, 0  , m 
2
1
 y  0   x  0
2
x
y
2
d
 4  0   2  0
2 5
2
1
 2
1   2 
x
 10  2 x
2
5 x  20
x4
2
 P  4, 2 
y2
f : R  R, f  x   e  k
x
Question 37
The tangent to the graph of f at the point where
x = a passes through the point (0, 0). Find the value
of k in terms of a.
f  x  e  k
x
f  x  e
x
x  a, f  a   e  k
a
The equation of the tangent is given by:
y  e  k   e
a
a
 x  a
y  e  k  xe  ae
a
a
a
y  xe  ae  e  k
a
a
a
As the tangent passes through the point
(0, 0), it follows that:
0  0e  ae  e  k
a
k  ae  e
a
a
a
k  e  a  1
a
a
Related Rates
dx dx
 
dt
dt
Question 39
dV
8
dt
V  4x
3
2
dV

 6x
dx
1
2
dx dx dV


dt dV dt
dx
1
 1 8
dt
6x 2
When x  4
dx
8

dt 6 4
2
 cm/s
3
Question 40
Analysis Question
f : 0,2   R, f  x   2sin  x 
a) i)
f   x   2cos  x 
a) ii)
2  f   x   2
 0  f  x  2
y
2

-2
2
x
b) i) Find the other value of x for which the
gradient of the curve is 1.
f  x  1
 2 cos  x   1
1
cos  x  
2
 5
x ,
3 3
5
x
3
ii) Find the equation of the tangent to the curve

at x 
3
 
f   3
3
y  y1  m  x  x1 
y1  3, m  1, x1 


y  3  1 x  
3

y  x 3

3

3
iii) Find the axes intercepts of the tangent found
in b) ii)
y  x 3

3
x int : y  0
x

 3
3
y int : x  0
y 3

3


   3, 0 
3



  0, 3  
3

c) A translation of m takes the tangent at A to
the tangent at B. Find the exact value of m.
The value of the translation m is the distance between
the x-intercepts of the tangents at points A and B.

Tangent at A:
y  x 3
Tangent at B:
5
y  x 3
3
3
5

Hence the x-intercepts are x  3  , x   3
3
3
5

x 3
, x  3
3
3
5  

m 3
  3
3 3

4
m2 3
3
d) Find the general solution, for x of the
equation 2 sin  x   1 .
h : R  R, h  x   2 sin  x 
2sin  x  sin  x   0
h
2sin  x  sin  x   0
h  x  1
 2sin  x   1,  2sin  x   1
1
When sin x 
2

5
x   2 k ,
 2 k , k  Z
6
6
1
When sin  x   
2

5
  2 k , 
 2 k , k  Z
6
6

5
 x   k ,
 k , k  Z
6
6
S
A
T
C
Antidifferentiation and Integral Calculus
1 n 1
x
dx

x

c
,
n


1

n 1
n
n 1
(ax  b)
 (ax  b) dx  a(n  1)  c
n
Rules of Antidifferentiation
Trigonometric Functions
d
sin( kx )  k cos( kx )
dx
1
 sin( kx ) dx  k cos( kx )  c
d
cos( kx )  k sin( kx )
dx
1
 cos( kx ) dx  k sin( kx )  c
Rules of Antidifferentiation
Exponential Functions
d x
x
e e
dx
d kx
kx
e  ke
dx
1 kx
 e dx  k e  c
kx
Rules of Antidifferentiation
Logarithmic Functions
d
f ( x)
loge f ( x) 
dx
f ( x)

f ( x)
dx  log e f ( x)  c
f ( x)
Example
4
f ( x)
 4 x  3 dx  f ( x) dx where f ( x)  4x  3
 loge 4 x  3  c
Question 41
f   x   g  x   3, f  0  2 and g  0  1
f  x     g   x   3 dx
 g  x   3x  c
f  0   2, g  0   1
 2  1 c
c 1
f  x   g  x   3x  1
ANSWER: A
VCAA 2006: 66%
Definite Integrals
b

f  x  dx   F  x  a
b
a
 F (b)  F (a)
Properties of Definite Integrals
3
Question 43
3
 f  x  dx  5
1
  2 f  x   3 dx
1
3
3
1
1
 2 f  x  dx   3dx
 2  5  3x 1
3
ANSWER: A
VCAA 2008: 49%
 10  9  3
4
Calculating Area
• Sketch a graph of the function, labelling all
x-intercepts.
• Shade in the region required.
• Divide the area into parts above the x-axis and
parts below the x-axis.
• Find the integral of each of the separate sections,
using the x-intercepts as the terminals of
integration.
• Subtract the negative areas from the positive
areas to obtain the total area.
Question 45
The total area of the shaded region is
given by:
y
ANSWER: D
y = f( x )
-1
VCAA 2006: 81%
6
1
x
4
1
4
6
1
1
4
  f  x  dx   f  x  dx   f  x  dx
Integration by recognition
d
 f ( x)  g ( x), then it follows
dx
g
(
x
)
dx

f
(
x
)

c

Question 46
f : R  R, f  x   x loge  x 

a) Find the derivative of x loge  x 
2
d 2
x log e  x  

dx
d 2
2 d
x
log e  x   log e  x   x 
dx
dx
21
 x    2 x log e  x 
x
 x  2 x log e  x 
b) Use your answer to part a. to find the area of the
shaded region in the form, a loge b  c
where a, b and c are non-zero real constants.
3
A   x loge  x  dx
1
d 2
x log e  x    x  2 x log e  x 

dx
   x  2 x log e  x   dx  x 2 log e  x 
  2 x log e  x  dx  x log e  x    xdx
2
2
x
 x log e  x  
2
2
2
2
x
x
 x log e  x  dx  log e  x  
2
4
Hence,
3
A   x log e  x  dx
1
3
x
x 
  log e  x   
4 1
2
9 1
1
9
  log e  3     log e 1  
4 2
4
2
9
9 1
 log e  3  
2
4 4
9
 log e  3  2
2
2
2
Area between curves
b
b
a
a
A   f ( x ) dx   g  x  dx
y
f(x)
b
   f ( x)  g ( x)  dx
a
g(x)
x
a
b
Method
• Sketch the curves, locating the points of
intersection.
• Shade in the required region.
• If the terminals of integration are not given – use
the points of intersection.
• Check to make sure that the upper curve remains
as the upper curve throughout the required
region. If this is not the case then the area must
be divided into separate sections.
• Evaluate the area.
Question 47
a)
x
2
f  x  e 1
1
 y  2 x  2
f  x  e
2
1
f   0 
mnormal  2
2
x  0, f  0   2
x
2
 y  2  2  x  0 
b)
x
2
f  x  e 1
g  x   2 x  2
 2x 

A    e  1   2 x  2   dx


0 

1
x
 2

   e  2 x  1 dx

0
x  int : 1, 0 
1
1


2
  2e  x  x 

0
x
2
 A  2e  2


  2e  1  1   2e0  0  0 


1
2
1
2
Numerical techniques for finding area
y
Question 48

 
  
A   sin    sin   
6 6
 3 
1
 1
3
  

62 2 




3 1
12
O

6

3

2
ANSWER: D
VCAA 2008: 73%
x
Average value of a function
Question 49
f  x   loge 3x 1 x 0,2
b
1
f  x  dx

ba a
2
1

log
3
x

1
dx




e
2  0 0
7 log e  7   6

6
ANSWER: D
VCAA 2008: 59%
Question 50
Analysis Question
7
f : R  R, f  x  
x
a >1
y

C
A
Ob
1
a
x
a)
i) Calculate the gradient of CA in terms of a.
7

7
f  a   f 1 a

a 1
a 1
7

a
a) ii) At what value of x does the gradient of the
tangent to the graph of f have the same
gradient as CA?
7
f  x   2
x
7 7
 2 
x
a
x2  a  0
x a
x a
x >1
e
b) i) Calculate
 f  x  dx
1
e

1
e
7
f  x  dx   dx
x
1
e
 7 log e x 1
 7 log e  e   7 log e 1
7
7
b  x  dx  7
1
b) ii) Find b such that
7
b  x  dx  7
1
1
7 log e x  b  7
log e 1  log e  b   1
log e  b   1
be
1
c) i) Express the area of the region bounded by
the line segment CA, the x-axis, the line x  1
and the line x  a in terms of a.
1
7
A   7    a  1
2
a
 a  1   7  a  1 



7 
a

a  2 
7
a
–
1

7  a 2  1
2a
c) ii) For what exact value of a does this area equal 7?
7  a  1
2
7
2a
7 a 2  7  14a
a  2a  1  0
2
a  1  2,1  2
 a  2 1
a >1
a
c) iii) Explain why
 f  x  dx  7
1
The area under the curve is less than the area of
the trapezium. Therefore,
a
 f  x  dx  7
1
e
We already know that
 f  x  dx  7
1
a
Therefore it follows that if
ae
 f  x  dx  7
1
then
d)
7
1  x  dx  3
m
n
mn
7
1  x  dx  2
mn
7 log e x 1  3
7 log e  mn   3
mn  e
3
7
1
m
n
1
7 log e x   2
m
7 log e    2
n
m
e
n
2
7
 2
Solving these equations simultaneously,
5
14
1
14
me , ne
5
14
1
14
or m  e , n  e
Probability – Assumed Knowledge
Law of Probability:
Pr  A  B  Pr  A  Pr  B  Pr  A  B
Mutually Exclusive Events:
Pr  A  B  0
Independent Events:
Pr  A  B  Pr  A Pr  B
Discrete Random Variables
A discrete random variable takes only distinct or
discrete values and nothing in between.
Discrete variables are treated using either discrete or
binomial distributions. These values are usually
obtained by counting.
A continuous random variable can take any value
within a given domain. These values are usually
obtained through measurement of a quantity.
Continuous variables are often treated using normal
distributions.
Expected value and expectation theorems
  E( X )
 x1 Pr(X  x1 )  x2 Pr(X  x2 )  .....xn Pr(X  xn )
  x Pr(X  x)
E(aX )  aE( X )
E(aX  b)  aE( X )  b
Variance and Standard Deviation
  Var ( X )
2
 E ( X )  E ( X )
2
2
SD ( X )  Var ( X )  
Var (aX )  a Var ( X )
2
Question 51
The discrete random variable, X, has the following
probability distribution. If the mean of X is 1 then
X
0
1
2
Pr  X  x 
a
b
0.4
E X  1
1  0  a  1 b  2  0.4
1  b  0.8
b  0.2
 a  0.4
ANSWER: C
VCAA 2010: 83%
Markov Chains
A Markov chain is a chain of events for which the
probabilities of outcomes or states depend on what
has happened previously.
Tree diagrams are a useful tool for solving
problems.
Question 53
If he goes to the Dandy one Friday, the probability he
goes to the Cino the next Friday is 0.5.
If he goes to the Cino one Friday, then the probability
he goes to the Dandy the next Friday is 0.6.
If he goes to the Cino one Friday, what is the
probability that he goes to the Cino on exactly two of
the next three Friday’s?
Pr  2 of next 3
 Pr  DCC   Pr  CDC   Pr  CCD 
 0.6  0.5  0.4  0.4  0.6  0.5  0.4  0.4  0.6
 0.12  0.12  0.096
 0.336
Question 55
A bag contains four white balls and six black balls.
Three balls are drawn without replacement. The
probability that they are all black is:
Pr( BBB)
6 5 4
  
10 9 8
1

6
ANSWER: A
VCAA 2010: 81%
The Binomial Distribution
X ~ Bi(n, p)
Pr(X  x) Cx ( p) (1  p)
n
x
n x
, x  0,1, 2, ......n
  E( X )  n p
  Var ( X )  n p q,
2
  SD ( X )  n p q
where q  1  p
Question 57
A soccer player has 15 attempts at goal. The
probability that the number of goals she scores is
less than 7 is:
3

X ~ Bi  n  15, p  
5

Pr  X  7   Pr  X  6 
ANSWER: B VCAA 2010: 44%
Question 58
Let X be a discrete random variable with a
binomial distribution. The mean of X is 1.2 and
the variance of X is 0.72.
Mean:
1.2  np
Var:
0.72  npq
0.72  1.2 q
 q  0.6  p  0.4
n3
ANSWER: E
VCAA 2008: 68%
Continuous Random Variables
Properties of probability density functions
a)
f  x   0 for all real numbers x

b)

f  x  dx  1

b
c)
Pr  a  X  b    f  x  dx
a
Continuous Random Variables
Mean:
  E X  

 x f  x  dx

Mode: the value for which Pr  X  x  is a maximum
m
Median: the value m such that

a
Variance:
  Var  X  
2
1
f  x  dx 
2

 x f  x  dx  
2

2
Question 59
0.5
  sin  2 x  dx > 0.2
a
 a  0.35
ANSWER: D
VCAA 2009: 61%
Question 60
3
a)
x
Pr  X  3   dx
12
1
3
x 
 
 24 1
9
1


24 24
1

3
2
x

f  x   12
0
1 x  5
otherwise
b)
5
Pr  X  a  
8
5
5
 x
a  12  dx  8
5
x 
5
 24   8
 a
2
25  a  15
2
a   10, 10
but a > 1,  a  10
The Normal Distribution


The mean, mode and median are the same.
The total area under the curve is one unit.
b
Pr(a  X  b)   f  x  dx
a
Using symmetry properties
z
Pr(Z > z )  1  Pr( Z  z)
-z
z
Pr  Z  z   Pr  Z > z 
a
b
Pr  a  Z  b  Pr  Z  b  Pr  Z  a 
Question 62
X is normally distributed with a mean of 72 and a standard
deviation of 8. Use the result that Pr  Z  1  0.84 to find:
a)
Pr  X > 80 
 Pr  Z > 1
 1  Pr  Z  1
 0.16
72
80
b)
Pr  64  X  72 
 Pr  1  Z  0 
 0.34
64
72
c)
Pr  X  64 \ X  72 
Pr  X  64  X  72 

Pr  X  72 
Pr  X  64 

Pr  X  72 
Pr  Z  1

Pr  Z  0 
 0.32
Conditional probability
Pr( A  B)
Pr( A \ B) 
Pr( B)
Solving normal distribution problems
• Draw a diagram, clearly labelling the mean.
• Shade the region required.
• Use either the appropriate symmetry
properties or a calculator to find the required
probability
• Remember that:
z
x

Question 63
The time (in minutes) taken for students to complete
a university test are normally distributed with a
mean of 200 minutes and a standard deviation of 10
minutes.
The proportion of students who complete the test in
less than 208 minutes is closest to
a) 0.200
b) 0.212
d) 0.788
e) 0.800
c) 0.758
208  200 

Pr( X  208)  Pr  Z 

10


Pr( Z  0.8)  0.788
ANSWER: D
VCAA 2006: 75%
Applications of the normal distribution
• Draw a diagram, clearly shading the region
that corresponds to the given probability.
• Use the symmetry properties of the curve to
write down the appropriate z value.
• Use the inverse normal function to find the
required probability and the corresponding
z value.
• Use the relationship z 
x

calculate the required x value.
to
Question 64
The heights of the children in a queue for an amusement
park ride are normally distributed with mean 130 cm and
standard deviation 2.7 cm. 35% of the children are not
allowed to go on the ride because they are too short. The
minimum acceptable height correct to the nearest
centimetre is
a) 126
b) 127
c) 128
d) 129
X ~ N 130, 2.7
2

Pr  X  a   0.35
e) 130
a
ANSWER: D
130
VCAA 2007: 71%
Question 65
Analysis Question
a) i)
Pr  GPPP   0.4  0.32
4
9
 
10 100
9

250
G
0.6
G
P
G
0.7
0.4
G
P
G
0.3
P
0.3
P
a) ii)
Pr  GGPP   0.6  0.4  0.3
Pr  GPGP   0.4  0.7  0.4
Pr  GPPG   0.4  0.3  0.7
67
 Pr  pool twice  
250
b)
1  Gym
S0   
0  Pool
0.6 0.7
T 

0.4
0.3


S100  T
100
 S0
100
0.6 0.7 


0.4 0.3
0.636 


0.364 
1 
0
 
Hence, the proportion of nights she goes to the pool is 0.364.
c)
y
(
1.42
,
1.54
)
O
(
1
,
0
)
(
2
,
0
)
t
d)
Pr T  1.25 
1.25
  4t
1
 0.191
3
 24t  44t  24  dt
2
e)
X
Bi  5, 0.809 
Pr  X  4   0.41
f)
m
3
2
4
t

24
t
 44t  24  dt  0.5

1
m  1.459 hours
m  88 minutes
THE FINAL RESULT