REVISION LECTURE
MATHEMATICAL METHODS
UNITS 3 AND 4
Exam Preparation
Examinations
EXAMINATION 1 - Facts, Skills –
Short-answer questions
EXAMINATION 2 - Multiple Choice and
Analysis Task
Examination Advice
General Advice
• Answer questions to the required degree of
accuracy.
• If a question asks for an exact answer then a
decimal approximation is not acceptable.
• 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 to questions
worth more than one mark if appropriate
working is not shown.
Examination Advice
Notes Pages
• Well-prepared and organised into topic
areas.
• Prepare notes throughout the year.
• Include process steps rather than just
specific examples of questions.
Examination Advice
Notes Pages
• Some worked examples can certainly be of
benefit.
• Include key steps for using your graphic
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 plan an approach
for the paper.
• Make sure that you answer each question.
There is no penalty for incorrect answers.
• It may be sensible to obtain the “working
marks”.
Examination Advice
Strategy - Examination 1
• Questions generally require only one or two
steps – however, you should still expect to
do some calculations.
Examination Advice
Strategy - Examination 2
• 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
• 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
6
7
8
9
Question 1
The derivative of
a) (cos x)e
sin x
b)
e
sin x
e
sin x
is equal to
c)
e
cos x
d) (cos x)ecos x e) (cos x)e x
A
Question 2
The range of the function with graph as shown is
a)  2, 6
f(x)
b)  5,  4   2, 4
5
4
3
2
1
c)  5, 4
d)  5,  4   2, 3
e)  2, 4  5, 6
-3 -2 -1
-1
-2
-3
-4
-5
B
1 2 3 4 5 6
x
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)
30
10
C10 (0.2) (0.8)
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
0
1
2
0
c)  f ( x) dx   f ( x) dx
x
-2
1
d)  f ( x) dx   f ( x) dx
e)
1
0
0
2
 f ( x) dx   f ( x) dx
D
-1
Question 5
What does V.C.A.A. stand for?
a) Vice-Chancellors Assessment Authority
b) Victorian Curriculum and Assessment Authority
c) Victorian Combined Academic Authority
d) Victorian Certificate of Academic Aptitude
e) None of the above
B
Question 6
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
Bonus Prize!!
Question 8
P( x) = ( x  a)( x  b)( x  c)
2
2
where a, b and c are three different positive
real numbers. The equation has exactly
a) 1 real solution
b) 2 distinct real solutions
c) 3 distinct real solutions
d) 4 distinct real solutions
e) 5 distinct real solutions
B
Question 9
For the equation 2 sin x  3 = 0 the sum of the
solutions on the interval 0 , 2π  is

5
b)
3
c)
2
7
d)
3
e)
3
a)
E
EXAMINATION 1 - FACTS, SKILLS
AND APPLICATIONS TASK
• Part A
– 27 multiple-choice questions (27 marks)
• Part B
– short-answer questions (23 marks)
• Time limit:
– 15 minutes reading time
– 90 minutes writing time
EXAMINATION 2 - ANALYSIS TASK
• Extended response questions
– 4 questions (55 marks)
• Time limit:
– 15 minutes reading time
– 90 minutes writing time
Question 1
The linear factors of the polynomial
x  x  3 x  3x are
4
3
2
x 4  x3  3x 2  3x
= x  x 3  x 2  3 x  3
= x  x  1  x 2  3


= x  x  1 x  3 x  3
ANSWER: B

Question 4
a) Expand 3x(2 x  5) fully
3

3 x (2 x)  3(2 x) (5)  3(2 x)( 5)  (5)
3
2
2
= 3 x(8 x  60 x  150 x  125)
3
2
= 24 x  180 x  450 x  375 x
4
3
2
3

b)
x  2 x  ax  2 is exactly divisible by
3
2
x  2.
Find the value of a.
P ( 2) = 0
0 = 2  2 ( 2)  2 a  2
0 = 14  2a
a = 7
3
2
Question 5
y = x  5 x  x  10
3
2
a)
y = x 3  5 x 2  x  10
x 2  3x  5
x  2 x 3  5 x 2  x  10
y
10
x3  2x 2
 3x 2  x
2
x
 3x 2  6 x
 5 x  10
 5 x  10
 ( x  2)( x 2  3 x  5)
y = x  5 x  x  10
3
2
y = ( x  2)( x  bx  c)
2
2  c = 10
 c = 5
2 x  bx = 5 x
 b = 3
2
y = ( x  2)( x  3 x  5)
2
2
2
b)
( x  2)( x  3 x  5) = 0
2
x = 2, x  3 x  5 = 0
2
3  3  4(1)( 5)
x=
2
3  29
x=
2
3  29 3  29
 x = 2,
,
2
2
2
Coefficient of x 3 in (2 x  a) 6
Question 6
6
C 3 (2) (a) = 4320
3
3
160a = 4320
3
a = 27
3
a=3
ANSWER: B
Question 7
Coefficient of x 2 in (3x  2) 7
7
C 2 (3) (2)
2
5
= 21(9)( 32)
= 6048
ANSWER: D
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
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
Question 9
The range of the function with graph as shown is
a)  2, 3
b)  3, 3
c)  3, 1   2, 3
d)  2, 1   2, 3
e)  2, 0   2, 3
f(x)
4
3
2
1
-3 -2 -1
-1
-2
-3
ANSWER: D
1
2
3
4
x
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.
Consider the function
f ( x) = 2 x  3
2x  3  0
So the maximal domain is:

x : x 

3  3

 or  ,  
2  2

Question 10
This question requires EVERY option
to be checked carefully.
b)
f ( x) = x( x 2  4)
2
f ( x) = x( x  2)( x  4)( x  1)
c)
f ( x) = (3  x)( x  16)
d)
f ( x) = ( x  x  6)( x  4)
e)
f ( x) = ( x  x  6)( x  x  12)
a)
4
2
2
2
f ( x) = ( x  x  6)( x  x  12)
= ( x  3)( x  2)( x  4)( x  3)
Cuts the axis in four places
2
ANSWER: E
2
Question 11
The graph shown could be that of the function f
whose rule is
a) f ( x) = ( x  a)( x  b)
b) f ( x) = ( x  a)( x  b)
y
2
c) f ( x) = ( x  a)( x  b)
2
d) f ( x) = ( x  a)( x  b)
2
e) f ( x) = ( x  a) ( x  b)
2
y = f(x)
2
x
a
b
ANSWER: A
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 - Translation, Dilation, Reflection
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 to the RIGHT.
If the solution for x is negative – move the graph x
units to the LEFT.
2. Dilations
a) Parallel to the y-axis – the dilation factor is the
number outside the brackets. This can also be
described as a dilation from the x-axis.
b) Parallel to the x-axis – the dilation factor is the
reciprocal of the coefficient of x. This can also be
described as a dilation from the y-axis.
Note: A dilation of a parallel to the y-axis is the
1
same as a dilation of
parallel to the x-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
y
x
Reflection about the x-axis
y
x
Reflection about the y-axis
y
x
Reflection about both axes
Question 13
y = f (x)
y
2
x
Determine the graph of
y = 1  f ( x)
y
y =  f (x)
-2
Reflection about the x-axis
x
y
y = 1  f ( x)
ANSWER: A
-1
x
Translation of 1 unit parallel to the y-axis
EXTRA QUESTION
The graph of the function f is obtained from the
graph of the function with equation y = x
by a reflection in the y-axis followed by a dilation
of 2 units from the x-axis. The rule for f is:
a)
f  x  = 2 x
b)
f  x  = 2 x
c)
f  x  = 0.5 x
d)
f  x  = 0.5 x
e)
f  x = 2 x
Reflection: f  x  =  x
Dilation:
f  x = 2 x
ANSWER: E
Transform f(x) to g(x)
Question 15
y
2
y = g(x)
1
y = f(x)

2

3
2
2
x
– 1
– 2
Dilation by a factor of 0.5 from the y-axis
Dilation by a factor of 2 from the x-axis
Graphs of Rational Functions
Question 16
The equations of the horizontal and vertical
asymptotes of the graph with equation
2
y=
3
x4
Vertical:
x4=0
x=4
Horizontal:
y=3
ANSWER: E
Inverse Functions
Key features:
Domain and range are interchanged
Reflection about the line y = x
The original function must be one-to-one
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 17
f : R  R, where f ( x) = e
F
f
f
1
(2 x)
1
D
R
R
 1,  
Inverse :
 1,  
R
x 1 = e
log e ( x  1) = 2 y
ANSWER: A
x=e
2y
1
2y
1
y = log e ( x  1)
2
Question 18
Graph of the inverse function
y
x
ANSWER: C
Question 20
f : (2, )  R, f ( x) = 4 log e ( x  2)
f(x)
8
6
4
2
-8 -6 -4 -2
-2
-4
-6
-8
2
4
6
8
x
f ( x) = 4 log e ( x  2)
(k , 2)  2 = 4 log e (k  2)
1
2
k =e 2
k = 3.649 (3 decimal places)
f(x)
(0, 3)
y =2
(3, 0)
Label asymptotes
Approach asymptotes
Label coordinates
x =2
x
Question 21
The equation relating x and y is most likely:
y
y = ax
1
2
ANSWER: E
x
Question 22
y
a) f ( x) = g ( x)  h( x)
b) f ( x)  g ( x) = h( x)
2
1.5
c) f ( x) = 2 g ( x)
d) g ( x) = h( x  1)
e) h( x) = 2 g ( x)
1
g(x)
f(x)
0.5
– 0.5
– 0.5
ANSWER: B
h(x)
– 1
0.5
1
1.5
2
2.5
3
x
Solving indicial equations
Step 1: Use appropriate index laws to reduce
both sides of the equation to one term.
Step 2: Manipulate the equation so that either
the bases or the powers are the same.
Step 3: Equate the bases or powers. If this is not
possible then take logarithms of both
sides to either base 10 or base e.
Question 23
= 1997
1997
2x
e =
3
1
 1997 
x = log e 

2
 3 
x = 3.25 (2 decimal places)
3e
2x
Solving logarithmic equations
Step 1: Use the logarithmic laws to reduce the
given equation to two terms – one on
each side of the equality sign.
Step 2: Convert the logarithmic equation to
indicial form.
Step 3: Manipulate the given equation so that
either the bases or the powers are the
same.
Step 4: Equate the bases or powers. If this is not
possible then take logarithms of both
sides to either base 10 or base e.
Step 5: Check to make sure that the solution
obtained does not cause the initial
function to be undefined.
Question 26
2 log a  x  = log a 16   4
log a  x
2
  log 16  = 4
a
x 
log a   = 4
 16 
ANSWER: A
2
x
4
a =
16
x 2 = 16a 4 ,
x = 4 a 2
2
but x > 0
 x = 4a
2
Question 27
  = 1  log
3 log 10 x  log 10 x
2
10
y
3 log 10 x  2 log 10 x = 1  log 10 y
log 10 x  log 10 y = 1
x
log 10   = 1
 y
x
= 10
y
x = 10 y
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 29
Amplitude: 2
2
Period:
=4
b
y
4
3
b=
2
Translation: 2 units up
2
1
1
– 1

2

y = 2sin 
2
3
4

x  2

x
ANSWER: C
Question 30
f : R  R, f ( x) = 2 cos(3x   )  1
Amplitude: 2
2

Period:
3
Range:
2 1 = 1
2  1 = 3
  3, 1
ANSWER: B
Question 31
f ( x) = p sin( 2 x)  q, where p > 0
y
a) q > 0
b)
p > q
c)
p<q
ANSWER: C
p
x
d)
p<q< p
e)
1
p> q
2
-p
y
Question 32
2
y = f (x)
1


2
3
2
2
x
ANSWER: C
– 1
– 2
y = g(x)
Dilation of factor 2 from the x-axis
Reflection in the x-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 33
a) t = 4
 4 
C = 4 cos 
  16
 12 
1
= 4   16
2
= 14
 t 
b) 20 = 4 cos 
  16
 12 
 t 
cos 
 = 1
 12 
t
=
12
t = 12
 t = 4 pm
Question 34
x  0, 4 
sin 2 x = 1
0  2 x  8
sin 2 x = 1
2x =
x=

2
, 2 

2
, 4 
 5 9 13
4
,
4
,
4
,

2
, 6 

2
4
ANSWER: E
Question 35
sin( 2 x) = a cos( 2 x)
tan( 2 x) = a
x=

6
 
tan   = a
3
a= 3
ANSWER: E
Question 36
T = 25  4 cos
Analysis Question
  t  3
12
, for 0  t  24
a) Maximum: 25  4 = 29 C
Minimum: 25  4 = 21 C
b)
cos  is maximum when  = 

  t  3
12
t  3 = 12
t = 15
 3pm
=
c)
25  4 cos
  t  3
12
  t  3 1
cos
=
12
2
  t  3  5
= ,
12
3 3
t  3 = 4, 20
t = 7, 23
t = 7 am, 11pm
= 23
d)
Maximum at t = 15
Interval: 15  2, 15  2
t = 13 T = 28.46 C
t = 17 T = 28.46 C
Minimum temp: 28.46 C
e)
i)
  t  3
dT 4
=
sin
dt 12
12
  t  3

= sin
3
12
e)
ii)

3
sin
sin
  t  3
12
  t  3
12
  t  3
=
 0.2
0.6

= 0.192, 2.949
12
t = 3.73, 14.27
Interval is: 3.73, 14.27 
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
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
Further Rules of Differentiation
Logarithmic Functions
y = log e x
dy 1
=
dx x
y = log e  f ( x)
dy f ( x)
=
dx
f ( x)
Examples:
y = log e (5x  7)
y = log e (sin x)
dy
5
=
dx 5 x  7
dy cos x
=
dx sin x
= cot x
Further Rules of Differentiation
Exponential Functions
y=e
x
y=e
f ( x)
dy
x
=e
dx
dy
f ( x)
= f ( x)  e
dx
Examples:
y=e
( x  5 x  3)
y=e
2
cos x
dy
( x 2 5 x  3)
= (2 x  5)e
dx
dy
cos x
=  sin x e
dx
Question 37
x
y = e ( x  4)
dy
d x
x d
3
3
=e
( x  4)  ( x  4) e
dx
dx
dx
x
2
3
x
= e (3x )  ( x  4)(e )
3
x
= e ( x  3x  4)
3
2
ANSWER: D
Question 39
cos(3t )
y=
2
t
d 2
2 d
t
cos(3t )  cos(3t ) t
dy
dt
dt
=
2
2
dt
t
 
dy  3t sin( 3t )  (2t ) cos(3t )
=
4
dt
t
2
ANSWER: A
Graphs of Derived Functions
f(x)
Question 40
1
– 2
– 1
ANSWER: A
1
– 1
2
3
x
Question 42
When x = 4,
dy
= 3x
dx
y = 16
1
2
dy
x = 4,
=6
dx
y  16 = 6  x  4 
y = 6x  8
ANSWER: C
Question 43
y =  x  x  2x  2
3
2
Positive gradient for (1.215, 0.548)
ANSWER: B
Approximations
Question 44
f  x  h   f  x   hf   x 
f (16) = 16
h = 0.04
ANSWER: B
f 16.04   f 16   0.04 f  16 
Question 46
Analysis Question
f : R  R, f  x  = e  2ke  3
2x
a)
a=3
b)
x
at x = 0, f ( x) = 0
0 = e  2ke  3
0 = 1  2k  3
2k = 4
k=2
0
0
c) Use CALCULUS to find the EXACT values of
the COORDINATES of the turning point.
y = e  4e  3
dy
2x
x
= 2e  4e
dx
2x
x
0 = 2e  4 e
2x
x
2e (e  2) = 0
x
x
 e = 2 as e  0
x = log e 2
x
x
f ( x) = e
2 loge 2
 4e
= 483
= 1
  log e 2,  1 
log e 2
3
d)
i)
When x = c, y = 0
0 = e  4e  3
2c
c
(e  3)(e  1) = 0
c
c
e = 3, e = 1
c = log e 3, 0
c
c
but c > 0
 c = log e 3
ii)
A =
log e 3

2x
x
e

4
e
 3 dx

0
log e 3
 1 2x

x
=  e  4e  3x 
2
0
 1 2 loge 3
 1 0
log e 3
0
= e
 4e
 3log e 3   e  4e 
2
 2

1 loge 9
1
= e
 4  3  3log e 3   4
2
2
= 4  3log e 3
y
y = a
y = g(x)
– c
y = f(x)
x
c
g ( x) = e
2 x
 4e
x
3
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
Examples
(5 x  3)
(
5
x

3
)
dx
=

c

7(5)
7
6
(5 x  3)
=
c
35
7

2
5
3
( 2 x  7)
2
5
dx = 3 (2 x  7) dx
3
5
( 2 x  7)
= 3
c
3
2
5
3
5
= 3   ( 2 x  7) 5  c
6
3
5
= ( 2 x  7) 5  c
2
Question 47
dy
=
dx
3
2
2  (4 x  1) dx
2
(4 x  1)
3
2
ANSWER: E
1 

 (4 x  1) 2 
= 2
c


1

4 
 2

= (4 x  1)
1
=
1
(4 x  1) 2
1
2
c
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)
log e f ( x) =
dx
f ( x)

f ( x)
dx = log e f ( x)  c
f ( x)
Examples
4
f ( x)
 4 x  3 dx =  f ( x) dx where f ( x) = 4 x  3
= log e (4 x  3)  c
x
1
2x
 x 2  5 dx = 2  x 2  5 dx
1 f ( x)
= 
dx where f ( x) = x 2  5
2 f ( x)
1
= log e ( x 2  5)  c
2
Definite Integrals

b
a
f ( x) dx = F ( x)
b
a
= F (b)  F (a)
Example
2
2
(
x
  3x  4) dx
0
2
x

3x
= 
 4x
2
3
0
 8 12

=    8   (0)
3 2

2
=
3
3
2
Properties of Definite Integrals
Question 49
4
 (2 f ( x)  1) dx
1
4
4
1
1
=  2 f ( x) dx   1 dx
4
= 2  f ( x) dx   x 1
4
1
4
= 2  f ( x) dx  3
ANSWER: D
1
EXTRA QUESTION
4
If

1
4
f ( x)dx = 2 then
  2 f ( x)  3 dx
is equal to:
1
4
4
4
1
1
1
  2 f  x   3 dx =  2 f  x  dx   3 dx
4
= 2  f  x  dx  3x 1
4
1
= 4  12  3
= 13
Integration by recognition
d
 f ( x) = g ( x), then it follows
dx
g
(
x
)
dx
=
f
(
x
)

c

Question 50
d
x log e x = 1  log e x
dx
  (1  log e x)dx = x log e x  d
 1 dx   log x dx = x log x  d
log
x
dx
=
x
log
x

1
dx

d


 log x dx = x log x  x  c
e
e
e
e
e
e
ANSWER: B
y
Question 52
On the interval (a, b)
the gradient of g(x) is
positive.
a
x
b O
y = f(x)
ANSWER: B
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 53
The total area of the shaded region is given by:
y
0

2
1
f ( x) dx   f ( x) dx
y = f(x)
2
0
-2
ANSWER: C
1
x
Question 54
The total area bounded by the curve and the x-axis is
y
given by:
y = f(x)
a
ANSWER: D
b
c
O

b
a
x
f  x  dx   f  x  dx
b
c
Question 55
a)
y = x log e 2 x  x
dy
 2 
= x    log e 2 x 1  1
dx
 2x 
= log e 2 x
b) Hence, find the exact area of the shaded region
e
2
 log
1
2
e
2 x dx =  x log 2 2 x  x 
e
2
1
2
e 1
1
e
=  log e  e      log e 1  
2 2
2
2
1
=
2
y
1
2
e
2
x
Area between curves
y
f(x)
g(x)
x
a
b
b
b
a
a
A =  f ( x) dx   g ( x) dx
=   f ( x)  g ( x)dx
b
a
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.
The area of the shaded region is given by:
y
y = g(x)
b
c
x
y = f(x)
c
 ( f ( x)  g ( x))dx
0
BOS 1997 CAT 2 Q. 18
Question 56
Find the exact area of the shaded region
y
y = cosx
1
– 1


4
2
3
4

5
4
3
2
7
4
2
y = sinx
x
A=
5
4
  sin x  cos x 
dx
4
5
4
=   cos x  sin x  
4
 5
=  cos 
 4
=2 2

 5
  sin 

 4

 
 
  cos    sin  

4
4
Numerical techniques for finding area
Question 57
y
A = f (0)  f (1)  f (2)
= e0  e1  e 2
= 1 e  e
1
2
f(2)
f(0)
ANSWER: A
f(1)
1
2
3
x
Question 58
a)
Analysis Question
1
4
3
2
y =  2 x  x  5 x  3x 
2
2
dy
3x
3
3
= 4x 
 5x 
dx
2
2
2
3x
3
3
4x 
 5x  = 0
2
2
b)
i)
2
dy
3x
3
3
= 4x 
 5x 
dx
2
2
dy
when x = 1,
= 1
dx
mnormal = 1
1
x = 1, y =
2
1
y  = 1 x  1
2
 y = x  1.5
b) ii)
x  1.5 = 0.5  2 x  x  5 x  3 x 
4
3
2
x  0.5 x  2.5 x  0.5 x  1.5 = 0
4
3
2
 x  1 x  1.5  x  1
2
=0
A repeated root at x = -1 indicates that the normal is
a tangent to the curve at this point.
5
When x = 1, y =
2
5 

 B  1,

2 

y
c) i)
x
A
B
c) i)
  0.5  2 x
1
A =
4
2
1
1
=
 x
1
4

 x  5 x  3x    x  1.5  dx
3
 0.5 x  2.5 x  0.5 x  1.5  dx
3
2
c) ii)
Discrete Random Variables
A discrete random variable takes only distinct or
discrete values and nothing in between.
Discrete variables are treated using either discrete,
binomial or hypergeometric distributions.
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 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 60
Melissa constructs a spinner that will fall onto one of
the numbers 1 to 5 with the following probabilities.
Number
Probability
1
2
3
4
5
0.3
0.2
0.1
0.1
0.3
The mean and standard deviation of the number that the
spinner falls onto are, correct to two decimal places,
x
Pr( X = x)
1
0.3
0.3
0.3
2
0.2
0.4
0.8
3
0.1
0.3
0.9
4
0.1
0.4
1.6
5
0.3
1.5
7.5
2.9
11.1
Var( X ) = E( X )   E( X ) 
2
2
x Pr( X = x) x 2 Pr( X = x)
= 11.1  (2.9) 2
= 2.69
SD( X ) = 1.64
ANSWER: E
The Binomial Distribution
X ~ Bi (n, p )
Pr( X = x)= C x ( p) (1  p)
n
x
n x
, x = 0,1, 2, ......n
 = E( X ) = n p
 2 = Var ( X ) = n p q,
 = SD( X ) = n p q
where q = 1  p
Question 61
In a two-week period of ten school days, the
probability that the traffic lights have been green
on exactly nine occasions is:
X
Bi 10, 0.4 
Pr( X = 9) = C9 (0.4) (0.6)
10
ANSWER: A
9
1
Question 63
Pr  X = 3 = C3  0.2   0.8 
10
X
3
Bi  n = 10, p = 0.2 
= np
=2
Variance = npq
= 1.6
Mean
ANSWER: A
7
The Hypergeometric Distribution
X ~ Hg (n, D, N )
Pr( X = x) =
D
Cx 
N D
N
Cn
C n x
,
nD
 = E( X ) =
N
 n D  D  N  n 
 = Var ( X ) = 
1  

 N  N  N  1 
2
 = SD( X ) = Var ( X )
Question 64
A team of four is selected from six women and four
men. What is the probability that the team consists
of exactly one woman and three men.
C1  C3
Pr( X = 1) = 10
C4
6
ANSWER: A
4
Question 65
A jar contains fifteen jellybeans of which twelve are
green. Four jelly beans are taken from the jar at
random and eaten, calculate Pr( X  3)
Pr( X  3) = Pr( X = 3)  Pr( X = 4)
C3  C1
C4  C0
= 15

15
C4
C4
12
= 0.846
3
12
3
Calculator program
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
Same 
Different 

Same 
Different 

1

2
Question 67
Which one of the following sets of statements
is true?
X ~ N (  ,2 )
1
1 1
1 < 2 , 1 <  2
X ~ N (  ,2 )
2
2 2
ANSWER: A
Method
• Draw a diagram, clearly labelling the
mean.
• Shade the region required.
• Determine the z value which
corresponds to the value of x by using
z=
x

• Use the cumulative normal distribution
table to find the required probability.
Using the cumulative normal distribution table
1
Pr( Z < 1) = 0.8413
1
Pr( Z > 1) = 1  Pr( Z  1)
= 1  0.8413
= 0.1587
-1
Pr( Z < 1) = Pr( Z > 1)
= 1  Pr( Z < 1)
= 1  0.8413
= 0.1587
-2
1
Pr( 2 < Z < 1) = Pr( Z < 1)  Pr( Z < 2)
= Pr( Z < 1)  Pr( Z > 2)
= Pr( Z < 1)  1  Pr( Z < 2)
= 0.8413  (1  0.9772)
= 0.8413  0.0228
= 0.8185
Question 68
The mass of fruit jubes, in a packet labelled as containing
200 grams, has been found to be normally distributed with a
mean of 205 grams and a standard deviation of 4 grams.
The percentage of packets that contain less than 200 grams
is, correct to one decimal place,
200  205 

Pr( X < 200) = Pr Z <

4


Pr( Z < 1.2) = Pr( Z > 1.2)
= 1  Pr( Z < 1.2)
= 1  0.8944
= 0.1056
ANSWER: C
Question 71
The eggs laid by a particular breed of chicken have a
mass which is normally distributed with a mean of 61 g
and a standard deviation of 2.5 g. The probability, correct
to four decimal places, that a single egg has a mass
between 60 g and 65 g is
Pr(60 < X < 65) = Pr  0.4 < Z < 1.6
= Pr( Z < 1.6)  Pr( Z < 0.4)
= Pr( Z < 1.6)  1  Pr( Z < 0.4)
= 0.9452  (1  0.6554)
= 0.6006
ANSWER: C
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 distribution table (or
graphic calculator) to find the required
probability and the corresponding z value.
• Use the relationship z =
x

calculate the required x value.
to
Question 72
Black Mountain coffee is sold in packets labeled as
being of 250 grams weight. The packing process
produces packets whose weight is normally distributed
with a standard deviation of 3 grams.
In order to guarantee that only 1% of packets are
under the labeled weight, the actual mean weight (in
grams) would be required to be closest to
a) 243
b) 247
c) 250
d) 254
e) 257
250

Pr  X < 250  = 0.01
250  
2.33 =
3
 = 257
ANSWER: E
Question 74
Pr( X > d ) = 0.25
 Pr( X < d ) = 0.75
d  80
= 0.6745
3
d = 82
Question 75
a)
Analysis Question
  2 = 84  2(12)
= 108
b)
94.2  84 

Pr( X < 94.2) = Pr Z <

12


= Pr( Z < 0.85)
= 0.802
c)
75  84 

Pr( X < 75) = Pr Z <

12 

= Pr( Z < 0.75)
= 1  Pr( Z < 0.75)
= 0.2266
= 23%
d)
Pr( X > a ) = 0.12
Pr( X < a ) = 0.88
a  84 

 Pr X <
 = 0.88
12 

a  84

= 1.175
12
a  84 = 14.1
a = 98
e)
Pr( X > 114 / Jumbo )
Pr( X > 114)
=
Pr( Jumbo )
0.0062
=
0.12
Conditional probability
= 0.052
Pr( A  B)
Pr( A / B) =
Pr( B)
f)
Small
Standard
Jumbo
Price
9
19
30
Prob
0.23
0.65
0.12
E ( X ) = 9  0.23  19  0.65  30  0.12
= 18.02
Income = 2000  18.02
= $ 360
g)
Pr( X  2) = 1   Pr( X = 0)  Pr( x = 1) 
= 1  C0 (0.12) (0.88)  C1 (0.12) (0.88)
6
0
= 1  0.4644  0.3780
= 0.156
6
6
1
5
THE FINAL RESULT