### File - VCE Solutions

```Mathematical
Methods (CAS)
Unit 3
VCE SOLUTIONS
Functions and
relations
Set Notation
[ or ]
‘Closed brackets’ – including that end-point
[2,3] means every real number between and including 2 and 3
( or )
‘Open brackets’ – excluding that end-point
When describing a set with end-point ∞ or -∞, make sure to use the ‘open brackets’ as opposed to the closed
brackets.
(2,3) means every real number between but not including 2 or 3
{…}
A set of numbers containing the specific terms as listed within the brackets
{1, 5, 19} means the set of numbers 1, 5 and 19
\
R\[2,3] means every real number excluding the set of numbers between 2 and 3, and also
excluding 2 and 3 themselves
R\{2,3} means every real number excluding 2 and 3
∩
Intersection of two sets – gives the set which contains all elements common to both of the two
original sets
[-5,7] ∩ (2,10) notates the intersection of the sets [-5,7] and (2,10) and is defined by the set (2,7]
U
Union of the two sets – gives the set which contains all elements contained in either of the two
original sets
[-5,7] U (2,10) notates the union of the sets [-5,7] and (2,10) and is defined by the set [-5,10)
∈
‘is an element of the set described following’
1 ∈ R means that 1 is an element of the set of all real numbers
⊆
‘is a subset of the set described following’ – i.e. all the elements of the set preceding the symbol
are contained in the set following the symbol
(1,5) ⊆ (-1,7) means that all the elements of the set (1,5) are contained within the set (-1,7)
Types of functions
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Linear functions
Cubic functions
Quartic functions
Inverse functions
Composite functions
Hybrid Functions
Modulus functions
Discriminant and Determinant
Sketching functions
Linear and Quartic – very rarely examined
‒ Know how the shapes of each quartic graph
Quadratic and Cubic – sketching mostly relates to transformations and
integration/differentiation
‒ Sketching quadratic and cubic functions are essential to know
‒ Know the shapes and how to find turning points
‒ Different ways to write the equations
Sketching functions
A few basic tips:
◦ When sketching any graph, a good place to start is always to find any x and y axis
intercepts.
◦ For example, If you need to sketch a linear graph, all you need to do is find the x
and y axis intercepts, and then draw a line through those points.
◦ As the degree (the degree is essentially the value of the highest power in a
polynomial function – i.e. x^3 + x^2 +2 would be of degree 3, and x + 2 would be
of degree 1) of a polynomial increases, polynomials generally require a few more
key points to be found.
◦ For instance, a quadratic function (parabola) has an turning point that needs to
be found. A cubic function may have two turning points that need to be found,
and a quartic may have three. But in each case, the first steps can simply be
finding the x and y ints. You will learn how to find turning points when you start
calculus – it’s not too hard, but for now all you need to do is find x & y ints!
◦ But to find x and y ints polynomials, you’ll need to know how to factorise
themwell
Factorising polynomials
What is the process of carrying out polynomial long division?
In maths methods, long division is used to divide a polynomial by another polynomial, where the
degree of the numerator is equal to or greater than the degree of the denominator.
Sometimes dividing through will result with a remainder
Example: (2x2-5x-1)/(x-3)
And sometimes it will result with no remainder
Example: (x2-3x-4)/(x-4)
When there is no remainder, it can be easier to use another method (instead of long division) to
find the unknown factor:
Example: (x2-3x-4)/(x-4)
What is the remainder theorem? What is the factor theorem?
if x-a if a factor, then P(a) = r (where r is the remainder)
What is the factor theorem?
if x-a if a factor, then P(a) = 0
- Whether there is a remainder or no remainder depends on the polynomial.
Sketching Polynomials
Now that we know how to factorise polynomials, we can go about sketching them.
Just as you learnt in year 10 and 11 that it is easier to sketch a parabola when it is
in the form (x-a)(x-b) as compared to ax^2 + bx + c, it’s the same for polynomials
of higher degree
This is because we can let y=0 to find the x-intercepts, and by the null factor law,
the x-ints are essentially the negative components of each factor; that is, for f(x) =
(x-a)(x-b)(x-c), the x-ints are x=a, x=b, x=c (as we put f(x) = 0).
And once we’ve found these x-ints, finding the y-ints is always a lot easier
Therefore, being able to sketch polynomials is more about being able to factorise
them than anything else. Of course, there are a few other considerations, such as
whether it is positive or negative, or translated, etc (another key consideration is
the turning points, but you learn this later on in calculus). But if you know how to
factorise polynomials well, then you can basically sketch their graph.
Note: this is why the quick method for factorising polynomials is so useful – to
sketch graphs faster.
Square root functions
Firstly, whenever you see a square root function, you must be very conscious that you can’t
take the square root of a negative number.
This is why the standard sq root functions only exist for positive x-values (and zero).
Knowing sq root functions only exist when the “inside” part is greater than zero, we can easily
establish our implied domain for the function
Example: f(x) =
As always, it’s a good idea to find x and y intercepts
Other things to know: if there’s a negative sign in front of the sq root, then the graph will be
flipped upside down, and if there’s a negative sign in front of the ‘x’ term inside the sq root,
then the graph will be flipped side ways. You will understand this better once you go through
transformations.
Then you can sketch. However, it also helps to find one other point to help guide your sketch
(i.e. you could substitute x=1 into f(x) to find the y-value, and then you’d have an extra point to
play around with)
Hyperbolae
With sq root functions, we established that you must always remember
that you can’t take the square root of a negative number. With
hyperbolae, the condition to remember is that you can’t divide by zero.
i.e. the denominator cannot equal zero
Example: f(x) =
This gives us the vertical asymptote.
However, we also know that no matter how large x becomes, the overall
quotient term will not equal zero. So we get a horizontal asymptote as
well; that is, y=0 is the horizontal asymptote (if the hyberbola was
translated +c units, the asymptote would y = c)
Trunci
Very similar to Hyperbolae – denominator cannot equal zero, so the
process is the same.
Since the denominator is squared, all negative components on the left
become positive, just like on the right. Hence, the graph is symmetrical
Matrices
An (m x n) matrix multiplied by an (n x r) is defined as the number of
columns in the first matrix equals the number of rows in the second matrix.
The dimension of the resulting matrix is (m x r)
That is (m x n) x (n x r) produces a (m x r) matrix.
Inverse matrices:
Firstly, the determinant of a matrix is det(A) = ad-bc for a matrix A=
-You can think of the determinant as the “magnitude” of a matrix, but this
magnitude will often turn out to be completely different to what you would
expect
-We need the determinant to be able to find the inverse of a matrix A, which
is denoted A^-1:
Using inverse matrices
The most common matrix equation is Ax=B where A has dimensions (2x2) and B has
dimensions (2x1). If we want to solve for x, we simply pre-multiply both sides of
the equation by A^-1 (which is why the inverse matrix is useful) -(explain further
on board)
Once you understand the inverse matrix, you can apply it to more difficult
problems. The most commonly examined application of the inverse matrix
resonates with the following example. For this example, it is important to realise
that solutions only exist when the inverse matrix exists, and the inverse on exists
when the determinant isn’t equal to zero (as the determinant is on the
denominator in the expression for the inverse – and you can’t divide by zero)
Find the values of m for which there is a unique solution, and then infinitely many
or no solutions.
Inverse functions
Only exists when function is one-to-one, that is it fulfils the
requirements of both the horizontal and vertical line test
Sketching these graphs – reflect about the line  =
To find the inverse functions must state: for inverse functions, switch
and
Common exam question asks you to find the maximal domain of ()
so that an inverse function exists
Be aware that an inverse function and its original function ALWAYS
intersect at  =
Example – Inverse functions
Consider the function : (, 3) → , () =  + 1
a)
Find the smallest value for
b) Find the inverse function
2
Composite functions
∘ () only exists when the range of  is a subset of or equal to the
domain of  (i.e. ran  ⊆ dom )
Common exam questions include finding the maximal domain of  for a
composite function to exist
Good technique is to draw the table to find the suitable range of  then
by extension domain of g that allows  ∘ () to exist
Example – Composite
Functions
Let :  → ,   = 4 −  2 and  be the set of all real values of  for
which () is deﬁned. Let :  → , where () =  2 + 1. Given  ∘
() exists:
a)
Find the maximum domain for (), .
b) Is  ∘ () defined for the () with domain ? State reasons
c)
Sketch  ∘ ()
Hybrid functions
Simply, more than one function sketched on one set of axis
‒ Can be tied with kinematics and integration
‒ Cannot be differentiated when function changes (or at the cusp)
Common questions asks you to sketch, find maximal domain and range
(examined most commonly in the first exam)
Sketch each part of the hybrid function as separate functions
Modulus function is a special type of hybrid function
Modulus functions
Modulus is essentially the magnitude – hence all negatives become
positive
A type of composite function where one of the functions is () = ||
First type – takes the modulus of the entire function –  ∘
Second type – takes the modulus of only the x value –  ∘ ()
Common questions asks you to sketch, find maximal domains and
ranges
Example – Modulus functions
Let () =  2 − 6 + 5 and () = ||
a)
Find  ∘   and sketch the graph and state the maximal domain
and range.
b) Find  ∘ () and sketch the graph and state the maximal domain
and range.
Discriminant and determinant
The discriminant is given by the expression Δ = 2 − 4
‒ When it equals 0, there is one solution
‒ When it is greater than 0, there are multiple solutions
‒ When it is less than 0, there are no solutions
The determinant of some matrix  is given by the expression det  =
=  −  (from matrix)
‒ det() = 0, no solutions or infinitely many solutions exist
‒ det() ≠ 0, a unique solution exists
Can use either to solve simultaneous equations
Example
Discriminant and determinant
Consider the simultaneous equations
+ 2 = 8 and 4 − (2 − ) = 2
Find the value(s) of  for which there are
a)
no solutions
b)
infinitely many solutions
c)
a unique solution
Index laws
×  = +

= −

=
×  =

1
=
=

1
− =
0 = 1
1 =

Log laws
log   = log  () + log  ()
log

= log  () − log  ()
log   =  log  ()
log

log  () = log
− log   = log
log  1 = 0
log   = 1
1

Transformations
Transformations
=×  + +
‒  is a dilation of factor  from the  axis (changing the  value)
‒  is a dilation of factor  from the  axis (changing the  value)
‒  is a translation of  units in the negative direction of the  axis
‒  is a translation of  units in the positive direction of the  axis
‒ If  is negative, it is also a reflection in the  axis
‒ If  is negative, it is also a reflection in the  axis
Transformations
Situation 1: Given a set of transformations to apply to a function, you
are asked to find the new transformed function.
1
Example:   = 2 is transformed by:

‒ a reflection in the  axis
‒ a translation of 3 units in the positive direction of the  axis
‒ a dilation of 2 units from the  axis
Write down the equation of the new transformed function.
Transformations
Situation 2: Given the transformed function and original function, you
are asked to find the transformations performed to change the original
function to the transformed function.
Example:
1

= is transformed to () =
2
−1
+3
State the sequence of transformations required.
Transformations
Situation 3: Given a “complex” transformed function and you are asked
to find the transformations you have to perform to map it back to the
original “simple” function.
Example:
() =
3
– 7 is
1− 2
transformed to () =
1
2
State the sequence of transformations required
Transformations using the
dash method
What are the steps to using the dash method?
Transformation using matrices
Remember top left box affects the  axis and so is therefore a
reflection/dilation in or from the  axis
Bottom right box affects the  axis and so is therefore a
reflection/dilation in or from the  axis
Transformations of the top right and bottom left boxes are rarely
examined in this course but is a reflection in the line  = .
Translations are graphed on the 2 × 1 matrix.
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