Math Methods 3/ 4 - unit 3, 4 syllabus by
topic/ sub topic
Lesson Theory / Topics covered
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Course requirements + work through booklets
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Revision
Functions & Relations
Linear Equations
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Solving them with numbers
Solving algebraic Questions
Simultaneous linear equations – go through all methods
 Graphical – substitution (the most useful) – elimination (not as useful but
assumed and solvable with matrices if that’s your thing) – the CAS solve function
 The CAS simultaneous equation solving function on the Math1 keyboard
Linear Graphs
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Equations of straight lines
All the basics, i.e.
1. 𝑦 = 𝑚𝑥 + 𝑐 know how to find the gradient, the y intercept and calculate the
x intercept
2. If given the gradient and one point, work out the equation
3. If given two points, work out the gradient and then work out the equation
4. From the equation, sketch the graph; From the graph work out the equation.
Transposing 𝑎𝑥 + 𝑏𝑦 = 𝑑 ↔ 𝑦 = 𝑚𝑥 + 𝑐
Midpoint – average out the x and y coordinates (and z if they go 3D) of the given
𝑥 +𝑥 𝑦 +𝑦
points (𝑥𝑚 , 𝑦𝑚 ) = ( 1 2 2 , 1 2 2 )
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Distance b/w points (pythag) *This has a LOT of use *
𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
Parallel lines (same gradient 𝑚1 = 𝑚2 )
−1
−1
Perpendicular lines 𝑚1 = 𝑚 𝑜𝑟 𝑚2 = 𝑚 or 𝑚1 × 𝑚2 = −1
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Applications of straight lines
2
1
Jac
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 Ch 1
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Cambridge
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 2A 1,
 2B Q1
 2A Q2, *3, *4,
*5
 2B*2, *3, *4
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 2C 1-7, 10-17,
18, *19 - *25
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 2D 2- 5, *6
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Simultaneous equations with unique solutions OR no unique solutions I.e. no
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solutions or infinite solutions
No unique solution with 2 equations means either:1. Infinite solutions i.e. the line is the same, the equation and its answer are the
same
2. No solutions i.e. the lines are parallel, the equation is the same but the answer is
different/ if in y = mx + c form, m is the same c is different
If a constant is involved, find x and y in terms of the constant (using any method) and
n.u.s. means the value that gives no real answer (usually a denominator = 0). Trial
and error required to see if this is a zero or infinite case
Must be able to use simultaneous solve function on CAS – on the Math1 keyboard,
column 3 row 4 it looks like { with two rectangles.
Eg. To solve 2x + 1 = y and x + 2y = 12, Enter
2𝑥 + 1 = 𝑦
{
|
} and the answer {x = 2, y = 5} should appear
𝑥 + 2𝑦 = 12 𝑥 , 𝑦
𝑇𝑜 𝑔𝑒𝑡 𝑎𝑛 𝑒𝑥𝑡𝑟𝑎 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑐𝑙𝑖𝑐𝑘 𝑜𝑛 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠 𝑎𝑛𝑑 𝑝𝑢𝑠ℎ 𝑡ℎ𝑒 𝑏𝑢𝑡𝑡𝑜𝑛 𝑎𝑔𝑎𝑖𝑛
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Domain & Range, relation – function – 1 to 1 function
Domain & Range -will need to revise some basic functions, straight line, parabola,
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cubic, circle, square root, hyperbola, truncus
Know what the domain and range of the “basic” functions are
Know the different forms of writing the domain
eg 𝑥𝜖 𝑅 = (−∞, ∞), 𝑥 > 3 = (3, ∞), 𝑥 ≥ 3 = [3, ∞),
2 < 𝑥 ≤ 5 = (2, 5] and so on
By revise, includes awareness of appearance and major features of the graphs,
including x and y intercepts, turning points, stationary points of inflection,
asymptotes, starting points
inequalities associated with the above.
Will need to revise how to graph restricted functions re full and half circles
Maths vocab i.e. 𝑦 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑡𝑜 𝑓(𝑥)𝑜𝑟 𝑔(𝑥) 𝑒𝑡𝑐, 𝑓 −1 (𝑥)𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑓(𝑥)
Revise all features of the methods basic shape collection
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1. To prepare for the year’s work later
2. To illustrate implied domain and range
 2F 1-3, 5-9
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 1B 1, 2, 3- 6,
7, 8, 10, 11,
15, 16, *12 *14

methodsbasicgraph
s.docx
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Will NEED to revise basic shape and transformations of
1 1
3
𝑥 2 , 𝑥 3 , 𝑥 𝑛 , 𝑥 , 𝑥 2 , 𝑎𝑛𝑑 √𝑥 , √𝑥 graphs, including major features (asymptotes, starting
points, turning points, SPI,) and what happens to these under translation.
As these things should / need to be known
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1 to 1 functions – revise relation and function
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A function, each x has a maximum of one y value
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One to one function, as well as being a function, each y value has only one x value
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Revise set and implied domains
Set = you are told eg 𝑦 = 2𝑥 + 5, 𝑥𝜖 (2, 5] this restricts x to the set values, and hence
restricts y to 𝑦𝜖 (9, 15]
Implied eg 𝑅 → 𝑅, 𝑓(𝑥) = √𝑥 + 3 + 4 where by implication 𝑥 ≥ −3 𝑎𝑛𝑑 𝑦 ≥ 4
because of the nature of the graph
 Set domains can be important eg 𝑦 = (𝑥 − 3)2 𝑥 𝜖 𝑅 is not a one to one function,
but 𝑦 = (𝑥 − 3)2 𝑥 ≥ 3 is.
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Hybrid functions A function is a combination (hybrid) of different rules for different x
values eg
2𝑥 + 5 𝑥 ≤ 2
𝑓(𝑥) = { 2
} which is a straight line for 𝑥 ≤ 2 and a parabola for x>2. To
𝑥
𝑥>2
find f(a), a must be substituted into the correct rule.
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Sums and products of functions
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Will NEED to revise basic shape and transformations of
1 1
𝑥 2 , 𝑥 3 , 𝑥 , 𝑥 2 , 𝑎𝑛𝑑 √𝑥 graphs, including major features (asymptotes, starting points,
turning points, SPI,) and what happens to these under translation.
 + go over hints for ordinate sketching graphs eg sketch the basic graphs
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Sum of functions f(x) = g(x) + h(x)
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𝐷𝑜𝑚𝑎𝑖𝑛 𝑓 = 𝐷𝑜𝑚 𝑔 ∩ 𝐷𝑜𝑚 ℎ i.e. for f to have a value, there must be a valid x value
for both f and g
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To sketch them, sketch the individual functions and add ordinates (y values) at
various points – good to chose known / easy or significant points eg x intercepts, y
intercepts, turning points
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Difference of functions f(x) = g(x) - h(x) gets converted to the sum of functions f(x) =
g(x) + -h(x) so it can done the same way
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Product of functions 𝑓(𝑥) = 𝑔(𝑥) × ℎ(𝑥)
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2.2 Q13
 1C 1 - 8
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 1C 9 - 17
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 1D All
 Except *1c
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 1D?
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 1D?
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𝐷𝑜𝑚𝑎𝑖𝑛 𝑓 = 𝐷𝑜𝑚 𝑔 ∩ 𝐷𝑜𝑚 ℎ i.e. for f to have a value, there must be a valid x value
for both f and g
To sketch them, sketch the individual functions and multiply ordinates (y values) at
various points – good to chose known / easy or significant points eg x intercepts, y
intercepts, turning points
Division of functions 𝑓(𝑥) = 𝑔(𝑥) ÷ ℎ(𝑥) gets converted to the product of functions
1
𝑓(𝑥) = 𝑔(𝑥) × ℎ(𝑥) so it can done the same way
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Power functions
What they are, how to get all basic shapes (text is pretty good)
Cover all scenarios, 𝑦 = 𝑥,
𝑦 = 𝑥 2, 𝑦 = 𝑥 ℎ𝑖𝑔ℎ𝑒𝑟 𝑒𝑣𝑒𝑛,
𝑦 = 𝑥 𝑜𝑑𝑑>1
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, 𝑦 = 𝑥 −𝑜𝑑𝑑 , 𝑦 = 𝑥 −𝑒𝑣𝑒𝑛
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𝑦 = 𝑥 𝑜𝑑𝑑 , 𝑦 = 𝑥 𝑒𝑣𝑒𝑛
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𝑦 = 𝑥 𝑜𝑑𝑑 , 𝑦 = 𝑥 𝑒𝑣𝑒𝑛 , 𝑦 = 𝑥 𝑜𝑑𝑑 𝑓𝑜𝑟 𝑡ℎ𝑒𝑠𝑒, 𝑠𝑘𝑒𝑡𝑐ℎ
1
 1G Q1 - 4
1
𝑒𝑣𝑒𝑛
1
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𝑜𝑑𝑑
𝑜𝑑𝑑
1
𝑡ℎ𝑒 𝑥 𝑜𝑑𝑑 𝑜𝑟 𝑥 𝑒𝑣𝑒𝑛 𝑓𝑖𝑟𝑠𝑡, 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑟𝑎𝑖𝑠𝑒 𝑖𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑝𝑜𝑤𝑒𝑟 (more in chapter 7)
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Composite functions
What they are, what f o g (x) and g o f (x) means
For f o g(x) to be defined, range of g must be a subset of the domain of f If this can’t
be done, it is not defined
Advanced, restrict domain of g to hence restrict range of g so that it is a subset of
range of f
i.e. 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑔 ⊂ 𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑓
Inverse functions 𝑓 −1 (𝑥)
To get an inverse, swap x and y values, and then transpose.
To get an inverse graph, swap x and y values / reflect across y = x line
To get an inverse function, the original function must be one to one
Advanced, to get an inverse function restrict the domain of the original so that it is
one to one
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 IE 1 -10, *11,
*12
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 1F 1 – 10
algebra
 1F 11, 12, 17
estimate from
graph
 1F 13- 16, 18
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The graphs of 𝑓(𝑥)𝑎𝑛𝑑 𝑓 −1 (𝑥)𝑚𝑒𝑒𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑦 = 𝑥
Domain and ranges of 𝑓(𝑥)𝑎𝑛𝑑 𝑓 −1 (𝑥) 𝑎𝑟𝑒 𝑠𝑤𝑎𝑝𝑝𝑒𝑑
Application Questions to reinforce above concepts
Transformations – translations, dilations, reflections across x axis and y axis,
reflection across y = x (inverse), rotation (not doing it)
𝑦 = 𝑓(𝑥) → 𝑦 = 𝐴 × 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛(𝑛(𝑥 − 𝐸)) + 𝐵 means
1
𝑥𝑁 = 𝑛 𝑥𝑂 + 𝐸 𝑎𝑛𝑑 𝑦𝑁 = 𝐴 𝑦𝑂 + 𝐵
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 1H 4, 6, 7
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1
𝑥𝑁
𝐸
0 𝑥𝑂
Or in matrix form as [𝑦 ] = [ 𝑛
] [𝑦 ] + [ ]
𝑁
𝐵
0 𝐴 𝑂
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“starting point” (original origin) is moved 𝐸 right and 𝐵 up. (this is the translation)
It is dilated away from the original 𝑥 axis (or the translated one) using the rule
𝑁𝑒𝑤 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 × 𝐴 aka dilated vertically aka dilated in the
direction of 𝑦 by a factor of 𝐴
−𝐴 means reflected vertically across the original 𝑥 axis.
eg what happens with the amplitude when you deal with sin and cos graphs
It is dilated away from the original 𝑦 axis (or the translated one) using the rule
𝑜𝑙𝑑
𝑂𝑙𝑑
𝑁𝑒𝑤 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 ÷ 𝑛 ↔ 𝑁𝑒𝑤 = 𝑛 ↔ 𝑛 = 𝑛𝑒𝑤 aka dilated
1
horizontally aka dilated in the direction of 𝑥 by a factor of 𝑛
−𝑛 means reflected horizontally across the original 𝑦 axis.
eg what happens with the wavelength when you deal with sin, cos and tan graphs
Eg 𝑦 = sin(𝑥) ℎ𝑎𝑠 𝑎 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑎𝑡 (0,0), 𝑎𝑛𝑑 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 1
𝑎𝑛𝑑 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 2𝜋
𝜋
𝜋
𝑦 = 4sin 3 (𝑥 − 6 ) + 7 ℎ𝑎𝑠 𝑎 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑎𝑡 ( 6 , 7) , 𝑎𝑛𝑑 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 4
𝑎𝑛𝑑 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓
2𝜋
3
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Translations of points by description
Eg 𝑥 ′ = 𝑥 + 3 𝑚𝑒𝑎𝑛𝑠 𝑛𝑒𝑤 𝑥 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑥 + 3
𝑚𝑒𝑎𝑛𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑥 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑎𝑙𝑙 𝑚𝑜𝑣𝑒𝑑 3 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡
Eg 𝑦 ′ = 𝑦 − 12 𝑚𝑒𝑎𝑛𝑠 𝑛𝑒𝑤 𝑦 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑦 − 12
𝑚𝑒𝑎𝑛𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑦 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑎𝑙𝑙 𝑚𝑜𝑣𝑒𝑑 12 𝑑𝑜𝑤𝑛
E right and B up (opposite for +E and –B)
Translations of functions
Dilations of points by description
Eg 𝑥 ′ = 2𝑥 𝑚𝑒𝑎𝑛𝑠 𝑛𝑒𝑤 𝑥 = 2 × 𝑜𝑙𝑑 𝑥
𝑚𝑒𝑎𝑛𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑥 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑎𝑙𝑙 𝑑𝑜𝑢𝑏𝑙𝑒𝑑
New distance up from original x axis (vertical dilation) = old distance x A
𝑜𝑙𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
New distance from original y axis (horizontal dilation) =
𝑛
Dilations of functions
Reflections across original x and y axis
Vertical reflection (across x axis) A is negative
Horizontal reflection (across y axis) n is negative
Combinations of transformations
Estimating from graphs
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Further combinations – algebra
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One way to work out the transformations from a complicated graph back to a simpler
original, is to work out the steps of how to do the simple original to the complicated
graph and then work backwards.
Further applications of sketching, using the know what the original looks like, get the 
equation into standard form, and then systematically transform technique
Transforming power rules
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1. Revise earlier material on power graphs (Ex 1G) and then 2. transform them using
the same techniques as everything else
Transformations of basic power functions of the form 𝑦 = 𝐴 (𝑥 − ℎ)𝑛 + 𝑐 both
equation ↔ graph
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 3A Q1
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 3A Q1-8
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 3B Q1 - 11
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 3C All
 3D 2, 3
 3D – the rest
are OK, a bit
repetitive
 3E 1, 2 a) – d),
4a) –e) *4f
 3F All
 3G Q1, 2, 4, 6,
7
 The rest are
OK
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Use concepts of start point (original x = 0), centreline (original y = 0) to
help with translations, dilations and reflections
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Graph → Equation using the knowledge of transformations
Matrix Transformations
 How to show transformations with matrices
𝑥
𝑐
𝑥′
𝑎 0 𝑥
𝑇 ([𝑦]) = [ ] = [
] [𝑦 ] + [ ]
𝑑
𝑦′
0 𝑏
𝑥𝑁
𝑐
𝑎 0 𝑥𝑂
Rewrite as [𝑦 ] = [
] [ ] + [ ] which can get written as
𝑑
𝑁
0 𝑏 𝑦𝑂
𝑥𝑁 = 𝑎 𝑥𝑂 + 𝑐 𝑎𝑛𝑑 𝑦𝑁 = 𝑏 𝑦𝑂 + 𝑑
Which can get substituted in, to find new rules in terms of xN and yN (x new and y
new) OR transposed
𝑥𝑁 − 𝑐
𝑦𝑁 − 𝑑
𝑥𝑂 =
𝑎𝑛𝑑 𝑦𝑂 =
𝑎
𝑏
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Quadratic functions and parabolas
𝑒𝑥𝑝𝑎𝑛𝑑𝑒𝑑 𝑓𝑜𝑟𝑚 ↔ 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑠𝑒𝑑 𝑓𝑜𝑟𝑚 ↔ 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑓𝑜𝑟𝑚 ↔ 𝑒𝑥𝑝𝑎𝑛𝑑𝑒𝑑 𝑓𝑜𝑟𝑚
What each form is useful for, and the algebra involved to convert them.
Exp form – y intercept, and form best to do calculus on, and use the quadratic
formula on.
Factorised form – best to find solutions to f(x) = 0 and x intercepts (SAME THING)
TP form – best to find max and min point, and use the standard transformation rules
on
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Quadratic formula
Use of the discriminant to determine how many solutions, and then what they are.
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
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 3H
 3I 7 – 17
 3J Q6 - 13

 4A Q1 – Tp
form to graph
 4A q2 convert
to TP form and
then graph
 4A 3a - d
factorise and
sketch *3e, *3f
 4A 4 complete
the square
and sketch
 4A – 8, 9
 *4A 6, 7,
 4A Q10 - 18
 A quadratic equation is an equation that can be written in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 where
𝑎, 𝑏 𝑎𝑛𝑑 𝑐 are constants, 𝑎 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 2 and 𝑏 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 and 𝑐 is
the constant
 If it can’t be solved easily by factorisation, or you are not told to use completing the square,
these can be solved using the quadratic formula
 The discriminant of the equation is given by the rule
𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡 = 𝑏 2 − 4 × 𝑎 × 𝑐 𝑜𝑟 𝑏 2 − 4𝑎𝑐


If 𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡 = 0, 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑜𝑛𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑤ℎ𝑒𝑟𝑒
−𝑏
𝑥 = 2𝑎
If 𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡 > 0, 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠: −
𝑥=
−𝑏+ √𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡
2𝑎
and 𝑥 =
𝑠𝑜𝑚𝑒𝑡𝑖𝑚𝑒𝑠 𝑐𝑜𝑚𝑏𝑖𝑛𝑒𝑑 𝑎𝑠 𝑥 =
𝑥=

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
−𝑏− √𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡
2𝑎
−𝑏± √𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡
2𝑎
or
−𝑏± √𝑏2 −4𝑎𝑐
2𝑎
If 𝑑𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡 < 0, 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑛𝑜 𝑟𝑒𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠
(Yes there are complex solutions – specialist maths only)
25
If the discriminant is a perfect square eg 16, 100, 4 , the solution can be rational.
Graph → Equation
May need to revise simultaneous equations with CAS and otherwise
Polynomials
What they are
The two main forms Expanded and Factorised
Expanded eg 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 − 10𝑥 + 24 is the form you need to easily see the y
intercept as well as calculate the derivative and anti derivative
Factorised eg 𝑓(𝑥) = (𝑥 − 2)(𝑥 − 4)(𝑥 + 3) is the form that enables you to find the x
intercepts, where they are and what they are like, as well as the solutions to
𝑓(𝑥) = 0 (exactly the same as the x intercepts)
Because both forms of the equation are useful, you may wish to expand the
factorised version AND / OR factorise the expanded form

 4B all
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The Basics f(a) means sub x = a into f(x)

Expanding
Expand brackets and simplify.- manually, and using the expand( ) function on CAS
Eg, with CAS, to expand (𝑥 + 3)2 type expand( (𝑥 + 3)2 )
Expand (𝑎 ± 𝑏)2 , (𝑎 ± 𝑏)3 𝑒𝑡𝑐 inc – when b power is odd, and + when it is even for
(𝑎 − 𝑏)𝑛 in the expanded section
Binomial expansion using formula and Pascal’s triangle
(𝑎 ± 𝑏)2 = 𝑎2 ± 2𝑎𝑏 + 𝑏 2
(𝑎 ± 𝑏)3 = 𝑎3 ± 3𝑎2 𝑏 + 3𝑎𝑏 2 ± 𝑏 3 and so on
𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 ℎ𝑒𝑟𝑒 − 𝑖𝑛𝑐 𝑃𝑎𝑠𝑐𝑎𝑙𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑎𝑛𝑑
𝑛
( 𝑟𝐶 𝑏𝑜𝑡ℎ 𝑤𝑖𝑡ℎ 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 [ 𝑠ℎ𝑖𝑓𝑡 ÷]𝑎𝑛𝑑 𝐶𝐴𝑆 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟 [𝑠𝑒𝑒 𝑎𝑙𝑝ℎ𝑎𝑏𝑒𝑡𝑖𝑐𝑎𝑙 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠)
𝑙𝑖𝑠𝑡 𝑢𝑛𝑑𝑒𝑟 𝑁]
Factorising
Polynomial division and related stuff
Polynomial long division
CAS will factorise, using the factor( ) instruction
Remainder theorem and factor theorem (ft = special case of rt where remainder = 0)
Using these, along with PLD, to factorise polynomial expressions
Special factorisations
𝑎2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏) 𝑎𝑘𝑎 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒𝑠
and
𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑇𝐻𝐸 𝑚𝑎𝑖𝑛 𝑠𝑝𝑒𝑐𝑖𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑎𝑛𝑑 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑠𝑎𝑡𝑖𝑜𝑛
𝑎3 + 𝑏 3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 )
𝑎𝑛𝑑 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 )
and these which are an application of the above
𝑎4 − 𝑏 4 = , 𝑎6 − 𝑏 6 = , 𝑎6 + 𝑏 6 =
Solving equations in factorised form
 4C All

 4D 1 -5

 4D 6 -14, *20,
21

 4D Q15

 4D Q 18 a – f
 18 g, h, I – sub
in x = 1 and x
= -2 first
 Q19a) b – try x
= 1, c – try x =
2, d try x = 1
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Factorise using CAS factor (…..)
(Make sure that they can expand as well) expand( expression )
Cubic sketching
Using intercepts (where and what sort) combined with short cuts

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Some quartic questions

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Finding equations - basic
Graph sketching of polynomials in factorised form, where the factor gave you the x
intercept and the type of intercept – n=1→ straight through, n = 2, 4, 6, etc
→parabolaish touch, n = 3, 5, 7, → cubey bend


Finding equations using simultaneous equations in lots of variables + CAS
Solving simultaneous equations between various polynomials and relations – a bit
advanced so may be hard for some
Indices and logarithms
Indices (plural of index) have many applications, as do logarithms (logs)
Indices and logarithms are opposites or inverse functions of each other
i.e. 23 = 8 𝐴𝑁𝐷 𝑙𝑜𝑔2 8 = 3 are equivalent statements
The most useful base for an index or log function is euler’s number “e”
Index graphs – important basic features + transformations

 2.2
7b,8,
10,
12,
18a, c, 
19,
21b,
22c*,
MORE

 4G 2, 5, 7, 8

 4H 1- 7, *8*21

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 4D Q22, 23

 4E Q1 – draw
graphs
 Find more
 4F Q1 – draw
graphs
 4F *2, *3, 4,
 4F 5, 6 using
CAS
 4G 1, 3, 4, 6,

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 5A 1 -8
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One big idea, the transformations are identical in terms of translations, dilations and
reflections as before
Eg 𝑦 = 2𝑥 𝑐𝑎𝑛 𝑏𝑒 𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑦 = 𝐴 × 2𝑛(𝑥−𝐸) + 𝐵
𝑒 𝑥 graphs
𝐻𝑜𝑤 𝑡𝑜 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝑎 𝑥 𝑖𝑛𝑡𝑜 𝑒 𝑘𝑥 and vice versa - will need later 𝑦 = 𝑎 𝑥 → 𝑦 =
𝑒 𝑙𝑜𝑔𝑒𝑎 ×𝑥 eg 𝑦 = 2𝑥 → 𝑦 = 𝑒 𝑙𝑜𝑔𝑒2 ×𝑥 (so that you can use deriv and anti deriv rules
among others)
Basic exponential rules and exact equations solving
Logarithms
Listing index laws, and log laws, show how they are related to themselves and each
other
𝑙𝑜𝑔𝑒 (𝑠𝑜𝑚𝑒𝑡𝑖𝑚𝑒𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑙𝑜𝑔𝑎𝑟𝑖𝑡ℎ𝑚) 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 ln 𝑜𝑟 𝑡ℎ𝑒 ln
𝑏𝑢𝑡𝑡𝑜𝑛 𝑜𝑛 𝑦𝑜𝑢𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟
How to work out logs / indices on a calc.
Change of base rule – This is important. Change of base log law
𝑙𝑜𝑔 (𝑥)
𝑙𝑜𝑔𝑎 (𝑥 ) = 𝑏(𝑎)
𝑙𝑜𝑔𝑏
𝑙𝑜𝑔𝑒 (𝑥)
1
𝑒𝑔 𝑙𝑜𝑔𝑎 (𝑥 ) =
=
× 𝑙𝑜𝑔𝑒 (𝑥 )
𝑙𝑜𝑔𝑒 (𝑎)
𝑙𝑜𝑔𝑒 (𝑎)
1
𝑒𝑔 𝑙𝑜𝑔2 (𝑥 ) =
× 𝑙𝑜𝑔𝑒 (𝑥 )
𝑙𝑜𝑔𝑒 (2)
Solving log equations using indices as an inverse
Applications of solving log and index equations
What they are – opposite of indices
Using log laws to simplify expressions
 Logarithms – what they mean
𝑒𝑔 23 = 8 ↔ 𝑙𝑜𝑔2 8 = 3
 Indices and logs are opposites i.e.
𝑙𝑜𝑔𝑒 (𝑒 𝑥 ) = 𝑥
𝑎𝑛𝑑 𝑒 (𝑙𝑜𝑔𝑒 𝑥) = 𝑥 and hence logs are used to solve index
equations, and indices are used to solve log equations

 5B Q1


 5C
 5D


𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛
𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛
(𝑎𝑚 )𝑛 = 𝑎𝑚×𝑛
𝑎1 = 𝑎
𝑎0 = 1 (𝑎 ≠ 0)
1
𝑎−𝑚 = 𝑚
𝑎−1 =
1.
2.
3.
4.
5.
6.
1
𝑛
𝑎

Index-log summary
sheet.docx
1
𝑎1
1
2
𝑛
7. 𝑎 = √𝑎
𝑎 = √𝑎
𝑚
𝑚
8. (𝑎 × 𝑏) = 𝑎 × 𝑏 𝑚
𝑎
9. ( )𝑚 = =
𝑏
10.
1.
2.
3.
4.
5.
6.



𝑎
( )−𝑚
𝑏
𝑎𝑚
𝑏𝑚
𝑏
= ( )𝑚
𝑎
=

𝑏𝑚
𝑎𝑚
𝑙𝑜𝑔𝑚 𝑎 + 𝑙𝑜𝑔𝑚 𝑏 = 𝑙𝑜𝑔𝑚 (𝑎 × 𝑏)
𝑎
𝑙𝑜𝑔𝑚 𝑎 − 𝑙𝑜𝑔𝑚 𝑏 = 𝑙𝑜𝑔𝑚 ( )
𝑏
𝑛
𝑛 × 𝑙𝑜𝑔𝑚 𝑎 = 𝑙𝑜𝑔𝑚 (𝑎 )
𝑙𝑜𝑔𝑚 𝑚 = 1
𝑙𝑜𝑔𝑚 1 = 0
𝑙𝑜𝑔𝑚 (𝑎−𝑛 ) = −𝑛 𝑙𝑜𝑔𝑚 (𝑎) … . 𝑙𝑜𝑔𝑚 (𝑎−1 ) = −𝑙𝑜𝑔𝑚 (𝑎) (
application of law 3)
Using logs to solve exponential equations, using the idea that logs
are opposite to indices
Demonstrate how to solve index equations using logs
Method 1 Eg 2 𝑥 = 10 → 𝑙𝑜𝑔𝑒 2𝑥 = 𝑙𝑜𝑔𝑒 10 → 𝑥 𝑙𝑜𝑔𝑒 2 =
𝑙𝑜𝑔𝑒 10
𝑙𝑜𝑔𝑒 10
→ 𝑥=
𝑙𝑜𝑔𝑒 2
OR any base will do

 5G1, 4, *5, **6
(use change of
base), 8, *9,
*10

Method 2 Eg 2 𝑥 = 10 → 𝑙𝑜𝑔2 2 𝑥 = 𝑙𝑜𝑔2 10 → 𝑥 𝑙𝑜𝑔2 2 =
𝑙𝑜𝑔2 10 → 𝑥 = 𝑙𝑜𝑔2 10 𝑤ℎ𝑖𝑐ℎ =
𝑙𝑜𝑔𝑒 10
𝑙𝑜𝑔𝑒 2
by the change of
base rule
 Scientific calculator only does 𝑙𝑜𝑔𝑒 the ln button and
𝑙𝑜𝑔10 the log button
CAS will do other bases

Log graphs – basic features and transformations

One big idea, the transformations are identical in terms of
translations, dilations and reflections as before
Eg 𝑦 = 𝑙𝑜𝑔𝑒 (𝑥) 𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑒𝑑 𝑡𝑜 𝑦 = 𝐴 × 𝑙𝑜𝑔𝑒 (𝑛(𝑥 − 𝐸)) + 𝐵

𝐸 𝑎𝑛𝑑 𝐵 𝑑𝑜 𝑡ℎ𝑒 𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑑 𝑢𝑝 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙
𝐴 𝑑𝑜𝑒𝑠 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑙𝑎𝑡𝑖𝑜𝑛 𝑢𝑝, 𝑎𝑤𝑎𝑦 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑥 𝑎𝑥𝑖𝑠
𝑁𝑒𝑤 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑜𝑙𝑑 × 𝐴
−𝐴 = 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑎𝑐𝑟𝑜𝑠𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑥 𝑎𝑥𝑖𝑠 = 𝑓𝑙𝑖𝑝𝑝𝑒𝑑 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙𝑙𝑦
 5E q1, 2, 3 *5
𝑛 𝑑𝑜𝑒𝑠 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑙𝑎𝑡𝑖𝑜𝑛 𝑟𝑖𝑔ℎ𝑡, 𝑎𝑤𝑎𝑦 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑦 𝑎𝑥𝑖𝑠
𝑜𝑙𝑑
𝑁𝑒𝑤 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑛
−𝑛 = 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑎𝑐𝑟𝑜𝑠𝑠 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑦 𝑎𝑥𝑖𝑠 = 𝑓𝑙𝑖𝑝𝑝𝑒𝑑 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙𝑙𝑦
 5E Q6, is an application of the change of base law for indices,
and can be proved using log laws
𝑎 𝑥 = 𝑒 𝑙𝑜𝑔𝑒𝑎 × 𝑥

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Solving log equations and exponential equations using CAS
Inverse functions / graphs using the fact that exponentials and logs
are opposites
The set of exercises also helps to revise index and log graph
drawing
Using log laws to help find the equation of index graphs
Using index laws to help find the equation of log graphs.
*Both of the above sometimes require use of simultaneous
equations, which you can solve with CAS, or at an advanced (but



 5E Q4, 5B Q5


 5F 1 -6, 9 Index
graphs
 5F 7, 8, 10 – 12
log graphs
 5H 1 – 5, 10
 The rest are
OK
still on the study design) level, particularly curious use of log and
index rules
(examples in text)
















Trigonometry
Degrees ↔ Radians
Basic sohcahtoa and its uses
Find sin,cos,tan on calculator
Exact values of some stuff including 30, 45 and 60 degrees
Sin & cos & tan in 4 quadrants using the unit circle
𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1 and where it’s from
𝑠𝑖𝑛𝜃
𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑠𝜃 and where it’s from
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑡𝑟𝑖𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
Transformations of sin, cos and tan graphs
Basic degrees ↔ radians, how radians come from unit circle


trigmindmap.pdf




Cos as x coordinate, sin as y coordinate and tan as intersection of θ 
line with x = 1 tangent of unit circle, and applying this to symmetry
and solutions in all 4 quadrants.
i.e. If you know the sin, cos and tan values for an angle in the first
quadrant, can you work out the sin cos and tan values for related
angle in the other quadrants.
Eg if you know sin, cos and tan for 25°, you also know them for 155,
205, 335, -25, -155, 385, etc
Exact trig ratios, for the special angles AND their related angles in
the 4 quadrants see the attached document
Complementary relationships and
Triangle drawing / Pythagoras to work out exact sin, cos, tan rules
eg if you know one trig value, and the quadrant, be able to work out
the others


trig summary sheet
12 methods.docx
 6A all

exact trig-v2.docx

 6B 1, 3, 6 all
tech free using
exact values in
each quadrant
 6B 4, 5 know
values for sin,
cos, tan in one
quadrant, work
out the rest
 6C Q1 all tech
free


See the explanations on p 264. The below rules are known by
everybody and will be demonstrated.
𝜋
sin ( 2 − 𝜃) = cos 𝜃

cos ( 2 − 𝜃) = sin 𝜃

tan ( 2 − 𝜃) = tan 𝜃
𝜋
𝜋
1
degrees
radians
sin
cos
tan
0
0
0
1
0
𝜋
1
1
√3
√3
𝑜𝑟
6
2
3
2
√3
𝜋
1
1
√2
√2
1
𝑜𝑟
𝑜𝑟
4
2
2
√2
√2
𝜋
1
√3
√3
3
2
2
𝜋
90
1
0
∞
2
Must know the above, and then be able to work out the related
angles eg if you know sin, cos, tan of 30, the values for 150, 210,
330, -30, etc are the same value but (maybe) different signs.
If you know one of sin, cos, tan for an angle, can you draw a
diagram, figure out the others, and then use quadrant knowledge to
work out if they are positive or negative.
 Solving trig equations in General form, and then narrowing the
domain, Cos and sin, use a unit circle diagram, tan – just +/wavelengths
 There is a link between solving trig equations, and backtracking
from a trig graph
 Solving trig equations for cos and sin
 Use calculator or exact value table to find one basic answer
 Use angle properties / symmetry / diagram to find the second
answer

30
45

60




 6C Q2 all tech
free
 6E Q1,2,3,7, 8
exact value
radians
 6E 6 Exact
value degrees
 6E 4, 9
Calculator
value radians

 6E 5 calculator
value degreesl
Add / subtract multiples of 2π
i.e. ±2𝑛𝜋 𝑤ℎ𝑒𝑟𝑒 𝑛 𝜖 𝑁 𝑖. 𝑒. 𝑛 = 0, 1, 2, … OR
+2𝑘𝜋 𝑤ℎ𝑒𝑟𝑒 𝑘 𝜖 𝑍 𝑖. 𝑒. 𝑘 = ⋯ − 2, −1, 0, 1, 2, …
If working in radians
or add subtract multiples of 360° i.e ±360𝑛, +360𝑘 if working in
degrees to your answers
 Add / subtract / divide as required to convert to solutions
 Show examples
Eg 4 sin 2(𝑥 − 11) + 5 = 8
→ 4 sin 2(𝑥 − 11) = 3
→ sin 2(𝑥 − 11) = 0.75 At this point, use a diagram to identify
where the two answers will be. Positive sin = positive y value = 1 st
and 2nd quadrants the calculator gives 48.6, checking the diagram
gives 180 – 48.6 = 131.4
→
→
2(𝑥 − 11) = 48.6 ± 360𝑛
(𝑥 − 11) = 24.3 ± 180𝑛
→ 𝑥 = 13.3 ± 180𝑛
𝐴𝑁𝐷
→ 2(𝑥 − 11) = 131.4 ± 360𝑛
→ (𝑥 − 11) = 65.7 ± 180𝑛
→ 𝑥 = 54.7 ± 180𝑛
*This works EXACTLY the same with radians, draw a diagram, use
a calculator to find one answer, work out the other (subtracting
from 𝜋 𝑜𝑟 2𝜋 rather than 180 or 360), and then adding ± 2𝑛𝜋

For tan,
1. use calc or exact value to find one answer.
2. Add subtract multiples of π / 180° to your answers
3. Add / subtract / divide as required to convert to solutions
Eg 4 tan 2(𝑥 − 11) + 5 = 7
→ 4 tan 2(𝑥 − 11) = 2
→ tan 2(𝑥 − 11) = 0.5

 ,These are a
mix of sin, cos,
tan 6K 2, 3, 6 9
→ 2(𝑥 − 11) = 26.6 ± 180𝑛
→ (𝑥 − 11) = 13.3 ± 90𝑛
→ 𝑥 = 23.3 ± 90𝑛
Note, Tan equations don’t really need diagrams to solve,
sin and cos equations do


As per comment on sin and cos, tan problems are done the
same with radians, but ±𝑛𝜋 rather than ±180𝑛

Some mixed sin and cos problems turn into tan problems
Eg sin 3(𝑥 − 1) = 2 cos 3(𝑥 − 1) 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑏𝑦 cos 3(𝑥 − 1) →
sin 3(𝑥−1)
cos 3(𝑥−1)







= 2 → tan 3(𝑥 − 1) = 2 solve as for a normal tan problem
General transformations for any graph of form 𝑦 = 𝑓(𝑥) into 𝑦 =
𝐴 × 𝑓(𝑛(𝑥 − 𝐵)) + 𝐷
Major features of trig graphs + the vocab ; max , min, centreline,
starting point , amplitude, period / wavelength (period and
wavelength are the same thing in maths, they are DIFFERENT in
physics)

Use of the above to transform sin, cos graphs calculate equations of
transformed sin , cos graphs.
See the link 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 ↔ 𝑔𝑟𝑎𝑝ℎ i.e. if the equation will give you the
key features of the transformed graph in terms of starting point,
centreline, amplitude and wavelength; you can also identify those
features from the graph and hence figure out the equation.
The exercises start easy, and get more intricate, may be able to skip
some
calculate equations of transformed sin , cos graphs.
Use of the above to transform tan graphs AND calculate equations
of transformed tan graphs.
 2.5 7,8, 9, 18,
19, 20
 2.6 5, 8, 13


 2.5 11 – 14,
21, 22
 6D All
(selection of –
a bit of
repetition)
 6F
 6G
 6I
 6J Lots

Same basic ideas as sin and cos, but a different natural wavelength
(180 or π) rather than 360 or 2π. There is also no amplitude, to
judge vertical dilation use the half way points (normally
𝜋
𝜋
(45, 1)𝑎𝑛𝑑 (−45, −1)𝑜𝑟 ( , 1) 𝑎𝑛𝑑 (− , − 1)
4
4

Combination trig / applications



Addition of ordinates for trig graphs
Difficult, but it involves sketching the two starting graphs (easy, and
a skill you should practice) and then working out key points on these
graphs eg max, min points and 0 values (easy, and a skill you
should practice) and then adding these y values (reinforcing skills
learned earlier)
More functions and relations

Fractional power functions

1
1

𝑦 = 𝑥 𝑜𝑑𝑑 , 𝑦 = 𝑥 𝑒𝑣𝑒𝑛

𝑦 = 𝑥 𝑜𝑑𝑑 , 𝑦 = 𝑥 𝑒𝑣𝑒𝑛 , 𝑦 = 𝑥 𝑜𝑑𝑑 𝑓𝑜𝑟 𝑡ℎ𝑒𝑠𝑒, 𝑠𝑘𝑒𝑡𝑐ℎ

𝑒𝑣𝑒𝑛
1





𝑜𝑑𝑑
1
 2.5, 23, 24*,
25, 26

 6L +MORE
 6H


𝑜𝑑𝑑
𝑡ℎ𝑒 𝑥 𝑜𝑑𝑑 𝑜𝑟 𝑥 𝑒𝑣𝑒𝑛 𝑓𝑖𝑟𝑠𝑡, 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑟𝑎𝑖𝑠𝑒 𝑖𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑝𝑜𝑤𝑒𝑟 (more in
chapter 7)
(the text has info / questions on “odd” and “even” functions, as far as
I can figure they are NOT part of the course)
Extra questions on composite functions and inverses (some are a

bit tricky)

Extra questions on sums, differences and products of functions (all

are a bit tricky)

Weird stuff – the book explanations are good for what is a minor and 
tricky part of the course
Extra advanced stuff on solving equations, both linear and

quadratic; exponential and log where there will be pronumerals
 7A Q1 – 4, *5
 *7B Do a
selection

 **7C a
selection

 **7D
 *7E 1- 13
(letters) in the answers. These are good advanced algebra skill
questions

Chapter 8 is revision on chapters 1 to 7


All the revision material in chapter 8 is fine, with the possible

exception of some of the very wordy parts of the extended response
questions







Basic Calculus
What it means, and shortcuts and rules to find it
for certain functions







Big idea, differentiation comes from doing a normal straight line

approximation between two points on a curved line, and then getting
the points closer and closer together.
𝑦 −𝑦
Average rate of change 𝑚 = 𝑥2 −𝑥1
2







 9A 1, 2 (av)

1
Big idea, differentiation comes from doing a normal straight line

approximation between two points on a curved line, and then getting
the points closer and closer together, until the distance between
them approaches (gets close to) 0.
First principles example
𝑓(𝑥) = 𝑥 2 + 5𝑥 + 6
Do this one as a thorough example.
Differentiation by 1st principles, showing how it is nothing more that
𝑦2 − 𝑦1
𝑚=
𝑤ℎ𝑒𝑟𝑒 𝑥2 − 𝑥1 = ℎ 𝑎𝑛𝑑 ℎ → 0
𝑥2 − 𝑥1
𝑑𝑦
∆𝑦
𝑑
Introduce the symbols 𝑦 ′ , 𝑓 ′ (𝑥), 𝑑𝑥 , ∆𝑥 , 𝑑𝑥 ( )
 9A 3, 5, 6, 7( a
couple) first
principals
The words differentiate, find the gradient(function), Find the gradient
of the tangent, Find the derivative, Find the rate of change of the
function, instantaneous rate of change ALL mean the same thing








Derivatives of xn functions using basic formula
1
Converting expressions such as √𝑥, 𝑥 into 𝑥 𝑛 form, differentiating
and then converting back
Taking constants out the front before diff.and then popping them
back in
Doing stuff separated by + and – separately and then recombining
Find the gradient at a point by substituting a known x value into the
𝑑𝑦
formula
𝑑𝑥
 Find basic
from
Heinemann
 Ex5.2 Q3, 4,
11,
Know what a positive, negative and 0 gradient look like
Find the value of the value of x for a given value of the gradient

Know that the gradient of any straight line = tan 𝛼 𝑤ℎ𝑒𝑟𝑒 𝛼 is the
angle above the direction of the positive x axis eg for 𝑦 = 𝑥, 𝛼 =
45°, 𝑓𝑜𝑟 𝑦 = − √3𝑥, 𝛼 = −60°

Strictly increasing. Part of a graph is strictly increasing, if when the x 
gets a tiny bit bigger, the y gets bigger. Eg a cos graph is strictly
increasing [180, 360) Even though the gradient is 0 at 180, because
the graph to the right of 180 is slightly higher (cos 180.001>
cos180), it is called strictly increasing. 360 is not included, because
the graph to the right of 360 is slightly lower (cos 360.001 < cos
360)

Simply decreasing is the opposite

Simply increasing means positive gradient, + stationary points
where it’s about to go up.

Simply decreasing means negative gradient + stationary points
where it’s about to go down
This stuff, in my opinion is a bit trivial, but it turns up on exams
 Ex9B 1 – 6, 12,
13 (expand
first) 14, 15,
 Ex 9C 4,
5(divide first), 6
 Ex 9F 2, 3
 9B Q7, 17

 9B Q 8 - 10

 9B 17 - 21




𝑑
TECHNOLOGY:- use the 𝑑∎ ( ) function on CAS
𝑛
Use of the chain rule with basic 𝑥 formula, must specify and write
𝑑𝑢
down both u and 𝑑𝑥 in margin.
If asked to “Use calculus to solve..” you MUST show these steps if
relevant (also vital when we get to product & quotient and
antidifferentiation in Spec.)


 6.2 Q1, 7a,
7c) 7e, 7h, 8b,
8d (explain
4/3)
Eg Chain is where you have composite functions, functions of
functions eg
𝑦 = 𝑠𝑖𝑛2 (𝑥) → 𝑦 = (sin(𝑥))2
𝑑𝑢
→ 𝑦 = (𝑢)2 𝑤ℎ𝑒𝑟𝑒 𝑢 = sin(𝑥) 𝑎𝑛𝑑
= cos 𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑦
𝑑𝑢
→ 𝑑𝑥 = 𝑑𝑢 × 𝑑𝑥
= 2𝑢 × cos(𝑥)
= 2sin(𝑥) × cos(𝑥)
 9E 1-5, *6
 9F 4 -7
𝑦 = (𝑥 3 + cos(3𝑥))5 →
→ 𝑦 = (𝑢)5
𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑥 3 + cos(3𝑥)
𝑑𝑢
𝑎𝑛𝑑 𝑑𝑥 = 3𝑥 2 − 3 sin(3𝑥)
𝑑𝑦
𝑑𝑦
𝑑𝑢
→ 𝑑𝑥 = 𝑑𝑢 × 𝑑𝑥
= 5(𝑢)4 × (3𝑥 2 − 3 sin(3𝑥))
= 5(𝑥 3 + cos(3𝑥))4 × (3𝑥 2 − 3 sin(3𝑥))

o
o
o
o
Brainstorm what is differentiation for
Finding the gradient at a point anywhere on a function
Using the above to find equations of tangent & normal
(perpendicular) lines
Using the above to find turning / stationary points (grad = 0)
Finding maximum / minimum values of functions (grad = 0)
 5.2 Q5, 6
 5.2 9, 15, 16,
18a, 21
Tangent
 5.2 17, 18b
normal

Continuity and limits (maths for lawyers)


 9L Q1, 2, 3
o















When a function is continuous
Estimating gradient from the graph
Sketching the gradient graph from the original (thus reinforcing that
gradient actually means something) **MUST BE ABLE TO DO
THESE**
When is a function differentiable?
1. function must exist – if the function does not exist, either because
of a given or implied domain, it has no derivative function there
2. It must be continuous at the point – no break
3. Not an endpoint
4. Not a sharp point (gradient different either side of point)
 5.2 Q7,8, 12
find more
 9D Q1, 2, 3



derivativesketchpra
phs1-2.docx
sketching of
derivative graphs-answers.pdf
 9M 1
 9M2,3,4
 *9M 5


Revise exponential rules
 5.3 Q1, 2, 3,
𝑥
𝑙𝑜𝑔𝑒 𝑎 ×𝑥
𝑥
4, 5, 6, 7 – 12
Including change of base 𝑦 = 𝑎 → 𝑦 = 𝑒
eg 𝑦 = 2 → 𝑦 =
a selection,
𝑒 𝑙𝑜𝑔𝑒2 ×𝑥
 9G 1, 3
*13, *14
Derivative of the exponential function using basic formula
 5.4 Q1,2,3,4,
6.2 2b,
Derivative of the exponential function using basic formula, combined 
 9G 2, 4, 5, 6
with chain
 Heinemann
Revise log rules
5.6 Q1-7
Derivative of the natural log function using basic formula
 Ex9.2 1a, 2a,
Derivative of the natural log function using basic formula combined
2c(separate
with chain
 9H 1a) 4a)
first), 3, 4, 5,
1/2 clever stuff using log combined with exponential rules
(without chain)
7a, 10a,
𝑙𝑜𝑔𝑏 (𝑥)
 9H 1- 7, *8
𝑙𝑜𝑔𝑎 (𝑥 ) =
10b,12,13,
𝑙𝑜𝑔𝑏 (𝑎)

𝑙𝑜𝑔𝑒 (𝑥)
1
(
)
𝑒𝑔 𝑙𝑜𝑔𝑎 (𝑥 ) =
=
×
𝑙𝑜𝑔
𝑥
𝑒
𝑙𝑜𝑔𝑒 (𝑎)
𝑙𝑜𝑔𝑒 (𝑎)
𝑙𝑜𝑔𝑒 (𝑥)
𝑙𝑜𝑔𝑒 (2)

𝑒𝑔 𝑦 = 𝑙𝑜𝑔2 (𝑥 ) → 𝑦 =



Derivative of sin, cos, tan by sketching
1
Know that secθ = 𝑐𝑜𝑠𝜃 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑖𝑡 𝑡𝑢𝑟𝑛𝑠 𝑢𝑝

=
1
𝑙𝑜𝑔𝑒 (2)
× 𝑙𝑜𝑔𝑒 (𝑥 )

Use of basic trig differentiation formulae


Use of basic trig diff. formulae combined with chain rule
𝑠𝑖𝑛2 𝜃 = (𝑠𝑖𝑛𝜃)2 + 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑓𝑜𝑟 𝑐𝑜𝑠, 𝑡𝑎𝑛


Product rule – what it is and how to use it
𝑑𝑢
𝑑𝑣
Specify both u and 𝑑𝑥 as well as v and 𝑑𝑥in margin – this is vital
𝑑𝑢
if one or both of 𝑑𝑥 and / or



𝑑
𝑑𝑣
Product rule 𝑑𝑥 (𝑢𝑣) = 𝑢 𝑑𝑥 + 𝑣
need calculating using chain
𝑑𝑢
𝑑𝑥

Quotient rule – the fact that it’s a pain, what it is and how to use it
𝑑𝑢
𝑑𝑣
Specify both u and 𝑑𝑥 as well as v, v2 and 𝑑𝑥in margin – this is vital
𝑑𝑢
if one or both of 𝑑𝑥 and / or
 Quotient rule

𝑑𝑣
𝑑𝑥
𝑑
𝑢
( )=
𝑑𝑥 𝑣
𝑑𝑣
𝑑𝑥
𝑣
need calculating using chain
𝑑𝑢
𝑑𝑣
−𝑢
𝑑𝑥
𝑑𝑥
𝑣2
 5.5 Q1a, c,e,
2, 5a,d,h, 6, 7,
8, 10, *11,

*12, *14
15CAS,
16CAS

 9I 1a, b, c, 2a,
b, c, f 3a, b, c
 5.5 1b, d, 5g
 6.2 2a, 3, 8a,  9I 1d- j, 2d, e,
3d, 5, 6 NOT4
8e, 10, *12a,
13b*,
 6.3 Q1, 2, 3,
 9J 1- 4, 6, 7.
4, 5, *6, 7
*5a, *8
selection,
8selection, 9,
11, 12, *16
 6.4 Q1, 2, 3,
 9K Q1 2d,f, 3c
4, 5(some),
 9K4 - 7
6a, **6b, 7, 8,
9, 10, *11, 12,
13, 14,

𝑑𝑢
Product AND chain eg set up your problem specifying u and 𝑑𝑥 as 
well as v and
𝑑𝑢

𝑑𝑥
𝑎𝑛𝑑 / 𝑜𝑟
𝑑𝑣
𝑑𝑥
𝑑𝑣
 9K 2a, b, c 3a,
b
, BUT you might need to use the chain rule to find
𝑑𝑥
Make sure that you can deal with Quotient AND chain as well
Applications of Basic Differential Calculus
Having found it, what can you do with it?








 6.7 Q5, 6, 7, 8
15CAS,
16Cas.
 6.7
disp/speed/
 10B 11 - 14
vel Q3, 11, 12
Applications to motion / kinematics
A bit specialist(ish) but OK questions
𝑥 𝑚𝑒𝑎𝑛𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛, 𝑣 𝑚𝑒𝑎𝑛𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑎 𝑚𝑒𝑎𝑛𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
Stationary means 𝑣 = 0
𝑑𝑥
o 𝑣 = 𝑑𝑡
𝑑𝑣
o 𝑎 = 𝑑𝑡
Eg 𝑥 = 5𝑡 3 − 12𝑡 2 − 3𝑡 + 4 → 𝑣 = 15𝑡 2 − 24𝑡 − 3 → 𝑎 = 30𝑡 − 24

Applications of finding derivatives,



Finding stationary points on a graph / function
Cover three types of stationary points – (local)Max, (local) Min and
SPI
Turning points, maximum and minimum
Stationary point of inflection – increasing and decreasing (Point
of inflection is NOT the correct term)
𝑑𝑦
They all occur when = 0
 6.5 Q1, 2, 3,
4, 7, 8, 10, 11,
12, 17, 18
 6.5 Q5, 6 , 14,
15 (max/ min)  10C 1 – 10 (A
large selection)







𝑑𝑥
 10B 1- 10, 15 –
17 (Do a
selection)
𝑑
TECHNOLOGY:- use the combination Solve (𝑑∎ ( 𝑓(∎) ) = 0, ∎)
functions on CAS
Types of stationary points, see notes above
Do sign diagram (using known points either side of your known
stationary points) to determine nature of points
+, 0 , - = maximum

 10 D 1, 2,3, 5,
6
 10D 8, 9 CAS
 10D *13


-, 0, + = minimum
-, 0, - Or +, 0, + = SPI
 10D 14, 15
 Lots of others
are a bit tricky /
time consuming

𝑑𝑦
Applications of 𝑑𝑥 = 0, in finding (local) maximum & (local) minimum  6.6 2, 3, 4, 6,
8, 11
values of functions
 6.6 Q7, 16

Local max , min (calculus) AND
 Absolute max and min – will be domain and range dependant and
these MIGHT be different to the local max and min points
 10E1 – 4, 6, 7
 10F 2, 3, 5
 10E *9 – *16
are more of the
same

𝑑𝑦
Eg1 𝑦 = 𝑥 2 − 4𝑥 + 7 𝑑𝑥 = 0 → 2𝑥 − 4 = 0 →
→ 𝑥 = 2 → 𝑙𝑜𝑐𝑎𝑙 min (2,3),
IF 0 ≤ 𝑥 ≤ 6, 𝑡𝑒𝑠𝑡 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 6, → (0, 7)𝑎𝑛𝑑 (6, 19)
→ 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 and local min 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑎𝑡 (2, 3)𝑎𝑛𝑑 absolute max 𝑎𝑡 (6, 19)
𝑑𝑦
Eg2 𝑦 = 𝑥 3 − 9𝑥 2 + 24𝑥 𝑑𝑥 = 0 → 3𝑥 2 − 18𝑥 + 24 = 0 →
𝑥 2 − 6𝑥 + 8 = 0 → 𝑥 = 2 𝑜𝑟 𝑥 = 4
→ 𝑙𝑜𝑐𝑎𝑙 max(2, 20), 𝑙𝑜𝑐𝑎𝑙 min(4, 16)
IF 0 ≤ 𝑥 ≤ 6, 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 min 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑎𝑡 (0, 0)𝑎𝑛𝑑 max 𝑎𝑡 (6, 36)

Applications where you draw a diagram first
 A cuboid = a rectangular based prism i.e. a box

Applications involving some written word interpretation



Questions where you need to know about cylinders
Cylinder TSA = 2 × 𝜋 𝑟 2 + 2𝜋𝑟ℎ (2 ends + curved surface)
Cylinder volume = 𝐵𝑎𝑠𝑒 𝐴𝑟𝑒𝑎 × ℎ𝑒𝑖𝑔ℎ𝑡 (𝑠𝑎𝑚𝑒 𝑎𝑠 𝑎𝑛𝑦 𝑝𝑟𝑖𝑠𝑚) =
𝜋𝑟 2 × ℎ
Distance questions using pythagoras


 10E 5, 8
 10F 1, 4, 6

 10F 5, 7, 8, 9

 10F 11
 Find more

Distance from a fixed point (𝑎, 𝑏) to a point (𝑥, 𝑦) on a function 𝑦 =
𝑓(𝑥) 𝑑 = √(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 or
𝑑 = √(𝑥 − 𝑎)2 + (𝑓(𝑥) − 𝑏)2 (useful if you want to find derivative,
set it to 0 to find x value that gives the minimum distance)
𝑒𝑔 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 (1, 3)𝑡𝑜 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒
𝑦 =𝑥+8
𝑑 = √(𝑥 − 1)2 + (𝑓(𝑥) − 3)2
→ 𝑑 = √(𝑥 − 1)2 + (𝑥 + 8 − 3)2
→ 𝑑 = √(𝑥 − 1)2 + (𝑥 + 5)2 = √𝑥 2 − 2𝑥 + 1 + 𝑥 2 + 10𝑥 + 25 =
√2𝑥 2 + 8𝑥 + 26
𝑇ℎ𝑖𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎𝑡 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚, 𝑤ℎ𝑒𝑛 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑢𝑛𝑑𝑒𝑟
𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑖𝑠 𝑎𝑡 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑑
(2𝑥 2 + 8𝑥 + 26) = 4𝑥 + 8
𝑑𝑥

𝑠𝑒𝑡(4𝑥 + 8 = 0) → 𝑥 = −2 → 𝑦 = 6 → 𝑑 = √18
These are common on exams, both with straight lines and
parabolas


Distance questions using pythag and trig

Using differentiation to find equations of tangents and normal at a
point on a curve
1. Find derivative of function (= gradient of tangent)
−1
1a gradient of normal = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡
2.
3.


Have / find out x and y coordinates of point
Sub answer to 1 and 2 into y = mx + c formula to find c, and hence
write down formula!
 10F *10, 12
 10F Q16, 17 +
more of these
 10A 1 – 6, 8 –
10 (all the
same – pick a
selection)
 10 A *11 – 17
 10A 7 – Use
CAS
tangentnormalprac
tice.pdf

Extension questions


 **10G**


Antidifferentiation aka Integration and some of its
applications



Brainstorm what is anti differentiation and what is it for

Knowing the gradient and working back to find the function
Knowing the gradient graph and working back to find the function
graph
o Because the derivative of f(x) + c = f’(x) {the c disappears}, hence
the antiderivative of f’(x) = f(x) + c
o From above point, general solution will have a +c in the answer, with
a specific solution you will be given an x and y value and / or point
on the function to help find c
o Antiderivative of f’(x) = f(x) + c; or Antiderivative of f(x) = F(x) + c, or
Antiderivative of f(x) with respect to x = ∫ 𝑓(𝑥) dx
o General antiderivative ∫ 𝑓(𝑥) = 𝐹(𝑥) + 𝑐
o An antiderivative ∫ 𝑓(𝑥) does not have a +c in it

Later on, anti differentiation can also be used to find exact areas
under graphs and between graphs

In methods, anti differentiation will be done using

1. 6 Basic formulae from the sheet (possibly after some reorganisation
into standard form using index and log laws)
1a) There will always be a +c
1b) ∫ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑥
2. Taking constants out the front before anti diff and then popping
them back in
3. Doing stuff separated by + and – separately and then recombining
4. Just like you use the change in base index rules to do
differentiation, you may need them to do anti differentiation.
𝑦 = 𝑎 𝑥 → 𝑦 = 𝑒 𝑙𝑜𝑔𝑒𝑎 ×𝑥
5. Using the (ax + b) techniques as shown below – In Specialist a
technique called anti differentiation by change of variable
(Reverse Chain rule) will be used




o
o


6.
7.

Anti differentiation by recognition
In specialist there will be more, in methods you can stop now
1
∫ 𝑥 𝑛 𝑑𝑥 = 𝑛+1 𝑥 𝑛+1 + 𝑐 𝑛 ≠ 1 i.e. opposite of differentiation, i.e. put
up power by one, and divide by new power

∫(𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 =

𝑛 ≠ 1 - show how it is derived using chain rule to find the derivative
of the right hand side
1
∫ 𝑒 𝑎𝑥 𝑑𝑥 = 𝑎 𝑒 𝑎𝑥 + 𝑐 i.e. opposite of differentiation using chain


1
𝑎(𝑛+1)
(𝑎𝑥 + 𝑏)𝑛+1 + 𝑐
1

Extra formula
∫ 𝑒 𝑎𝑥+𝑏 𝑑𝑥 = 𝑎 𝑒 𝑎𝑥+𝑏 + 𝑐
AND
1
Extra f.
∫ 𝑒 𝑘(𝑎𝑥+𝑏) 𝑑𝑥 = 𝑘𝑎 𝑒 𝑘(𝑎𝑥+𝑏) + 𝑐

∫ sin(𝑎𝑥) 𝑑𝑥 = − 𝑎 cos(𝑎𝑥) + 𝑐

∫ cos(𝑎𝑥) 𝑑𝑥 = 𝑎 sin(𝑎𝑥) + 𝑐
The formulae below may NOT be used, but they can be derived
using chain rule and show that the derivative of the right hand side
is the same as the left hand side
1
Extra f
∫ sin(𝑎𝑥 + 𝑏) 𝑑𝑥 = − 𝑎 cos( 𝑎𝑥 + 𝑏) + 𝑐



1
1
1

Extra f
∫ cos(𝑎𝑥 + 𝑏) 𝑑𝑥 = 𝑎 sin(𝑎𝑥 + 𝑏) + 𝑐
AND
1
Extra f. ∫ sin 𝑘(𝑎𝑥 + 𝑏) 𝑑𝑥 = − 𝑘𝑎 cos 𝑘(𝑎𝑥 + 𝑏) + 𝑐

Extra f ∫ cos 𝑘(𝑎𝑥 + 𝑏) 𝑑𝑥 =
1
𝑘𝑎
sin 𝑘(𝑎𝑥 + 𝑏) + 𝑐
 7.2Q1a,b,
2a,3, 4, 9, 10,
 11B 1 – 4, 5
11a,b,c,f, 12c,
(expand or
12e, 12f, 13,
divide first), 6,7
15
*8
 Heinemann
7.1

 11C Q1, 4c, f
 Heinemann
7.2
 7.3 Q 3, 5, 6,
10,
 11D 1 -4, *5, *6
 Heinemann
7.3
 7.3 Q 1,2, 13a
 11G 1a – h, j
2a –f, h, j
 k(ax + b) NB
as mentioned,
not on the
course 1i, 2g, i

mixed exponent And trig

∫ 𝑥 𝑑𝑥 = 𝑙𝑜𝑔𝑒 (𝑥) + 𝑐 , 𝑥 > 0

Extra f.
1
1
1
𝑎
1
∫ 𝑎𝑥 𝑑𝑥 =
1


𝑙𝑜𝑔𝑒 (𝑥) + 𝑐1 𝑂𝑅
𝑎
𝑙𝑜𝑔𝑒 (𝑎 × 𝑥) + 𝑐2 = 𝑎 𝑙𝑜𝑔𝑒 (𝑎) +
= 𝑎 𝑙𝑜𝑔𝑒 (𝑥) +

1
1
1
1
𝑎
1
𝑎
𝑙𝑜𝑔𝑒 (𝑎𝑥) + 𝑐2 =
𝑙𝑜𝑔𝑒 (𝑥) + 𝑐2 =
 7.3 4, 9, 11,
12, *18,
 Heinemann
7.4
 9.3 q1, 2, 9,
10

 Ex 11C 2, 3, 5,
6, 8 -10
𝑙𝑜𝑔𝑒 (𝑎) + 𝑐2 (same thing)
𝑎
1
1
∫ 𝑎𝑥+𝑏 𝑑𝑥 = 𝑎 𝑙𝑜𝑔𝑒 (𝑥) + 𝑐1 𝑂𝑅 𝑎 𝑙𝑜𝑔𝑒 (𝑎𝑥 + 𝑏) + 𝑐2
Both of the above can be shown to be true using the chain rule
𝑇ℎ𝑒𝑟𝑒 are other formulas in the book etc, but these are NOT on the
formula sheet – you are expected to use chain rule to do them





1
More ∫ 𝑥 𝑑𝑥 = 𝑙𝑜𝑔𝑒 (𝑥) + 𝑐 using chain mostly, but divide first
𝑑
Anti differentiation by recognition (Hence) i.e. if 𝑑𝑥
𝑓(𝑥) = 𝑔(𝑥) ↔
∫ 𝑔(𝑥)𝑑𝑥 = 𝑓(𝑥) + 𝑐
𝑑
Part a find
𝑓(𝑥) = 𝑓 ′ (𝑥)
𝑑𝑥

Part b Hence ∫ 𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓(𝑥) + 𝑐
𝑑
Eg a) Find 𝑑𝑥 𝑥 sin(𝑥) = 𝑥 cos(𝑥) + sin 𝑥 (using product rule)
b) Hence find ∫ 𝑥 cos(𝑥) 𝑑𝑥 =
∫ 𝑥 cos(𝑥) + 𝑠𝑖𝑛𝑥 − sin 𝑥 𝑑𝑥 (so that part of the question looks
EXACTLY like the previous answer)
∫ 𝑥 cos(𝑥) + 𝑠𝑖𝑛𝑥 𝑑𝑥 − ∫ sin 𝑥 𝑑𝑥=
𝑥 sin(𝑥) − ∫ sin 𝑥 𝑑𝑥 = 𝑥 sin(𝑥 ) + cos(𝑥) + 𝑐

 Anti diff by CAS

Antiderivative of f’(x) = f(x) + c; or Antiderivative of f(x) = F(x) +
c, or Find the antiderivative of f(x) with respect to x or ∫ 𝑓(𝑥)𝑑𝑥

 7.2 7,8, 16,
17, 18,
 Heinemann
7.5
 7.3 Q7, 8, 14,
15, 16
 9.3 3, 4, *12,
13, 14, 15
 7.2 19, 20
 7.3 Q19, 20
 7.4 24

 11C 4a, b,
d(expand as
well), f, 7
 11H 3 – 6, 10,
13


or integrate f(x) with respect to x ALL mean the same thing and
will have a +c in the answer
1
eg ∫ 2𝑥 − sin(3𝑥) + 4 𝑑𝑥 = 𝑥 2 + 3 cos(3𝑥) + 4𝑥 + 𝑐
𝒅𝒚
𝒅𝒙
= {𝑤ℎ𝑎𝑡𝑒𝑣𝑒𝑟} → 𝐹𝑖𝑛𝑑 𝒚 will have a +c
𝑑𝑦
1
eg 𝑑𝑥 = 2𝑥 − sin(3𝑥) + 4 → 𝑦 = 𝑥 2 + 3 cos(3𝑥) + 4𝑥 + 𝑐
General antiderivative ∫ 𝑓(𝑥) = 𝐹(𝑥) + 𝑐
 An antiderivative ∫ 𝑓(𝑥) does not have a +c in it, the “c” has a
fixed value, usually 0
eg An antiderivative of 2𝑥 − sin(3𝑥) + 4 with respect to 𝑥 =
1
𝑥 2 + 3 cos(3𝑥) + 4𝑥

The antiderivative ∫ 𝑓(𝑥) does have a +c in it, BUT there will be
clues to find what c is
𝑑𝑦
Eg The antiderivative of 𝑑𝑥 = 2𝑥 − sin(3𝑥) + 4 , given that y(0) = 2,
is 𝑥 2 +
→𝑐=

5
3
1
3
1
cos(3𝑥) + 4𝑥 + 𝑐 → 2 = 0 + 3 + 0 + 𝑐
→ 𝑦 = 𝑥2 +
1
Later on, anti differentiation can also be used to find definite
integrals which can be used to find exact areas under graphs
and between graphs, this will not have a +c either (reasons
given later)
4
Eg ∫1 2𝑥 − sin(3𝑥) + 4 𝑑𝑥 = [𝑥 2 +


5
cos(3𝑥) + 4𝑥 + 3
3
1
cos(3𝑥) + 4𝑥]
3
Getting from gradient graph to the function graph
4
1




Ex7.4 7, 8, 17
Extras
Ex 6.2 Q10
Ex 7.10 All

good

 Other spec
stuff


sketching
antiderivative graphs.pdf
Finding Areas under graphs




Using left and right rectangles as an approximation
The narrower the rectangles, the more accurate
If you get the rectangles close to zero, the approximation gets better
– very much like antidifferentiation by first principles
very carefully can be called upper & lower rectangles IF you know
which is the underestimate and which is the over estimate as an
approximation
 8.2 Q1, 2, 7,
8, 9,
Finding areas using rectangle or triangle or trapezium rules
𝑏
Fundamental theorem of calculus, ie demonstrate that ∫𝑎 𝑦 𝑑𝑥 =






 11A 1, 2, 3a),
4, 6
 11A Q8
𝑏


∫𝑎 𝑓(𝑥) 𝑑𝑥 means the sum of all the areas of the rectangles between

x = a and x = b of height y (aka f(x)) and width dx
𝑏
𝑥=𝑏
For Spec purposes∫𝑎 𝑓(𝑥) 𝑑𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎𝑠 ∫𝑥= 𝑎 𝑓(𝑥) 𝑑𝑥
Definite integrals, how to use them, the basic rules, the fact that DIs  8.2, 3, 4, 5, 6,
do not give you an area straight away, but intelligently used they
10, 11, 12, 13,
can help you find it
14
Eg area bounded between the lines x=2, x = 5, the y axis and the
5

line 𝑓(𝑥) = 𝑥 2 + 3 = ∫2 𝑥 2 + 3 𝑑𝑥 =
1
5
125
2
3
[3 𝑥 3 + 3𝑥] = [




1.
2.
8
+ 15] − [3 + 6] = 48
DIs do not have a +c, more to the point they do but it cancels out
More definite integral practice (NOT areas)
Application of definite integral using CAS
Using DI s (Definite integrals) to find areas between function and x
axis
Dividing up areas into above and below x axis (b/c above areas will
be + and below will be -) – will need to find x intercepts first
Any negative DIs to be converted to + (because areas are +) before
adding them up because areas are positive

 8.2 18

 11E

 Ex 11F 1- 8
 11G 4, 6
 11F *9 -*12

**Assorted further **


1.
2.


Using DIs to find areas between function f(x) and function g(x)
𝑏
Show where Area = ∫𝑎 𝑓(𝑥) − 𝑔(𝑥) 𝑑𝑥 formula comes from
Dividing up areas into bits where f(x) on top and g(x) at the bottom,
and vice versa (b/c above areas will be + and below will be -) – will
need to find intersection points first so can do anti diff. separately
Any negative DIs to be converted to + (because areas are +) before
adding them up because areas are positive
Area between two curves, find the intersection points first
One piece areas – do with one anti derivative and you are told the x
limits
Areas where you need to calculate limits or intersection points first

Using CAS

Average values – Integrate between a and b to get the area, and
divide by x difference (=b-a). In effect, turning the integrated shape
into a rectangle, with the same area, the same base (b –a) and
height = y average
Sometimes written as “If the same area was in the form of a
rectangle of width “a” what would be the height?
These have been turning up in exams most years
𝑑𝑥
𝑑𝑣
Kinematics but use 𝑣 =
,𝑎 =
and




3.
4.







𝑑𝑡
𝑑𝑡
 11I 1 -6
 8.3 Q1, 2, 5,
6, 7, 13
 Heinemann
7.8 6, 7, *13
8.3 Q16
8.4 21
9.3 19, 20
8.4 Q5, 6, 7
 11I Q 7 –
NEED MORE

 11J 1,2, 3 4, 5,
6 (choose
some)*11 - *19
 8.4 9, 10, 17
hence 𝑥 = ∫ 𝑣 𝑑𝑡 𝑎𝑛𝑑 𝑣 = ∫ 𝑎 𝑑𝑡

Chapter 11 revision
 *11H 1, 2, 11,
12, 15, 18, 21a,
b, e, f
 11J 7 - 10

 Tech free
1, 5, 6, 9, 10,
12, 14, 15, 16,
17, 18, 19.

SAC 2 Modelling or Problem Solving Task on calculus and
integration

SAC 3 Modelling or Problem Solving Task on probability and
statistics

Probability
Probability is a measure of how likely an event / combination of events
is to occur.
Lesson
Enduring Understandings / Key Questions










*2, 4, 7, 8, 11,
13
 Multi choice
1 – 8, 10
 Extended
answer
1, 2, 7
*5, *6, *8

/8
/10
/7
 See p 79
course advice

/7
/10
/8
 See p 79
course advice


Extra resources Exercises
Questions from
Jacaranda 12
Cambridge
and extras

What is probability? A measure of the likelihood of an event or
combination of events occurring. Between 0 (impossible) and 1
(certain). Hence usually written as a fraction or decimal or
percentage
 Ways of calculating probability are either:1. By measurement (collecting data)
2. Predicting mathematically



What are examples of each type of probability calculation? 1.
Probability that the next car on a road is yellow, Probability that a
student will be > 180cm tall., probability that a voter will vote
Labor or Liberal or Green. 2. Probability that 2 dice have a total
>8, chance of two kings being taken from a pack
 There are two types of probability distributions
1. DISCRETE
2. CONTINUOUS
 What are examples of the two types of distribution?
Discrete – anything with fixed discrete values eg no of children in a
family, cost of petrol at a petrol station Continuous – heights,
weights, volume of petrol at a petrol station
 Things common to all probability distributions
1. They are a measure of likelihood, not a guarantee
2. With numeric probabilities, an Expected Value can be
calculated (aka Average score) and a variance / SD (measure
of expected spread)
3. Probabilities distributions add up to 1 – This may involve the
total probabilities adding up to one in a discrete distribution
eg a binomial distribution OR the total area under the
probability distribution graph totalling to one in a continuous
distribution graph eg a normal distribution




There are two types of DISCRETE distribution studied in
Math Methods
1. Made up / defined according to a given rule
Can you graph, and then calculate max, min, modal value,
mean and median of a made up distribution?





Ex 10.1

probsheet1ANSWE
RS.docx
probwheel.docx
 prob wheel
exercise

Can you use a tree diagram to calculate probabilities of events
and combinations of events?



What is a probability distribution?
Can you construct / interpret a probability table?



Introduction to the binomial (probability) distribution
The first two sheets on this show a binomial distribution, the
third sheet shows a hypergeometric distribution.
10.2 Q1
– 10
covers basic
probability,
prob
distribution
tables, sum of
p = 1, as well
as 𝑬(𝒙) and
𝑬(𝒇(𝒙)) (its
really good)
 13A 1, 2
 13B 1, 5, 6, 8,
11, *13, 14,
*17, 19
 13C 1,2
(discrete vs
continuous)
 13C 4, 5,8, 10
,11, 13, 18,
*19
 6,7,9, 16 are
dice questions,
and are a bit
long


binomial1ANSWERS
.docx
Binomial intro.docx
 Can you calculate Exp(x)? = expected value = average score
̅ = 𝑬(𝒙) = ∑ 𝒙 × 𝒑(𝒙)
𝑴𝒆𝒂𝒏 𝝁 = 𝒙


Can you calculate Exp(f(x))?
𝑬(𝒇(𝒙)) = ∑ 𝒇(𝒙) × 𝒑(𝒙)

And, for Exp (f(x)) 𝑬(𝒂 𝑿 + 𝒃) = 𝒂 𝑬(𝑿) + 𝒃

Var(x) and SD(x) – standard deviation of x – are measures of
how spread out you would expect your possible range of values
to be.
Can you calculate Var(x), SD(x) using exp(x2) etc
𝟐
𝑽𝒂𝒓(𝑿) = 𝑬(𝑿𝟐 ) − (𝑬(𝑿)) 𝒐𝒓𝑬(𝑿𝟐 ) − 𝝁𝟐 𝒐𝒓


Binomial intro.docx


𝑬((𝑿 − 𝝁)𝟐 ) = ∑(𝒙 − 𝝁)𝟐 × 𝒑(𝒙)

𝑽𝒂𝒓(𝑿) = 𝝈 , 𝑺𝑫(𝒙) = √𝑽𝒂𝒓(𝑿) = 𝝈


Can you solve problems using a Venn Diagram?
What is the difference between independent and mutually
exclusive events?
What formulae can you use to help if you know that a scenario
is independent or mutually exclusive?
Can you use the various formulae to find probabilities,
especially for discrete distributions?
Can you use a Venn diagram to show an explanation for the
formula
Pr(𝐴′ ) = 1 − Pr(𝐴) 𝑓𝑜𝑟 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑒𝑣𝑒𝑛𝑡𝑠
Can you use a Venn diagram to show an explanation for the
formula
Pr(𝐴 ∪ 𝐵) = Pr(𝐴) + Pr(𝐵) − Pr(𝐴 ∩ 𝐵)




Ex 10.3
Q1 -16,
20


 13D, 1, 2, 3, 5,
7a,b, ,d , *8,
13, *14, *15,
*17, *18,





binomial1ANSWERS
.docx
𝟐

 13A 5 – 9, 15
– 17
 13B 2, 3, 4, 10

VENN12methods.d
ocx










Can you use a Venn diagram to show an explanation for the
formula
Pr(𝐴 ∩ 𝐵)
Pr(𝐴|𝐵) =
Pr(𝐵)
Can you solve worded problems involving the above formulae?
2. Binomial Distribution
What is a Bernoulli trial?
What is a binomial distribution?
What is an example of a binomial distribution?
A distribution involving a number of identical independent
trials each with only two outcomes. (The probability does
not change between trials)
Can you use tree diagrams to solve simple binomial distribution
problems?
Do you know the difference b/w binomial (independent and / or
with replacement) and hypergeometric (w/o replacement) ?
Can you solve a simple conditional probability question using a
tree diagram? Eg sampling without replacement.

binomialintropart2.
docx
binomialintropart2
ANSWERS.docx
binomialintropart3.
docx

binomialintropart3
answers.docx
binomialintropart4.
docx
binomial4ANSWERS
.docx

 Binomial intro
sheets which
go from basic
through to use
of CAS to find
pdf and cdf






Some phrasing in binomial questions
eg if possible results are 0, 1, 2, 3, 4, 5, 6
At least two = two or more = no less than two = greater than or
equal to two
𝑥 ≥ 2, 𝑥 = 2, 3, 4, 5, 6
At most two = two or less = no more than two = less than or
equal to two
𝑥 ≤ 2, 𝑥 = 0, 1, 2
Less than four = three or less 𝑥 < 4, 𝑥 = 0, 1, 2, 3
More than four = five or more 𝑥 > 4, 𝑥 = 5, 6
*This going up or down to the next value for > or < works for
discrete distributions eg binomial, NOT continuous distributions.
 If the value is between two values, the question needs to be
clear whether the values are included or not
Eg greater than two, but less than 5, 2 < 𝑥 < 5, 𝑥 = 3, 4
Eg greater than two, but no more than than 5, 2 < 𝑥 ≤ 5, 𝑥 =
3, 4, 5
 If you know that a distribution is binomial, what formulae
shortcuts can be used to calculate probabilities instead of
having to draw up tree diagrams?
 Can you define p, q, n and x in a binomial problem?
 p is the probability of a success in a single trial, q is the
probability of a failure in a single trial (q = 1 – p), n is the total
number of trials and x is the number of successes that you are
interested in
 Define p, q, n and x and find probabilities and cumulative
probabilities (probs b/w given values) both manually and using
the CAS (using binomPDf and binomCDf functions)
 CAS – can access through main screen Interactive –
Distribution / Inv.Dist - Discrete
 Eg to find 𝑝𝑟 𝑥 = 2 𝑤𝑖𝑡ℎ 6 𝑡𝑟𝑖𝑎𝑙𝑠 𝑎𝑛𝑑 𝑝 = 0.2
(Casio) CAS binomialPDf(2,6,0.2)




11.3
11.4
 14A 2,3,4, 6 –
11, *12, 14 –
18, *19 - *22

Eg to find 𝑝𝑟 2 ≤ 𝑥 ≤ 5, 𝑤𝑖𝑡ℎ 6 𝑡𝑟𝑖𝑎𝑙𝑠 𝑎𝑛𝑑 𝑝 = 0.2
(Casio) CAS binomialCDf(2,5,6,0.2)
*If you use this to solve a problem on an exam, you need to
specify Binomial, your n value, your p value and your x
value or x limits

A handy shortcut - to list all 𝒑𝑟
𝑓𝑟𝑜𝑚 𝑥 = 0 𝑡𝑜 𝑥 = 6 𝑤𝑖𝑡ℎ 6 𝑡𝑟𝑖𝑎𝑙𝑠 𝑎𝑛𝑑 𝑝 = 0.2
(Casio) CAS binomialPDf(6,0.2)

Finding probabilities (binomial pdf) using the
𝒏!
𝒏
𝒓 𝒏−𝒓
function where 𝒏𝒓𝑪 = (𝒏−𝒓)!𝒓!
𝒓𝑪 𝒑 𝒒

𝒏
𝒓𝑪






feature is in your CAS menu (see alphabetic list) or shift ÷
on your fx 82 will do it as well eg 5 shift ÷ 2 will display as 5 C 2
= 10
Sketching binomial graphs, and noticing that they look a bit like
normal graphs (hold that thought)
Use of binomial shortcuts to find Exp(x) = np ( 𝝁 ) and
Var(x) = npq ( 𝝈𝟐 ) and SD(x) = √𝒗𝒂𝒓(𝒙) = ( 𝝈 ) for binomial
distributions, these are used mainly for large values of n,
where the binomial distribution gets close to the normal
distribution (see below)
Advanced binomial, including Pr(𝑥 𝑖𝑠 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒) =
Pr(𝑥 ≥ 1) = 1 − Pr(𝑥 = 0)
Link between confidence limits and 1, 1.96, 3 SD (as per normal
distribution)
1SD = 68.3 % confidence interval
1.96 SD = 95.0 % confidence interval – You must know this,
a different value to further, and the pre 2016 methods
course design, and some text questions
3SD = 99.7 % confidence interval

 14B 1, 2, 3

 14B 4 - 10

 14C 1a, b, *c,
2, *5, *6


Extra resource on binomial distribution questions


binomial extra
Jac.pdf
CONTINUOUS (The second type of distribution)
 What are the similarities / differences between discrete and
continuous distributions?
Differences
 Diff b/w discrete and continuous eg Pr (X = x) = 0
(impossibility of getting an exact value)
Eg In a large group of students, with heights ranging from
135cm to 204cm, with a mean of 171cm, the probability that
a particular student is exactly 171cm tall is close to 0.
Similarities
 They are a measure of likelihood, not a guarantee
 With numeric probabilities, an Expected Value can be
calculated (aka Average score) and a variance / SD
(measure of expected spread)
 Probabilities distributions add up to 1 – with continuous
this means the area under the pr(x) graph – calculated as
either a basic shape or using integration
 You can find the probability that an event is between two values





Two types of continuous probability studied in Math Methods
are
1. “Made up” and
2. Normal
 MADE UP
 Can you find probabilities?
Lots of integration / finding areas by other means.
 12.2 (assume
that all
functions are
prob density
functions)
 Ex15A 1 – 4, 6
-15 do a
selection
 12.3 same as
12.2 really, bit
more
advanced and
using lots of
CAS solving




















For made up, probabilities, are there formulae / concepts to help
calculate mean, median, mode, var SD?
Methods of calculating probability domains and ranges,
mean, Var / Sd, Median (set cumulative pr to 0.5) and mode
(max point on pr graph)
∞
Mean 𝝁 = ∫−∞ 𝒙 × 𝒇(𝒙)𝒅𝒙

12.4
 Ex 15 B Lots
𝒏
Median Solve ∫−∞ 𝒇(𝒙)𝒅𝒙 = 𝟎. 𝟓
(use similar sums to find quartiles and percentiles)
∞
Variance 𝝈𝟐 = ∫−∞(𝒙 − 𝝁)𝟐 × 𝒇(𝒙)𝒅𝒙
This stuff, is NOT math methods, it is Specialist
What is a combined event?
How do you find expected values of combined events?
Can you think of examples of the below combined events?
Combined events can be defined as X and Y
𝑬(𝒂 𝑿 + 𝒃) = 𝒂 𝑬(𝑿) + 𝒃
Combined events 𝑬(𝑿 + 𝒀) = 𝑬(𝑿) + 𝑬(𝒀)
𝑬(𝒂𝑿 + 𝒃𝒀) = 𝒂𝑬(𝑿) + 𝒃𝑬(𝒀)
𝑽𝒂𝒓(𝒂 𝑿 + 𝒃) = 𝒂𝟐 𝑽𝒂𝒓(𝑿)
𝑽𝒂𝒓(𝒂𝑿 + 𝒃𝒀) = 𝒂𝟐 𝑽𝒂𝒓(𝑿) + 𝒃𝟐 𝑽𝒂𝒓(𝒀)
NORMAL
Big Ideas
 With many distributions eg heights, values are more common
close to the mean and drop away. Eg. If the mean height in a
group is 170cm, there will be more people of height 170 to
 15 C 1 -10 lots
*11


12.5




Ex 13.2
3,4, 9 14, 16
 16A 1, 2
 16B 1, 3-9 (1,
1.96, 3 SD)
 16B 10 – 14 z
scores
175 than there will be from 175 to 180, and there will be more
of those than people 180 to 185 etc. NORMAL distribution is
one of these.
 NORMAL distribution is also symmetrical about the mean.
Eg. If the mean height in a group is 170cm, there will be as
many people of height 170 to 175 as there will be 165 to 170.
There will be as many from 170 to 178 as there are 162 to 170.
 With Normal distributions, mean µ is the same as the median
and is the central average value.
 With Normal distributions, standard deviation σ is the main
(and most useful) measure of how spread out the values are









What is a normal distribution, what does its graph look like, what
is an example?
What symbols get used?
What sort of problems can be solved using it?
Very useful, mean and SD usually determined by experiment
Highest probability centred around the mean, and then drops off
either side. i.e. things are more likely to be closer to the mean
than far away from it.
Written as 𝑋 ~ 𝑁 (𝜇, 𝜎 2 ) where 𝜇 is the mean and 𝜎 2 is the
variance → 𝜎 as the standard deviation is the useful measure.
Calculations using 1, 1.96, 3 SD / confidence limits similar to
Further Maths.
WARNING – Your textbook uses +- 2 SD for 95% confidence,
the methods course design specifies 1.96 SD
Standardised value = z score = no. of standard deviations a
𝑥−𝜇
score is above or below the mean 𝑧 = 𝜎
NB I have avoided the questions using the basic?
Normal equation formula = unnecessary IMHO.

Can you calculate probabilities apart from these with a
calculator? i.e. Standard normal =all SD values?

13.3 1 13
 16C 1, 2, ,3, 4,
12, 13, 17
a,b,c, 18 a,b

normCDf with calc to find probability of scores within a
range
 eg the probability that a student gets a study score of
between25 and 32, where the mean 𝜇 is 30 and SD 𝜇 = 7 is
normCDf(25, 32, 7, 30) = 0.3749 to 4dp. If the question is worth
more than one mark, you MUST specify that it is normal, and
what the mean, SD, max and min are.
 Eg the probability that x is less than 35 when mean = 30 SD =
7 is normCDf(−∞, 35, 7, 30) = 0.7625
 NB because you cannot get an exact value in a continuous
distribution, normPDf is fairly useless and < 𝑎𝑛𝑑 ≤
𝑚𝑒𝑎𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑡ℎ𝑖𝑛𝑔 𝑒𝑔 Pr(𝑥 < 35) = 𝑃𝑟 (𝑥 ≤ 35)
similarly > 𝑎𝑛𝑑 ≥ 𝑚𝑒𝑎𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑡ℎ𝑖𝑛𝑔
eg Pr(26 < 𝑥 < 35) = 𝑃𝑟 (26 ≤ 𝑥 ≤ 35) This is different to a
discrete distribution like the binomial
 What does standard normal mean, and how do you convert
from normal ↔ standard normal?
Standard norm means 𝜇 = 0 𝑎𝑛𝑑 𝜎 = 1
 Be able to convert normal ↔ standard normal using the
formula AND the sketch graph.
 Meaning of standard normal 𝑿 ~ 𝑵 (𝟎, 𝟏) and how to convert
to & from the given question and showing that the answers
are the same, eg
normCDf(15, 30, 5, 20) = 0.8186 and normCDf(−1,2, 1, 0) = 0.8186
 If you know the probabilities, can you work backwards and find
the x scores?
 Inverse normal distribution is the opposite to finding the
probabilities. The calculator uses the invNormCDf function
to do it. Interactive → dist /inverse dist → inverse →
invNormCDf
 If the probability that a student gets a study score less than a is
70%, where the mean 𝜇 is 30 and SD 𝜇 = 7 , a is calculated by
o 𝒊𝑛𝑣𝑁𝑜𝑟𝑚𝐶𝐷𝑓("L", 0.7, 7, 30) ≈ 33.67 “L” means left tail

13.4 1 16
 16C 5 -11, 14,
15, 16, 17d,e,
18 c,d,e
 If the probability that a student gets a study score more than a is
70%, where the mean 𝜇 is 30 and SD 𝜇 = 7 , a is calculated by
o 𝒊𝑛𝑣𝑁𝑜𝑟𝑚𝐶𝐷𝑓("R", 0.7, 7, 30) ≈ 26.33 “R” means right tail
 If the probability that a student gets a study score within the
central 70%, where the mean 𝜇 is 30 and SD 𝜇 = 7 , a (the lower
limit) is calculated by
o 𝒊𝑛𝑣𝑁𝑜𝑟𝑚𝐶𝐷𝑓("C", 0.7, 7, 30) ≈ 22.74 “C” means Center,
the upper limit is worked out because the upper and lower
are symmetrical about the mean. 30 − 22.74 = 7.26,
𝑠𝑜 𝑢𝑝𝑝𝑒𝑟 𝑤𝑖𝑙𝑙 𝑒𝑞𝑢𝑎𝑙 30 + 7.26 = 37.26
.






You may need to convert problem to a Z one (standard
normal) if problem is too complex.
Eg If they tell you a probability, and a value and the mean and
you need to calculate SD
Can you apply normal distribution, in combination with other
techniques, esp binomial, to solve worded questions? (Common
high level questions on exams)
Eg the probability that a given student score x is less than 35,
when mean = 30 SD = 7 is normCDf(−∞, 35, 7, 30) = 0.7625
The chance that a group of 8 students will have more than 2
with scores of less than 35 is binomial, given by
binomialCDf(3,8,8,0.7625)
Extra mixed questions resources on normal, inverse normal and
mixed

13.5
 16D lots



normal and inv
normal extra q's Jac.pdf




Sample Probability
What is a sample probability?
̂ is a probability found experimentally by
a sample probability 𝒑
running a number of trials. It serves as an estimate of the real,
whole population probability 𝒑. The more trials, the better the
estimate.
̂
Sample Statistics to determine a sample probability 𝒑
𝑿
̂=
𝒑
𝒘𝒉𝒆𝒓𝒆 𝑿 = 𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒅 𝒔𝒖𝒄𝒄𝒆𝒔𝒔𝒆𝒔,
𝒏
̂
𝒏 = 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒊𝒛𝒆. 𝑿 𝒘𝒊𝒍𝒍 𝒐𝒃𝒗𝒊𝒐𝒖𝒔𝒍𝒚 𝒗𝒂𝒓𝒚 𝒂 𝒃𝒊𝒕, 𝒉𝒆𝒏𝒄𝒆 𝒔𝒐 𝒘𝒊𝒍𝒍 𝒑
̂ is the probability found if you actually did the experiment and used
𝒑
the result.
Eg for 4 binomial trials, the number of possible successes (x) will be
0, 1, 2, 3, 4,
Hence the value of possible experimental probabilities found will be
0 1 2 3 4
, , , ,
4 4 4 4 4

̂ problems are very similar to x questions
Some 𝒑
Eg for a binomial distribution with n = 4, and p = 0.6 the distribution table
for x is as follows (You’ve seen this before)
x
0
1
2
3
4
Pr (X = x)
0.0256
0.1536
0.3456
0.3456
0.1296
for a binomial distribution with n = 4, and p = 0.6 the probability
̂ is as follows
distribution table for 𝒑
̂
𝒑
0
0.25
0.5
0.75
1
̂)
Pr (P = 𝒑
0.0256
0.1536
0.3456
0.3456
0.1296




14.3
14.4
 17A 12, 13
 17B Lots
̂ can be pretty much anything, but some are more
Measured values of 𝒑
̂ will be close to 𝒑 and
likely than others. Clearly, the expected value of 𝒑
will have a variance depending on the sample size. The larger the
sample size, the smaller the variance i.e. the better approximation for 𝒑 it
is

 Once you have a sample probability, can you use it to get an
estimate of the real probability?
̂(𝟏−𝒑
̂)
𝒑

̂, 𝑺𝑫 = √
𝝁= 𝒑

̂
𝒄𝒐𝒏𝒇𝒊𝒅𝒆𝒏𝒄𝒆 𝒍𝒊𝒎𝒊𝒕𝒔 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒕𝒓𝒖𝒆 𝒑, 𝒚𝒐𝒖 𝒖𝒔𝒆 𝒑


̂(𝟏−𝒑
̂)
𝒑
̂ ± 𝒛√
𝒑= 𝒑
 17C 1-8, *9,
*10
 17D lots
𝒏
𝒏

𝒇𝒐𝒓 𝟗𝟓% 𝒖𝒔𝒆 𝒛 = 𝟏. 𝟗𝟔
̂±
i.e. the real probability value 𝒑 𝒘𝒊𝒍𝒍 𝒃𝒆 𝒑
𝟏. 𝟗𝟔 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏𝒔 𝒇𝒐𝒓 𝒂 𝟗𝟓% 𝒄𝒆𝒓𝒕𝒂𝒊𝒏𝒕𝒚
 SAC 3 probability exercise – 2 classes, 2.5 hour total.
Technology allowed (Cas and standard scientific calc.) /
summary book allowed.
Exam Practice for multi choice / analysis. The main points are when
doing the exam: Read through whole exam quickly
 Read through again slowly thinking about how to do each
Question and referring to formula & cheat sheets.
 Do whole exam from front to back doing the easy questions
and/or attempting them all but stopping if they get tricky. (leave
a note on the question paper and move on) It is stupid to
spend 5 minutes getting a 1 point multi choice wrong,
when you could be spending 3 minutes getting a 3 point
short question right.
 Then come back and do the harder Questions.
 Then guess any multi choice Questions that you’ve left blank.




Exam Practice for analysis tests is definitely worth doing. The main
points are when doing the exam: Read through whole exam quickly
 Read through again slowly thinking about how to do each
Question and referring to formula & cheat sheets.
One method of doing the test is as follows: Do whole exam from front to back doing the first bit/ bits of
each Question. This gives you a start on each Question, and
gets your brain going on the harder ones, and your spirits up
on the easy ones(‘cos you know you can come back and clean
up some easy points).
 Then come back and complete (as far as possible) the easy
questions
 Then have a go at the rest.

Exam Practice for tech free. The main points are when doing the
exam: Read through whole exam quickly
 Read through again slowly thinking about how to do each
Question and referring to formula sheets.
 Do whole exam from front to back doing the easy questions
and/or attempting them all but stopping if they get tricky. (leave
a note on the question paper and move on) It is stupid to
spend 5 minutes getting a 1 point section wrong, when




Analysis
practice
exams
Do some 1
lesson
efforts
(including 15
minutes
reading
time) to get
used to
reading &
starting and
see how far
they get.
(self esteem
improved by
knowing that
if they can
get 12 marks
in 30
minutes they
can get
more in 90)


you could be spending 3 minutes getting a 3 point short
question right.
Then come back and do the harder Questions / parts of
questions.