IB Math AI HL Revision Notes

Formulas, conditions, key connections and common exam traps across the five AI HL topics. Use it alongside the formula booklet and regular question practice.

Written by IB Math tutor Abdulhamid

IB Math Master · Updated November 2026

For the current AI HL syllabus, for exams up to November 2028.

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How to use these notes

Each topic is split into short sections. Every section keeps its formulas, conditions and diagrams together.

Connection (blue dot): links between formulas that the formula booklet does not state.
Exam note (gold diamond): short reminders and common traps.
Video: a link to my explanation of the topic.

Topic 1 Number and Algebra

Numbers and Approximation

Standard form and accuracy

Name	Formula	Condition
Standard form	$a \times 10^{k}$	$1 \leq a < 10$ , $k \in ℤ$

Connection: Standard form products: multiply the $a$ parts, add the powers of 10, then rescale to $1 \leq a < 10$ .
Exam note: The GDC shows 4.52E−4. Write $4.52 \times 10^{- 4}$ .
Exam note: Give answers exactly or to 3 significant figures, unless the question says otherwise.
Exam note: Keep full calculator values in your working. Round only the final answer.

Percentage error

Name	Formula	Condition
Percentage error	$ε = \| \frac{v_{A} - v_{E}}{v_{E}} \| \times 100 %$	$v_{E}$ = exact value $v_{A}$ = approximate value

Exam note: Percentage error divides by the exact value $v_{E}$ .

Back to topics

Sequences and Series

Arithmetic and geometric sequences

Item	Arithmetic common difference $d$	Geometric common ratio $r$
$n$ th term	$u_{n} = u_{1} + (n - 1) d$	$u_{n} = u_{1} r^{n - 1}$
Sum of $n$ terms	$\begin{aligned} S_{n} & = \frac{n}{2} (2 u_{1} + (n - 1) d) \\ = \frac{n}{2} (u_{1} + u_{n}) \end{aligned}$	$S_{n} = \frac{u_{1} (r^{n} - 1)}{r - 1} = \frac{u_{1} (1 - r^{n})}{1 - r}, r \neq 1$
Finding $d$ or $r$	$u_{2} - u_{1} = u_{3} - u_{2} = \dots = d$	$\frac{u_{2}}{u_{1}} = \frac{u_{3}}{u_{2}} = \dots = r$

Exam note: $n$ counts terms, so $n \in ℤ^{+}$ .

Sums and sigma notation

Name	Formula	Condition
Sigma notation	$\sum_{i = 1}^{n} u_{i} = u_{1} + u_{2} + \dots + u_{n}$	$i$ runs through whole numbers

Connection: For any sequence:
$S_{1} = u_{1}$
$S_{2} = u_{1} + u_{2}$
$S_{3} = u_{1} + u_{2} + u_{3}$
$u_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$
Exam note: In sigma notation, check the first and last values of $i$ .

Back to topics

Financial Mathematics

Video: Financial mathematics (opens YouTube in a new tab)

Compound interest and depreciation

Name	Formula	Condition
Compound interest	$F V = P V \times {(1 + \frac{r}{100 k})}^{k n}$ $F V$ = future value, $P V$ = present value, $n$ = number of years, $k$ = number of compounding periods per year, $r %$ = nominal annual rate of interest	geometric, ratio $1 + \frac{r}{100 k}$ per period
Depreciation	$V_{n} = V_{0} \times {(1 - \frac{r}{100})}^{n}$ $V_{0}$ = starting value, $V_{n}$ = value after $n$ years, $r %$ = annual rate of depreciation	geometric, ratio $1 - \frac{r}{100}$ per year

Exam note: Number of years to reach a target: the answer is not a whole number. Round up to the next whole compounding period (to whole years when compounding is yearly).

Loans and annuities

TVM solver entry	Meaning
$N$	number of payments
$I %$	annual interest rate
$P V$	present value: the amount borrowed or invested at the start
$P M T$	payment each period
$F V$	future value: the amount at the end
$P / Y$	payments per year
$C / Y$	compounding periods per year

Connection: Loan fully repaid: $F V = 0$
Connection: Total repaid: add up all the payments. That is the number of payments times the amount of each payment.
Connection: Interest paid: the total repaid minus the amount borrowed.
Exam note: Money paid out is negative and money received is positive.
Exam note: Write down the values you enter in the TVM solver. They earn method marks.
Exam note: Monthly payments: $P / Y = 12$ . Set $C / Y$ from the question.

Back to topics

Exponents and Logarithms

Video: Logarithms (opens YouTube in a new tab)

Exponent laws

Name	Formula	Condition
Same base	$a^{m} \times a^{n} = a^{m + n}$ $\frac{a^{m}}{a^{n}} = a^{m - n}$ ${(a^{m})}^{n} = a^{m n}$	$a > 0$ for rational $m, n$
Zero and negative	$a^{0} = 1$ $a^{- n} = \frac{1}{a^{n}}$	$a \neq 0$
Rational exponents	$a^{1 / n} = \sqrt[n]{a}$ $a^{m / n} = {(\sqrt[n]{a})}^{m} = \sqrt[n]{a^{m}}$	$a > 0$

Connection: $\sqrt{a} = a^{1 / 2}$ and $\frac{1}{\sqrt{a}} = a^{- 1 / 2}$

Logarithms

Name	Formula	Condition
Definition of a log	$a^{x} = b ⟺ x = \log_{a} b$	$a > 0$ , $b > 0$ , $a \neq 1$
Laws of logarithms	$\log_{a} x y = \log_{a} x + \log_{a} y$ $\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$ $\log_{a} x^{m} = m \log_{a} x$	$a, x, y > 0$
Base 10 and base $e$	$\log x = \log_{10} x$ $\ln x = \log_{e} x$	$e \approx 2.718$

Connection: $\log_{a} 1 = 0$ and $\log_{a} a = 1$
Connection: $a^{\log_{a} x} = x$ and $\log_{a} a^{x} = x$
Connection: $a^{x}$ $= b$ $\Rightarrow x$ $= \log_{a} b$ $= \frac{\ln b}{\ln a}$
Exam note: No law exists for $\log (x + y)$ , $(\log x)^{2}$ or $(\log x) (\log y)$ . Leave them as they are.
Exam note: On the GDC, log is base 10 and ln is base $e$ .
Exam note: To find an unknown power, take ln of both sides or use the GDC solver.

Back to topics

Complex Numbers

Cartesian form and the Argand diagram

Name	Formula	Condition
Cartesian form	$z = a + b i, a, b \in ℝ$ $R e (z) = a$ $I m (z) = b$ $i^{2} = - 1$
Conjugate and modulus	$z^{*} = a - b i$ $\| z \| = r = \sqrt{a^{2} + b^{2}}$
Argument	$\arg z = θ$ $\tan θ = \frac{b}{a}$ $a = r \cos θ$ $b = r \sin θ$

z^{*}

is the reflection of

z

in the real axis.

Connection: $z z^{*} = | z |^{2}$
$z + z^{*} = 2 R e (z)$
$z - z^{*} = 2 i I m (z)$
Connection: Real $a$ , $b$ , $c$ and $Δ = b^{2} - 4 a c < 0$ : the roots of $a x^{2} + b x + c = 0$ are a conjugate pair:
$x$ $= \frac{- b \pm i \sqrt{4 a c - b^{2}}}{2 a}$
Exam note: $\tan^{- 1} (\frac{b}{a})$ gives the correct $θ$ only in quadrants 1 and 4. Sketch $z$ first.
Exam note: To divide, multiply the top and bottom by the conjugate of the bottom.

Arguments on the axes

$z$ lies on the	$z$	$\arg z$	All its arguments
positive real axis	$a$	$0$	$0 + 2 k π$
negative real axis	$- a$	$π$	$π + 2 k π$
positive imaginary axis	$b i$	$\frac{π}{2}$	$\frac{π}{2} + 2 k π$
negative imaginary axis	$- b i$	$- \frac{π}{2}$	$- \frac{π}{2} + 2 k π$

Here $a > 0$ , $b > 0$ and $k \in ℤ$ . The $\arg z$ column uses $- π < θ \leq π$ .

Exam note: Adding a multiple of $2 π$ gives the same point. Give the argument in the range the question asks: with $0 \leq θ < 2 π$ , $- b i$ has $\arg z = \frac{3 π}{2}$ .

Measure from the positive real axis.
Anticlockwise is positive. Clockwise is negative.

z = 0

has no argument.

Polar and exponential forms

Name	Formula	Condition
Polar and exponential	$z = r (\cos θ + i \sin θ) = r e^{i θ} = r cis θ$
Product	$z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})} = r_{1} r_{2} cis (θ_{1} + θ_{2})$
Quotient	$\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})} = \frac{r_{1}}{r_{2}} cis (θ_{1} - θ_{2}), z_{2} \neq 0$
Powers	$z^{n} = r^{n} e^{i n θ} = r^{n} cis n θ, n \in ℤ$

| z_{1} z_{2} | = r_{1} r_{2}

and

\arg (z_{1} z_{2})

= θ_{1} + θ_{2}

Connection: $| z_{1} z_{2} |$ $= | z_{1} | | z_{2} |$ and $\arg (z_{1} z_{2})$ $= \arg z_{1} + \arg z_{2}$
$| \frac{z_{1}}{z_{2}} |$ $= \frac{| z_{1} |}{| z_{2} |}$ and $\arg \frac{z_{1}}{z_{2}}$ $= \arg z_{1} - \arg z_{2}$
Adjust arguments by $2 π$ if needed.
Connection: Multiplying by $i$ turns $z$ through $\frac{π}{2}$ anticlockwise.
Exam note: Use radians for $θ$ unless the question uses degrees.
Exam note: Check answers with the GDC in complex mode.

Back to topics

Matrices

Matrix operations

Name	Formula	Condition
Order and elements	An $m \times n$ matrix has $m$ rows and $n$ columns. $a_{i j}$ is the element in row $i$ , column $j$ .
Add, subtract	Element by element	same order only
Scalar multiple	Multiply every element by the scalar
Product	$(m \times n) (n \times p)$ gives an $m \times p$ matrix	columns of $A$ = rows of $B$
Identity	$A I = I A = A$

Connection: $A B \neq B A$ in general.

Determinant and inverse

Name	Formula	Condition
Determinant	$A = (\begin{array}{cc} a & b \\ c & d \end{array}) \Rightarrow \det A = \| A \| = a d - b c$
Inverse	$A^{- 1} = \frac{1}{\det A} (\begin{array}{cc} d & - b \\ - c & a \end{array})$ $A A^{- 1} = A^{- 1} A = I$	$a d \neq b c$

Connection: $(A B)^{- 1} = B^{- 1} A^{- 1}$ and $\det (A B) = \det A \times \det B$
Connection: $\det A = 0$ : $A$ has no inverse.
Exam note: $A X = B$ gives $X = A^{- 1} B$ .
$X A = B$ gives $X = B A^{- 1}$ .
Exam note: $3 \times 3$ determinants and inverses: use the GDC.

Systems of linear equations

Name	Formula	Condition
Matrix form	$A 𝐱 = 𝐛 \Rightarrow 𝐱 = A^{- 1} 𝐛, \det A \neq 0$
$\det A \neq 0$	One solution. With 2 unknowns, the two lines meet at one point.
$\det A = 0$	No unique solution. With 2 unknowns, the lines are parallel (no solution) or the same line (infinitely many solutions).

Exam note: Systems of up to 3 equations and polynomial equations: solve on the GDC, but write the equations first.

Back to topics

Eigenvalues and Eigenvectors

Video: Eigenvalues and eigenvectors (opens YouTube in a new tab)

Eigenvalues and eigenvectors

Name	Formula	Condition
Definition	$A 𝐯 = λ 𝐯, 𝐯 \neq 𝟎$
Characteristic equation	$\det (A - λ I) = 0$

Connection: Any non-zero multiple of an eigenvector is also an eigenvector.
Exam note: For $2 \times 2$ , the two rows of $(A - λ I) 𝐯$ $= 𝟎$ are multiples. Use one of them.
Exam note: Check your answer: $A 𝐯 = λ 𝐯$ .

A

= (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})

A 𝐯

stays on the line of

𝐯

A 𝐮

changes direction.

Diagonalisation and powers

Name	Formula	Condition
Diagonalisation	$A = P D P^{- 1}$	distinct real eigenvalues
Power formula	$M^{n} = P D^{n} P^{- 1}$ $P$ = matrix of eigenvectors, $D$ = diagonal matrix of eigenvalues

Connection: $D^{n}$ $= (\begin{array}{cc} λ_{1}^{n} & 0 \\ 0 & λ_{2}^{n} \end{array})$
Connection: $λ_{1} = 1$ and $| λ_{2} | < 1$ : $λ_{2}^{n} \to 0$ , so $M^{n}$ tends to a limit. Used for Markov chains.
Exam note: Put the eigenvectors in $P$ in the same order as the eigenvalues in $D$ .

Back to topics

Topic 2 Functions

Straight Lines

Equations and gradients

Name	Formula	Condition
Straight line	$y = m x + c$ $a x + b y + d = 0$ $y - y_{1} = m (x - x_{1})$
Gradient	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$	$x_{1} \neq x_{2}$
Parallel lines	$m_{1} = m_{2}$
Perpendicular lines	$m_{1} \times m_{2} = - 1$ $m_{2} = - \frac{1}{m_{1}}$

Connection: In a linear model, $m$ is the change in $y$ for each unit increase in $x$ and $c$ is the value at $x = 0$ .
Exam note: $a x + b y + d = 0$ usually needs whole numbers $a$ , $b$ and $d$ .
Exam note: A vertical line is $x = k$ . Its gradient is undefined.

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Functions, Composites and Inverses

Functions and composites

Name	Formula	Condition
Function	Each $x$ in the domain gives exactly one $y$
Domain, range	Domain: the allowed inputs. Range: the outputs.
Composite	$(f \circ g) (x) = f (g (x))$	apply $g$ first

Connection: In general $f \circ g \neq g \circ f$ .
Exam note: In a model, the domain comes from the context, for example $t \geq 0$ .

Inverse functions

Name	Formula	Condition
Inverse	$y = f (x)$ : swap $x$ and $y$ , then make $y$ the subject. $f$ must be one-to-one.
Identity	$(f \circ f^{- 1}) (x) = (f^{- 1} \circ f) (x) = x$

Connection: The graph of $f^{- 1}$ is the reflection of $f$ in $y = x$ .
Connection: Domain of $f^{- 1}$ = range of $f$
Range of $f^{- 1}$ = domain of $f$
Exam note: $f^{- 1} (x)$ is not $\frac{1}{f (x)}$ .
Exam note: Not one-to-one? Restrict the domain first, usually to one side of a turning point.
Exam note: Always state the domain of $f^{- 1}$ .

(1, 2)

f

and

(2, 1)

f^{- 1}

Back to topics

Transformations

Video: Transformations of graphs: translations, stretches and reflections (opens YouTube in a new tab)

Transformations of $y = f (x)$

New graph	Transformation	Point $(x, y)$ moves to
$y = f (x) + b$	translation $b$ units up	$(x, y + b)$
$y = f (x) - b$	translation $b$ units down	$(x, y - b)$
$y = f (x - a)$	translation $a$ units to the right	$(x + a, y)$
$y = f (x + a)$	translation $a$ units to the left	$(x - a, y)$
$y = - f (x)$	reflection in the $x$ -axis	$(x, - y)$
$y = f (- x)$	reflection in the $y$ -axis	$(- x, y)$
$y = p f (x)$	vertical stretch, scale factor $p$	$(x, p y)$
$y = f (q x)$	horizontal stretch, scale factor $\frac{1}{q}$	$(\frac{x}{q}, y)$
$y = f (a x + b)$	translation $b$ units to the left, then horizontal stretch, scale factor $\frac{1}{a}$	$(\frac{x - b}{a}, y)$

Exam note: In this table $a, b, p, q > 0$ . So $f (x) + b$ always moves up and $f (x) - b$ always moves down.
Connection: Inside the bracket: changes $x$ and works the opposite way to how it looks.
Connection: Outside the bracket: changes $y$ and works the way it looks.
Connection: $y = p f (x) + b$ : stretch first, then translate.
Connection: As a vector:
$b$ units up: $(\begin{array}{c} 0 \\ b \end{array})$ $a$ units right: $(\begin{array}{c} a \\ 0 \end{array})$
Connection: A vertical stretch keeps points on the $x$ -axis fixed. A horizontal stretch keeps points on the $y$ -axis fixed.
Exam note: Describe each transformation fully: type, direction and scale factor or vector.
Exam note: For a composite transformation, apply the steps in the order the question gives.

Dashed

y = x^{2}

. Left:

y = (x - 2)^{2}

moves right 2. Right:

y = (2 x)^{2}

divides each

x

by 2.

Back to topics

Polynomial and Variation Models

Video: Graphing quadratics (opens YouTube in a new tab)

Quadratic models

Form	Equation	What it shows
Standard	$f (x) = a x^{2} + b x + c,$ $a \neq 0$	$y$ -intercept $(0, c)$ axis of symmetry $x = - \frac{b}{2 a}$ vertex $(- \frac{b}{2 a}, f (- \frac{b}{2 a}))$
Factor	$f (x) = a (x - p) (x - q)$	zeros $p$ and $q$ axis of symmetry $x = \frac{p + q}{2}$
Vertex	$f (x) = a (x - h)^{2} + k$	vertex $(h, k)$ axis of symmetry $x = h$

Connection: $a > 0$ : minimum at the vertex. $a < 0$ : maximum at the vertex.
Exam note: Check the maximum or minimum lies inside the domain of the model.

f (x)

= x^{2} - 6 x + 5

= (x - 1) (x - 5)

= (x - 3)^{2} - 4

Cubic models

Name	Formula	Condition
Cubic	$f (x) = a x^{3} + b x^{2} + c x + d, a \neq 0$ At most 2 turning points

Exam note: Zeros, turning points and intersections: use the GDC.

Direct and inverse variation

Name	Formula	Condition
Direct variation	$y$ is directly proportional to $x^{n}$ : $y = a x^{n}, n \in ℤ^{+}, a \neq 0$ The graph passes through the origin. With $n = 1$ it is a straight line.	$y = 2 x$ $(n = 1)$
Inverse variation	$y$ is inversely proportional to $x^{n}$ : $y = \frac{a}{x^{n}}, n \in ℤ^{+}, a \neq 0$ Asymptotes $x = 0$ and $y = 0$ .	$y = \frac{4}{x}$ $(n = 1)$

Name

Formula

Condition

Direct variation

$y$ is directly proportional to $x^{n}$ :

y = a x^{n}, n \in ℤ^{+}, a \neq 0

The graph passes through the origin. With $n = 1$ it is a straight line.

y = 2 x

(n = 1)

Inverse variation

$y$ is inversely proportional to $x^{n}$ :

y = \frac{a}{x^{n}}, n \in ℤ^{+}, a \neq 0

Asymptotes $x = 0$ and $y = 0$ .

y = \frac{4}{x}

(n = 1)

Exam note: “ $y$ is proportional to $x^{2}$ ” means $y = a x^{2}$ . “ $y$ is inversely proportional to $x$ ” means $y = \frac{a}{x}$ . Find $a$ from one data point.

Back to topics

Exponential and Logistic Models

Exponential and logarithmic models

Name	Formula	Condition
Exponential	$f (x) = k a^{x} + c, a > 0, a \neq 1$ $f (x) = k e^{r x} + c$ asymptote $y = c$
Logarithmic	$f (x) = a + b \ln x, x > 0$

Connection: The value at $x = 0$ is $k + c$ .
Connection: $k a^{x} = k e^{x \ln a}$
Connection: $k e^{r x}$ : $r > 0$ growth and $r < 0$ decay
Exam note: In $f (t) = k e^{r t} + c$ , the initial value is $k + c$ , not $k$ .
Exam note: Long-term value: let $x \to \infty$ . Exponential decay tends to $c$ .

y = 4 e^{- 0.5 x} + 1

: starts at

k + c = 5

and tends to

c = 1

Logistic models

Name	Formula	Condition
Logistic	$f (x) = \frac{L}{1 + C e^{- k x}}, L, C, k > 0$ asymptotes $y = 0$ and $y = L$

Connection: $L$ is the upper limit, the carrying capacity.
Connection: $f (0) = \frac{L}{1 + C}$
Connection: Fastest growth where $f (x) = \frac{L}{2}$
Exam note: Long-term value: the logistic model tends to $L$ .

y = \frac{L}{1 + C e^{- k x}}

with

L = 10

C = 9

k = 1

Back to topics

Sinusoidal and Piecewise Models

Sinusoidal models

Name	Formula	Condition
Sinusoidal	$f (x) = a \sin (b (x - c)) + d$ $f (x) = a \cos (b (x - c)) + d$	$b > 0$

Feature	From the function	From the graph
Amplitude	$\| a \|$	$\frac{max - min}{2}$
Period	radians: $\frac{2 π}{b}$ degrees: $\frac{360^{\circ}}{b}$	from a maximum to the next maximum, or from a minimum to the next minimum then $b = \frac{2 π}{period}$ or $b$ $= \frac{360^{\circ}}{period}$
Principal axis	$y = d$	$y = \frac{max + min}{2}$
Maximum, minimum	$d + \| a \|, d - \| a \|$	the highest and lowest $y$ -values
Horizontal shift	$c$ to the right ( $c < 0$ : to the left)	sine with $a > 0$ : the $x$ where the graph crosses $y = d$ going up cosine with $a > 0$ : the $x$ of a maximum point

Connection: A cos model is a sin model shifted by a quarter period.
Exam note: Set the GDC to degrees or radians to match the question.

y

= 2 \sin (\frac{π}{6} (x - 1)) + 3

a = 2

b = \frac{π}{6}

c = 1

d = 3

Piecewise models

Name	Formula	Condition
Piecewise	A different rule on each interval of the domain	Both rules give 5 at $x = 2$ .

Exam note: Check where the pieces meet. Use the correct rule on each side.

Back to topics

Modelling and Linearising Data

Linearising with logarithms

Model	Take logs	Straight line when	Graph
Exponential $y = k a^{x}$	$\ln y = \ln k + x \ln a$	$\ln y$ against $x$	semi-log
Power $y = a x^{n}$	$\ln y = \ln a + n \ln x$	$\ln y$ against $\ln x$	log-log

Connection: From $\ln y = m x + c$ : $k = e^{c}$ and $a = e^{m}$
Connection: From $\ln y = m \ln x + c$ : $a = e^{c}$ and $n = m$
Connection: On a log scale, each equal step multiplies by the same factor, often 10.

y = 3 (1.5)^{x}

becomes a straight line when

\ln y

is plotted against

x

Choosing and using a model

Exam note: Choose a model from the shape of the data and the context. Give a reasonable domain.
Exam note: Predictions far outside the data are not reliable.
Exam note: “Sketch”: show the shape and key features. “Draw”: plot accurately to scale.
Exam note: Label intercepts, turning points, asymptotes and end points of the domain.

Back to topics

Topic 3 Geometry and Trigonometry

3D Geometry and Triangles

Points in 3D

Name	Formula	Condition
Distance	$d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}}$
Midpoint	$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$

Solids

Name	Formula	Condition
Right pyramid	$V = \frac{1}{3} A h,$ $A$ = area of base, $h$ = height
Right cone	$V = \frac{1}{3} π r^{2} h$ curved surface $A = π r l$ total surface $= π r l + π r^{2}$ $l^{2} = r^{2} + h^{2}$ , $l$ = slant height
Sphere	$V = \frac{4}{3} π r^{3}$ $A = 4 π r^{2}$
Hemisphere	$V = \frac{2}{3} π r^{3}$ curved surface $2 π r^{2}$

Right-angled triangles

Name	Formula	Condition
Ratios	$\sin θ = \frac{opp}{hyp}$ $\cos θ = \frac{adj}{hyp}$ $\tan θ = \frac{opp}{adj}$
Pythagoras	$a^{2} + b^{2} = c^{2},$ $c$ = hypotenuse

Opposite and adjacent are named from

θ

Angle between a line and a plane

Connection: Drop a perpendicular from a point on the line to the plane. This gives the projection of the line on the plane.
Connection: $θ$ is the angle between the line and its projection. Find it in that right-angled triangle.

θ

is measured to the projection on the plane.

Any triangle

Name	Formula	Condition
Sine rule	$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Cosine rule	$c^{2} = a^{2} + b^{2} - 2 a b \cos C$ $\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
Area	$A = \frac{1}{2} a b \sin C$

Side

a

is opposite angle

A

, and so on.

Use the sine rule when you know	Use the cosine rule when you know
two angles and one side	two sides and the angle between them
two sides and an angle not between them (this can be the ambiguous case)	all three sides

Exam note: Draw and label a diagram from the words first.

The ambiguous case of the sine rule

Exam note: It happens only when you know two sides and an angle not between them, and you use the sine rule to find an angle.
Exam note: $\sin θ = k$ gives two angles: $θ$ and $180^{\circ} - θ$ . Check both. Keep $180^{\circ} - θ$ only if the angles of the triangle still add up to $180^{\circ}$ with every angle positive.

Ambiguous case only: two triangles fit the same

A

c

and

a

Elevation, depression and bearings

Name	Formula	Condition
Elevation, depression	Measured from the horizontal
Bearings	Measured clockwise from north, with three figures, for example $065^{\circ}$

Back to topics

Circles, Radians and the Unit Circle

Radians, arcs and sectors

Name	Formula	Condition
Radians	$π radians = 180^{\circ}$
Arc length	$l = \frac{θ}{360} \times 2 π r$ $l = r θ$	degrees, then radians
Sector area	$A = \frac{θ}{360} \times π r^{2}$ $A = \frac{1}{2} r^{2} θ$	degrees, then radians

Connection: Degrees to radians: $\times \frac{π}{180}$ . Radians to degrees: $\times \frac{180}{π}$ .
Exam note: $l = r θ$ and $A = \frac{1}{2} r^{2} θ$ need $θ$ in radians.

The unit circle

Name	Formula	Condition
Unit circle	$\cos θ = x$ -coordinate $\sin θ = y$ -coordinate
Identity	$\cos^{2} θ + \sin^{2} θ = 1$
Tan	$\tan θ = \frac{\sin θ}{\cos θ}, \cos θ \neq 0$

Connection: $\sin θ$ and $\cos θ$ : period $2 π$ or $360^{\circ}$ , values from $- 1$ to $1$
Connection: $\sin (180^{\circ} - θ)$ $= \sin θ$ . This is why the sine rule can give two angles.

Trig graphs and solving trig equations

Exam note: Solve on the GDC: graph both sides, then find every intersection in the interval.
Exam note: Check the GDC mode: degrees or radians.

y = \sin x

and

y = \cos x

: period

2 π

\sin x = 0.4

has 2 solutions for

0 \leq x \leq 2 π

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Voronoi Diagrams

Perpendicular bisectors

Name	Formula	Condition
Perpendicular bisector of $A B$	Through the midpoint $M$ of $A B$ , gradient $- \frac{1}{m_{A B}}$ Every point on it is equidistant from $A$ and $B$ .

Name

Formula

Condition

Perpendicular bisector of

A B

Through the midpoint $M$ of $A B$ , gradient $- \frac{1}{m_{A B}}$

Every point on it is equidistant from $A$ and $B$ .

Connection: Equation: $y - y_{M}$ $= - \frac{1}{m_{A B}} (x - x_{M})$
Connection: $A B$ horizontal: the bisector is $x = x_{M}$ . $A B$ vertical: the bisector is $y = y_{M}$ .

Any point

P

on the bisector:

P A = P B

Voronoi diagrams

Name	Formula	Condition
Cell	All points closer to its site than to any other site
Edge	Part of the perpendicular bisector of two sites
Vertex	Where at least three edges meet. Equidistant from 3 or more sites.
Nearest neighbour interpolation	A point takes the value of the site whose cell contains it

Connection: Adding a site: draw the bisectors between the new site and its neighbours, then remove the old edges inside the new cell.
Connection: The bisectors of $A B$ and $B C$ meet at a point equidistant from $A$ , $B$ and $C$ , so the bisector of $A C$ passes through it too.

Sites in blue, vertices in gold.

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Matrix Transformations

Transformation matrices

Name	Formula	Condition
Reflection	$(\begin{array}{cc} \cos 2 θ & \sin 2 θ \\ \sin 2 θ & - \cos 2 θ \end{array})$	in the line $y = (\tan θ) x$
Reflection in an axis or $y = x$	$x$ -axis: $(\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ $y$ -axis: $(\begin{array}{cc} - 1 & 0 \\ 0 & 1 \end{array})$ $y = x$ : $(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$
Horizontal stretch	$(\begin{array}{cc} k & 0 \\ 0 & 1 \end{array})$	scale factor $k$
Vertical stretch	$(\begin{array}{cc} 1 & 0 \\ 0 & k \end{array})$	scale factor $k$
Enlargement	$(\begin{array}{cc} k & 0 \\ 0 & k \end{array})$	scale factor $k$ , centre $(0, 0)$
Rotation anticlockwise	$(\begin{array}{cc} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{array})$	by $θ$ about $(0, 0)$ , $θ > 0$
Rotation clockwise	$(\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$	by $θ$ about $(0, 0)$ , $θ > 0$

Connection: Rotation: $\det = 1$ . Reflection: $\det = - 1$ .

Columns of a matrix

Connection: The first column of $A$ is where the point $(1, 0)$ goes.
Connection: The second column of $A$ is where the point $(0, 1)$ goes.

Name	Formula	Condition
Example	$A = (\begin{array}{cc} 2 & 1 \\ 0 & 1 \end{array})$ $A (\begin{array}{c} 1 \\ 0 \end{array}) = (\begin{array}{c} 2 \\ 0 \end{array}),$ the first column of $A$ $A (\begin{array}{c} 0 \\ 1 \end{array}) = (\begin{array}{c} 1 \\ 1 \end{array}),$ the second column of $A$
Finding a matrix	Rotation by $90^{\circ}$ anticlockwise about $(0, 0)$ : $(1, 0)$ goes to $(0, 1)$ and $(0, 1)$ goes to $(- 1, 0)$ So $A = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$

(1, 0)

goes to

(2, 0)

and

(0, 1)

goes to

(1, 1)

.
The gold images are the columns of

A

.
The unit square becomes a parallelogram with area

| \det A | = 2

Composites and area

Name	Formula	Condition
With translation	$𝐱'$ $= A 𝐱 + 𝐛$ : apply $A$ , then translate by $𝐛$
Area	Area of image $= \| \det A \| \times$ area of object

Connection: $A$ then $B$ : the single matrix is $B A$ .
Connection: $A^{- 1}$ undoes the transformation $A$ .
Exam note: Area scale factor uses $| \det A |$ , so ignore a negative sign.
Exam note: The matrix multiplies the column vector: $A 𝐱$ , not $𝐱 A$ .
Exam note: The order of a composite matters. The first transformation goes on the right.

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Vectors

Video: Vector product (opens YouTube in a new tab)

Vector basics

Name	Formula	Condition
Components	$𝐯 = v_{1} 𝐢 + v_{2} 𝐣 + v_{3} 𝐤 = (\begin{array}{c} v_{1} \\ v_{2} \\ v_{3} \end{array})$
Magnitude	$\| 𝐯 \| = \sqrt{v_{1}^{2} + v_{2}^{2} + v_{3}^{2}}$
Unit vector	$\frac{𝐯}{\| 𝐯 \|}$
Displacement	$\vec{A B} = 𝐛 - 𝐚$

Connection: A vector of length $L$ in the direction of $𝐯$ : $L \frac{𝐯}{| 𝐯 |}$

Scalar product

Name	Formula	Condition
Scalar product	$𝐯 \cdot 𝐰 = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} = \| 𝐯 \| \| 𝐰 \| \cos θ$	$θ$ = angle between $𝐯$ and $𝐰$
Angle between vectors	$\cos θ = \frac{v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3}}{\| 𝐯 \| \| 𝐰 \|}$

Connection: Perpendicular: $𝐯 \cdot 𝐰 = 0$
Connection: $𝐯 \cdot 𝐯$ $= | 𝐯 |^{2}$
Exam note: $𝐯 \cdot 𝐰$ is a number.
Exam note: For the angle at a vertex, both vectors must start at that vertex.

Vector product

Name	Formula	Condition
Vector product	$𝐯 \times 𝐰 = (\begin{array}{c} v_{2} w_{3} - v_{3} w_{2} \\ v_{3} w_{1} - v_{1} w_{3} \\ v_{1} w_{2} - v_{2} w_{1} \end{array})$ $\| 𝐯 \times 𝐰 \| = \| 𝐯 \| \| 𝐰 \| \sin θ$
Area of parallelogram	$A = \| 𝐯 \times 𝐰 \|,$ $𝐯$ , $𝐰$ adjacent sides

Connection: $𝐯 \times 𝐰$ is perpendicular to both $𝐯$ and $𝐰$ .
Connection: Parallel: $𝐰 = k 𝐯$ or $𝐯 \times 𝐰$ $= 𝟎$
Connection: Area of triangle $= \frac{1}{2} | 𝐯 \times 𝐰 |$
Exam note: $𝐯 \times 𝐰$ is a vector.

Parallelogram area

| 𝐯 \times 𝐰 |

. Triangle: half of it.

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Vector Lines and Kinematics

Vector lines

Name	Formula	Condition
Vector equation of a line	$𝐫 = 𝐚 + λ 𝐛$ $(\begin{array}{c} x \\ y \\ z \end{array}) = (\begin{array}{c} x_{0} \\ y_{0} \\ z_{0} \end{array}) + λ (\begin{array}{c} l \\ m \\ n \end{array})$	$𝐚$ $= (\begin{array}{c} x_{0} \\ y_{0} \\ z_{0} \end{array})$ : a point on the line $𝐛$ $= (\begin{array}{c} l \\ m \\ n \end{array})$ : direction vector
Parametric form	$x = x_{0} + λ l, y = y_{0} + λ m, z = z_{0} + λ n$
Angle between lines	$\cos θ = \frac{\| 𝐛_{1} \cdot 𝐛_{2} \|}{\| 𝐛_{1} \| \| 𝐛_{2} \|}$	direction vectors only

Name

Formula

Condition

Vector equation of a line

𝐫 = 𝐚 + λ 𝐛

(\begin{array}{c} x \\ y \\ z \end{array}) = (\begin{array}{c} x_{0} \\ y_{0} \\ z_{0} \end{array}) + λ (\begin{array}{c} l \\ m \\ n \end{array})

$𝐚$ $= (\begin{array}{c} x_{0} \\ y_{0} \\ z_{0} \end{array})$ : a point on the line

$𝐛$ $= (\begin{array}{c} l \\ m \\ n \end{array})$ : direction vector

Parametric form

x = x_{0} + λ l, y = y_{0} + λ m, z = z_{0} + λ n

Angle between lines

\cos θ = \frac{| 𝐛_{1} \cdot 𝐛_{2} |}{| 𝐛_{1} | | 𝐛_{2} |}

direction vectors only

Exam note: The modulus gives the acute angle $θ$ .
Exam note: Without the modulus you may get the obtuse angle. The acute angle is then $180^{\circ} - θ$ .

Name	Formula	Condition
Parallel	$𝐛_{1} = k 𝐛_{2}$
Intersecting	Solve $𝐚_{1} + λ 𝐛_{1}$ $= 𝐚_{2} + μ 𝐛_{2}$ . Use two components, then check the third.
Skew	Not parallel and they do not meet

Kinematics

Name	Formula	Condition
Constant velocity	$𝐫 = 𝐫_{0} + t 𝐯$ $(\begin{array}{c} x \\ y \end{array}) = (\begin{array}{c} x_{0} \\ y_{0} \end{array}) + t (\begin{array}{c} v_{x} \\ v_{y} \end{array})$ $(\begin{array}{c} x_{0} \\ y_{0} \end{array})$ = position at $t = 0$
Speed	$\| 𝐯 \| = \sqrt{v_{x}^{2} + v_{y}^{2}}$
Variable velocity	$𝐫 = (\begin{array}{c} x (t) \\ y (t) \end{array})$ $𝐯 = \frac{d 𝐫}{d t} = (\begin{array}{c} x' (t) \\ y' (t) \end{array})$ $𝐚 = \frac{d 𝐯}{d t} = (\begin{array}{c} x ″ (t) \\ y ″ (t) \end{array})$ Differentiate or integrate one component at a time.

Connection: Position of $B$ relative to $A$ : $𝐫_{B} - 𝐫_{A}$
Connection: Closest approach: minimise $| 𝐫_{B} - 𝐫_{A} |$
Exam note: A collision needs the same position at the same time $t$ . Paths that cross may not collide.
Exam note: Speed is a number. Velocity is a vector.

𝐫_{A}

= t (\begin{array}{c} 1 \\ 2 \end{array})

𝐫_{B}

= (\begin{array}{c} 10 \\ 0 \end{array}) + t (\begin{array}{c} - 2 \\ 1 \end{array})

.
The paths cross at

(2, 4)

at different times. Closest at

t = 3

: distance

\sqrt{10}

Back to topics

Graph Theory

Graphs and degrees

Name	Formula	Condition
Degree	Number of edges at a vertex	a loop counts 2
Handshaking	Sum of degrees $= 2 \times$ number of edges
Simple graph	No loops and no multiple edges
Directed graph	Each vertex has an in-degree and an out-degree

Connection: The number of odd vertices is always even.

Complete graphs and trees

Name	Formula	Condition
Complete graph $K_{n}$	Every pair of vertices joined: $\frac{n (n - 1)}{2}$ edges	$K_{4}$ : 6 edges
Tree	Connected with no cycles: $n - 1$ edges for $n$ vertices	5 vertices, 4 edges

Walks, trails, paths and cycles

Name	Formula	Condition
Walk	A sequence of edges joined end to end. Vertices and edges may repeat. Example: $A C D A C$ (edge $A C$ is used twice)
Trail	A walk with no repeated edge. Example: $A B C D E C$ (vertex $C$ repeats)
Path	A walk with no repeated vertex. Example: $A B C D$
Circuit	A trail that starts and ends at the same vertex. Example: $C A B C D E C$
Cycle	A circuit with no repeated vertex, apart from the start and end. Example: $A B C A$
Hamiltonian path	Visits every vertex exactly once. Example: $B A D C E$
Hamiltonian cycle	Visits every vertex exactly once and returns to the start. Example: $A C D E B A$

Blue: the route. Gold: a vertex or an edge used more than once.

Adjacency matrices

Name	Formula	Condition
Adjacency matrix $M$	The entry in row $i$ , column $j$ is the number of edges joining vertex $i$ to vertex $j$ . For an undirected graph, $M$ is symmetric.
Walks of length exactly $k$	Raise $M$ to the power $k$ . Look in the row of the start vertex and the column of the end vertex. That entry is the number of walks. Notation: ${(M^{k})}_{i j}$ . Here $i$ is the start and $j$ is the end.
Walks of length $k$ or less	Add the powers: $M + M^{2} + \dots + M^{k}$ . Then read the same entry: row of the start, column of the end. Notation: ${(M + M^{2} + \dots + M^{k})}_{i j}$

Connection: The length of a walk is the number of edges in it.
Connection: Weighted adjacency table: the entries are the weights. Leave a gap where there is no edge.
Exam note: Keep the vertices in the same order in the rows and columns.
Exam note: Find the powers of $M$ on the GDC.

M^{2}

, row

A

, column

D

is 2 (in gold).
So there are 2 walks of length 2 from

A

D

A B D

and

A C D

Transition matrices for random walks

Name	Formula	Condition
Random walk	At each step, move along one edge. Each edge at the vertex is equally likely.
Transition matrix $T$	Each column is the vertex you leave. Each row is the vertex you move to. Fill in one column at a time. Share 1 equally between the edges at that vertex. Put 0 in the other rows.
Example	$B$ has 3 edges, to $A$ , $C$ and $D$ . So column $B$ has $\frac{1}{3}$ in rows $A$ , $C$ and $D$ , and 0 in row $B$ .

Connection: Each column adds to 1.
Connection: Probabilities after $n$ steps: $T^{n} 𝐬_{0}$ , as for Markov chains.
Connection: Directed graph: share 1 equally between the edges leaving the vertex.
Connection: Strongly connected: every vertex can be reached from every other vertex.

The graph above. Column = from, row = to.

A

and

D

have 2 edges: their columns use

\frac{1}{2}

B

and

C

have 3 edges: their columns use

\frac{1}{3}

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Graph Algorithms

Eulerian trails and the Chinese postman

Name	Formula	Condition
Eulerian circuit	Uses every edge once and returns to the start connected, all degrees even
Eulerian trail	Uses every edge once connected, exactly 2 odd vertices
Chinese postman	The shortest closed route that uses every edge at least once and returns to the start. Total weight + the shortest way to pair up the odd vertices

Connection: An Eulerian trail starts at one odd vertex and ends at the other.
Connection: 2 odd vertices: repeat the shortest path between them.
4 odd vertices: compare the 3 pairings.

Odd vertices

B

and

D

. Repeat

B D

: route

= 30 + 5 = 35

Minimum spanning trees

Name	Formula	Condition
Kruskal	Add the shortest edge that makes no cycle, until there are $n - 1$ edges
Prim	Start at any vertex. Add the shortest edge joining the tree to a new vertex.
Prim on a table	Choose a start vertex. Cross out its row and label its column. In the labelled columns, circle the smallest entry that is not crossed out. Its row gives the next vertex. Cross out that row and label that column. Repeat until every row is crossed out. The circled entries are the edges of the tree.

Exam note: Show the order in which you add edges in Kruskal and Prim.

Minimum spanning tree in blue: weight 13.

Travelling salesman problem

Name	Formula	Condition
Upper bound	Nearest neighbour: go to the nearest unvisited vertex, then return to the start
Lower bound	Delete a vertex. Minimum spanning tree of the rest + the 2 shortest edges from the deleted vertex

Connection: Lower bound $\leq$ optimal route $\leq$ upper bound
Connection: Best upper bound: the smallest.
Best lower bound: the largest.
Exam note: Nearest neighbour: start at the vertex the question gives.
Exam note: TSP needs a table of least distances first if the graph is not complete.

Upper bound
Nearest neighbour from

A

A B C D A

5 + 4 + 9 + 6 = 24

Lower bound: delete

A

Blue: minimum spanning tree
of

B

C

D

4 + 7 = 11

Gold: 2 shortest from

A

11 + 5 + 6 = 22

22 \leq

optimal route

\leq 24

Back to topics

Topic 4 Statistics and Probability

Sampling and Data

Data and sampling methods

Name	Formula	Condition
Types of data	Discrete data is counted. Continuous data is measured.
Simple random	Every possible sample of that size is equally likely
Systematic	Every $k$ th member from a list	$k$ $= \frac{population}{sample size}$
Stratified	Each group in proportion: $\frac{group size}{population} \times sample size$
Quota	A fixed number from each group, not chosen at random
Convenience	The members easiest to reach

Connection: Random sampling reduces bias. Quota and convenience samples are not random.

Collecting data: reliability and validity

Name	Formula	Condition
Reliability tests	Test-retest, parallel forms	same results again?
Validity tests	Content, criterion-related	measures what it should?

Exam note: Questionnaires: clear questions, not leading.

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Descriptive Statistics

Mean and spread

Name	Formula	Condition
Mean	$\bar{x} = \frac{\sum f_{i} x_{i}}{n}$	$n = \sum f_{i}$
Grouped data	Mid-interval values as $x_{i}$ . Modal class: highest frequency.	gives an estimate
Standard deviation	$σ$ from the GDC. Variance $= σ^{2}$

Connection: Add $k$ to every value: $mean + k$ . $σ$ and IQR unchanged.
Connection: Multiply every value by $k$ : $mean \times k$ , $σ \times | k |$ , $σ^{2} \times k^{2}$
Exam note: Use $σ_{x}$ from the GDC for the standard deviation of the data.

Quartiles, box plots and histograms

Name	Formula	Condition
Interquartile range	$I Q R = Q_{3} - Q_{1}$
Outlier	below $Q_{1} - 1.5 \times I Q R$ or above $Q_{3} + 1.5 \times I Q R$
Cumulative frequency	read across at $\frac{n}{4}$ for $Q_{1}$ , $\frac{n}{2}$ for the median, $\frac{3 n}{4}$ for $Q_{3}$

Exam note: Remove an outlier only if it is an error.
Exam note: Histograms of continuous data have no gaps between bars.

Box plot data: 1, 3, 4, 6, 7, 8, 10, 11, 13, 14, 16. Read each value from the scale.

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Correlation and Regression

Video: Regression line (opens YouTube in a new tab)

Correlation

Name	Formula	Condition
Pearson’s $r$	$- 1 \leq r \leq 1$	linear correlation
Spearman’s $r_{s}$	First rank the data for each variable. Then find Pearson’s $r$ of the ranks.	tied ranks: use the average

Connection: $r$ close to $\pm 1$ : strong linear correlation. $r$ close to 0: weak or no linear correlation.
Connection: $r_{s} = 1$ : always increasing, not always linear. $r_{s}$ is less affected by outliers than $r$ .
Connection: Scatter diagram: plot each pair $(x, y)$ as a point. A line of best fit drawn by eye passes through the mean point $(\bar{x}, \bar{y})$ .
Exam note: Use $r$ for linear data and $r_{s}$ for non-linear but monotonic data.
Exam note: Correlation does not mean causation.

r

close to 1: strong positive

r

close to 0: no linear correlation

r_{s} = 1

: always increasing,

r < 1

Regression line

Name	Formula	Condition
Regression line of $y$ on $x$	$y = a x + b$ : predicts $y$ from $x$

Connection: For each increase of 1 in $x$ , $y$ changes by $a$ .
When $x = 0$ , $y = b$ .
Connection: The line passes through $(\bar{x}, \bar{y})$ .
Exam note: Interpolation is reliable when the correlation is strong ( $r$ is close to $\pm 1$ ). Extrapolation is not reliable.

The line passes through

(\bar{x}, \bar{y})

How well a model fits

Name	Formula	Condition
Sum of squared residuals	$S S_{r e s} = \sum (y_{i} - f (x_{i}))^{2}$	smaller is a better fit
Coefficient of determination	$R^{2}$ : the proportion of the variation in $y$ explained by the model	$0 \leq R^{2} \leq 1$

Connection: For a straight line model, $R^{2} = r^{2}$ .
Exam note: Non-linear regression on the GDC: exponential, power, quadratic, cubic, sine. Compare $R^{2}$ .

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Probability

Video: Basic theoretical probability (opens YouTube in a new tab)

Basic rules

Name	Formula	Condition
Probability of $A$	$P (A)$ $= \frac{n (A)}{n (U)}$ , equally likely outcomes
Complementary events	$P (A) + P (A') = 1$
Combined events	$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Mutually exclusive	$P (A \cap B) = 0$
Expected number	$n p$ for $n$ trials with probability $p$

Connection: $P (A)$ $= P (A \cap B) + P (A \cap B')$

P (A) + P (B)

counts

A \cap B

twice, so subtract

P (A \cap B)

once.

Conditional probability and independence

Name	Formula	Condition
Conditional	$P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$
Independent	$P (A \cap B) = P (A) P (B)$

Connection: Independent $⟺ P (A ∣ B)$ $= P (A)$
Connection: Mutually exclusive events with non-zero probabilities are never independent.
Exam note: “Given that” means conditional: divide by the probability of the given event.

Tree diagrams

Connection: Multiply along branches and add the end results.
Connection: A conditional probability from a tree: the branch you want over all branches that give the condition.
Exam note: Without replacement, the second-stage probabilities change.

P (B ∣ A)

= \frac{0.01 \times 0.95}{0.01 \times 0.95 + 0.99 \times 0.08}

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Binomial and Poisson Distributions

Discrete random variables

Name	Formula	Condition
Discrete distribution	$\sum P (X = x) = 1$ $E (X) = \sum x P (X = x)$

Connection: Fair game: $E (gain) = 0$

Binomial distribution

Name	Formula	Condition
Binomial	$X \sim B (n, p)$ $E (X) = n p$ $Var (X) = n p (1 - p)$

Connection: Conditions: fixed $n$ , two outcomes, constant $p$ , independent trials
Connection: $P (X \geq k)$ $= 1 - P (X \leq k - 1)$
$P (X > k)$ $= 1 - P (X \leq k)$
Exam note: GDC: pdf for $P (X = x)$ and cdf for $P (X \leq x)$ .
Exam note: “At least”, “more than” and “fewer than”: rewrite with $\leq$ before using the cdf.

X \sim B (6, 0.4)

P (X \geq 3)

= 1 - P (X \leq 2)

Poisson distribution

Name	Formula	Condition
Poisson	$X \sim P o (m)$ $E (X) = m$ $Var (X) = m$
Sum of Poissons	$X \sim P o (m_{1}), Y \sim P o (m_{2}) \Rightarrow X + Y \sim P o (m_{1} + m_{2})$	$X, Y$ independent

Connection: Conditions: events occur at random, independently, at a constant average rate
Connection: The mean scales with the interval: 3 per hour gives $P o (1.5)$ per 30 minutes.

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Normal and Linear Combinations

The normal distribution

Name	Formula	Condition
Normal	$X \sim N (μ, σ^{2})$ , $σ^{2} =$ variance
Standardised value	$z = \frac{x - μ}{σ}$

Exam note: $N (μ, σ^{2})$ gives the variance. Take the square root for $σ$ .
Exam note: Inverse normal: check which tail the GDC uses.

Within

1 σ

: about 68%. Within

2 σ

: about 95%. Within

3 σ

: about 99.7%.

Linear transformations and combinations

Name	Formula	Condition
Linear transformation	$E (a X + b) = a E (X) + b$ $Var (a X + b) = a^{2} Var (X)$
Linear combination	$E (a_{1} X_{1} \pm a_{2} X_{2}) = a_{1} E (X_{1}) \pm a_{2} E (X_{2})$ $Var (a_{1} X_{1} \pm a_{2} X_{2}) = a_{1}^{2} Var (X_{1}) + a_{2}^{2} Var (X_{2})$	$X_{1}, X_{2}$ independent

Connection: Variances always add, even for $X_{1} - X_{2}$ .
Connection: $X_{1} + X_{2}$ is not $2 X$ : $Var (X_{1} + X_{2})$ $= 2 σ^{2}$ and $Var (2 X)$ $= 4 σ^{2}$
Connection: A linear combination of independent normal variables is normal.

Sample means and unbiased estimates

Name	Formula	Condition
Sample mean	If $X \sim N (μ, σ^{2})$ , then $\bar{X} \sim N (μ, \frac{σ^{2}}{n})$ for any $n$
Unbiased estimate of the population mean $μ$	$\bar{x}$
Unbiased estimate of the population variance $σ^{2}$	$s_{n - 1}^{2} = \frac{n}{n - 1} s_{n}^{2}$

Connection: Central limit theorem: for large $n$ (often $n \geq 30$ ), $\bar{X}$ is approximately normal for any distribution.
Exam note: The GDC gives $s_{n - 1}$ as $s_{x}$ and $s_{n}$ as $σ_{x}$ .

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Hypothesis Tests and Errors

Setting up a test

Name	Formula	Condition
Hypotheses	$H_{0}$ : no effect or no change. $H_{1}$ : what you are testing for.
One-tailed, two-tailed	One-tailed: $H_{1}$ uses $<$ or $>$ . Two-tailed: $H_{1}$ uses $\neq$ .
Decision	$p$ -value $<$ significance level: reject $H_{0}$
Critical region	the values of the test statistic that lead to rejecting $H_{0}$

Exam note: Conclude in context: “There is sufficient evidence at the 5% level that…”
Exam note: Never say $H_{0}$ is proved. Say “do not reject $H_{0}$ ”.

Choosing the test

You are testing	Test	$H_{0}$	$H_{1}$	Use when
one mean, $σ$ known	$z$ -test	$μ = μ_{0}$	$μ \neq μ_{0}$	normal population or large $n$
one mean, $σ$ unknown	$t$ -test, $n - 1$ degrees of freedom	$μ = μ_{0}$	$μ \neq μ_{0}$	normal population
paired data	$t$ -test on the differences	$μ_{d} = 0$	$μ_{d} \neq 0$	same items measured twice, differences normal
two means	two-sample $t$ -test	$μ_{1} = μ_{2}$	$μ_{1} \neq μ_{2}$	independent samples from normal populations. Pooled: equal variances.
a proportion	binomial test, $B (n, p)$	$p = p_{0}$	$p \neq p_{0}$	successes in $n$ trials
a Poisson mean	Poisson test, $P o (m)$	$m = m_{0}$	$m \neq m_{0}$	events in an interval
correlation	test on Pearson’s $r$	$ρ = 0$	$ρ \neq 0$	bivariate normal data
independence	$χ^{2}$ test for independence	independent	not independent	a contingency table
a distribution	$χ^{2}$ goodness of fit	the data fit	the data do not fit	frequencies in classes

Type I and Type II errors

Decision	$H_{0}$ is true	$H_{0}$ is false
Reject $H_{0}$	Type I error	correct
Do not reject $H_{0}$	correct	Type II error

Connection: Type I: you reject $H_{0}$ , but $H_{0}$ was true.
Connection: Type II: you do not reject $H_{0}$ , but $H_{0}$ was false.

Probability	What it means
$P (Type I)$	landing in the critical region when $H_{0}$ is true
continuous test	$z$ , $t$ or normal: equals the significance level, for example 0.05
discrete test	binomial or Poisson: the values go up in whole numbers, so 5% is rarely exact. Take the critical region with probability as close to 5% as possible without going over. $P (Type I)$ is that probability.
$P (Type II)$	not landing in the critical region when $H_{1}$ is true. It needs a given value of the parameter, for example $p = 0.7$ .

Connection: Example: $X \sim B (20, 0.5)$ under $H_{0}$ , $H_{1}$ : $p > 0.5$ at 5%. $P (X \geq 14) = 0.0577$ is above 5%, but $P (X \geq 15) = 0.0207$ is not. So the critical region is $X \geq 15$ and $P (Type I)$ $= 0.0207$ .
Connection: Same test with $p = 0.7$ in fact: $P (Type II)$ $= P (X \leq 14)$ with $X \sim B (20, 0.7)$ , which is 0.584.
Exam note: Find the critical region first, then its probability.

Back to topics

Confidence Intervals and $χ^{2}$ Tests

Confidence intervals

Name	Formula	Condition
$σ$ known	$(\bar{x} - z \frac{σ}{\sqrt{n}}, \bar{x} + z \frac{σ}{\sqrt{n}})$	$z = 1.96$ for 95%
$σ$ unknown	$(\bar{x} - t \frac{s_{n - 1}}{\sqrt{n}}, \bar{x} + t \frac{s_{n - 1}}{\sqrt{n}})$	$t$ with $n - 1$ degrees of freedom

Connection: The interval is centred on $\bar{x}$ . The GDC gives both ends.
Connection: A larger sample gives a narrower interval: the width is proportional to $\frac{1}{\sqrt{n}}$ .
Connection: Higher confidence gives a wider interval.

$χ^{2}$ tests

Test	$H_{0}$	Expected frequency $f_{e}$	Degrees of freedom $ν$
Independence	the two variables are independent	$f_{e} = \frac{row total \times column total}{grand total}$	$ν = (r - 1) (c - 1)$ $r$ rows, $c$ columns
Goodness of fit	the data fit the given distribution	$f_{e} =$ total frequency $\times$ probability of the class	$ν =$ number of classes $- 1$ , then subtract 1 more for each parameter estimated from the data

Connection: $f_{o} =$ observed frequency: the counts in the data. $f_{e} =$ expected frequency: the counts you would expect if $H_{0}$ were true.
Connection: The GDC gives $χ_{c a l c}^{2}$ and the $p$ -value. Reject $H_{0}$ when the $p$ -value $<$ significance level.
Exam note: Every expected frequency must be at least 5. If not, combine classes and reduce $ν$ .

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Markov Chains

Transition matrices and steady states

Name	Formula	Condition
Transition matrix	$T$ : each column adds to 1 column = from, row = to
State after $n$ steps	$T^{n} 𝐬_{0} = 𝐬_{n}$ , $𝐬_{0} =$ initial state
Steady state	$T 𝐬 = 𝐬$ , entries of $𝐬$ add to 1

Connection: The steady state is the eigenvector of $T$ for $λ = 1$ , scaled so its entries add to 1.
Connection: Regular chain: some power of $T$ has all entries positive. Then $T^{n} 𝐬_{0}$ tends to the steady state for any $𝐬_{0}$ .
Connection: Long-term probabilities: the columns of $T^{n}$ tend to $𝐬$ .
Exam note: Check that each column of $T$ adds to 1.
Exam note: Steady state by hand: write $𝐬$ $= (\begin{array}{c} a \\ b \end{array})$ , then solve $T 𝐬 = 𝐬$ with $a + b = 1$ .

T

= (\begin{array}{cc} 0.9 & 0.2 \\ 0.1 & 0.8 \end{array})

Columns are “from”, rows are “to”. Steady state

(\begin{matrix} 2 / 3 \\ 1 / 3 \end{matrix})

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Topic 5 Calculus

Derivatives, Tangents and Normals

The derivative

Name	Formula	Condition
Derivative	$\frac{d y}{d x} = f' (x)$ : the gradient function and the rate of change of $y$ with respect to $x$
Power rule	$f (x) = a x^{n} \Rightarrow f' (x) = a n x^{n - 1}$
Second derivative	$\frac{d^{2} y}{d x^{2}} = f ″ (x)$ : the derivative of $f' (x)$

Connection: The derivative is the limit of the gradient of a chord as the chord shrinks.
Connection: Rewrite first: $\frac{1}{x^{2}} = x^{- 2}$ and $\sqrt{x} = x^{1 / 2}$

h \to 0

the chord turns into the tangent.

Tangents and normals

Name	Formula	Condition
Tangent at $(x_{1}, y_{1})$	$y - y_{1} = m (x - x_{1}), m = f' (x_{1})$
Normal at $(x_{1}, y_{1})$	$y - y_{1} = - \frac{1}{m} (x - x_{1}), m \neq 0$

Connection: The normal is perpendicular to the tangent.
Exam note: Find $y_{1}$ from $f (x_{1})$ , not from $f' (x_{1})$ .
Exam note: GDC: the derivative at a point and the tangent line can be checked.

y = x^{3} - 2 x

(1, - 1)

tangent

y = x - 2

, normal

y = - x

Back to topics

Differentiation Rules

Standard derivatives

$f (x)$	$f' (x)$
$\sin x$	$\cos x$
$\cos x$	$- \sin x$
$\tan x$	$\frac{1}{\cos^{2} x}$

$f (x)$	$f' (x)$
$e^{x}$	$e^{x}$
$\ln x$	$\frac{1}{x}$

Connection: $\frac{d}{d x} \sin (k x)$ $= k \cos (k x)$ and $\frac{d}{d x} e^{k x} = k e^{k x}$

Chain, product and quotient rules

Name	Formula	Condition
Chain rule	$y = g (u)$ , where $u = f (x) \Rightarrow \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$
Product rule	$y = u v \Rightarrow \frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$
Quotient rule	$y = \frac{u}{v} \Rightarrow \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$

Related rates

Name	Formula	Condition
Related rates	$\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$

Connection: Sphere: $\frac{d V}{d t}$ $= 4 π r^{2} \frac{d r}{d t}$
Exam note: A decreasing quantity has a negative rate.

Back to topics

Stationary Points and Optimisation

Shape of a graph

If	Then
$f' (x) > 0$	$f$ is increasing
$f' (x) < 0$	$f$ is decreasing
$f' (a) = 0$	stationary point at $x = a$
$f' (a) = 0$ and $f ″ (a) < 0$	local maximum at $x = a$
$f' (a) = 0$ and $f ″ (a) > 0$	local minimum at $x = a$
$f' (a) = 0$ and $f ″ (a) = 0$	the test fails: check the sign of $f'$ on each side of $a$
$f' (a) = 0$ , $f ″ (a) = 0$ and $f ″$ changes sign at $a$	horizontal (stationary) point of inflection at $x = a$
$f ″ (x) > 0$	concave up
$f ″ (x) < 0$	concave down
$f ″ (a) = 0$ and $f ″$ changes sign at $a$	point of inflection at $x = a$

Connection: Second derivative test: first solve $f' (x) = 0$ to find the stationary point $x = a$ . Then test the sign of $f ″ (a)$ .
Connection: $f$ turns where $f' = 0$ and $f'$ changes sign. $f$ has an inflection where $f'$ turns.
Connection: Horizontal (stationary) inflection: the tangent is horizontal, $f' = 0$ there. Non-stationary inflection: $f' \neq 0$ there.
Exam note: $f ″ (a) = 0$ alone does not prove an inflection. Check the sign change.

f (x) = x^{3} - 3 x

. Step 1:

f' (x) = 3 x^{2} - 3 = 0

gives the turning points

a = - 1

and

a = 1

.
Step 2: test each one with

f ″ (x) = 6 x

f

has an inflection where

f'

turns.

f (x) = x^{3} - 3 x^{2} + 4

f ″ (x) = 6 x - 6

: concave down for

x < 1

, concave up for

x > 1

, inflection at

(1, 2)

f (x) = x^{3}

: a horizontal point of inflection at the origin. The tangent there is horizontal.

Optimisation

Connection: Write the quantity in one variable, set the derivative to 0, then confirm max or min.
Exam note: Check the end points of the domain too.
Exam note: Give the value of the quantity as well as the $x$ where it occurs, if asked.

Back to topics

Integration

Video: Integration by substitution (opens YouTube in a new tab)

Standard integrals

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C, n \neq - 1

\int \frac{1}{x} d x = \ln | x | + C

\int \sin x d x = - \cos x + C

\int \cos x d x = \sin x + C

\int \frac{1}{\cos^{2} x} d x = \tan x + C

\int e^{x} d x = e^{x} + C

Connection: A boundary condition finds $C$ .
Connection: Integration reverses differentiation: check by differentiating.
Exam note: Add $+ C$ to every indefinite integral.

Recognising the form

Name	Formula	Condition
Linear inside	$\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C$	$F' = f$
By inspection	$\int \frac{g' (x)}{g (x)} d x = \ln \| g (x) \| + C$ $\int g' (x) e^{g (x)} d x = e^{g (x)} + C$

Substitution and definite integrals

Name	Formula	Condition
Substitution	$\int f (g (x)) g' (x) d x$ : let $u = g (x)$ , $d u = g' (x) d x$
Definite integral	$\int_{a}^{b} f (x) d x = F (b) - F (a)$	the GDC gives the value

Exam note: Substitution in a definite integral: change the limits too.

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Kinematics

Displacement, velocity and acceleration

Name	Formula	Condition
Velocity, acceleration	$v = \frac{d s}{d t}$ $a = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$
Going back	$s = \int v d t$ $v = \int a d t$ use the starting values to find $C$
Displacement from $t_{1}$ to $t_{2}$	$\int_{t_{1}}^{t_{2}} v (t) d t$
Distance travelled from $t_{1}$ to $t_{2}$	$\int_{t_{1}}^{t_{2}} \| v (t) \| d t$
Speed	$\| v \|$

v = 3 (t - 1) (t - 3)

0 \leq t \leq 4

: integrals

4

- 4

4

displacement

4 - 4 + 4 = 4

, distance

4 + 4 + 4 = 12

Moving along a straight line

Connection: $v > 0$ : moving in the positive direction.
$v < 0$ : moving in the negative direction.
Connection: At rest: $v = 0$ . Changes direction: $v$ changes sign.
Connection: Speeding up: $v$ and $a$ have the same sign. Slowing down: opposite signs.

Motion along a straight line:

s

is the displacement from 0.

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Areas and Volumes

Areas

Name	Formula	Condition
With the $x$ -axis	$A = \int_{a}^{b} \| y \| d x$ $A = \int_{a}^{b} y d x$ when $y > 0$
With the $y$ -axis	$A = \int_{a}^{b} \| x \| d y$
Between curves	$A = \int_{a}^{b} (f (x) - g (x)) d x, f (x) \geq g (x)$

Connection: Below the $x$ -axis, the integral is negative. Use $| y |$ or split at the roots.
Exam note: Area between curves: find where they meet for the limits.

y = x (x - 2) (x - 3)

: area

= \frac{8}{3} + \frac{5}{12}

, integral

= \frac{8}{3} - \frac{5}{12}

Volumes of revolution

Name	Formula	Condition
About the $x$ -axis	$V = \int_{a}^{b} π y^{2} d x$
About the $y$ -axis	$V = \int_{a}^{b} π x^{2} d y$

Connection: Square first: $\int π y^{2} d x \neq π {(\int y d x)}^{2}$
Connection: About the $y$ -axis: write $x^{2}$ in terms of $y$ , and use $y$ limits.

Each slice is a disc of area

π y^{2}

Trapezoidal rule

Name	Formula	Condition
Trapezoidal rule	$\int_{a}^{b} y d x \approx \frac{1}{2} h ((y_{0} + y_{n}) + 2 (y_{1} + y_{2} + \dots + y_{n - 1}))$ $h = \frac{b - a}{n}$

Connection: Concave up: the trapezoidal rule overestimates.
Concave down: it underestimates.
Connection: More strips give a better estimate.
Exam note: $n$ strips need $n + 1$ values of $y$ .

h = 0.5

: estimate 4.75, exact

\frac{14}{3} \approx 4.67

. Concave up, so an overestimate.

Back to topics

Differential Equations

Setting up and solving

Name	Formula	Condition
Separation of variables	$\frac{d y}{d x} = f (x) g (y) \Rightarrow \int \frac{1}{g (y)} d y = \int f (x) d x$
Exponential model	$\frac{d y}{d t} = k y \Rightarrow y = A e^{k t}$

Connection: “Rate proportional to $y$ ” means $\frac{d y}{d t} = k y$ .
Connection: An initial condition finds the constant.
Exam note: Add the constant before you divide or take exponentials.

Slope fields

Name	Formula	Condition
Slope field	At each point $(x, y)$ , a short line with gradient $f (x, y)$

Connection: A solution curve follows the slope field lines.

\frac{d y}{d x} = x + y

: the solution through

(0, 1)

follows the field.

Euler’s method

Name	Formula	Condition
Euler’s method	$y_{n + 1} = y_{n} + h \times f (x_{n}, y_{n})$ $x_{n + 1} = x_{n} + h,$ $h$ = step length
Coupled systems	$x_{n + 1} = x_{n} + h \times f_{1} (x_{n}, y_{n}, t_{n})$ $y_{n + 1} = y_{n} + h \times f_{2} (x_{n}, y_{n}, t_{n})$ $t_{n + 1} = t_{n} + h$

Connection: A smaller $h$ gives a better approximation.
Exam note: Coupled Euler: use the old $x_{n}$ and $y_{n}$ in both lines.
Exam note: Keep full accuracy at every step. Use the GDC table.

\frac{d y}{d x} = x + y

y (0) = 1

h = 0.5

Back to topics

Phase Portraits

Video: Coupled differential equations: algebraic solutions (opens YouTube in a new tab)

Coupled linear systems

Name	Formula	Condition
Coupled linear system	$\frac{d x}{d t} = a x + b y$ $\frac{d y}{d t} = c x + d y$ $\frac{d 𝐱}{d t} = (\begin{array}{cc} a & b \\ c & d \end{array}) 𝐱$
Exact solution	$𝐱 = A e^{λ_{1} t} 𝐩_{1} + B e^{λ_{2} t} 𝐩_{2}$	$λ_{1}$ , $λ_{2}$ distinct real $𝐩_{1}$ , $𝐩_{2}$ eigenvectors
Equilibrium point	$\frac{d x}{d t} = 0$ and $\frac{d y}{d t} = 0$

Eigenvalues and the phase portrait

Eigenvalues	Behaviour near the origin	Sketch
Real, both negative Example: $λ_{1} = - 1$ , $λ_{2} = - 3$	Stable: all paths move into the origin
Real, both positive Example: $λ_{1} = 2$ , $λ_{2} = 5$	Unstable: all paths move away
Real, opposite signs Example: $λ_{1} = - 2$ , $λ_{2} = 3$	Saddle point: in along $𝐩$ for $λ < 0$ , out along $𝐩$ for $λ > 0$
Complex $p \pm q i$ , $p < 0$ Example: $λ = - 1 \pm 2 i$	Stable spiral: paths spiral in
Complex $p \pm q i$ , $p > 0$ Example: $λ = 1 \pm 2 i$	Unstable spiral: paths spiral out
Imaginary $\pm q i$ Example: $λ = \pm 2 i$	Closed loops round the origin

Connection: Paths that start on an eigenvector stay on that line.
Connection: As $t \to \infty$ , the term with the largest $λ$ dominates. The path becomes parallel to its eigenvector.
Connection: Direction of rotation: at $(1, 0)$ , $\frac{d y}{d t} = c$ . $c > 0$ : anticlockwise. $c < 0$ : clockwise.
Exam note: Sketch: draw the eigenvector lines first, then add arrows.
Exam note: Non-linear systems: find every equilibrium point, then use Euler or the GDC for paths.

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Last updated: November 2026.

How to use these notes

Topic 1 Number and Algebra

1Numbers and Approximation

Standard form and accuracy

Percentage error

2Sequences and Series

Arithmetic and geometric sequences

Sums and sigma notation

3Financial Mathematics

Compound interest and depreciation

Loans and annuities

4Exponents and Logarithms

Exponent laws

Logarithms

5Complex Numbers

Cartesian form and the Argand diagram

Arguments on the axes

Polar and exponential forms

6Matrices

Matrix operations

Determinant and inverse

Systems of linear equations

7Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors

Diagonalisation and powers

Topic 2 Functions

1Straight Lines

Equations and gradients

2Functions, Composites and Inverses

Functions and composites

Inverse functions

3Transformations

Transformations of y=f(x)

4Polynomial and Variation Models

Quadratic models

Cubic models

Direct and inverse variation

5Exponential and Logistic Models

Exponential and logarithmic models

Logistic models

6Sinusoidal and Piecewise Models

Sinusoidal models

Piecewise models

7Modelling and Linearising Data

Linearising with logarithms

Choosing and using a model

Topic 3 Geometry and Trigonometry

13D Geometry and Triangles

Points in 3D

Solids

Right-angled triangles

Angle between a line and a plane

Any triangle

The ambiguous case of the sine rule

Elevation, depression and bearings

2Circles, Radians and the Unit Circle

Radians, arcs and sectors

The unit circle

Trig graphs and solving trig equations

3Voronoi Diagrams

Perpendicular bisectors

Voronoi diagrams

4Matrix Transformations

Transformation matrices

Columns of a matrix

Composites and area

5Vectors

Vector basics

Scalar product

Vector product

6Vector Lines and Kinematics

Vector lines

Kinematics

7Graph Theory

Graphs and degrees

Complete graphs and trees

Walks, trails, paths and cycles

Adjacency matrices

Transition matrices for random walks

8Graph Algorithms

Numbers and Approximation

Sequences and Series

Financial Mathematics

Exponents and Logarithms

Complex Numbers

Matrices

Eigenvalues and Eigenvectors

Straight Lines

Functions, Composites and Inverses

Transformations

Transformations of $y = f (x)$

Polynomial and Variation Models

Exponential and Logistic Models

Sinusoidal and Piecewise Models

Modelling and Linearising Data

3D Geometry and Triangles

Circles, Radians and the Unit Circle

Voronoi Diagrams

Matrix Transformations

Vectors

Vector Lines and Kinematics

Graph Theory

Graph Algorithms

Sampling and Data

Descriptive Statistics

Correlation and Regression

Probability

Binomial and Poisson Distributions

Normal and Linear Combinations

Hypothesis Tests and Errors

Confidence Intervals and $χ^{2}$ Tests

$χ^{2}$ tests

Markov Chains

Derivatives, Tangents and Normals

Differentiation Rules

Stationary Points and Optimisation

Integration

Kinematics

Areas and Volumes

Differential Equations

Phase Portraits