HL Applications & Interpretation Revision Notes

Unit 1 · Number & Algebra Unit 2 · Functions Unit 3 · Geometry & Trig Unit 4 · Stats & Probability Unit 5 · Calculus

Unit 1 · Number & Algebra

HL AI revision notes

›

Part 1

IB MATHEMATICS AI HL
UNIT 1: NUMBER & ALGEBRA
Comprehensive Notes (Part 1 of 2)

ibblueComplete Syllabus Coverage

SL 1.1 & 1.6: Scientific Notation, Approximation, Bounds & Percentage Error.
SL 1.2 & 1.3: Arithmetic and Geometric Sequences & Series.
AHL 1.9 & 1.10: Laws of Logarithms and Rational Exponents.
AHL 1.11: Infinite Convergent Geometric Series.

ibmathrevision.com

SECTION 1: APPROXIMATION & ERROR (SL)

Scientific Notation & Percentage Error (SL 1.1, 1.6) 1. Standard Form (Scientific Notation): Written as $a \times 10^{k}$ , where $1 \leq a < 10$ and $k$ is an integer. Useful for very large or small numbers.
2. Bounds: If a value $x$ is rounded to the nearest unit, the absolute error can be up to half that unit. The true value lies in the range: $Lower Bound \leq x < Upper Bound$ .
3. Percentage Error: Used to evaluate the accuracy of an estimation ( $v_{E}$ ) against an exact theoretical value ( $v_{A}$ ). Formula: $ϵ = | \frac{v_{A} - v_{E}}{v_{E}} | \times 100 %$

Worked Examples: Error & Bounds Example 1: Calculating Percentage Error
The exact area of a rectangle is $125.4 {cm}^{2}$ . A student measures it and estimates the area to be $120 {cm}^{2}$ . Calculate the percentage error.
$v_{A} = 125.4$ and $v_{E} = 120$ .
$ϵ = | \frac{125.4 - 120}{120} | \times 100 % = | \frac{5.4}{120} | \times 100 % = 0.045 \times 100 % = 4.5 %$ .
Example 2: Establishing Bounds
The length of a fence is given as $45$ m, correct to the nearest metre. Write down the upper and lower bounds.
Since it is to the nearest $1$ m, the maximum error is $\pm 0.5$ m.
Lower Bound = $44.5$ m, Upper Bound = $45.5$ m. ( $44.5 \leq L < 45.5$ ).

SECTION 2: EXPONENTS & LOGARITHMS (AHL)

Rational & Negative Exponents (AHL 1.10) HL students must simplify expressions involving rational (fractional) and negative exponents. The fundamental rules connecting radicals and reciprocals to exponents are: $x^{\frac{m}{n}} = \sqrt[n]{x^{m}} = (\sqrt[n]{x})^{m} and x^{- n} = \frac{1}{x^{n}}$

Worked Examples: Advanced Exponents Example 1: Numerical Evaluation
Evaluate $27^{\frac{2}{3}}$ without a calculator.
$27^{\frac{2}{3}} = (\sqrt[1]{27})^{2} = (3)^{2} = 9$ .
Example 2: Solving Exponential Equations
Solve the equation $x^{- \frac{1}{2}} = \frac{1}{5}$ for $x$ .
Using the negative exponent rule: $\frac{1}{x^{1 / 2}} = \frac{1}{5} ⟹ \frac{1}{\sqrt{x}} = \frac{1}{5}$ .
Therefore, $\sqrt{x} = 5$ . Squaring both sides gives $x = 25$ .

The Laws of Logarithms (AHL 1.9) Logarithms are the inverse operations to exponents. In IB AI HL, the base $a$ will almost exclusively be $10$ ( $\log x$ ) or $e$ (Natural Logarithm, $\ln x$ ). For any strictly positive values $x > 0$ and $y > 0$ :

Product Law: $\log_{a} (x y) = \log_{a} x + \log_{a} y$
Quotient Law: $\log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y$
Power Law: $\log_{a} (x^{m}) = m \log_{a} x$

Worked Examples: Logarithm Laws Example 1: Expanding and Condensing
Express $\ln (\frac{\sqrt{x}}{y^{3}})$ in terms of $\ln x$ and $\ln y$ .
Using the Quotient Law: $\ln (\sqrt{x}) - \ln (y^{3})$ .
Using rational exponents and the Power Law: $\frac{1}{2} \ln x - 3 \ln y$ .
Example 2: Solving Exponential Equations with Logarithms
Solve the equation $5 \times 3^{x} = 40$ exactly.
Isolate the base: $3^{x} = 8$ .
Apply the natural logarithm to both sides: $\ln (3^{x}) = \ln (8)$ .
Use the Power Law: $x \ln (3) = \ln (8) ⟹ x = \frac{\ln (8)}{\ln (3)}$ .

SECTION 3: SEQUENCES & SERIES (SL & AHL)

Arithmetic & Geometric Sequences (SL 1.2, 1.3) 1. Arithmetic Sequences: The terms change by a constant difference, $d$ .
$n$ -th term: $u_{n} = u_{1} + (n - 1) d$
Sum to $n$ terms ( $S_{n}$ ): $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ or $\frac{n}{2} (u_{1} + u_{n})$ .
2. Geometric Sequences: The terms change by a constant ratio, $r$ .
$n$ -th term: $u_{n} = u_{1} r^{n - 1}$
Sum to $n$ terms ( $S_{n}$ ): $S_{n} = \frac{u_{1} (r^{n} - 1)}{r - 1} = \frac{u_{1} (1 - r^{n})}{1 - r}$ .

Worked Examples: Finite Sequences Example 1: Arithmetic Sum
An arithmetic sequence has $u_{1} = 4$ and a common difference $d = 5$ . Find the sum of the first $15$ terms.
Using the sum formula: $S_{15} = \frac{15}{2} (2 (4) + (15 - 1) (5))$ .
$S_{15} = 7.5 (8 + 14 (5)) = 7.5 (8 + 70) = 7.5 (78) = 585$ .
Example 2: Geometric Term
A geometric sequence has $u_{1} = 3$ and $u_{2} = 12$ . Find the 5th term ( $u_{5}$ ).
Find $r$ : $r = \frac{12}{3} = 4$ .
Find $u_{5}$ : $u_{5} = 3 (4)^{5 - 1} = 3 (4)^{4} = 3 (256) = 768$ .

Infinite Geometric Series (AHL 1.11) If a geometric sequence has a common ratio $r$ that falls strictly between $- 1$ and $1$ ( $| r | < 1$ ), the terms decay towards zero. We can find the exact sum of an infinite number of terms (a convergent series). $S_{\infty} = \frac{u_{1}}{1 - r} for | r | < 1$

Worked Examples: Infinite Series Example 1: Finding the Sum to Infinity
An infinite geometric series is given by $18 + 6 + 2 + \frac{2}{3} + \dots$ Find its exact sum.
Find the common ratio ( $r$ ): $r = \frac{6}{18} = \frac{1}{3}$ .
Check convergence: Since $| 1 / 3 | < 1$ , the series converges.
Calculate $S_{\infty}$ : $S_{\infty} = \frac{18}{1 - 1 / 3} = \frac{18}{2 / 3} = 18 \times \frac{3}{2} = 27$ .
Example 2: Evaluating Sigma Notation to Infinity
Evaluate exactly: $\sum_{k = 1}^{\infty} 5 (0.2)^{k - 1}$
This is an infinite geometric series. The first term ( $k = 1$ ) is $u_{1} = 5 (0.2)^{0} = 5$ .
The common ratio is the base of the exponent: $r = 0.2$ .
Since $| 0.2 | < 1$ , $S_{\infty} = \frac{5}{1 - 0.2} = \frac{5}{0.8} = 6.25$ .

Part 2

IB MATHEMATICS AI HL
UNIT 1: NUMBER & ALGEBRA
Comprehensive Notes (Part 2 of 2)

ibblueComplete Syllabus Coverage

SL 1.4 & 1.7: Financial Math (Compound Interest, Depreciation, Amortization).
SL 1.8: Systems of Linear Equations using technology.
AHL 1.12 & 1.13: Complex Numbers (Cartesian, Polar, and Euler forms).
AHL 1.14 & 1.15: Matrices, Eigenvalues, Eigenvectors, and Diagonalization.

ibmathrevision.com

SECTION 4: FINANCE & LINEAR SYSTEMS (SL)

CG50 Tip: The TVM Solver Do not use the standard compound interest formula by hand! Go to MENU $\to$ Financial $\to$ Compound Interest (F2). N = total periods, I% = interest rate, PV = present value (negative if depositing), PMT = regular payments, FV = future value, P/Y & C/Y = periods/compounds per year.

Compound Interest & Amortization (SL 1.4, 1.7) Compound Interest: Interest is earned on both the initial principal and the accumulated interest. Formula: $F V = P V \times {(1 + \frac{r}{100 k})}^{k n}$ .
Amortization (Annuities): When dealing with loans or retirement funds, regular payments (PMT) are made alongside the accumulating compound interest. This must be solved using the GDC’s TVM solver.

Worked Examples: Financial Mathematics Example 1: Basic Compound Interest
$5000 is invested at $4 %$ p.a., compounded quarterly. Find the value after 5 years.
Using the formula: $F V = 5000 \times {(1 + \frac{4}{100 \times 4})}^{4 \times 5} = 5000 (1.01)^{20} \approx $ 6100.95$ .
Using GDC: N=20, I%=4, PV=-5000, PMT=0, P/Y=4, C/Y=4 $\to$ Solve FV.
Example 2: Loan Repayment (Amortization)
A $20,000 car loan is charged $6 %$ p.a. compounded monthly. The buyer wants to pay it off in 3 years with monthly payments. Calculate the monthly payment.
Using GDC: N=36 (months), I%=6, PV=20000 (loan received), FV=0 (paid off), P/Y=12, C/Y=12. Solve for PMT.
PMT $\approx - 608.44$ . The monthly payment is $608.44.

Systems of Linear Equations (SL 1.8) HL students must solve systems of up to $3$ linear equations with $3$ unknowns ( $x, y, z$ ). You should always use your GDC to solve these via matrices or the simultaneous equation solver.

Worked Examples: Linear Systems Example 1: Solving a 3x3 System
Solve the system:
$x + 2 y - z = 5$
$3 x - y + 2 z = 8$
$2 x + y + z = 7$
Using GDC: MENU $\to$ Equation $\to$ Simultaneous (F1) $\to$ Unknowns: 3 (F2).
Enter the matrix coefficients: $[1, 2, - 1, 5]$ , $[3, - 1, 2, 8]$ , $[2 - 4]$ .
Solve: $x = 1, y = 3, z = 2$ .

SECTION 5: ADVANCED COMPLEX NUMBERS (AHL)

Cartesian, Polar, and Euler Forms (AHL 1.12, 1.13) A complex number extends the number line into 2D space using $i = \sqrt{- 1}$ .
1. Cartesian Form: $z = a + b i$ (Real part $a$ , Imaginary part $b$ ).
2. Polar Form: $z = r (\cos θ + i \sin θ) = r cis θ$ (Modulus $r = \sqrt{a^{2} + b^{2}}$ , Argument $\tan θ = \frac{b}{a}$ ).
3. Euler (Exponential) Form: $z = r e^{i θ}$ .

Worked Examples: Complex Forms and Conversions Example 1: Cartesian to Polar/Euler Conversion
Convert $z = 1 + i$ into Euler form.
Find $r$ : $r = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$ .
Find $θ$ : $\tan θ = \frac{1}{1} = 1 ⟹ θ = \frac{π}{4}$ .
Write in Euler form: $z = \sqrt{2} e^{i \frac{π}{4}}$ .
Example 2: Polar to Cartesian Conversion
Convert $w = 4 cis (\frac{5 π}{6})$ into Cartesian form $a + b i$ .
Expand using sine and cosine: $w = 4 (\cos \frac{5 π}{6} + i \sin \frac{5 π}{6})$ .
Substitute exact trig values: $w = 4 (- \frac{\sqrt{3}}{2} + i \frac{1}{2}) = - 2 \sqrt{3} + 2 i$ .

Multiplying & Dividing in Euler Form (AHL 1.13) To multiply complex numbers in Euler form, multiply their moduli ( $r$ ) and add their arguments ( $θ$ ). To divide, divide their moduli and subtract their arguments. $z_{1} \times z_{2} = (r_{1} r_{2}) e^{i (θ_{1} + θ_{2})} and \frac{z_{1}}{z_{2}} = (\frac{r_{1}}{r_{2}}) e^{i (θ_{1} - θ_{2})}$

Worked Examples: Exponential Arithmetic Example 1: Multiplying and Dividing
Given $z_{1} = 4 e^{i \frac{π}{2}}$ and $z_{2} = 2 e^{i \frac{π}{6}}$ , find $z_{1} z_{2}$ and $\frac{z_{1}}{z_{2}}$ .
Product: $z_{1} z_{2} = (4 \times 2) e^{i (\frac{π}{2} + \frac{π}{6})} = 8 e^{i \frac{4 π}{6}} = 8 e^{i \frac{2 π}{3}}$ .
Quotient: $\frac{z_{1}}{z_{2}} = (\frac{4}{2}) e^{i (\frac{π}{2} - \frac{π}{6})} = 2 e^{i \frac{2 π}{6}} = 2 e^{i \frac{π}{3}}$ .

SECTION 6: MATRICES & MATRIX ALGEBRA (AHL)

Matrix Definition, Addition & Multiplication (AHL 1.14) A matrix is an array of numbers. Its size is described by its order: $m \times n$ (Rows $\times$ Columns). Addition is performed element-by-element. To multiply two matrices $A$ and $B$ , you multiply the Rows of the first matrix by the Columns of the second matrix.

Worked Examples: Matrix Algebra Example 1: Addition and Scalar Multiplication
Let $A = (\begin{matrix} 4 & 1 \\ 2 & 0 \end{matrix})$ and $B = (\begin{matrix} 1 & - 2 \\ 3 & 5 \end{matrix})$ . Find $2 A - B$ .
Scalar multiply $A$ : $2 A = (\begin{matrix} 8 & 2 \\ 4 & 0 \end{matrix})$ .
Subtract $B$ element-by-element: $(\begin{matrix} 8 - 1 & 2 - (- 2) \\ 4 - 3 & 0 - 5 \end{matrix}) = (\begin{matrix} 7 & 4 \\ 1 & - 5 \end{matrix})$ .
Example 2: Matrix Multiplication
Let $A = (\begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix})$ and $B = (\begin{matrix} 5 \\ - 1 \end{matrix})$ . Find $A B$ .
Top Row $\times$ Column: $(2 \times 5) + (3 \times - 1) = 10 - 3 = 7$ .
Bottom Row $\times$ Column: $(1 \times 5) + (4 \times - 1) = 5 - 4 = 1$ .
Resulting Matrix: $A B = (\begin{matrix} 7 \\ 1 \end{matrix})$ .

Finding Eigenvalues, Eigenvectors & Diagonalization (AHL 1.15) Eigenvalues ( $λ$ ): Satisfy the characteristic equation $det (A - λ I) = 0$ .
Eigenvectors ( $\vec{v}$ ): Satisfy the system $(A - λ I) \vec{v} = \vec{0}$ .
Diagonalization: A $2 \times 2$ matrix $A$ can be written as $A = P D P^{- 1}$ , where $P$ contains the eigenvectors and $D$ contains the eigenvalues diagonally.

Worked Examples: Eigenvalues & Diagonalization Example 1: Finding Eigenvalues
Find the exact eigenvalues of $A = (\begin{matrix} 4 & 2 \\ 1 & 3 \end{matrix})$ .
Set up characteristic equation: $| \begin{matrix} 4 - λ & 2 \\ 1 & 3 - λ \end{matrix} | = 0$ .
Calculate determinant: $(4 - λ) (3 - λ) - (2 \times 1) = 0$ .
Expand and Factorize: $λ^{2} - 7 λ + 10 = 0 ⟹ (λ - 5) (λ - 2) = 0$ .
The exact eigenvalues are $λ_{1} = 5$ and $λ_{2} = 2$ .
Example 2: Finding Eigenvectors and Diagonalizing
Find the eigenvector for $λ_{1} = 5$ and construct $P$ and $D$ . (Assume $λ_{2} = 2$ gives ${\vec{v}}_{2} = (\begin{matrix} - 1 \\ 1 \end{matrix})$ ).
Substitute $λ = 5$ : $(\begin{matrix} 4 - 5 & 2 \\ 1 & 3 - 5 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) ⟹ (\begin{matrix} - 1 & 2 \\ 1 & - 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ .
Create equation from top row: $- 1 x + 2 y = 0 ⟹ x = 2 y$ . Let $y = 1 ⟹ x = 2$ .
Eigenvector ${\vec{v}}_{1} = (\begin{matrix} 2 \\ 1 \end{matrix})$ .
Construct $P$ and $D$ : $P = (\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix})$ , $D = (\begin{matrix} 5 & 0 \\ 0 & 2 \end{matrix})$ .
Diagonalized Form: $A = (\begin{matrix} 2 & - 1 \\ 1 & 1 \end{matrix}) (\begin{matrix} 5 & 0 \\ 0 & 2 \end{matrix}) P^{- 1}$ .

Practice Unit 1 questions → See HL AI plans

Unit 2 · Functions

HL AI revision notes

›

Part 1

IB MATHEMATICS AI HL
UNIT 2: FUNCTIONS
Comprehensive Notes (Part 1 of 2)

ibblueComplete Syllabus Coverage

SL 2.1: Linear Functions, Gradients, and Equations of Lines.
SL 2.2: Concept of a function, Domain, Range, and Notation.
AHL 2.7: Composite Functions and Inverse Functions (with domain restriction).
SL 2.3 & 2.4: Key features of graphs (Intercepts, Asymptotes, and Extrema).

ibmathrevision.com

SECTION 1: LINEAR FUNCTIONS (SL 2.1)

Equations of Straight Lines & Gradients (SL 2.1) Linear models have a constant rate of change called the gradient ( $m$ ). $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ 1. Gradient-Intercept Form: $y = m x + c$ (where $c$ is the $y$ -intercept).
2. Point-Gradient Form: $y - y_{1} = m (x - x_{1})$ (used when a point and gradient are known).
3. General Form: $a x + b y + d = 0$ (where $a, b, d$ are integers).
Parallel Lines: Have the exact same gradient ( $m_{1} = m_{2}$ ).
Perpendicular Lines: Have negative reciprocal gradients ( $m_{1} \times m_{2} = - 1$ ).

Worked Examples: Linear Equations Example 1: Finding an Equation from Two Points
A straight line passes through $A (2, 5)$ and $B (6, 17)$ . Find its equation in the form $y = m x + c$ .
Find the gradient $m$ : $m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3$ .
Use point-gradient form with point $A$ : $y - 5 = 3 (x - 2)$ .
Expand and solve for $y$ : $y - 5 = 3 x - 6 ⟹ y = 3 x - 1$ .
Example 2: Perpendicular Lines
Line $L_{1}$ has equation $y = - \frac{1}{4} x + 2$ . Line $L_{2}$ is perpendicular to $L_{1}$ and passes through $(1, 3)$ . Find the equation of $L_{2}$ in general form $a x + b y + d = 0$ .
The gradient of $L_{1}$ is $- \frac{1}{4}$ . The perpendicular gradient is the negative reciprocal: $m_{2} = 4$ .
Use point $(1, 3)$ : $y - 3 = 4 (x - 1) ⟹ y - 3 = 4 x - 4 ⟹ y = 4 x - 1$ .
Rearrange to general form (integers): $4 x - y - 1 = 0$ .

SECTION 2: DOMAIN AND RANGE (SL 2.2)

Function Basics, Domain & Range (SL 2.2) A function maps each input ( $x$ ) to exactly one output ( $y$ ).
Domain: The set of all possible valid input ( $x$ ) values.
Range: The set of all possible output ( $y$ ) values.
To algebraically determine a domain restriction, check the DEN rule:
1. Denominators cannot equal zero.
2. Even roots (like square roots) must have non-negative arguments ( $\geq 0$ ).
3. Natural Logarithms ( $\ln x$ ) must have strictly positive arguments ( $> 0$ ).

Worked Examples: Domain and Range Restrictions Example 1: Denominator Restriction
Find the maximal domain of $f (x) = \frac{3}{2 x - 8}$ .
The denominator cannot be zero: $2 x - 8 \neq 0 ⟹ 2 x \neq 8 ⟹ x \neq 4$ .
Domain: $x \in R, x \neq 4$ .
Example 2: Square Root Restriction
Find the domain of $g (x) = \sqrt{10 - 2 x}$ .
The inside of the root must be $\geq 0$ : $10 - 2 x \geq 0 ⟹ 10 \geq 2 x ⟹ 5 \geq x$ .
Domain: $x \leq 5$ .
Example 3: Finding Range Algebraically
Find the range of the quadratic function $h (x) = x^{2} - 4 x + 7$ .
Find the $x$ -coordinate of the vertex: $x = - \frac{b}{2 a} = - \frac{- 4}{2 (1)} = 2$ .
Find the $y$ -coordinate: $h (2) = (2)^{2} - 4 (2) + 7 = 4 - 8 + 7 = 3$ .
Since $a > 0$ , the parabola opens upwards, making the vertex a minimum.
Range: $y \geq 3$ .

SECTION 3: COMPOSITE & INVERSE FUNCTIONS (AHL 2.7)

Composite Functions (AHL 2.7) A composite function applies one function to the result of another: $(f \circ g) (x) = f (g (x))$ . You work from the inside out (right to left).

Worked Examples: Composite Functions Example 1: Algebraic Composition
Let $f (x) = 2 x - 3$ and $g (x) = x^{2} + 1$ . Find an expression for $(f \circ g) (x)$ .
$(f \circ g) (x) = f (g (x)) = f (x^{2} + 1)$ .
Substitute $(x^{2} + 1)$ into $f (x)$ : $2 (x^{2} + 1) - 3 = 2 x^{2} + 2 - 3 = 2 x^{2} - 1$ .
Example 2: Evaluating Composites
Using the same functions, evaluate $(g \circ f) (4)$ .
First find $f (4)$ : $f (4) = 2 (4) - 3 = 8 - 3 = 5$ .
Substitute the result into $g (x)$ : $g (5) = 5^{2} + 1 = 25 + 1 = 26$ .

Inverse Functions & Domain Restriction (AHL 2.7) An inverse function $f^{- 1} (x)$ reverses the operation of $f (x)$ . The graph of $f^{- 1} (x)$ is a perfect reflection of $f (x)$ across the diagonal line $y = x$ .
Crucially, a function only has an inverse if it is one-to-one (passes the horizontal line test). If it is many-to-one (like a quadratic), its domain must be restricted before an inverse can be found.

Worked Examples: Inverse Functions Example 1: Finding an Inverse Function Algebraically
Let $f (x) = \frac{3 x + 1}{x - 2}$ . Find $f^{- 1} (x)$ .
Step 1: Set $y = f (x) ⟹ y = \frac{3 x + 1}{x - 2}$ .
Step 2: Swap $x$ and $y$ : $x = \frac{3 y + 1}{y - 2}$ .
Step 3: Multiply out the denominator: $x (y - 2) = 3 y + 1 ⟹ x y - 2 x = 3 y + 1$ .
Step 4: Collect all $y$ terms on one side: $x y - 3 y = 2 x + 1$ .
Step 5: Factor out $y$ : $y (x - 3) = 2 x + 1$ .
Step 6: Divide to isolate $y$ : $y = \frac{2 x + 1}{x - 3}$ .
Answer: $f^{- 1} (x) = \frac{2 x + 1}{x - 3}$ .
Example 2: Restricting the Domain
Explain why $h (x) = x^{2} - 6 x$ does not have an inverse on $x \in R$ , and state a restricted domain that would allow an inverse to exist.
$h (x)$ is a parabola, which fails the horizontal line test (it is a many-to-one function).
To make it one-to-one, we restrict it to one half of the parabola by finding the vertex:
Vertex $x = - b / 2 a = - (- 6) / 2 = 3$ .
A valid restricted domain is $x \geq 3$ (or $x \leq 3$ ).

CG50 Tip: Verifying Inverses Because $f (f^{- 1} (x)) = x$ , you can check your inverse function on a GDC! Enter the original function into Y1 and your calculated inverse into Y2. In Y3, enter Y1(Y2(x)). If you graphed it correctly, Y3 will be a perfectly straight diagonal line ( $y = x$ )!

Part 2

IB MATHEMATICS AI HL
UNIT 2: FUNCTIONS
Comprehensive Notes (Part 2 of 2)

ibblueComplete Syllabus Coverage

AHL 2.8: Transformations of Graphs (Translations, Stretches, and Reflections).
SL 2.5 & 2.6: Polynomial and Exponential Modelling Skills.
AHL 2.9: Logistic Models and Piecewise Functions.
AHL 2.10: Scaling Laws and Logarithmic Linearisation.

ibmathrevision.com

SECTION 4: TRANSFORMATIONS OF GRAPHS (AHL 2.8)

Translations, Stretches & Reflections (AHL 2.8) HL students must be able to perform graphical transformations on any generic function $y = f (x)$ .
1. Translations: $y = f (x - a) + b$ shifts the graph right by $a$ units, and up by $b$ units.
2. Stretches: $y = p f (x)$ stretches vertically by a scale factor of $p$ . $y = f (q x)$ stretches horizontally by a scale factor of $\frac{1}{q}$ .
3. Reflections: $y = - f (x)$ reflects across the $x$ -axis. $y = f (- x)$ reflects across the $y$ -axis.

Worked Examples: Graph Transformations Example 1: Identifying Transformations
Describe the full sequence of geometric transformations that maps $f (x) = x^{2}$ onto the graph $g (x) = - 3 (x - 4)^{2} + 5$ .
1. Translation: $4$ units RIGHT and $5$ units UP.
2. Stretch: A vertical stretch by a scale factor of $3$ .
3. Reflection: A reflection in the $x$ -axis (due to the negative sign in front of the $3$ ).
Example 2: Composite Coordinate Shifts
The point $(2, 6)$ lies on the curve $y = f (x)$ . Find the coordinates of the corresponding point on the curve $y = f (2 x) - 1$ .
The $f (2 x)$ implies a horizontal stretch by scale factor $\frac{1}{2}$ . The $x$ -coordinate halves: $2 \times \frac{1}{2} = 1$ .
The $- 1$ implies a vertical translation DOWN $1$ . The $y$ -coordinate drops: $6 - 1 = 5$ .
New coordinate: $(1, 5)$ .

SECTION 5: ADVANCED MODELLING (AHL 2.9)

Logistic Models (AHL 2.9) Populations rarely grow exponentially forever. They are often restricted by a "carrying capacity". The IB AI syllabus models this using the Logistic Function: $f (x) = \frac{L}{1 + C e^{- k x}}$

$L$ is the carrying capacity (the absolute maximum horizontal asymptote).
$C$ and $k$ are constants that dictate the curve’s starting point and growth rate.

Worked Examples: Logistic Modelling Example 1: Interpreting the Logistic Formula
The population of fish in a lake $t$ years after introduction is modelled by $P (t) = \frac{5000}{1 + 24 e^{- 0.3 t}}$ .
(a) Find the initial population.
Substitute $t = 0$ : $P (0) = \frac{5000}{1 + 24 e^{0}} = \frac{5000}{1 + 24 (1)} = \frac{5000}{25} = 200$ fish.
(b) State the carrying capacity.
The numerator dictates the maximum limit. As $t \to \infty$ , the term $e^{- 0.3 t} \to 0$ , leaving $\frac{5000}{1 + 0}$ .
The carrying capacity is exactly $5000$ fish.

Piecewise Functions (AHL 2.9) A piecewise function uses different algebraic rules for different parts of its domain. This is highly applicable for models like mobile phone tariffs or tiered income taxes.

Worked Examples: Piecewise Functions Example 1: Evaluating Piecewise Boundaries
A taxi fare $C (x)$ for travelling $x$ kilometres is given by:
$C (x) = {\begin{cases} 5 & for 0 \leq x \leq 2 \\ 5 + 1.5 (x - 2) & for x > 2 \end{cases}$
Calculate the cost of a $10$ km journey.
Because $10 > 2$ , we must use the second equation in the piecewise definition:
$C (10) = 5 + 1.5 (10 - 2) = 5 + 1.5 (8) = 5 + 12 = $ 17$ .

SECTION 6: LOGARITHMIC LINEARISATION (AHL 2.10)

Scaling Laws & Linearisation (AHL 2.10) In data science, curved relationships (like exponential or power models) are difficult to analyse. By taking the logarithm of both sides, HL students must convert these curves into straight lines!
1. Power Models ( $y = a x^{n}$ ):
$\log y = \log (a x^{n}) ⟹ \log y = \log a + n \log x$ .
If you plot $\log y$ against $\log x$ , the gradient is $n$ and the y-intercept is $\log a$ .
2. Exponential Models ( $y = a \times b^{x}$ ):
$\ln y = \ln (a \times b^{x}) ⟹ \ln y = \ln a + x \ln b$ .
If you plot $\ln y$ against $x$ , the gradient is $\ln b$ and the y-intercept is $\ln a$ .

Worked Examples: Linearising Data Example 1: Extracting Exponential Parameters from a Linear Graph
The mass of bacteria $M$ after $t$ hours is modelled by $M = a \times b^{t}$ .
A scientist plots a graph of $\ln M$ against $t$ . The graph forms a perfect straight line with a $y$ -intercept of $4$ and a gradient of $1.5$ . Find the exact values of the constants $a$ and $b$ .
Step 1: Write out the linearised formula.
$\ln M = \ln a + t \ln b$ .
Step 2: Match the graph features to the formula.
The $y$ -intercept is $\ln a$ . Therefore: $\ln a = 4 ⟹ a = e^{4}$ .
The gradient is $\ln b$ . Therefore: $\ln b = 1.5 ⟹ b = e^{1.5}$ .
Step 3: State the final model.
$M = e^{4} \times (e^{1.5})^{t} = e^{4} \times e^{1.5 t}$ .
Example 2: Formulating a Power Model
Variables $P$ and $Q$ obey a power law $P = k Q^{n}$ . When $\log_{10} P$ is plotted on the vertical axis against $\log_{10} Q$ on the horizontal axis, the line passes through $(2, 7)$ and $(5, 13)$ . Find $n$ .
Step 1: Linearise.
$\log P = \log k + n \log Q$ .
Here, the gradient of the line is exactly $n$ .
Step 2: Calculate the gradient.
$n = m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{13 - 7}{5 - 2} = \frac{6}{3} = 2$ .
The power $n$ is exactly $2$ .

Practice Unit 2 questions → See HL AI plans

Unit 3 · Geometry & Trig

HL AI revision notes

›

Part 1

IB MATHEMATICS AI HL
UNIT 3: GEOMETRY & TRIGONOMETRY
Comprehensive Notes (Part 1 of 2)

ibblueComplete Syllabus Coverage

SL 3.1 & 3.2: 3D Coordinate Geometry, Midpoints, Distance, and Right-Angled Trigonometry (SOH CAH TOA).
SL 3.3 & 3.4: Non-Right Trigonometry (Sine/Cosine Rules) and Circles (Degrees).
AHL 3.7: Radian Measure, Arc Length, and Sector Area in Radians.
AHL 3.8 & 3.9: The Unit Circle, Pythagorean Identity, and Trigonometric Functions.

ibmathrevision.com

SECTION 1: 3D GEOMETRY & SOH CAH TOA (SL 3.1, 3.2)

3D Coordinate Geometry (SL 3.1) In three-dimensional space, coordinates are given as $(x, y, z)$ .
1. Distance between two points $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ : $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}$ 2. Midpoint of a 3D line segment: $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$

Worked Examples: 3D Coordinates Example 1: Finding 3D Distance and Midpoint
Given points $A (1, 2, 3)$ and $B (4, - 2, 15)$ , find the exact distance $A B$ and the midpoint $M$ .
Distance: $A B = \sqrt{(4 - 1)^{2} + (- 2 - 2)^{2} + (15 - 3)^{2}} = \sqrt{3^{2} + (- 4)^{2} + 12^{2}}$ .
$A B = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ units.
Midpoint: $M = (\frac{1 + 4}{2}, \frac{2 - 2}{2}, \frac{3 + 15}{2}) = (2.5, 0, 9)$ .

Right-Angled Trigonometry (SL 3.2) For right-angled triangles, use SOH CAH TOA:
$\sin θ = \frac{Opposite}{Hypotenuse}, \cos θ = \frac{Adjacent}{Hypotenuse}, \tan θ = \frac{Opposite}{Adjacent}$
*Note: Angles of elevation are measured upwards from the horizontal, and angles of depression are measured downwards from the horizontal. They are alternate interior angles and are equal.*

Worked Examples: SOH CAH TOA Example 1: Angle of Elevation
A person stands $50$ m from the base of a building. The angle of elevation to the top of the building is $31^{\circ}$ . Calculate the height of the building.
We have the Adjacent ( $50$ ) and want the Opposite ( $h$ ). Use Tangent:
$\tan (31^{\circ}) = \frac{h}{50} ⟹ h = 50 \tan (31^{\circ}) \approx 30.0$ m.

SECTION 2: NON-RIGHT TRIG & RADIANS (SL 3.3, AHL 3.7)

The Sine Rule, Cosine Rule & Triangle Area (SL 3.3) Used for any triangle where the sides are $a, b, c$ and corresponding opposite angles are $A, B, C$ .
1. Cosine Rule: $c^{2} = a^{2} + b^{2} - 2 a b \cos C$ (Use when given 3 sides, or 2 sides + included angle).
2. Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ (Use when you have a known "matching pair" of a side and its opposite angle).
3. Area of a Triangle: $A r e a = \frac{1}{2} a b \sin C$ .

Worked Examples: Advanced Triangle Rules Example 1: Finding an Angle with the Cosine Rule
A triangle has sides $a = 5$ , $b = 7$ , and $c = 8$ . Find the largest angle.
The largest angle ( $C$ ) is opposite the largest side ( $c = 8$ ).
Rearranged Cosine Rule: $\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{5^{2} + 7^{2} - 8^{2}}{2 (5) (7)} = \frac{25 + 49 - 64}{70} = \frac{10}{70} = \frac{1}{7}$ .
$C = \cos^{- 1} (\frac{1}{7}) \approx {81.8}^{\circ}$ .
Example 2: Finding Area
Using the triangle above, calculate its exact area using $\sin C$ .
Since $\cos C = \frac{1}{7}$ , we use $\sin^{2} C + \cos^{2} C = 1 ⟹ \sin C = \sqrt{1 - (1 / 7)^{2}} = \sqrt{\frac{48}{49}} = \frac{4 \sqrt{3}}{7}$ .
Area $= \frac{1}{2} a b \sin C = \frac{1}{2} (5) (7) (\frac{4 \sqrt{3}}{7}) = 10 \sqrt{3}$ .

Radian Measure, Arcs & Sectors (AHL 3.7) HL students must work seamlessly in radians. $π radians = 180^{\circ}$ .
When the angle $θ$ is in radians, the formulas simplify significantly compared to degrees:
Arc Length: $l = r θ$
Sector Area: $A = \frac{1}{2} r^{2} θ$

Worked Examples: Radians Example 1: Conversions and Sector Area
A circle has a radius of $12$ cm. A sector is formed by an angle of $150^{\circ}$ .
(a) Convert the angle to radians exactly.
$150^{\circ} \times \frac{π}{180^{\circ}} = \frac{150 π}{180} = \frac{5 π}{6}$ radians.
(b) Calculate the exact area of the sector.
$A = \frac{1}{2} r^{2} θ = \frac{1}{2} (12^{2}) (\frac{5 π}{6}) = \frac{1}{2} (144) (\frac{5 π}{6}) = 72 (\frac{5 π}{6}) = 60 π {cm}^{2}$ .

SECTION 3: UNIT CIRCLE & TRIG FUNCTIONS (AHL 3.8, 3.9)

CG50 Tip: Radian vs Degree Mode A massive source of lost marks in HL is having your calculator in the wrong angle setting. Always press SHIFT $\to$ SETUP and check Angle. Use Deg for SL triangle geometry, but switch to Rad for Trig Functions, Vectors, and Calculus!

The Unit Circle & Identities (AHL 3.8) The unit circle has a radius of $1$ . For any point $(x, y)$ on the circle at an angle $θ$ :
$x = \cos θ$ and $y = \sin θ$ . $\tan θ = \frac{\sin θ}{\cos θ}$ .
By Pythagoras’ Theorem on the unit circle, we get the fundamental identity: $\cos^{2} θ + \sin^{2} θ = 1$

Worked Examples: The Unit Circle Example 1: Using the Pythagorean Identity
Given that $\sin θ = \frac{3}{5}$ and $\frac{π}{2} < θ < π$ (obtuse angle), find the exact value of $\cos θ$ .
Use $\cos^{2} θ + \sin^{2} θ = 1 ⟹ \cos^{2} θ + {(\frac{3}{5})}^{2} = 1 ⟹ \cos^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ .
$\cos θ = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$ .
Because $θ$ is in the second quadrant ( $\frac{π}{2} < θ < π$ ), cosine is negative. $\cos θ = - \frac{4}{5}$ .

Trigonometric Functions & Modelling (AHL 3.9) Periodic data (like tides, daylight hours, pendulums) is modelled by $f (t) = a \sin (b (t - c)) + d$ .
Principal Axis: $d = \frac{Max + Min}{2}$
Amplitude: $a = \frac{Max - Min}{2}$
Period: The time for one full cycle. $Period = \frac{2 π}{b}$ (if in radians) or $\frac{360^{\circ}}{b}$ (if in degrees).

Worked Examples: Trig Functions Example 1: Extracting Graph Parameters
The height of water in a harbour is given by $h (t) = 4.2 \sin (\frac{π}{6} t) + 6.8$ metres, where $t$ is hours after midnight.
(a) Find the maximum and minimum heights.
Max $= d + a = 6.8 + 4.2 = 11.0$ m.
Min $= d - a = 6.8 - 4.2 = 2.6$ m.
(b) Find the period of the tides.
$b = \frac{π}{6}$ . Period $= \frac{2 π}{b} = \frac{2 π}{π / 6} = 12$ hours.

Part 2

IB MATHEMATICS AI HL
UNIT 3: GEOMETRY & TRIGONOMETRY
Comprehensive Notes (Part 2 of 2)

ibblueComplete Syllabus Coverage

SL 3.5 & 3.6: Perpendicular Bisectors and Voronoi Diagrams (Site location and Nearest Neighbour).
AHL 3.10: Vectors, Magnitude, Scalar (Dot) Product, and Angles between Vectors.
AHL 3.11: Vector equation of a line ( $r = a + λ b$ ) in 2D and 3D.
AHL 3.12: Vector Kinematics (Position, Velocity, Constant/Variable Acceleration).

ibmathrevision.com

SECTION 4: VORONOI DIAGRAMS (SL 3.5, 3.6)

Perpendicular Bisectors & Voronoi Diagrams (SL 3.5, 3.6) A Voronoi diagram divides a plane into regions based on the distance to a specific set of points called "sites". Every point in a region is closest to the site within that region.
The boundary between two adjacent sites $A$ and $B$ is perfectly formed by their perpendicular bisector.
Steps to find a perpendicular bisector:
1. Find the midpoint $M$ of $A$ and $B$ .
2. Find the gradient $m$ of the line segment $A B$ .
3. Find the perpendicular gradient $m_{⊥} = - \frac{1}{m}$ .
4. Substitute $M$ and $m_{⊥}$ into $y - y_{1} = m_{⊥} (x - x_{1})$ .

Worked Examples: Voronoi Boundaries Example 1: Finding the Equation of a Boundary
Sites $A (2, 5)$ and $B (6, 1)$ dictate two adjacent Voronoi regions. Find the exact equation of the boundary separating them.
1. Midpoint: $M = (\frac{2 + 6}{2}, \frac{5 + 1}{2}) = (4, 3)$ .
2. Gradient $A B$ : $m = \frac{1 - 5}{6 - 2} = \frac{- 4}{4} = - 1$ .
3. Perpendicular Gradient: $m_{⊥} = \frac{- 1}{- 1} = 1$ .
4. Equation: $y - 3 = 1 (x - 4) ⟹ y = x - 1$ .

SECTION 5: VECTORS & THE DOT PRODUCT (AHL 3.10)

Vectors, Magnitude & Dot Product (AHL 3.10) A vector has both magnitude and direction. It is written as a column vector $v = (\begin{matrix} x \\ y \\ z \end{matrix})$ .
Magnitude (Length): $| v | = \sqrt{x^{2} + y^{2} + z^{2}}$ .
Unit Vector: A vector with a length of exactly $1$ . Found by calculating $\frac{1}{| v |} v$ .
Scalar (Dot) Product: $v \cdot w = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}$ .
Angle Between Vectors: Found using the formula $\cos θ = \frac{v \cdot w}{| v | | w |}$ .
*If the dot product is $0$ , the vectors are perfectly perpendicular ( $90^{\circ}$ ).*

Worked Examples: Vector Mathematics Example 1: Dot Product and Angles
Let $p = (\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix})$ and $q = (\begin{matrix} 4 \\ 5 \\ - 1 \end{matrix})$ . Find the angle between $p$ and $q$ .
1. Dot Product: $p \cdot q = (2) (4) + (- 1) (5) + (3) (- 1) = 8 - 5 - 3 = 0$ .
2. Conclusion: Since $p \cdot q = 0$ , $\cos θ = 0$ . The angle is exactly $90^{\circ}$ (perpendicular).
Example 2: Unit Vectors
Find a unit vector in the same direction as $v = (\begin{matrix} 3 \\ 0 \\ - 4 \end{matrix})$ .
Magnitude $| v | = \sqrt{3^{2} + 0^{2} + (- 4)^{2}} = \sqrt{9 + 16} = 5$ .
Unit Vector $= \frac{1}{5} (\begin{matrix} 3 \\ 0 \\ - 4 \end{matrix}) = (\begin{matrix} 0.6 \\ 0 \\ - 0.8 \end{matrix})$ .

SECTION 6: VECTOR LINES & KINEMATICS (AHL 3.11, 3.12)

Vector Equations of Lines (AHL 3.11) A straight line in 2D or 3D is defined by a starting position vector ( $a$ ) and a direction vector ( $b$ ). $r = a + t b$ Where $t$ is a scalar parameter (often representing time).
This can be split into parametric equations: $x = x_{0} + t b_{x}, y = y_{0} + t b_{y}, z = z_{0} + t b_{z}$ .

Worked Examples: Vector Lines Example 1: Intersecting Lines
Line $L_{1}$ has equation $r_{1} = (\begin{matrix} 2 \\ 5 \end{matrix}) + t (\begin{matrix} 1 \\ - 2 \end{matrix})$ . Line $L_{2}$ has equation $r_{2} = (\begin{matrix} - 4 \\ 2 \end{matrix}) + s (\begin{matrix} 3 \\ 1 \end{matrix})$ . Find their point of intersection.
Equate the $x$ and $y$ components to form simultaneous equations:
$x$ -eq: $2 + t = - 4 + 3 s ⟹ t - 3 s = - 6$
$y$ -eq: $5 - 2 t = 2 + s ⟹ - 2 t - s = - 3$
Solve via GDC or substitution. Let’s substitute $t = 3 s - 6$ into the $y$ -equation:
$- 2 (3 s - 6) - s = - 3 ⟹ - 6 s + 12 - s = - 3 ⟹ - 7 s = - 15 ⟹ s = \frac{15}{7}$ .
Substitute $s$ into $r_{2}$ to find the intersection coordinate:
$x = - 4 + 3 (15 / 7) = 17 / 7$ , and $y = 2 + (15 / 7) = 29 / 7$ . Point is $(17 / 7, 29 / 7)$ .

Vector Kinematics (AHL 3.12) In kinematics, the vector equation of a line perfectly models an object moving with constant velocity.
$r (t) = r_{0} + v t$

$r (t)$ is the position at time $t$ .
$r_{0}$ is the initial position (when $t = 0$ ).
$v$ is the velocity vector (direction and rate of movement).
Speed is the magnitude of the velocity vector: $Speed = | v |$ .

Worked Examples: Kinematics Example 1: Position, Velocity, and Speed
A ship leaves a port at $O (0, 0)$ and moves with a constant velocity vector $v = (\begin{matrix} 12 \\ - 5 \end{matrix})$ km/h.
(a) Find the speed of the ship.
Speed $= | v | = \sqrt{12^{2} + (- 5)^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13$ km/h.
(b) Find the position vector of the ship after 3 hours.
$r (3) = r_{0} + 3 v = (\begin{matrix} 0 \\ 0 \end{matrix}) + 3 (\begin{matrix} 12 \\ - 5 \end{matrix}) = (\begin{matrix} 36 \\ - 15 \end{matrix})$ .
Example 2: Closest Distance Between Two Moving Objects
If Object A is at $r_{A} (t) = (\begin{matrix} 0 \\ 10 \end{matrix}) + t (\begin{matrix} 4 \\ 0 \end{matrix})$ and Object B is at $r_{B} (t) = (\begin{matrix} 20 \\ 0 \end{matrix}) + t (\begin{matrix} 0 \\ 5 \end{matrix})$ , find the distance between them at $t = 2$ .
Find position of A at $t = 2$ : $r_{A} (2) = (8, 10)$ .
Find position of B at $t = 2$ : $r_{B} (2) = (20, 10)$ .
Distance $= \sqrt{(20 - 8)^{2} + (10 - 10)^{2}} = \sqrt{12^{2} + 0} = 12$ units.

Practice Unit 3 questions → See HL AI plans

Unit 4 · Stats & Probability

HL AI revision notes

›

Part 1

IB MATHEMATICS AI HL
UNIT 4: STATISTICS & PROBABILITY
Comprehensive Notes (Part 1 of 2)

ibblueComplete Syllabus Coverage

SL 4.1 & 4.3: Central Tendency, Dispersion, Outliers, and Sampling Methods.
AHL 4.12: Reliability vs Validity in Data Collection.
SL 4.5 & 4.6: Probability, Venn/Tree Diagrams, Conditional & Independent Events.
SL 4.7 & 4.8: Discrete Random Variables and the Binomial Distribution.
AHL 4.14: Linear Transformations of Random Variables ( $E (a X + b)$ & $V a r (a X + b)$ ).

ibmathrevision.com

SECTION 1: DATA ANALYSIS & SAMPLING (SL & AHL)

Sampling, Outliers & Data Quality (SL 4.1, AHL 4.12) 1. Sampling Methods: Simple Random, Systematic (every $k$ -th subject), Convenience, Quota, and Stratified (proportional to subgroup sizes).
2. Outliers: An outlier is formally defined as any data point that is more than $1.5 \times I Q R$ below the Lower Quartile ( $Q_{1}$ ) or above the Upper Quartile ( $Q_{3}$ ).
3. Reliability vs Validity (AHL):
Reliability refers to consistency (would a repeat test yield the same result?).
Validity refers to accuracy (does the test actually measure what it claims to measure?).

Worked Examples: Outliers & Sampling Example 1: Identifying Outliers
A dataset of test scores has a lower quartile of $45$ and an upper quartile of $71$ . A student scored $105$ . Is this score a mathematical outlier?
1. Find the IQR: $I Q R = Q_{3} - Q_{1} = 71 - 45 = 26$ .
2. Calculate Upper Boundary: $Q_{3} + 1.5 (I Q R) = 71 + 1.5 (26) = 71 + 39 = 110$ .
3. Conclusion: Since $105 < 110$ , the score is strictly within the boundary and is not an outlier.
Example 2: Stratified Sampling
A school has $400$ juniors and $600$ seniors. A sample of $50$ students is needed. How many juniors should be sampled using stratified sampling?
1. Total Population: $400 + 600 = 1000$ .
2. Junior Proportion: $\frac{400}{1000} \times 50 = 20$ juniors.

SECTION 2: PROBABILITY & EVENTS (SL 4.5, 4.6)

Probability Laws & Conditional Events (SL 4.6) 1. Mutually Exclusive Events: Cannot happen at the same time. $P (A \cap B) = 0$ .
Addition Rule simplifies to: $P (A \cup B) = P (A) + P (B)$ .
2. Independent Events: The outcome of one does not affect the other.
Multiplication Rule: $P (A \cap B) = P (A) \times P (B)$ .
3. Conditional Probability: The probability of $A$ given that $B$ has already occurred. $P (A | B) = \frac{P (A \cap B)}{P (B)}$

Worked Examples: Advanced Probability Example 1: Using Conditional Probability Formulas
Given that $P (A) = 0.6$ , $P (B) = 0.5$ , and $P (A \cup B) = 0.8$ , find $P (A | B)$ .
1. Find $P (A \cap B)$ : Use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
$0.8 = 0.6 + 0.5 - P (A \cap B) ⟹ P (A \cap B) = 1.1 - 0.8 = 0.3$ .
2. Find Conditional: $P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{0.3}{0.5} = 0.6$ .
*(Note: Because $P (A | B) = P (A)$ , events A and B are mathematically independent!)*

SECTION 3: RANDOM VARIABLES & TRANSFORMATIONS

CG50 Tip: Binomial Distributions To calculate binomial probabilities, go to MENU 2 (Stat) $\to$ DIST (F5) $\to$ BINOMIAL (F5). Use Bpd for an exact exact number of successes ( $P (X = x)$ ) and Bcd for cumulative inequalities ( $P (X \leq x)$ ).

Discrete Random Variables & Binomial Distribution (SL 4.7, 4.8) 1. Expected Value (Mean): For a discrete distribution, $E (X) = \sum x P (X = x)$ .
2. Binomial Distribution $X \sim B (n, p)$ : Used when there are a fixed number of independent trials ( $n$ ), two outcomes (success/fail), and a constant probability of success ( $p$ ).
Mean: $E (X) = n p Variance: V a r (X) = n p (1 - p)$ .

Worked Examples: The Binomial Distribution Example 1: Binomial Probabilities and Expected Value
A biased coin has a $0.6$ probability of landing on heads. It is flipped $20$ times.
(a) Find the probability of getting exactly $14$ heads.
Let $X \sim B (20, 0.6)$ . We need $P (X = 14)$ .
Using GDC (Bpd): $x = 14, Numtrial = 20, p = 0.6 ⟹ P (X = 14) \approx 0.124$ .
(b) Find the expected number of heads and the variance.
$E (X) = n p = 20 \times 0.6 = 12$ .
$V a r (X) = n p (1 - p) = 20 \times 0.6 \times 0.4 = 4.8$ .

Linear Transformations of Random Variables (AHL 4.14) HL students must be able to algebraically scale and shift random variables. If you multiply a random variable by $a$ and add $b$ :
Expected Value (Mean): Scales and shifts exactly as you would expect. $E (a X + b) = a E (X) + b$ Variance (Spread): Shifts ( $+ b$ ) do absolutely nothing to the spread of data. Multiplication scales the variance by the square of the factor. $V a r (a X + b) = a^{2} V a r (X)$

Worked Examples: Linear Transformations Example 1: Transforming Mean and Variance
A factory produces bags of flour. The weight of a bag, $X$ , has an expected value $E (X) = 500$ g and a variance $V a r (X) = 16$ g $^{2}$ . The factory changes the packaging process. The new weight $Y$ is modelled by the transformation $Y = 1.2 X - 5$ .
(a) Find the new expected weight, $E (Y)$ .
$E (Y) = E (1.2 X - 5) = 1.2 E (X) - 5$ .
$E (Y) = 1.2 (500) - 5 = 600 - 5 = 595$ g.
(b) Find the new variance and standard deviation of $Y$ .
$V a r (Y) = V a r (1.2 X - 5) = (1.2)^{2} V a r (X)$ . (The $- 5$ is ignored for spread!)
$V a r (Y) = 1.44 \times 16 = 23.04$ g $^{2}$ .
The standard deviation is $\sqrt{23.04} = 4.8$ g.

Part 2

IB MATHEMATICS AI HL
UNIT 4: STATISTICS & PROBABILITY
Comprehensive Notes (Part 2 of 2)

ibblueComplete Syllabus Coverage

SL 4.4 & 4.10: Pearson’s $r$ and Spearman’s Rank ( $r_{s}$ ).
AHL 4.13: Non-Linear Regression, $R^{2}$ , and Sum of Square Residuals ( $S S_{r e s}$ ).
SL 4.9 & AHL 4.17: Normal Distribution and the Poisson Distribution.
SL 4.11 & AHL 4.18: $χ^{2}$ Tests, t-Tests, Critical Regions, and $p$ -values.
AHL 4.16 & 4.19: Confidence Intervals and Markov Chains (Transition Matrices).

ibmathrevision.com

SECTION 4: BIVARIATE DATA & REGRESSION

Correlation & Non-Linear Regression (SL 4.4, 4.10, AHL 4.13) Pearson’s $r$ (SL 4.4): Measures the strength of a linear relationship.
Spearman’s Rank $r_{s}$ (SL 4.10): Measures monotonic (constantly increasing/decreasing) relationships. Evaluates the correlation of the ranks of the data. $r_{s} = 1 - \frac{6 \sum d^{2}}{n (n^{2} - 1)}$ Sum of Square Residuals, $S S_{r e s}$ (AHL 4.13): A residual is the vertical distance between an actual data point and the regression model. $S S_{r e s}$ is the sum of the squares of all these errors. The best model will minimize the $S S_{r e s}$ (Least Squares Regression).
Coefficient of Determination ( $R^{2}$ ): Gives the proportion of variability accounted for by the chosen model. An $R^{2}$ closer to $1$ implies a highly accurate model.

Worked Examples: Regression Analysis Example 1: Spearman’s Rank Calculations
Two judges rank 5 competitors (A to E). Judge 1 ranks them: 1, 2, 3, 4, 5. Judge 2 ranks them: 2, 1, 4, 3, 5. Find Spearman’s Rank manually.
1. Calculate differences squared ( $d^{2}$ ):
A: $1 - 2 = - 1 ⟹ d^{2} = 1$
B: $2 - 1 = 1 ⟹ d^{2} = 1$
C: $3 - 4 = - 1 ⟹ d^{2} = 1$
D: $4 - 3 = 1 ⟹ d^{2} = 1$
E: $5 - 5 = 0 ⟹ d^{2} = 0$
Total $\sum d^{2} = 4$ .
2. Formula: $r_{s} = 1 - \frac{6 (4)}{5 (25 - 1)} = 1 - \frac{24}{120} = 1 - 0.2 = 0.8$ .
Example 2: Evaluating $S S_{r e s}$
A quadratic model is proposed for a dataset yielding an $S S_{r e s}$ of 2.45. An exponential model is tested on the exact same dataset yielding an $S S_{r e s}$ of 0.82. Which model is a better fit?
The exponential model is a better fit because its Sum of Square Residuals is much smaller, indicating the predicted curve passes much closer to the true data points.

SECTION 5: NORMAL, POISSON & INTERVALS

Continuous & Discrete Distributions (SL 4.9, AHL 4.17) Normal Distribution $X \sim N (μ, σ^{2})$ : A continuous, symmetrical bell curve defined by mean $μ$ and standard deviation $σ$ . Find probabilities using the GDC (Norm CD or InvNorm).
Poisson Distribution $X \sim P o (m)$ (AHL 4.17): Models the number of independent events occurring in a fixed interval of time or space at a uniform average rate ( $m$ ).
Poisson Sums: If $X \sim P o (m_{1})$ and $Y \sim P o (m_{2})$ are independent, then their sum $X + Y$ strictly follows $P o (m_{1} + m_{2})$ .

Worked Examples: Distributions Example 1: Normal Probabilities
Bicycle stopping distances are modelled by $N (6.76, {0.12}^{2})$ metres. Find the probability a randomly chosen bicycle stops in less than $6.5$ m.
Using GDC (Norm CD): Lower = $- 1 \times 10^{99}$ , Upper = $6.5$ , $σ = 0.12$ , $μ = 6.76$ .
$P (X < 6.5) \approx 0.0151$ .
Example 2: Poisson Distribution Sums
The number of emails received per hour follows $X \sim P o (4)$ . The number of texts follows $Y \sim P o (6)$ . Find the probability of receiving exactly $12$ messages in total in one hour.
Let $T = X + Y$ . Therefore $T \sim P o (4 + 6) ⟹ T \sim P o (10)$ .
We need $P (T = 12)$ . Using GDC (Poisson PD): $λ = 10$ , $x = 12$ .
$P (T = 12) \approx 0.0948$ .

Confidence Intervals for the Mean (AHL 4.16) Used to estimate a population mean $μ$ from a sample mean $\bar{x}$ . Always use your GDC: STAT $\to$ INTR $\to$ Z or t.
Use the Z-interval when the true population standard deviation $σ$ is known.
Use the t-interval when $σ$ is unknown, regardless of sample size (the GDC uses the sample standard deviation $s_{n - 1}$ as an unbiased estimate).

SECTION 6: HYPOTHESIS TESTS & MARKOV CHAINS

Hypothesis Testing (SL 4.11, AHL 4.18) A hypothesis test compares a Null Hypothesis ( $H_{0}$ , default state) against an Alternative ( $H_{1}$ ).
If the calculated $p$ -value is less than the significance level (e.g. $0.05$ ), we Reject $H_{0}$ .
If the Test Statistic is greater than the Critical Value, we fall into the critical region and Reject $H_{0}$ .
Types of Tests:

$χ^{2}$ Test for Independence: Tests if two categorical variables are linked.
$χ^{2}$ Goodness of Fit: Tests if data follows a specific mathematical distribution.
t-Test for Means: Tests if a sample mean significantly differs from a population mean.

Worked Examples: Hypothesis Conclusions Example 1: Interpreting Test Output
A $χ^{2}$ test for independence evaluates gender and career choice at a $5 %$ significance level. The GDC gives $χ^{2} = 8.54$ and $p = 0.0736$ . The critical value is $9.49$ . State the conclusion.
Method 1 ( $p$ -value): Since $p = 0.0736 > 0.05$ , there is insufficient evidence to reject $H_{0}$ .
Method 2 (Critical Value): Since the test statistic $8.54$ is less than the critical value $9.49$ , it is not in the critical region. Do not reject $H_{0}$ .

Markov Chains & Transition Matrices (AHL 4.19) Markov chains model systems that transition between discrete states (e.g. Sunny vs Rainy) over time.
The state matrix after $n$ transitions is $s_{n} = T^{n} s_{0}$ , where $T$ is the transition matrix and $s_{0}$ is the initial state matrix.
Steady State: A regular Markov chain eventually settles into a long-term equilibrium probability called the steady state, $s$ . At this point, further transitions do not change the probabilities: $T s = s$

Worked Examples: Markov Chains Example 1: Setting up and finding the Steady State
Customers buy either Brand A or Brand B. If they buy A, the probability they buy A next week is $0.8$ . If they buy B, the probability they switch to A is $0.3$ .
1. Set up the Transition Matrix $T$ (Columns = "From", Rows = "To"):
$T = (\begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix})$ (Notice columns sum to 1).
2. Find the steady state algebraically:
Let $s = (\begin{matrix} a \\ b \end{matrix})$ . We know $T s = s$ and $a + b = 1 ⟹ b = 1 - a$ .
$(\begin{matrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{matrix}) (\begin{matrix} a \\ 1 - a \end{matrix}) = (\begin{matrix} a \\ 1 - a \end{matrix})$ .
Top row equation: $0.8 a + 0.3 (1 - a) = a$ .
$0.8 a + 0.3 - 0.3 a = a ⟹ 0.5 a + 0.3 = a ⟹ 0.3 = 0.5 a ⟹ a = 0.6$ .
Since $a = 0.6$ , $b = 0.4$ . The long-term steady state is $60 %$ Brand A and $40 %$ Brand B.

Practice Unit 4 questions → See HL AI plans

Unit 5 · Calculus

HL AI revision notes

›

Part 1

IB MATHEMATICS AI HL
UNIT 5: CALCULUS
Comprehensive Notes (Part 1 of 2)

ibblueComplete Syllabus Coverage

SL 5.1 & 5.3: Limits, Gradients, and the Power Rule for polynomials.
SL 5.4: Equations of Tangents and Normals.
AHL 5.9: Derivatives of $e^{x}, \ln x, \sin x, \cos x$ , and rational powers.
AHL 5.9: The Chain, Product, and Quotient Rules.
SL 5.6, 5.7 & AHL 5.10: Optimization, Second Derivatives, and Points of Inflexion.

ibmathrevision.com

SECTION 1: DIFFERENTIATION FUNDAMENTALS

The Power Rule, Tangents & Normals (SL 5.1, 5.3, 5.4) Calculus finds the instantaneous rate of change (gradient) of a curve.
The Power Rule: If $f (x) = a x^{n}$ , then the derivative is $f^{'} (x) = a n x^{n - 1}$ .
Tangents: A line that touches the curve at a point. Its gradient $m_{T}$ is exactly equal to $f^{'} (x)$ at that point.
Normals: A line perpendicular to the tangent. Its gradient is $m_{N} = - \frac{1}{m_{T}}$ .
Use $y - y_{1} = m (x - x_{1})$ to find the linear equations for both!

Worked Examples: Tangents and Normals Example 1: Using the Power Rule
Find the derivative of $f (x) = 4 x^{3} - \frac{2}{x^{2}} + 5 x$ .
First, rewrite with negative exponents: $f (x) = 4 x^{3} - 2 x^{- 2} + 5 x^{1}$ .
Apply the power rule: $f^{'} (x) = 12 x^{2} - (- 2) (2) x^{- 3} + 5 = 12 x^{2} + \frac{4}{x^{3}} + 5$ .
Example 2: Equation of a Normal Line
Find the equation of the normal to the curve $y = x^{2} - 4 x$ at the point where $x = 3$ .
1. Find the $y$ -coordinate: $y = (3)^{2} - 4 (3) = 9 - 12 = - 3$ . Point is $(3, - 3)$ .
2. Find the tangent gradient ( $m_{T}$ ): $f^{'} (x) = 2 x - 4$ . At $x = 3$ , $m_{T} = 2 (3) - 4 = 2$ .
3. Find the normal gradient ( $m_{N}$ ): $m_{N} = - \frac{1}{2}$ .
4. Equation: $y - (- 3) = - \frac{1}{2} (x - 3) ⟹ y + 3 = - 0.5 x + 1.5 ⟹ y = - 0.5 x - 1.5$ .

SECTION 2: ADVANCED RULES (AHL 5.9)

Chain, Product & Quotient Rules (AHL 5.9) At HL, you must differentiate composite, multiplied, and divided functions.
1. Chain Rule (Function inside a function): $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ .
2. Product Rule (Two functions multiplied): $(u v)^{'} = u^{'} v + u v^{'}$ .
3. Quotient Rule (Two functions divided): ${(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}$ .
Standard Derivatives: $(\sin x)^{'} = \cos x$ , $(\cos x)^{'} = - \sin x$ , $(e^{x})^{'} = e^{x}$ , $(\ln x)^{'} = \frac{1}{x}$ .

Worked Examples: Advanced Rules Example 1: The Chain Rule
Differentiate $y = \cos (x^{3} + 2 x)$ .
Let $u = x^{3} + 2 x ⟹ u^{'} = 3 x^{2} + 2$ .
The derivative of $\cos (u)$ is $- \sin (u)$ .
Using Chain Rule: $y^{'} = - (3 x^{2} + 2) \sin (x^{3} + 2 x)$ .
Example 2: The Product Rule
Differentiate $f (x) = x^{2} e^{3 x}$ .
Let $u = x^{2} ⟹ u^{'} = 2 x$ .
Let $v = e^{3 x} ⟹ v^{'} = 3 e^{3 x}$ (chain rule required here!).
Product Rule: $f^{'} (x) = u^{'} v + u v^{'} = (2 x) (e^{3 x}) + (x^{2}) (3 e^{3 x}) = e^{3 x} (2 x + 3 x^{2})$ .
Example 3: The Quotient Rule
Differentiate $g (x) = \frac{\ln x}{x}$ .
Let $u = \ln x ⟹ u^{'} = \frac{1}{x}$ .
Let $v = x ⟹ v^{'} = 1$ .
Quotient Rule: $g^{'} (x) = \frac{u^{'} v - u v^{'}}{v^{2}} = \frac{(\frac{1}{x}) (x) - (\ln x) (1)}{x^{2}} = \frac{1 - \ln x}{x^{2}}$ .

SECTION 3: OPTIMIZATION & CONCAVITY

CG50 Tip: Finding Max/Min Graphically To check your algebraic optimization, go to MENU 5 (Graph), plot the function, and press F5 (G-Solv). Press F2 (MAX) or F3 (MIN) to instantly verify the exact coordinates of the turning points!

Optimization & The Second Derivative (SL 5.6, AHL 5.10) 1. Turning Points (SL 5.6): Local maximums and minimums occur when the gradient is strictly zero: $f^{'} (x) = 0$ .
2. The Second Derivative (AHL 5.10): $f^{″} (x)$ measures the rate of change of the gradient, defining the curve’s concavity.
If $f^{″} (x) > 0$ , the curve is concave up ( $\cup$ ), meaning a turning point is a Minimum.
If $f^{″} (x) < 0$ , the curve is concave down ( $\cap$ ), meaning a turning point is a Maximum.
3. Point of Inflexion: Occurs where the concavity changes, requiring $f^{″} (x) = 0$ (and a confirmed sign change either side).

Worked Examples: Optimization & Concavity Example 1: Classifying Turning Points
Given $f (x) = 2 x^{3} - 9 x^{2} + 12 x$ , find the coordinates of the turning points and classify them using the second derivative test.
1. Set first derivative to zero: $f^{'} (x) = 6 x^{2} - 18 x + 12 = 0$ .
Divide by 6: $x^{2} - 3 x + 2 = 0 ⟹ (x - 1) (x - 2) = 0$ .
Turning points are at $x = 1$ and $x = 2$ .
2. Find $y$ -coordinates: $f (1) = 2 - 9 + 12 = 5 ⟹ (1, 5)$ .
$f (2) = 16 - 36 + 24 = 4 ⟹ (2, 4)$ .
3. Find second derivative: $f^{″} (x) = 12 x - 18$ .
4. Classify:
At $x = 1$ , $f^{″} (1) = 12 - 18 = - 6$ . Since $f^{″} (x) < 0$ , $(1, 5)$ is a Local Maximum.
At $x = 2$ , $f^{″} (2) = 24 - 18 = 6$ . Since $f^{″} (x) > 0$ , $(2, 4)$ is a Local Minimum.
Example 2: Finding Points of Inflexion
Find the coordinates of the point of inflexion for $f (x)$ from Example 1.
1. Set second derivative to zero: $f^{″} (x) = 12 x - 18 = 0$ .
$12 x = 18 ⟹ x = 1.5$ .
2. Find $y$ -coordinate: $f (1.5) = 2 (1.5)^{3} - 9 (1.5)^{2} + 12 (1.5) = 4.5$ .
The point of inflexion is at $(1.5, 4.5)$ .

Part 2

IB MATHEMATICS AI HL
UNIT 5: CALCULUS
Comprehensive Notes (Part 2 of 2)

ibblueComplete Syllabus Coverage

SL 5.5, 5.8 & AHL 5.11: Definite Integrals, Area, Substitution, and the Trapezoidal Rule.
AHL 5.13: Kinematics (Displacement, Velocity, Acceleration, and Total Distance).
AHL 5.14 - 5.18: Differential Equations (Separation of Variables, Integrating Factors, Euler’s Method, and Phase Portraits).
AHL 5.19: Maclaurin Series Expansions.

ibmathrevision.com

SECTION 4: INTEGRATION & AREA

Integration Rules, Area & Substitution (SL 5.5, AHL 5.11) Integration is the reverse process of differentiation. The arbitrary constant $+ C$ must be included for indefinite integrals.
Standard Rules: $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ . $\int e^{x} d x = e^{x} + C$ . $\int \frac{1}{x} d x = \ln | x | + C$ .
Area: The definite integral $\int_{a}^{b} f (x) d x$ calculates the area bounded by the curve and the $x$ -axis. Area between two curves is $\int_{a}^{b} (Upper - Lower) d x$ .
Integration by Substitution: Used when a function and its derivative are both present. Let $u$ equal the "inner" function.

Worked Examples: Integration Example 1: Definite Integrals & Area
Find the exact area enclosed by the curve $y = 3 x^{2} - 2$ and the $x$ -axis between $x = 1$ and $x = 3$ .
Area $= \int_{1}^{3} (3 x^{2} - 2) d x$ .
Find the antiderivative: $[x^{3} - 2 x]_{1}^{3}$ .
Evaluate at bounds: $((3)^{3} - 2 (3)) - ((1)^{3} - 2 (1)) = (27 - 6) - (1 - 2) = 21 - (- 1) = 22$ .
Example 2: Integration by Substitution
Evaluate the indefinite integral $\int 2 x e^{x^{2}} d x$ .
Let $u = x^{2}$ . Then $\frac{d u}{d x} = 2 x ⟹ d x = \frac{d u}{2 x}$ .
Substitute $u$ and $d x$ : $\int 2 x e^{u} \frac{d u}{2 x}$ .
The $2 x$ terms cancel: $\int e^{u} d u = e^{u} + C$ .
Substitute back for $x$ : $e^{x^{2}} + C$ .

SECTION 5: KINEMATICS (AHL 5.13)

Displacement, Velocity & Acceleration (AHL 5.13) Calculus links the three kinematic vectors:
To move down the chain (Differentiate): $v = \frac{d s}{d t}$ and $a = \frac{d v}{d t}$ .
To move up the chain (Integrate): $v = \int a d t$ and $s = \int v d t$ .
Total Distance Travelled: Evaluated as $\int_{t_{1}}^{t_{2}} | v (t) | d t$ . Absolute value brackets are essential because an object may change direction (where $v (t) = 0$ ).

Worked Examples: Kinematics Example 1: Integrating Acceleration
A particle starts from rest ( $v = 0, t = 0$ ). Its acceleration is $a (t) = 6 t$ . Find its velocity $v (t)$ .
$v (t) = \int 6 t d t = \frac{6 t^{2}}{2} + C = 3 t^{2} + C$ .
Use initial condition $v (0) = 0 ⟹ 0 = 3 (0)^{2} + C ⟹ C = 0$ .
Velocity is $v (t) = 3 t^{2}$ .
Example 2: Total Distance Travelled (With Direction Change)
A particle has velocity $v (t) = 2 t - 4$ . Find the total distance travelled from $t = 0$ to $t = 4$ .
The particle stops when $v (t) = 0 ⟹ 2 t - 4 = 0 ⟹ t = 2$ .
We must split the integral: Distance $= | \int_{0}^{2} (2 t - 4) d t | + | \int_{2}^{4} (2 t - 4) d t |$ .
$s (t) = [t^{2} - 4 t]$ .
$s (0) = 0$ . $s (2) = 4 - 8 = - 4$ . $s (4) = 16 - 16 = 0$ .
Distance $= | - 4 - 0 | + | 0 - (- 4) | = 4 + 4 = 8$ metres.

SECTION 6: DIFFERENTIAL EQUATIONS (AHL 5.14 - 5.18)

Differential Equations & Integrating Factors (AHL 5.15) 1. Separation of Variables: Get all $y$ ’s and $d y$ on one side, and $x$ ’s and $d x$ on the other, then integrate both sides.
2. Integrating Factor (IF): Used for linear DEs in the form $\frac{d y}{d x} + P (x) y = Q (x)$ .
The Integrating Factor is $I (x) = e^{\int P (x) d x}$ . Multiply the entire equation by this factor to collapse the left hand side into a single product-rule derivative.

Worked Examples: Solving DEs Example 1: Separation of Variables
Solve $\frac{d y}{d x} = 2 x y$ given $y (0) = 3$ .
Separate: $\frac{1}{y} d y = 2 x d x$ .
Integrate: $\int \frac{1}{y} d y = \int 2 x d x ⟹ \ln | y | = x^{2} + C$ .
Convert to exponential: $y = e^{x^{2} + C} = A e^{x^{2}}$ .
Use initial condition $y (0) = 3 ⟹ 3 = A e^{0} ⟹ A = 3$ .
Solution: $y = 3 e^{x^{2}}$ .
Example 2: Euler’s Method (AHL 5.16)
Given $\frac{d y}{d x} = x + y$ , $y (0) = 1$ , use Euler’s method with step $h = 0.1$ to find $y (0.1)$ .
Formula: $y_{1} = y_{0} + h \times f (x_{0}, y_{0})$ .
$y_{1} = 1 + 0.1 \times (0 + 1) = 1 + 0.1 = 1.1$ .

SECTION 7: MACLAURIN SERIES (AHL 5.19)

Maclaurin Series Expansions (AHL 5.19) A Maclaurin series approximates any function as an infinite polynomial centered at $x = 0$ . $f (x) = f (0) + x f^{'} (0) + \frac{x^{2}}{2!} f^{″} (0) + \frac{x^{3}}{3!} f^{‴} (0) + \dots$ HL students must generate these manually, or construct them by substituting into the standard expansions (for $e^{x}, \sin x, \cos x, \ln (1 + x)$ ) given in the formula booklet.

Worked Examples: Maclaurin Series Example 1: Manual Generation
Find the first three non-zero terms of the Maclaurin series for $f (x) = e^{2 x}$ .
$f (x) = e^{2 x} ⟹ f (0) = 1$ .
$f^{'} (x) = 2 e^{2 x} ⟹ f^{'} (0) = 2$ .
$f^{″} (x) = 4 e^{2 x} ⟹ f^{″} (0) = 4$ .
Substitute into formula: $f (x) = 1 + x (2) + \frac{x^{2}}{2} (4) = 1 + 2 x + 2 x^{2}$ .
Example 2: Substitution Method
The formula booklet gives $\sin x \approx x - \frac{x^{3}}{3!} + \dots$
Find the series for $\sin (x^{2})$ .
Substitute $x^{2}$ for $x$ : $\sin (x^{2}) \approx (x^{2}) - \frac{(x^{2})^{3}}{6} = x^{2} - \frac{x^{6}}{6}$ .

Practice Unit 5 questions → See HL AI plans

Summer Revision — €20, everything included

HL Applications & Interpretation Revision Notes

Unit 1 · Number & Algebra

Unit 2 · Functions

Unit 3 · Geometry & Trig

Unit 4 · Stats & Probability

Unit 5 · Calculus