HL Analysis & Approaches 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 AA revision notes

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Part 1

IB MATHEMATICS AA HL
UNIT 1 REVISION NOTES (PART 1)
Exhaustive Core Algebra, Sequences & Binomial Theorem

Unit 1 Part 1 Overview: Real-Valued Algebra

Scientific Notation & Logs: Operations with $a \times 10^{k}$ , laws of exponents, natural logarithms ( $\ln x$ ), and solving exponential equations.
Sequences & Finance: Arithmetic and geometric sequences/series, sigma notation, infinite convergent series, and compound interest/depreciation.
Counting & Polynomials: Permutations ( $^{n} P_{r}$ ), combinations ( $^{n} C_{r}$ ), and the standard Binomial Theorem for positive integer powers.
AHL Extensions: Partial fractions decomposition, and the Extended Binomial Theorem for fractional and negative indices.

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SECTION 1: Scientific Notation, Exponents & Logs

1.1 Scientific Notation Scientific notation allows us to express very large or very small numbers compactly: $a \times 10^{k}$ Where $1 \leq a < 10$ and $k \in Z$ . Note that calculator notation (e.g., $5.2 E 30$ ) is never accepted in IB examinations; you must write it out fully as $5.2 \times 10^{30}$ .

Worked Example: Scientific Notation Calculations Given $x = 4.5 \times 10^{4}$ and $y = 3.0 \times 10^{5}$ , calculate the exact value of $x \times y$ . Give your answer in the form $a \times 10^{k}$ .
Solution:
Multiply the constants and add the exponents using exponent laws:
$x \times y = (4.5 \times 10^{4}) \times (3.0 \times 10^{5}) = (4.5 \times 3.0) \times 10^{4 + 5} = 13.5 \times 10^{9}$ .
Since $13.5$ is not strictly between 1 and 10, we must adjust the decimal and the exponent:
$13.5 \times 10^{9} = 1.35 \times 10^{10}$ .

1.2 Laws of Exponents and Logarithms Logarithms are the mathematical inverse operations of exponents. The fundamental relationship is: $a^{x} = b ⟺ x = \log_{a} b$ The natural logarithm uses the mathematical constant $e$ (Euler’s number) as its base: $\ln x = \log_{e} x$ .

Key Logarithm Laws:

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: $\log_{a} (x^{n}) = n \log_{a} x$

Worked Example: Solving Exponential Equations Solve the equation $e^{2 x} - 5 e^{x} + 4 = 0$ exactly.
Solution:
This is a "hidden quadratic" equation. Let $u = e^{x}$ .
Substitute $u$ into the equation: $u^{2} - 5 u + 4 = 0$ .
Factorise the quadratic: $(u - 1) (u - 4) = 0 ⟹ u = 1 or u = 4$ .
Substitute $e^{x}$ back in for $u$ :
$e^{x} = 1 ⟹ x = \ln 1 ⟹ x = 0$ .
$e^{x} = 4 ⟹ x = \ln 4$ .
The exact solutions are $x = 0$ and $x = \ln 4$ .

SECTION 2: Sequences, Series & Finance

2.1 Arithmetic & Geometric Sequences Arithmetic Sequences: A sequence where each term increases by a constant common difference, $d$ .

$n$ -th term: $u_{n} = u_{1} + (n - 1) d$
Sum to $n$ terms: $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$

Geometric Sequences: A sequence where each term is multiplied by a constant common ratio, $r$ .

$n$ -th term: $u_{n} = u_{1} r^{n - 1}$
Sum to $n$ terms: $S_{n} = \frac{u_{1} (r^{n} - 1)}{r - 1} = \frac{u_{1} (1 - r^{n})}{1 - r}$
Sum to infinity: $S_{\infty} = \frac{u_{1}}{1 - r}$ , provided $| r | < 1$ .

Worked Example: Infinite Geometric Series Find the exact sum to infinity of the geometric sequence $18, 6, 2, \frac{2}{3}, \dots$
Solution:
Identify the first term $u_{1} = 18$ .
Calculate the common ratio $r$ : $r = \frac{6}{18} = \frac{1}{3}$ .
Check for convergence: Since $| r | = \frac{1}{3} < 1$ , the series converges and the sum to infinity exists.
Substitute into the formula: $S_{\infty} = \frac{u_{1}}{1 - r} = \frac{18}{1 - \frac{1}{3}} = \frac{18}{2 / 3}$ .
$S_{\infty} = 18 \times \frac{3}{2} = 27$ .

2.2 Financial Mathematics Compound Interest Formula: $F V = P V \times {(1 + \frac{r}{100 k})}^{k n}$ Where $F V$ is Future Value, $P V$ is Present Value, $r$ is the nominal annual interest rate (%), $n$ is the number of years, and $k$ is the number of compounding periods per year.

Worked Example: Compound Interest Calculate the future value if $$ 5000$ is invested for $3$ years at a nominal annual interest rate of $4 %$ , compounded quarterly.
Solution:
Identify the variables: $P V = 5000$ , $r = 4$ , $k = 4$ (quarterly means 4 times a year), and $n = 3$ .
Substitute into the compound interest formula:
$F V = 5000 {(1 + \frac{4}{100 \times 4})}^{4 \times 3} = 5000 {(1 + \frac{4}{400})}^{12}$ .
$F V = 5000 (1.01)^{12}$ .
$F V = 5000 \times 1.126825 \approx $ 5634.13$ .

SECTION 3: Algebra & The Binomial Theorem

3.1 AHL Partial Fractions Rational algebraic expressions can be split into partial fractions. This is highly useful for simplifying expressions before utilizing the extended binomial theorem or calculating integrals. For up to two distinct linear factors in the denominator: $\frac{p x + q}{(a x + b) (c x + d)} \equiv \frac{A}{a x + b} + \frac{B}{c x + d}$

Worked Example: Partial Fraction Decomposition Express $\frac{2 x + 1}{x^{2} + x - 2}$ in partial fractions.
Solution:
First, factorise the denominator: $x^{2} + x - 2 = (x - 1) (x + 2)$ .
Set up the algebraic identity: $\frac{2 x + 1}{(x - 1) (x + 2)} \equiv \frac{A}{x - 1} + \frac{B}{x + 2}$ .
Multiply through by the common denominator $(x - 1) (x + 2)$ :
$2 x + 1 \equiv A (x + 2) + B (x - 1)$ .
To find $A$ , let $x = 1$ (which eliminates $B$ ):
$2 (1) + 1 = A (1 + 2) ⟹ 3 = 3 A ⟹ A = 1$ .
To find $B$ , let $x = - 2$ (which eliminates $A$ ):
$2 (- 2) + 1 = B (- 2 - 1) ⟹ - 3 = - 3 B ⟹ B = 1$ .
Therefore, the partial fraction decomposition is: $\frac{1}{x - 1} + \frac{1}{x + 2}$ .

3.2 Standard & Extended Binomial Theorem Standard Binomial Theorem ( $n \in Z^{+}$ ): Used for positive integer expansions. $(a + b)^{n} = a^{n} + (\binom{n}{1}) a^{n - 1} b + \dots + (\binom{n}{r}) a^{n - r} b^{r} + \dots + b^{n}$

AHL Extended Binomial Theorem ( $n \in Q$ ): For fractional or negative powers, the expansion is infinite and only mathematically valid for values where $| x | < 1$ : $(1 + x)^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots$

Worked Example: Extended Binomial Expansion Find the first three terms of the Maclaurin/binomial expansion of $(1 - 2 x)^{- 1 / 2}$ .
Solution:
Use the extended binomial formula with the exponent $n = - 1 / 2$ and substitute $(- 2 x)$ for $x$ .
$= 1 + (- \frac{1}{2}) (- 2 x) + \frac{(- \frac{1}{2}) (- \frac{3}{2})}{2!} (- 2 x)^{2}$
$= 1 + x + \frac{3 / 4}{2} (4 x^{2})$
$= 1 + x + \frac{3}{8} (4 x^{2}) = 1 + x + \frac{3}{2} x^{2}$ .

Part 2

IB MATHEMATICS AA HL
UNIT 1 REVISION NOTES (PART 2)
Exhaustive Complex Numbers, Formal Proof & Systems

Unit 1 Part 2 Overview: Abstract & Systems

AHL Complex Numbers: The number $i$ , Cartesian ( $a + b i$ ), Polar ( $r cis θ$ ), and Euler ( $r e^{i θ}$ ) forms. Representing complex numbers on the Argand diagram.
AHL Advanced Complex: De Moivre’s theorem for fractional and integer powers. Finding the $n$ -th roots of complex numbers, and complex conjugate root pairs.
AHL Formal Proof: Logical deduction, mathematical induction, proof by contradiction, and proof by counterexample.
AHL Systems of Equations: Algebraic elimination of $3 \times 3$ linear systems, dealing with general parameters for infinite solutions, and the geometric interpretation of intersecting 3D planes.

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SECTION 4: Complex Numbers

4.1 The Argand Diagram & Forms of Complex Numbers A complex number $z$ contains the imaginary unit $i$ , where $i^{2} = - 1$ . It can be expressed in three primary formats:

Cartesian Form: $z = a + b i$
Polar Form: $z = r (\cos θ + i \sin θ) = r cis θ$
Euler Form: $z = r e^{i θ}$

The Modulus is $r = | z | = \sqrt{a^{2} + b^{2}}$ , representing the physical distance from the origin on the complex plane.
The Argument $θ = \arg (z)$ is the angle measured counter-clockwise from the positive real axis.

Worked Example: Converting Complex Forms Express the Cartesian complex number $z = - 1 + i \sqrt{3}$ in Euler form ( $r e^{i θ}$ ).
Solution:
1. Find the modulus $r$ : $r = \sqrt{a^{2} + b^{2}} = \sqrt{(- 1)^{2} + (\sqrt{3})^{2}} = \sqrt{1 + 3} = 2$ .
2. Find the argument $θ$ : The point $(- 1, \sqrt{3})$ lies in the 2nd quadrant of the Argand plane.
Calculate the acute reference angle $α = \arctan (\frac{\sqrt{3}}{1}) = \frac{π}{3}$ .
Because it’s in the 2nd quadrant, the argument is $θ = π - \frac{π}{3} = \frac{2 π}{3}$ .
3. State the Euler form: $z = 2 e^{i \frac{2 π}{3}}$ .

4.2 De Moivre’s Theorem & Roots De Moivre’s theorem provides a powerful way to raise complex numbers to integer and fractional powers. It states that for any complex number $z = r cis θ$ and any rational number $n \in Q$ : $[r (\cos θ + i \sin θ)]^{n} = r^{n} (\cos (n θ) + i \sin (n θ))$ Or in Euler form: $(r e^{i θ})^{n} = r^{n} e^{i n θ}$ .

Worked Example: Roots of a Complex Number Find the three cube roots of $z = 8 i$ . Express your answers in Euler form.
Solution:
1. Convert $8 i$ to polar/Euler form: Modulus $r = 8$ . Argument $θ = \frac{π}{2}$
$8 i = 8 cis (\frac{π}{2})$ 2. Set up the general formula for the $n$ -th roots:
$w_{k} = r^{1 / n} cis (\frac{θ + 2 π k}{n})$ for $k = 0, 1, 2, \dots, n - 1$ .
$w_{k} = 8^{1 / 3} cis (\frac{\frac{π}{2} + 2 π k}{3}) = 2 cis (\frac{π}{6} + \frac{2 π k}{3})$
3. Substitute $k = 0, 1, 2$ :
$k = 0 : w_{0} = 2 cis (\frac{π}{6}) = 2 e^{i π / 6}$
$k = 1 : w_{1} = 2 cis (\frac{π}{6} + \frac{4 π}{6}) = 2 cis (\frac{5 π}{6}) = 2 e^{i 5 π / 6}$
$k = 2 : w_{2} = 2 cis (\frac{π}{6} + \frac{8 π}{6}) = 2 cis (\frac{9 π}{6}) = 2 cis (\frac{3 π}{2}) = 2 e^{i 3 π / 2}$

SECTION 5: Formal Mathematical Proof

5.1 Proof by Mathematical Induction Mathematical Induction is a formal method used to logically prove that a statement $P (n)$ is true for all positive integers $n \in Z^{+}$ .

Base Case: Prove that the statement holds true for $n = 1$ .
Assumption: Assume that $P (k)$ is true for some positive integer $k$ .
Inductive Step: Using the assumption $P (k)$ , algebraically prove that the next term $P (k + 1)$ must also be true.
Conclusion Statement: "Since $P (1)$ is true, and $P (k) ⟹ P (k + 1)$ , by the principle of mathematical induction, $P (n)$ is true for all $n \in Z^{+}$ ."

Worked Example: Mathematical Induction Prove by mathematical induction that $\sum_{i = 1}^{n} i = \frac{n (n + 1)}{2}$ for all $n \in Z^{+}$ .
Solution:
Base Case ( $n = 1$ ): LHS $= 1$ . RHS $= \frac{1 (1 + 1)}{2} = \frac{2}{2} = 1$ . LHS = RHS, so the statement is true for $n = 1$ .
Assumption: Assume the statement is true for $n = k$ .
Therefore, $\sum_{i = 1}^{k} i = \frac{k (k + 1)}{2}$ .
Inductive Step ( $n = k + 1$ ):
We must prove that $\sum_{i = 1}^{k + 1} i = \frac{(k + 1) (k + 2)}{2}$ .
$\sum_{i = 1}^{k + 1} i = \sum_{i = 1}^{k} i + (k + 1)$
$= \frac{k (k + 1)}{2} + (k + 1)$ (Using our assumption)
$= (k + 1) (\frac{k}{2} + 1)$ (Factorising out the $(k + 1)$ )
$= (k + 1) (\frac{k + 2}{2}) = \frac{(k + 1) (k + 2)}{2}$ . This is exactly the formula for $n = k + 1$ .
Conclusion: Since it holds for $n = 1$ , and $P (k) ⟹ P (k + 1)$ , it is true for all $n \in Z^{+}$ .

5.2 Proof by Contradiction & Counterexamples Proof by Contradiction: Assume the opposite of what you are trying to prove is true. Use logical deduction to reach a mathematical impossibility (e.g., $1 = 0$ ). Conclude that the initial assumption was false, so the original statement is true.
Proof by Counterexample: Find one single case where a universal rule fails to show that the statement is false.

Worked Example: Irrationality of $\sqrt{2}$ Prove by contradiction that $\sqrt{2}$ is an irrational number.
Solution:
1. Assumption: Assume $\sqrt{2}$ is a rational number.
If it is rational, it can be written as a fraction in simplest form: $\sqrt{2} = \frac{p}{q}$ , where $p$ and $q$ share no common factors.
2. Deduction: Square both sides: $2 = \frac{p^{2}}{q^{2}} ⟹ p^{2} = 2 q^{2}$ .
Since $p^{2}$ is a multiple of 2, $p^{2}$ is an even number, meaning $p$ is also even.
Let $p = 2 k$ . Substitute back: $(2 k)^{2} = 2 q^{2} ⟹ 4 k^{2} = 2 q^{2} ⟹ 2 k^{2} = q^{2}$ .
This means $q^{2}$ is even, so $q$ is also even.
3. Contradiction: If $p$ is even and $q$ is even, they share a common factor of 2. This directly contradicts our initial assumption that $\frac{p}{q}$ was in simplest form.
4. Conclusion: Since the assumption led to a contradiction, $\sqrt{2}$ must be irrational.

SECTION 6: Systems of Linear Equations

6.1 Solving $3 \times 3$ Systems of Equations A system of three linear equations with three variables ( $x, y, z$ ) represents three planes in 3-dimensional space. $a_{1} x + b_{1} y + c_{1} z = d_{1}$ $a_{2} x + b_{2} y + c_{2} z = d_{2}$ $a_{3} x + b_{3} y + c_{3} z = d_{3}$

Geometric Interpretations of Solutions:

Unique Solution: You find exactly one value for $x, y,$ and $z$ . Geometrically, the planes intersect at a single distinct point.
Infinite Solutions: Algebraic elimination yields a statement like $0 = 0$ . The planes intersect along a common line. You must define a general parametric solution by letting $z = λ$ and solving for $x$ and $y$ in terms of $λ$ .
No Solution: Elimination yields a logical contradiction (e.g., $0 = 5$ ). Geometrically, planes are parallel or form a triangular prism with no common intersection.

Worked Example: Infinite Solutions & Parameters Consider the system of equations:
(1) $x + 2 y - z = 4$
(2) $2 x - y + 3 z = 3$
(3) $4 x + 3 y + z = 11$
Show that the system has an infinite number of solutions, and find the general solution in terms of a parameter $λ$ .
Solution:
Eliminate $y$ from (1) and (2). Multiply (2) by 2 and add it to (1):
$x + 2 y - z = 4$
$4 x - 2 y + 6 z = 6$
——————-
$5 x + 5 z = 10 ⟹ x + z = 2$ (Eq A)
Eliminate $y$ from (2) and (3). Multiply (2) by 3 and add it to (3):
$6 x - 3 y + 9 z = 9$
$4 x + 3 y + z = 11$
——————-
$10 x + 10 z = 20 ⟹ x + z = 2$ (Eq B)
Attempting to eliminate $x$ from (Eq A) and (Eq B) yields $0 = 0$ . Since this is a true mathematical statement, the planes intersect along a single line, creating an infinite number of solutions.
Parametric Solution:
Let $z = λ$ .
From (Eq A): $x + λ = 2 ⟹ x = 2 - λ$ .
Substitute $x$ and $z$ into Equation (1) to find $y$ :
$(2 - λ) + 2 y - λ = 4 ⟹ 2 y - 2 λ = 2 ⟹ 2 y = 2 + 2 λ ⟹ y = 1 + λ$ .
The general solution for the infinite points of intersection is: $(2 - λ, 1 + λ, λ)$ .

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Unit 2 · Functions

HL AA revision notes

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Part 1

IB MATHEMATICS AA HL
UNIT 2 REVISION NOTES (PART 1)
Core Functions, Quadratics & Polynomials

Unit 2 Part 1 Overview: Core Functions

SL Core Functions: Concept of domain and range, linear functions, and parallel/perpendicular gradients [2].
SL Quadratics: The three forms of a quadratic function (standard, factored, and vertex forms) and properties of parabolas [3].
SL Equations: Solving quadratics and utilizing the discriminant ( $Δ = b^{2} - 4 a c$ ) to determine the nature of roots [4].
AHL Polynomials: Polynomial functions, the Factor and Remainder theorems, and finding the sum and product of polynomial roots [5].
AHL Symmetry: Identifying algebraically and graphically whether a function is odd or even [5].

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SECTION 1: Quadratic Functions & The Discriminant

1.1 Forms of a Quadratic Function A quadratic graph (parabola) can be represented in three equivalent forms, each revealing different key features [3, 4]:

Standard Form: $f (x) = a x^{2} + b x + c$ . The $y$ -intercept is $(0, c)$ . The axis of symmetry is $x = - \frac{b}{2 a}$ .
Factored (Intercept) Form: $f (x) = a (x - p) (x - q)$ . The $x$ -intercepts (roots) are at $(p, 0)$ and $(q, 0)$ .
Vertex Form: $f (x) = a (x - h)^{2} + k$ . The turning point (vertex) is at exactly $(h, k)$ .

Worked Example: The Discriminant Find the exact values of $k$ for which the equation $3 x^{2} + 2 x + k = 0$ has two equal real roots [4].
Solution:
For two equal real roots, the discriminant must be zero: $Δ = b^{2} - 4 a c = 0$ [4].
Identify the coefficients: $a = 3$ , $b = 2$ , $c = k$ .
Substitute into the discriminant: $(2)^{2} - 4 (3) (k) = 0$ .
$4 - 12 k = 0 ⟹ 12 k = 4 ⟹ k = \frac{1}{3}$ .

SECTION 2: Polynomials (AHL)

2.1 The Factor and Remainder Theorems For any polynomial $P (x)$ :

Remainder Theorem: If $P (x)$ is divided by $(a x - b)$ , the remainder is exactly equal to $P (\frac{b}{a})$ [5].
Factor Theorem: If $(a x - b)$ is a perfect factor of $P (x)$ , then the remainder is zero, meaning $P (\frac{b}{a}) = 0$ [5].

Worked Example: Using the Remainder Theorem The polynomial $P (x) = x^{3} + a x^{2} - 3 x + 4$ leaves a remainder of $14$ when divided by $(x - 2)$ . Find the value of the constant $a$ .
Solution:
By the Remainder Theorem, $P (2) = 14$ .
Substitute $x = 2$ into the polynomial:
$(2)^{3} + a (2)^{2} - 3 (2) + 4 = 14$
$8 + 4 a - 6 + 4 = 14$
$4 a + 6 = 14 ⟹ 4 a = 8 ⟹ a = 2$ .

2.2 Sum and Product of Roots For any general polynomial equation $\sum_{r = 0}^{n} a_{r} x^{r} = 0$ of degree $n$ , the roots obey the following relationships [5]:

Sum of the roots: $\sum α = \frac{- a_{n - 1}}{a_{n}}$ (The negative of the second coefficient divided by the leading coefficient) [5].
Product of the roots: $\prod α = \frac{(- 1)^{n} a_{0}}{a_{n}}$ [5].

GDC Steps: Polynomial Root Finder If you need to find the roots of a polynomial up to degree 6 on Paper 2:
Go to MENU A (Equation) $\to$ POLY (F2). Select the degree (e.g., F2 for a cubic). Enter the coefficients $a, b, c, d$ in descending order of powers. Press SOLV (F1) to output all real and complex roots instantly [6]!

SECTION 3: Odd & Even Functions (AHL)

3.1 Algebraic Symmetry Functions can exhibit specific algebraic symmetries, known as being "odd" or "even" [5].

Even Functions: Satisfy $f (- x) = f (x)$ for all $x$ in the domain. They are geometrically symmetric (reflected) across the $y$ -axis [5]. Examples include $y = \cos x$ and $y = x^{2}$ .
Odd Functions: Satisfy $f (- x) = - f (x)$ for all $x$ in the domain. They exhibit rotational symmetry of $180^{\circ}$ about the origin $(0, 0)$ [5]. Examples include $y = \sin x$ and $y = x^{3}$ .

Worked Example: Proving Function Symmetry Determine algebraically whether $f (x) = x^{4} - 3 x^{2} + 5$ is odd, even, or neither.
Solution:
To test for symmetry, substitute $(- x)$ into the function:
$f (- x) = (- x)^{4} - 3 (- x)^{2} + 5$
Because negative numbers raised to even powers become positive:
$f (- x) = x^{4} - 3 x^{2} + 5$
Since $f (- x)$ is identically equal to the original $f (x)$ , the function is even.

Part 2

IB MATHEMATICS AA HL
UNIT 2 REVISION NOTES (PART 2)
Advanced Functions, Rationals & Transformations

Unit 2 Part 2 Overview: Advanced Concepts

SL Composite & Inverse: The identity function, composite chains $(f \circ g) (x)$ , and finding inverse functions $f^{- 1} (x)$ [7].
SL Rational Functions: Graphing $f (x) = \frac{a x + b}{c x + d}$ , finding vertical/horizontal asymptotes, and axes intercepts [4].
AHL Rational Extensions: Rationals with quadratic denominators and oblique asymptotes [5, 8].
AHL Transformations: Advanced modulus graphs ( $y = | f (x) |$ and $y = f (| x |)$ ), reciprocal graphs $y = \frac{1}{f (x)}$ , and solving modulus inequalities [9].

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SECTION 4: Composite & Inverse Functions

4.1 Composite and Inverse Properties Composite Functions: $(f \circ g) (x) = f (g (x))$ . The output of $g (x)$ becomes the direct input for $f (x)$ [7].
Inverse Functions: An inverse function $f^{- 1} (x)$ "undoes" the original function.

Geometrically, $f^{- 1} (x)$ is a perfect reflection of $f (x)$ across the line $y = x$ [7].
The domain of $f$ becomes the range of $f^{- 1}$ , and the range of $f$ becomes the domain of $f^{- 1}$ .
A function only has a valid inverse if it is "one-to-one" (passes the horizontal line test) [7]. If it fails, the domain must be restricted (AHL) [5].
Self-Inverse: A function is self-inverse if $f (x) = f^{- 1} (x)$ [5].

Worked Example: Finding an Inverse Function Given $f (x) = \frac{2 x + 1}{x - 3}$ , find $f^{- 1} (x)$ .
Solution:
Step 1: Write $y = \frac{2 x + 1}{x - 3}$ .
Step 2: Swap the $x$ and $y$ variables: $x = \frac{2 y + 1}{y - 3}$ .
Step 3: Solve algebraically for the new $y$ :
$x (y - 3) = 2 y + 1$
$x y - 3 x = 2 y + 1$
$x y - 2 y = 3 x + 1$
$y (x - 2) = 3 x + 1 ⟹ y = \frac{3 x + 1}{x - 2}$ .
Therefore, $f^{- 1} (x) = \frac{3 x + 1}{x - 2}$ .

SECTION 5: Rational Functions & Asymptotes

5.1 SL Rational Functions Rational functions of the form $f (x) = \frac{a x + b}{c x + d}$ feature horizontal and vertical asymptotes (boundary lines the curve approaches but never touches) [4].

Vertical Asymptote (VA): Occurs when the denominator equals zero: $c x + d = 0 ⟹ x = - \frac{d}{c}$ [10].
Horizontal Asymptote (HA): Found by looking at the ratio of the leading coefficients: $y = \frac{a}{c}$ [10].

Worked Example: Graphing Rational Functions Find the asymptotes and axis intercepts of $f (x) = \frac{3 x - 6}{x + 2}$ .
Solution:
VA: Set denominator to 0. $x + 2 = 0 ⟹$ Vertical Asymptote is $x = - 2$ .
HA: Ratio of leading coefficients. $y = \frac{3}{1} ⟹$ Horizontal Asymptote is $y = 3$ .
$x$ -intercept: Set numerator to 0. $3 x - 6 = 0 ⟹ x = 2$ . Coordinate is $(2, 0)$ .
$y$ -intercept: Set $x = 0$ . $f (0) = \frac{- 6}{2} = - 3$ . Coordinate is $(0, - 3)$ .

SECTION 6: Transformations & Modulus Equations

6.1 AHL Modulus Transformations The modulus (absolute value) function strictly outputs non-negative values [9]. When applied to graphs:

$y = | f (x) |$ : Any part of the graph that lies below the $x$ -axis is reflected upwards to become positive [9].
$y = f (| x |)$ : The entire graph on the left side of the $y$ -axis ( $x < 0$ ) is erased, and the right side of the graph ( $x \geq 0$ ) is reflected across the $y$ -axis to replace it [9].

Worked Example: Solving Modulus Equations Solve the modulus equation $| 2 x - 5 | = 7$ algebraically.
Solution:
A modulus equation splits into a positive and a negative case [9]:
Case 1 (Positive):
$2 x - 5 = 7 ⟹ 2 x = 12 ⟹ x = 6$ .
Case 2 (Negative):
$2 x - 5 = - 7 ⟹ 2 x = - 2 ⟹ x = - 1$ .
The solutions are $x = 6$ and $x = - 1$ .

GDC Steps: Graphing Modulus (Absolute Value) To graph absolute value equations like $y = | 2 x - 5 |$ on the CG50 on Paper 2:
Go to MENU 5 (Graph). In the function line, press OPTN $\to$ NUMERIC (F5) $\to$ Abs (F1). This will insert the modulus bars $| |$ into your equation line! You can use this to quickly find intercepts using G-Solv.

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Unit 3 · Geometry & Trig

HL AA revision notes

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Part 1

IB MATHEMATICS AA HL
UNIT 3 REVISION NOTES (PART 1)
Exhaustive Trigonometry & Circular Functions

Unit 3 Part 1 Overview: Trigonometry

SL Radians & Sectors: Radian measure, calculating arc length ( $l = r θ$ ), and sector area ( $A = \frac{1}{2} r^{2} θ$ ).
SL Unit Circle: Exact trigonometric ratios ( $\sin, \cos, \tan$ ) for standard angles ( $\frac{π}{6}, \frac{π}{4}, \frac{π}{3}$ ) and exploring symmetries across the four quadrants.
SL Trig Equations: Solving equations like $\sin (2 x) = 0.5$ over bounded domains.
AHL Identities: Pythagorean identity ( $\cos^{2} θ + \sin^{2} θ = 1$ ), compound angle identities, and double angle identities ( $\sin (2 θ) = 2 \sin θ \cos θ$ ).
AHL Functions: Graphing $y = a \sin (b (x - c)) + d$ and determining amplitude, period, phase shift, and principal axis.

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SECTION 1: Radians, Arcs & Sectors

1.1 Radian Measure A radian is a standard unit of angular measure. There are $2 π$ radians in a full $360^{\circ}$ circle.

To convert degrees to radians: Multiply by $\frac{π}{180}$ .
To convert radians to degrees: Multiply by $\frac{180}{π}$ .

When an angle $θ$ is measured strictly in radians, we can use two powerful formulas:

Arc Length: $l = r θ$
Area of a Sector: $A = \frac{1}{2} r^{2} θ$

Worked Example: Arcs and Sectors A circle with centre $O$ has radius $4 cm$ . The points $P, Q,$ and $R$ lie on the circumference such that the arc length $P Q R$ is $10 cm$ . Find the angle $P \hat{O} R$ in radians, and calculate the exact area of the shaded sector.
Solution:
Use the arc length formula to find the angle $θ$ :
$l = r θ ⟹ 10 = 4 θ ⟹ θ = 2.5 radians$ .
Now substitute $θ$ into the area formula:
$A = \frac{1}{2} r^{2} θ = \frac{1}{2} (4)^{2} (2.5) = \frac{1}{2} (16) (2.5) = 20 {cm}^{2}$ .

SECTION 2: The Unit Circle & Exact Values

2.1 The Unit Circle The Unit Circle is defined on the Cartesian plane with a radius of exactly 1. For any point $(x, y)$ on the unit circle at an angle $θ$ from the positive $x$ -axis, the coordinates define the fundamental trigonometric ratios: $x = \cos θ and y = \sin θ$ Furthermore, $\tan θ = \frac{\sin θ}{\cos θ}$ .

You must memorize the exact values for the first quadrant:

$\frac{π}{6} (30^{\circ}) : \sin = \frac{1}{2}, \cos = \frac{\sqrt{3}}{2}, \tan = \frac{1}{\sqrt{3}}$
$\frac{π}{4} (45^{\circ}) : \sin = \frac{\sqrt{2}}{2}, \cos = \frac{\sqrt{2}}{2}, \tan = 1$
$\frac{π}{3} (60^{\circ}) : \sin = \frac{\sqrt{3}}{2}, \cos = \frac{1}{2}, \tan = \sqrt{3}$

Worked Example: Finding Exact Values Without using a calculator, evaluate exactly: $\cos (\frac{5 π}{6}) - \sin^{2} (\frac{π}{4})$ .
Solution:
1. Recognize that $\frac{5 π}{6}$ is in the 2nd quadrant, where cosine is negative. Its reference angle is $\frac{π}{6}$ .
$\cos (\frac{5 π}{6}) = - \cos (\frac{π}{6}) = - \frac{\sqrt{3}}{2}$ .
2. The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ . Squaring this yields:
${(\frac{\sqrt{2}}{2})}^{2} = \frac{2}{4} = \frac{1}{2}$ .
3. Subtract the two values:
Result $= - \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{- \sqrt{3} - 1}{2}$ .

SECTION 3: Identities & Trig Equations

3.1 Trigonometric Identities (AHL) Trigonometric identities allow us to rewrite complex equations into solvable forms.

Pythagorean Identity: $\cos^{2} θ + \sin^{2} θ = 1$
Double Angle (Sine): $\sin (2 θ) = 2 \sin θ \cos θ$
Double Angle (Cosine): $\cos (2 θ) = \cos^{2} θ - \sin^{2} θ$
*(This can also be written as $2 \cos^{2} θ - 1$ or $1 - 2 \sin^{2} θ$ by substituting the Pythagorean identity).*

Worked Example: Solving Trig Equations with Domains Solve the equation $\sin (2 x) = \frac{1}{2}$ for the domain $0 \leq x \leq 2 π$ .
Solution:
1. First, adjust the domain for the argument $(2 x)$ . If $0 \leq x \leq 2 π$ , then $0 \leq 2 x \leq 4 π$ .
This means we must take two full rotations around the unit circle.
2. Identify where sine is equal to positive $\frac{1}{2}$ . This occurs at reference angle $\frac{π}{6}$ .
Since sine is positive in the 1st and 2nd quadrants, our base angles are $\frac{π}{6}$ and $\frac{5 π}{6}$ .
3. Find all solutions for $2 x$ within the adjusted $4 π$ domain:
$2 x = \frac{π}{6}, \frac{5 π}{6}, (\frac{π}{6} + 2 π), (\frac{5 π}{6} + 2 π)$
$2 x = \frac{π}{6}, \frac{5 π}{6}, \frac{13 π}{6}, \frac{17 π}{6}$ .
4. Divide all solutions by 2 to solve for $x$ :
$x = \frac{π}{12}, \frac{5 π}{12}, \frac{13 π}{12}, \frac{17 π}{12}$ .

GDC Steps: Graphical Solutions to Trig Equations If a trigonometric equation is too complex to solve analytically on Paper 2 (or involves mixing trig functions with polynomials like $\cos x = x^{2}$ ), you must use a graphical approach!
Ensure your CG50 is in Radians. In MENU 5 (Graph), plot the left side of the equation as Y1 and the right side as Y2. Adjust your V-Window X-min and X-max to precisely match the domain given in the question. Press G-Solv (F5) $\to$ ISCT (F5) to find all points of intersection!

Part 2

IB MATHEMATICS AA HL
UNIT 3 REVISION NOTES (PART 2)
Exhaustive Vectors, Lines & Kinematics (AHL)

Unit 3 Part 2 Overview: Vectors & Kinematics

AHL Vector Basics: Position vectors, displacement vectors, magnitude ( $| v |$ ), and determining unit vectors.
AHL Scalar Product: Using the dot product $v \cdot w = | v | | w | \cos θ$ to find the angle between two vectors. Perpendicular vectors have a scalar product of $0$ .
AHL Lines in 3D Space: The vector equation of a line $r = a + λ b$ . Converting between vector form and parametric form.
AHL Intersecting Lines: Solving simultaneous equations to determine if two lines intersect, are parallel, or are skew in 3-dimensional space.
AHL Vector Kinematics: Modelling linear motion using $r (t) = a + t b$ , interpreting $a$ as initial position, $b$ as constant velocity, and $| b |$ as speed.

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SECTION 4: Vector Fundamentals & Scalar Product

4.1 Magnitude and Unit Vectors A vector represents both magnitude (length) and direction. A unit vector is any vector that has a magnitude of exactly 1 unit. If we have a vector $v = (\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix})$ , its magnitude is $| v | = \sqrt{v_{1}^{2} + v_{2}^{2} + v_{3}^{2}}$ .
To find the unit vector in the same direction as $v$ , divide the vector by its own magnitude: $\frac{1}{| v |} v$ .

4.2 The Scalar (Dot) Product The scalar product is used to find the exact angle between two intersecting vectors.
Algebraically: $v \cdot w = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3}$
Geometrically: $v \cdot w = | v | | w | \cos θ$ .
By equating these two formulas, we can easily isolate and solve for $\cos θ$ .

Crucial Property: If two non-zero vectors are perpendicular (orthogonal), the angle between them is $90^{\circ}$ . Since $\cos (90^{\circ}) = 0$ , their scalar product must equal zero. ( $v \cdot w = 0$ ).

Worked Example: Angle Between Vectors Find the acute angle $θ$ between the vectors $a = (\begin{matrix} - 1 \\ 1 \\ - 2 \end{matrix})$ and $b = (\begin{matrix} 2 \\ - 1 \\ 3 \end{matrix})$ .
Solution:
1. Calculate the scalar product algebraically:
$a \cdot b = (- 1) (2) + (1) (- 1) + (- 2) (3) = - 2 - 1 - 6 = - 9$ .
2. Calculate the magnitudes of both vectors:
$| a | = \sqrt{(- 1)^{2} + 1^{2} + (- 2)^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6}$ .
$| b | = \sqrt{2^{2} + (- 1)^{2} + 3^{2}} = \sqrt{4 + 1 + 9} = \sqrt{14}$ .
3. Substitute into the geometric formula:
$- 9 = \sqrt{6} \sqrt{14} \cos θ ⟹ \cos θ = \frac{- 9}{\sqrt{84}}$ .
$θ = \arccos (\frac{- 9}{\sqrt{84}}) \approx {169.1}^{\circ}$ .
Because the question asks for the acute angle between the lines, we subtract this from $180^{\circ}$ :
Acute Angle $= 180^{\circ} - {169.1}^{\circ} = {10.9}^{\circ}$ .

SECTION 5: Lines in 3D Space

5.1 Vector and Parametric Equations of a Line A line in 3D space is uniquely defined by a known position point it passes through ( $a$ ) and a direction vector it runs parallel to ( $b$ ). The vector equation is given by: $r = a + λ b$ Where $λ$ is a scalar parameter. This can be split into three parametric equations: $x = x_{0} + λ b_{1}, y = y_{0} + λ b_{2}, z = z_{0} + λ b_{3}$

Worked Example: Intersecting Lines Determine if the line $L_{1} : r_{1} = (\begin{matrix} 1 \\ - 1 \\ 4 \end{matrix}) + s (\begin{matrix} 2 \\ - 2 \\ 3 \end{matrix})$ intersects with the line $L_{2} : r_{2} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + u (\begin{matrix} 2 \\ - 1 \\ 7 \end{matrix})$ . If so, find the point of intersection.
Solution:
Equate the $x, y,$ and $z$ parametric equations from both lines:
1) $1 + 2 s = 1 + 2 u ⟹ 2 s = 2 u ⟹ s = u$ .
2) $- 1 - 2 s = - u$ .
3) $4 + 3 s = 7 u$ .
Substitute $s = u$ into equation 2:
$- 1 - 2 (u) = - u ⟹ - 1 = u$ .
Since $s = u$ , $s = - 1$ as well.
Crucial Check Step: You must verify these values in the unused 3rd equation to ensure intersection!
$4 + 3 (- 1) = 4 - 3 = 1$ .
$7 (- 1) = - 7$ .
Since $1 \neq - 7$ , the third equation fails. The lines do not intersect. Because their direction vectors are not multiples of each other, they are not parallel either. They are skew lines.

SECTION 6: Vector Kinematics

6.1 Modelling Motion If an object moves with constant velocity $v$ , its position vector at time $t$ is given by: $r (t) = r_{0} + t v$ Where:

$r_{0}$ is the initial position vector at $t = 0$ .
$v$ is the velocity vector (representing speed and direction).
$| v |$ is the speed of the object (a scalar quantity).

Worked Example: Kinematics and Position A remote-controlled boat is initially at the coordinates $(2, 3)$ . It travels with a constant velocity vector of $v = (\begin{matrix} 4 \\ - 5 \end{matrix}) {km h}^{- 1}$ .
a) Write down the vector equation of the boat’s path.
b) Find the boat’s exact speed.
c) Find the boat’s position after $90 minutes$ .
Solution:
a) The initial position is $r_{0} = (\begin{matrix} 2 \\ 3 \end{matrix})$ . The path equation is:
$r (t) = (\begin{matrix} 2 \\ 3 \end{matrix}) + t (\begin{matrix} 4 \\ - 5 \end{matrix})$ .
b) Speed is the magnitude of the velocity vector:
$| v | = \sqrt{4^{2} + (- 5)^{2}} = \sqrt{16 + 25} = \sqrt{41} {km h}^{- 1}$ .
c) Time must be in hours to match the velocity units. $90 minutes = 1.5 hours$ .
Substitute $t = 1.5$ into the equation:
$r (1.5) = (\begin{matrix} 2 \\ 3 \end{matrix}) + 1.5 (\begin{matrix} 4 \\ - 5 \end{matrix}) = (\begin{matrix} 2 \\ 3 \end{matrix}) + (\begin{matrix} 6 \\ - 7.5 \end{matrix}) = (\begin{matrix} 8 \\ - 4.5 \end{matrix})$ .
The boat’s coordinates are $(8, - 4.5)$ .

GDC Steps: Scalar Product and Magnitudes You can use the CG50 to quickly calculate the dot product and vector magnitudes on Paper 2!
Go to MENU 1 (Run-Matrix). Press F3 (MAT/VCT). Enter your vectors into Vct A and Vct B.
Press OPTN $\to$ MAT/VCT (F2).
To find the scalar product: Press DotP( and type Vct A, Vct B).
To find the magnitude (length): Press Norm( and type Vct A).

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Unit 4 · Stats & Probability

HL AA revision notes

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Part 1

IB MATHEMATICS AA HL
UNIT 4 REVISION NOTES (PART 1)
Exhaustive Descriptive Statistics, Bivariate Data & Probability

Unit 4 Part 1 Overview: Data & Probability

SL Descriptive Statistics: Concepts of population, sample, and discrete/continuous data [2]. Calculating mean, median, mode, variance, and standard deviation [3].
SL Outliers: Interpreting and identifying mathematical outliers using the Interquartile Range ( $1.5 \times I Q R$ ) [2].
SL Bivariate Data: Scatter diagrams, Pearson’s product-moment correlation coefficient ( $r$ ), and lines of best fit [4]. Using regression equations for prediction and understanding the dangers of extrapolation [5].
SL Probability: Combined events, mutually exclusive events, independent events, and conditional probability [6]. Solving problems using Venn diagrams and tree diagrams.

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SECTION 1: Descriptive Statistics & Outliers

1.1 Central Tendency, Dispersion & Outliers Data is classified as discrete (exact counted values) or continuous (measured values in intervals) [2, 7].

Measures of Central Tendency: Mean ( $μ$ ), Median ( $Q_{2}$ ), Mode.
Measures of Dispersion: Variance ( $σ^{2}$ ), Standard Deviation ( $σ$ ), Interquartile Range ( $I Q R = Q_{3} - Q_{1}$ ) [3].
Outliers: An outlier is defined as any data item that is more than $1.5 \times I Q R$ from the nearest quartile [2, 7].
- Lower Bound for Outliers: $x < Q_{1} - 1.5 \times I Q R$
- Upper Bound for Outliers: $x > Q_{3} + 1.5 \times I Q R$

Worked Example: Calculating Outliers The test scores of a class have a lower quartile ( $Q_{1}$ ) of $45$ and an upper quartile ( $Q_{3}$ ) of $65$ . A student scores $98$ on the test. Determine mathematically if this score is an outlier.
Solution:
1. Calculate the Interquartile Range ( $I Q R$ ):
$I Q R = Q_{3} - Q_{1} = 65 - 45 = 20$ .
2. Calculate the upper outlier boundary:
Upper Bound $= Q_{3} + 1.5 \times I Q R = 65 + 1.5 (20) = 65 + 30 = 95$ .
3. Compare the score to the boundary:
Since $98 > 95$ , the score of $98$ is mathematically an outlier.

SECTION 2: Bivariate Data & Linear Regression

2.1 Correlation and Regression When exploring the relationship between two variables, we use a scatter diagram.

Pearson’s correlation coefficient ( $r$ ): Measures the strength and direction of a linear relationship ( $- 1 \leq r \leq 1$ ) [4, 8].
Regression Line ( $y = a x + b$ ): The line of best fit. It always passes exactly through the mean point $(\bar{x}, \bar{y})$ [8].
Prediction: The regression line can be used for interpolation (reliable predictions within the given data range), but it is generally unreliable for extrapolation (predicting outside the data range) [5, 8].

Worked Example: Linear Regression & Predictions The regression line of $y$ on $x$ for a set of bivariate data is $y = 1.2 x + 3.4$ . The mean of the $x$ -values is $\bar{x} = 5$ . a) Find the mean of the $y$ -values, $\bar{y}$ . b) Predict the value of $y$ when $x = 10$ , and state whether this prediction is reliable given that the original $x$ -values range from $2$ to $8$ [8].
Solution:
a) The regression line always passes through the mean point $(\bar{x}, \bar{y})$ [8]. Substitute $\bar{x} = 5$ into the equation:
$\bar{y} = 1.2 (5) + 3.4 = 6.0 + 3.4 = 9.4$ .
b) Substitute $x = 10$ into the equation to predict $y$ :
$y = 1.2 (10) + 3.4 = 12 + 3.4 = 15.4$ .
This prediction is unreliable because $x = 10$ lies outside the original data range ( $2 \leq x \leq 8$ ), making it an extrapolation [8].

SECTION 3: Probability Rules & Events

3.1 Advanced Probability Rules Probability calculates the theoretical likelihood of events. The formal rules are:

Combined Events: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ [6, 8].
Mutually Exclusive Events: Events that cannot happen at the same time. Therefore, $P (A \cap B) = 0$ [6, 8].
Independent Events: The outcome of one event does not affect the other. $P (A \cap B) = P (A) P (B)$ [6, 8].
Conditional Probability: The probability of $A$ occurring given that $B$ has already occurred: $P (A | B) = \frac{P (A \cap B)}{P (B)} ⟹ P (A \cap B) = P (B) P (A | B) [6, 8]$

Worked Example: Conditional Probability Events $A$ and $B$ are such that $P (A) = 0.4$ , $P (A | B) = 0.25$ , and $P (A \cup B) = 0.55$ [9, 10]. Find the exact value of $P (B)$ .
Solution:
1. Use the conditional probability formula to express the intersection:
$P (A | B) = \frac{P (A \cap B)}{P (B)} ⟹ P (A \cap B) = 0.25 \times P (B)$ .
2. Substitute this into the combined events formula:
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$0.55 = 0.4 + P (B) - 0.25 P (B)$
$0.15 = 0.75 P (B)$
3. Solve for $P (B)$ :
$P (B) = \frac{0.15}{0.75} = \frac{1}{5} = 0.2$ .

GDC Steps: Calculating Pearson’s $r$ To find the equation of a regression line and the correlation coefficient on Paper 2:
Enter your $x$ -values into List 1 and $y$ -values into List 2 via MENU 2 (Stat).
Press CALC (F2) $\to$ REG (F3) $\to$ X (F1) $\to$ ax+b (F1) [8].
The screen will instantly display the gradient ( $a$ ), the $y$ -intercept ( $b$ ), and Pearson’s correlation coefficient ( $r$ ).

Part 2

IB MATHEMATICS AA HL
UNIT 4 REVISION NOTES (PART 2)
Exhaustive Discrete, Binomial & Normal Distributions

Unit 4 Part 2 Overview: Probability Distributions

SL Discrete Random Variables: Defining valid probability distributions where $\sum P (X = x) = 1$ . Calculating Expected Value $E (X) = \sum x P (X = x)$ , and interpreting fair games [6, 11].
AHL Discrete Random Variables: Calculating the variance of a discrete random variable [12].
SL/AHL Binomial Distribution: Applying the binomial model $X \sim B (n, p)$ for independent trials with constant probability. Finding the mean and variance of binomial distributions [11, 13].
SL/AHL Normal Distribution: Analysing continuous data using the normal curve $X \sim N (μ, σ^{2})$ . Calculating probabilities, inverse normal values, and $Z$ -score standardizations [13, 14].

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SECTION 4: Discrete Random Variables

4.1 Expected Value and Variance A discrete random variable ( $X$ ) has specific, countable outcomes, each with an associated probability. The sum of all probabilities in the sample space must equal exactly 1 [11].

Expected Value $E (X)$ : The theoretical long-run average (mean) of the distribution. $E (X) = \sum x P (X = x)$ Note: In the context of gambling, if $E (X) = 0$ , the game is mathematically "fair" [6, 11].
Variance (AHL): Measures the spread of the discrete random variable [12].

Worked Example: Expected Value & Constants A discrete random variable $X$ has a probability distribution given by $P (X = n) = \frac{k}{n}$ , where $n \in {1, 2, 3, 4}$ and $k$ is a positive constant. Find $k$ , and calculate $E (X)$ [11].
Solution:
1. The sum of all probabilities must equal 1:
$P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1$
$\frac{k}{1} + \frac{k}{2} + \frac{k}{3} + \frac{k}{4} = 1$
$k (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = 1$
$k (\frac{12 + 6 + 4 + 3}{12}) = 1 ⟹ k (\frac{25}{12}) = 1 ⟹ k = \frac{12}{25} = 0.48$ [11].
2. Calculate Expected Value $E (X)$ :
$E (X) = \sum x P (X = x) = 1 (\frac{k}{1}) + 2 (\frac{k}{2}) + 3 (\frac{k}{3}) + 4 (\frac{k}{4})$
$E (X) = k + k + k + k = 4 k$
$E (X) = 4 (0.48) = 1.92$ [11].

SECTION 5: The Binomial Distribution

5.1 Binomial Variables $X \sim B (n, p)$ The binomial distribution models situations with a fixed number of independent trials ( $n$ ), where each trial has only two possible outcomes (Success or Failure) and a constant probability of success ( $p$ ) [11].

Mean (Expected Value): $E (X) = n p$ [13].
Variance: $Var (X) = n p (1 - p)$ [13].

Worked Example: Binomial Probabilities A factory produces cereal boxes. The heaviest $3 %$ of the cereal boxes produced are separated and classified as "Premium" boxes. A random sample of $12$ boxes is selected from the factory floor. Find the probability that exactly 2 of these boxes are classified as Premium [15, 16].
Solution:
This is a Binomial Distribution problem. Let $Y$ be the number of premium boxes.
$n = 12$ (number of trials)
$p = 0.03$ (probability of success)
$Y \sim B (12, 0.03)$ [16].
We need to find exactly 2 successes: $P (Y = 2)$ .
Using technology, $P (Y = 2) = 0.0435$ (or $4.35 %$ ) [16].

SECTION 6: The Normal Distribution

6.1 Normal Variables $X \sim N (μ, σ^{2})$ The normal distribution is a continuous, symmetrical bell-shaped curve defined by its mean ( $μ$ ) and standard deviation ( $σ$ ).

Approximately $68 %$ of data lies between $μ \pm σ$ , $95 %$ lies between $μ \pm 2 σ$ , and $99.7 %$ lies between $μ \pm 3 σ$ [13].
Inverse Normal: Used to find the boundary value ( $k$ ) when the probability area is already known [14].
Standardization (AHL): Using $Z$ -values to calculate unknown means or standard deviations. The $Z$ -score represents the number of standard deviations from the mean: $Z = \frac{x - μ}{σ}$ [14].

Worked Example: Normal & Inverse Normal The weights of cereal boxes are normally distributed with a mean of $502 g$ and a standard deviation of $2 g$ . Let $X \sim N (502, 2^{2})$ . a) Find the probability that a randomly chosen box weighs more than $504 g$ . b) Any box weighing less than $k$ grams is destroyed. The probability of a box being destroyed is $1.5 %$ . Find the value of $k$ [15, 16].
Solution:
a) We require $P (X > 504)$ .
Using a GDC Normal CDF with Lower = $504$ , Upper = $10^{99}$ , $μ = 502$ , and $σ = 2$ :
$P (X > 504) \approx 0.159$ ( $15.9 %$ ) [16].
b) We require $P (X < k) = 0.015$ [16].
Using a GDC Inverse Normal with Area = $0.015$ (Left tail), $μ = 502$ , and $σ = 2$ :
$k = 497.66 \dots \approx 498 g$ [16].

GDC Steps: Distributions on the CG50 For all distribution calculations on Paper 2, use the Stat menu!
Go to MENU 2 (Stat) $\to$ DIST (F5).

Binomial: Press BINOMIAL (F5). Use Bpd (F1) for exact $P (X = x)$ or Bcd (F2) for inequalities like $P (X \leq x)$ [16].
Normal: Press NORM (F1). Use Ncd (F2) for finding probabilities between two boundaries, and InvN (F3) for finding an unknown boundary $x$ when given the probability area [16].

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Unit 5 · Calculus

HL AA revision notes

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Part 1

IB MATHEMATICS AA HL
UNIT 5 REVISION NOTES (PART 1)
Exhaustive Differential Calculus & Applications

Unit 5 Part 1 Overview: Differential Calculus

SL Limits & Rules: Estimation of limits, standard differentiation of polynomials $f^{'} (x) = a n x^{n - 1}$ , and tangents/normals [2, 3].
AHL First Principles: Formal limits (convergence/divergence) and finding the derivative from first principles [4].
AHL Advanced Rules: The Chain, Product, and Quotient rules for composite functions [5]. Implicit differentiation and related rates of change [6].
AHL Extended Functions: Derivatives of trigonometric, exponential, logarithmic, and inverse trigonometric functions ( $\arcsin x$ , $\arctan x$ ) [7].
SL/AHL Applications: Local maximum/minimum points, points of inflexion, 2nd derivative concavity tests, optimization, and kinematics ( $s, v, a$ ) [8, 9].

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SECTION 1: Limits & First Principles

1.1 Formal Limits & Derivative from First Principles Calculus mathematically describes instantaneous rates of change [10]. The derivative of a function $f (x)$ at a specific point can be found formally using limits (First Principles) [4]: $f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ This represents the gradient of the chord connecting $(x, f (x))$ and $(x + h, f (x + h))$ as the horizontal distance $h$ becomes infinitesimally small [4].

Worked Example: First Principles (AHL) Use the first principles formula to find the instantaneous rate of change for $f (x) = x^{2} + 2 x$ at the exact point where $x = 5$ [11].
Solution:
1. Evaluate the function at $x = 5$ : $f (5) = 5^{2} + 2 (5) = 35$ [11].
2. Substitute into the limit definition:
$f^{'} (5) = lim_{h \to 0} \frac{f (5 + h) - f (5)}{h}$ [11]
$= lim_{h \to 0} \frac{(5 + h)^{2} + 2 (5 + h) - 35}{h}$ [11]
$= lim_{h \to 0} \frac{25 + 10 h + h^{2} + 10 + 2 h - 35}{h}$
$= lim_{h \to 0} \frac{h^{2} + 12 h}{h}$ [11]
3. Factorise and cancel $h$ (since $h \neq 0$ ):
$= lim_{h \to 0} (h + 12) = 12$ [12].
The instantaneous rate of change is $12$ [12].

SECTION 2: Differentiation Rules & Tangents

2.1 Advanced Differentiation Rules (AHL) For complex functions, we use specific differentiation rules [5, 7]:

Chain Rule (Composite functions): $y = g (u)$ and $u = h (x) ⟹ \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ [13].
Product Rule: $y = u v ⟹ y^{'} = u v^{'} + v u^{'}$ [5].
Quotient Rule: $y = \frac{u}{v} ⟹ y^{'} = \frac{v u^{'} - u v^{'}}{v^{2}}$ [5].
Implicit Differentiation: Used when $y$ cannot be easily isolated. Differentiate both sides with respect to $x$ , applying the chain rule to attach $\frac{d y}{d x}$ whenever differentiating a $y$ term [6].

A tangent is a straight line that touches the curve with the same gradient ( $m_{T} = f^{'} (x)$ ) [14]. A normal intersects perpendicularly ( $m_{T} \times m_{N} = - 1$ ) [14].

Worked Example: Finding a Normal Equation Find the exact equation of the normal to the curve $f (x) = e^{2 x}$ at the point where $x = 0$ [15].
Solution:
1. Find the full coordinate: $f (0) = e^{0} = 1 ⟹ (0, 1)$ [15].
2. Find the gradient of the tangent ( $m_{T}$ ) using the chain rule [15]:
$f^{'} (x) = 2 e^{2 x} ⟹ f^{'} (0) = 2 e^{0} = 2$ [15].
3. Find the perpendicular gradient of the normal ( $m_{N}$ ):
$m_{N} = - \frac{1}{m_{T}} = - \frac{1}{2}$ [15].
4. Substitute into the linear equation formula $y - y_{1} = m (x - x_{1})$ [15]:
$y - 1 = - \frac{1}{2} (x - 0) ⟹ y = - 0.5 x + 1$ [15].

SECTION 3: Curve Sketching, Optimization & Kinematics

3.1 Stationary Points & Optimization Stationary points (turning points) occur exactly when $f^{'} (x) = 0$ [15]. We classify them using the second derivative ( $f^{″} (x)$ ) [8]:

Local Minimum: $f^{″} (x) > 0$ (The curve is "concave-up") [8, 16].
Local Maximum: $f^{″} (x) < 0$ (The curve is "concave-down") [8, 16].
Point of Inflexion: Occurs where the concavity changes sign, meaning $f^{″} (x) = 0$ AND changes from positive to negative (or vice versa) [8].

We use these principles to solve Optimization problems to find maximum profits, areas, or minimum costs [8].

Worked Example: Optimization and Profit A company’s profits (in thousands) are modeled by $P (x) = - 0.1 x^{3} + 12 x^{2} - 60 x + 100$ , where $x$ is the number of sets sold in hundreds [17, 18]. Find the number of sets required to maximize profit.
Solution:
1. Differentiate and set to zero to find the stationary points [17, 18]:
$P^{'} (x) = - 0.3 x^{2} + 24 x - 60 = 0$ [18].
Using a GDC Polynomial solver, $x \approx 77.4$ or $x \approx 2.58$ [17]. Since they sell whole items, we test $x \approx 77$ [17].
2. Use the second derivative to verify it is a maximum [17]:
$P^{″} (x) = - 0.6 x + 24$ [17].
$P^{″} (77.4) = - 0.6 (77.4) + 24 = - 22.44$ [17].
Since $P^{″} (x) < 0$ , the curve is concave down, strictly confirming a maximum [17].

3.2 Kinematics (Motion in a Straight Line) Calculus connects displacement ( $s$ ), velocity ( $v$ ), and acceleration ( $a$ ) [19].

Differentiating: $v (t) = \frac{d s}{d t}$ and $a (t) = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$ [9].
Change of Direction: An object changes direction when $v (t) = 0$ [20].
Speed: The magnitude (absolute value) of velocity $| v (t) |$ [9].

GDC Steps: Finding Turning Points To instantly find turning points of a polynomial on Paper 2 without manual differentiation [17]:
Go to MENU A (Equation) $\to$ Polynomial (F2) $\to$ Select Degree.
For $f^{'} (x) = - 0.3 x^{2} + 24 x - 60$ , enter coefficients a = -0.3, b = 24, c = -60 and press EXE to solve for $x$ instantly [17, 18]!

Part 2

IB MATHEMATICS AA HL
UNIT 5 REVISION NOTES (PART 2)
Exhaustive Integration, Series & Differential Equations

Unit 5 Part 2 Overview: Integration & Advanced Calculus

SL Integration & Area: Anti-differentiation, definite integrals, and finding the area between curves [9, 21].
AHL Integration Techniques: Integration by substitution, integration by parts ( $\int u d v = u v - \int v d u$ ), and using partial fractions to rearrange integrands [7, 22].
AHL Volumes of Revolution: Calculating volumes generated by rotating a bounded region $360^{\circ}$ about the $x$ -axis [23].
AHL Maclaurin Series & L’Hôpital’s Rule: Evaluating limits of indeterminate forms ( $\frac{0}{0}, \frac{\infty}{\infty}$ ) [6]. Using Maclaurin series to approximate complex functions [24].
AHL Differential Equations: Solving first-order DEs via separation of variables, homogeneous substitutions ( $y = v x$ ), integrating factors, and Euler’s Method.

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SECTION 4: Integration Techniques & Area/Volume

4.1 Advanced Integration Methods (AHL) Integration is the reverse process of differentiation [19].

Integration by Substitution: Used for integrals like $\int k g^{'} (x) f (g (x)) d x$ [25]. Let $u = g (x)$ .
Integration by Parts: Used for products of functions [22]. Formula: $\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$ .
Partial Fractions: Splitting complex rational functions to integrate them as natural logarithms [7]. Example: $\int \frac{1}{x^{2} + 3 x + 2} d x = \int (\frac{1}{x + 1} - \frac{1}{x + 2}) d x = \ln | x + 1 | - \ln | x + 2 | + C$ [7].

Definite Integrals: $\int_{a}^{b} f (x) d x$ evaluates to a numerical value [19]. Geometrically, this calculates the area under a curve, or the area enclosed between two curves $\int_{a}^{b} (Upper - Lower) d x$ [26]. Rotating this area $360^{\circ}$ around the $x$ -axis generates a Volume of Revolution [23]: $V = π \int_{a}^{b} (f (x))^{2} d x$ .

Worked Example: Area Between Curves Find the exact area enclosed between the parabola $y = x^{2}$ and the line $y = x + 2$ [26].
Solution:
1. Find the points of intersection by equating the functions [26]:
$x^{2} = x + 2 ⟹ x^{2} - x - 2 = 0 ⟹ (x - 2) (x + 1) = 0$ .
The intersections are at $x = - 1$ and $x = 2$ [26].
2. Determine which curve is on top. Between $- 1$ and $2$ , the line $x + 2$ is the upper curve [26].
3. Set up the definite integral:
Area $= \int_{- 1}^{2} ((x + 2) - x^{2}) d x$ [26].
$= {[\frac{x^{2}}{2} + 2 x - \frac{x^{3}}{3}]}_{- 1}^{2}$
$= (\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = (6 - \frac{8}{3}) - (- \frac{7}{6}) = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = 4.5$ .
The exact area is $4.5$ square units [26].

SECTION 5: L’Hôpital’s Rule & Maclaurin Series

5.1 Indeterminate Forms and Maclaurin Expansions (AHL) L’Hôpital’s Rule: Used to evaluate limits that result in indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ [6]. $lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ Maclaurin Series: Used to express transcendental functions as infinite polynomials [24]. You are expected to use simple substitutions, products, integration, and differentiation to manipulate standard series provided in the formula booklet [24].

Worked Example: L’Hôpital’s Rule Evaluate $lim_{θ \to 0} \frac{\sin θ}{θ}$ [6].
Solution:
1. Direct substitution yields $\frac{\sin 0}{0} = \frac{0}{0}$ , which is an indeterminate form [6].
2. Apply L’Hôpital’s rule by differentiating the numerator and the denominator separately:
$\frac{d}{d θ} (\sin θ) = \cos θ$
$\frac{d}{d θ} (θ) = 1$
3. Re-evaluate the limit:
$lim_{θ \to 0} \frac{\cos θ}{1} = \frac{\cos 0}{1} = \frac{1}{1} = 1$ [6].

SECTION 6: Differential Equations

6.1 First-Order Differential Equations (AHL) The AA HL syllabus requires mastery of several methods to solve first-order differential equations:

Separable Variables: Rearranging to integrate $y$ terms on one side and $x$ terms on the other.
Homogeneous DEs: Using the substitution $y = v x ⟹ \frac{d y}{d x} = v + x \frac{d v}{d x}$ to transform the DE into a separable form.
Integrating Factor (IF): For linear DEs in the form $\frac{d y}{d x} + P (x) y = Q (x)$ , calculate IF $= e^{\int P (x) d x}$ and multiply the entire equation by it.
Euler’s Method: A numerical step-by-step approximation method: $y_{n + 1} = y_{n} + h \times f (x_{n}, y_{n})$ .

Worked Example: Separation of Variables Find the general solution to the differential equation $\frac{d y}{d x} = 4 x e^{- y}$ . Express your answer in the form $y = f (x)$ .
Solution:
1. Separate the variables by moving all $y$ terms to the left and $x$ terms to the right:
$\frac{1}{e^{- y}} d y = 4 x d x ⟹ e^{y} d y = 4 x d x$ .
2. Integrate both sides:
$\int e^{y} d y = \int 4 x d x$
$e^{y} = 2 x^{2} + C$ .
3. Take the natural logarithm of both sides to isolate $y$ :
$y = \ln (2 x^{2} + C)$ .

GDC Steps: Definite Integrals & Area To calculate exact definite integrals or areas on Paper 2 without manual integration [19]:
Run-Matrix Mode: Press MATH (F4) $\to$ \int dx (F6 -> F1). Fill in the function and the lower/upper bounds to instantly calculate the numerical value [19].
Graph Mode: Draw the curve. Press G-Solv (SHIFT F5) $\to$ \int dx (F6 -> F3) $\to$ ISCT (F3) if finding area between intersections! [26]

Practice Unit 5 questions → See HL AA plans

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HL Analysis & Approaches Revision Notes

Unit 1 · Number & Algebra

Unit 2 · Functions

Unit 3 · Geometry & Trig

Unit 4 · Stats & Probability

Unit 5 · Calculus