Composite Functions & Order of Operations
- Definition of Composition: The composite function $(f \circ g)(x)$ is defined by: $$(f \circ g)(x) = f(g(x))$$ The inside function $g$ acts first on $x$, producing an output $g(x)$. The outside function $f$ then acts on that output.
- Non-Commutativity: In general, composition is not commutative: $$(f \circ g)(x) \ne (g \circ f)(x)$$ Always maintain strict adherence to the specified order.
- Domain of a Composite Function: For $(f \circ g)(x)$ to exist, two conditions must be satisfied:
- $x$ must belong to the domain of the inside function: $x \in \text{dom } g$.
- The output $g(x)$ must lie within the domain of the outside function: $g(x) \in \text{dom } f$.
(a) Find the numerical value of $(f \circ g)(8)$.
(b) Find an expression for $(f \circ g)(x)$ and state its maximum domain.
Part (a): Evaluate $(f \circ g)(8)$
First find $g(8)$:
$$g(8) = \sqrt{8 + 1} = \sqrt{9} = 3$$Now evaluate $f(3)$:
$$f(3) = \frac{2}{3 - 3} = \frac{2}{0} \quad (\text{undefined})$$Therefore, $(f \circ g)(8)$ is undefined (8 is not in the domain of $f \circ g$).
Part (b): Expression and Domain
$$(f \circ g)(x) = f(g(x)) = \frac{2}{\sqrt{x + 1} - 3}$$Condition 1 (inside function): $x + 1 \ge 0 \implies x \ge -1$.
Condition 2 (denominator non-zero): $\sqrt{x + 1} - 3 \ne 0 \implies \sqrt{x + 1} \ne 3 \implies x + 1 \ne 9 \implies x \ne 8$.
Domain: $\{x \in \mathbb{R} \mid x \ge -1, x \ne 8\}$ or $[-1, 8) \cup (8, \infty)$.
Inverse Functions & The One-to-One Condition
- One-to-One Condition (Horizontal Line Test): A function $f$ possesses an inverse function $f^{-1}$ if and only if it is one-to-one (bijective onto its range).
- If a horizontal line intersects the graph of $f$ more than once, $f$ is many-to-one and does not have an inverse on that domain.
- To invert many-to-one functions (such as parabolas), the domain must be restricted (e.g., $f(x) = (x-h)^2 + k$ restricted to $x \ge h$ or $x \le h$).
- Algebraic Protocol to Find $f^{-1}(x)$:
- Write $y = f(x)$.
- Swap the variables $x$ and $y$: write $x = f(y)$.
- Rearrange the equation to make $y$ the subject.
- Replace $y$ with $f^{-1}(x)$.
- Critical Notation Alert: $f^{-1}(x) \ne \frac{1}{f(x)}$. The exponent $-1$ denotes an inverse mapping, NOT an algebraic reciprocal.
Step 1: Write $y = f(x)$ and swap variables.
$$x = \frac{3y + 1}{2y - 5}$$Step 2: Multiply by denominator to clear fraction.
$$x(2y - 5) = 3y + 1 \implies 2xy - 5x = 3y + 1$$Step 3: Collect all terms involving $y$ on one side.
$$2xy - 3y = 5x + 1$$Step 4: Factor out $y$ and divide.
$$y(2x - 3) = 5x + 1 \implies y = \frac{5x + 1}{2x - 3}$$Answer: $f^{-1}(x) = \frac{5x + 1}{2x - 3}$, where $x \ne \frac{3}{2}$.
Domain-Range Inversion & Reflection Across $y = x$
- Domain and Range Swap: The domain of $f^{-1}$ is identical to the range of $f$, and the range of $f^{-1}$ is identical to the domain of $f$: $$\text{dom } f^{-1} = \text{range } f \qquad \text{and} \qquad \text{range } f^{-1} = \text{dom } f$$
- Geometric Reflection Property: The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ across the line $y = x$. $$\text{If } (a, b) \text{ lies on } f \iff (b, a) \text{ lies on } f^{-1}$$
- Intersections Along $y = x$: For strictly increasing functions, if $f(x)$ and $f^{-1}(x)$ intersect, their points of intersection must lie on the line of symmetry $y = x$. Hence, solving $f(x) = f^{-1}(x)$ is equivalent to solving the simpler equation $f(x) = x$.
(a) Find an expression for $f^{-1}(x)$.
(b) State the domain and range of $f^{-1}$.
Part (a): Algebraic inversion
Swap $x$ and $y$:
$$x = (y - 2)^2 + 3$$ $$x - 3 = (y - 2)^2$$Take square roots: $y - 2 = \pm\sqrt{x - 3}$.
Since the original domain is $x \ge 2$, the range of $f^{-1}$ must satisfy $y \ge 2$, so we select the positive branch:
$$y = 2 + \sqrt{x - 3}$$ $$f^{-1}(x) = 2 + \sqrt{x - 3}$$Part (b): Domain and range of $f^{-1}$
Since $x \ge 2$, the minimum value of $f(x)$ is $f(2) = 3$, so $\text{range } f = [3, \infty)$.
Domain of $f^{-1}$ $= \text{range } f = [3, \infty)$ (or $x \ge 3$).
Range of $f^{-1}$ $= \text{dom } f = [2, \infty)$ (or $y \ge 2$).
Identity & Self-Inverse Functions
- The Identity Function Property: Composing a function with its inverse in either order yields the identity function $I(x) = x$: $$(f \circ f^{-1})(x) = x \quad \text{for all } x \in \text{dom } f^{-1}$$ $$(f^{-1} \circ f)(x) = x \quad \text{for all } x \in \text{dom } f$$
- Self-Inverse Functions: A function $f$ is self-inverse if it equals its own inverse: $$f(x) = f^{-1}(x) \iff (f \circ f)(x) = x$$ Geometrically, the graph of a self-inverse function is its own reflection across the line $y = x$. Common examples: $f(x) = \frac{1}{x}$, $f(x) = -x + c$, and linear rationals where the top-left and bottom-right coefficients are opposite in sign ($f(x) = \frac{ax + b}{cx - a}$).
Step 1: Set up the composition $(f \circ f)(x) = f(f(x))$.
$$(f \circ f)(x) = \frac{4\left(\frac{4x + 1}{x - 4}\right) + 1}{\left(\frac{4x + 1}{x - 4}\right) - 4}$$Step 2: Multiply numerator and denominator by $(x - 4)$ to clear fractions.
$$(f \circ f)(x) = \frac{4(4x + 1) + 1(x - 4)}{(4x + 1) - 4(x - 4)}$$Step 3: Expand and simplify.
$$\text{Numerator: } 16x + 4 + x - 4 = 17x$$ $$\text{Denominator: } 4x + 1 - 4x + 16 = 17$$ $$(f \circ f)(x) = \frac{17x}{17} = x$$Since $(f \circ f)(x) = x$ for all $x \ne 4$, $f(x) = f^{-1}(x)$. Thus, $f$ is a self-inverse function. $\blacksquare$
Inverse-Identity Intersections & Domain Restrictions
(a) State the range of $f$. [2 marks]
(b) Find an expression for $f^{-1}(x)$ and state its domain. [4 marks]
(c) Solve the equation $f(x) = f^{-1}(x)$. [3 marks]
Part (a): Find the vertex and range of $f$:
$$f(x) = (x - 2)^2 - 4 + 6 = (x - 2)^2 + 2 \quad [M1]$$Since the minimum occurs at $x = 2$ where $f(2) = 2$, the range is $y \ge 2$ (or $f(x) \ge 2$) [A1].
Part (b): Find $f^{-1}(x)$ by swapping $x$ and $y$:
$$x = (y - 2)^2 + 2 \implies (y - 2)^2 = x - 2 \quad [M1]$$Take the positive square root because the original domain specifies $y \ge 2$:
$$y - 2 = \sqrt{x - 2} \implies y = 2 + \sqrt{x - 2} \quad [A1]$$ $$f^{-1}(x) = 2 + \sqrt{x - 2} \quad [A1]$$Domain of $f^{-1} = \text{Range of } f \implies x \ge 2$ [A1].
Part (c): Solve $f(x) = f^{-1}(x)$ using line of symmetry $y = x$:
Since the graphs of $f$ and $f^{-1}$ reflect across $y = x$, any intersection must lie on $y = x$ [R1]:
$$f(x) = x \implies x^2 - 4x + 6 = x \implies x^2 - 5x + 6 = 0 \quad [M1]$$ $$(x - 2)(x - 3) = 0 \implies x = 2 \quad \text{or} \quad x = 3 \quad [A1]$$Both roots satisfy $x \ge 2$.
Answers: (a) $y \ge 2$; (b) $f^{-1}(x) = 2 + \sqrt{x - 2}$, domain $x \ge 2$; (c) $x = 2$ or $x = 3$
10 Progressive Practice Exercises
Tier 1 tests core composition and linear inverting; Tier 2 introduces rational inverses, restricted quadratics, and parameter matching; Tier 3 challenges multi-step self-inverse proofs and real-world modeling.
Tier 1 Core Fluency (Problems 1–3)
Let $f(x) = 2x - 3$ and $g(x) = x^2 + 4$. Find the exact values of $(f \circ g)(2)$ and $(g \circ f)(2)$.
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Part (a): Find $(f \circ g)(2)$
$$g(2) = 2^2 + 4 = 8$$ $$(f \circ g)(2) = f(g(2)) = f(8) = 2(8) - 3 = 16 - 3 = 13$$Part (b): Find $(g \circ f)(2)$
$$f(2) = 2(2) - 3 = 1$$ $$(g \circ f)(2) = g(f(2)) = g(1) = 1^2 + 4 = 5$$Answer: $(f \circ g)(2) = 13$ and $(g \circ f)(2) = 5$. (Notice that $13 \ne 5$).
Find the inverse function $f^{-1}(x)$ for $f(x) = \frac{4x - 5}{3}$.
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Step 1: Swap $x$ and $y$:
$$x = \frac{4y - 5}{3}$$Step 2: Solve for $y$:
$$3x = 4y - 5 \implies 4y = 3x + 5 \implies y = \frac{3x + 5}{4}$$Answer: $f^{-1}(x) = \frac{3x + 5}{4}$ (or $f^{-1}(x) = 0.75x + 1.25$).
Let $f(x) = \sqrt{2x - 1}$ for $x \ge 0.5$ and $g(x) = \frac{x^2 + 1}{2}$ for $x \ge 0$. Algebraically verify that $(g \circ f)(x) = x$.
View Full Worked Solution
Substitute $f(x)$ into $g$:
$$(g \circ f)(x) = g(f(x)) = \frac{(\sqrt{2x - 1})^2 + 1}{2}$$Since $x \ge 0.5$, $2x - 1 \ge 0$, so $(\sqrt{2x - 1})^2 = 2x - 1$:
$$(g \circ f)(x) = \frac{(2x - 1) + 1}{2} = \frac{2x}{2} = x$$Since $(g \circ f)(x) = x$, $g$ is the inverse of $f$ on this domain. $\blacksquare$
Tier 2 Exam Modeling (Problems 4–7)
Consider the rational function $f(x) = \frac{2x - 1}{x + 3}$, where $x \ne -3$.
(a) Find $f^{-1}(x)$.
(b) State the domain and range of $f^{-1}$.
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Part (a): Find $f^{-1}(x)$
$$x = \frac{2y - 1}{y + 3} \implies x(y + 3) = 2y - 1 \implies xy + 3x = 2y - 1$$ $$xy - 2y = -3x - 1 \implies y(x - 2) = -(3x + 1) \implies y = \frac{3x + 1}{2 - x}$$ $$f^{-1}(x) = \frac{-3x - 1}{x - 2} = \frac{3x + 1}{2 - x}$$Part (b): Domain and Range
The horizontal asymptote of $f(x)$ is $y = \frac{2}{1} = 2$, so $\text{range } f = \mathbb{R} \setminus \{2\}$.
Therefore, $\text{dom } f^{-1} = \mathbb{R} \setminus \{2\}$ (or $x \ne 2$).
The vertical asymptote of $f(x)$ is $x = -3$, so $\text{range } f^{-1} = \text{dom } f = \mathbb{R} \setminus \{-3\}$ (or $y \ne -3$).
Let $f(x) = (x - 3)^2 + 4$ for $x \ge 3$. Find an expression for $f^{-1}(x)$ and state its domain.
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Step 1: Invert algebraically:
$$x = (y - 3)^2 + 4 \implies (y - 3)^2 = x - 4$$Since $x \ge 3$, $y \ge 3$, so we take the positive square root:
$$y - 3 = \sqrt{x - 4} \implies y = 3 + \sqrt{x - 4}$$ $$f^{-1}(x) = 3 + \sqrt{x - 4}$$Step 2: Determine domain of $f^{-1}$:
For $x \ge 3$, the minimum value of $f(x)$ is $f(3) = 4$, so $\text{range } f = [4, \infty)$.
Answer: $f^{-1}(x) = 3 + \sqrt{x - 4}$ with domain $\{x \in \mathbb{R} \mid x \ge 4\}$.
Let $f(x) = 3x + 2$ and $g(x) = \frac{6}{x - 1}$ for $x \ne 1$. Solve the equation $(f \circ g)(x) = 8$.
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Method 1: Direct expansion
$$(f \circ g)(x) = f(g(x)) = 3\left(\frac{6}{x - 1}\right) + 2 = \frac{18}{x - 1} + 2$$ $$\frac{18}{x - 1} + 2 = 8 \implies \frac{18}{x - 1} = 6$$ $$18 = 6(x - 1) \implies 3 = x - 1 \implies x = 4$$Method 2: Using inverse $f^{-1}$
$$f(g(x)) = 8 \implies g(x) = f^{-1}(8)$$ $$f^{-1}(x) = \frac{x - 2}{3} \implies f^{-1}(8) = \frac{8 - 2}{3} = 2$$ $$g(x) = 2 \implies \frac{6}{x - 1} = 2 \implies x - 1 = 3 \implies x = 4$$Answer: $x = 4$.
Given $f(x) = 2x + k$ and $g(x) = 3x - 4$, find the value of the constant $k$ such that $(f \circ g)(1) = (g \circ f)(1)$.
View Full Worked Solution
Step 1: Evaluate LHS $(f \circ g)(1)$:
$$g(1) = 3(1) - 4 = -1$$ $$(f \circ g)(1) = f(-1) = 2(-1) + k = k - 2$$Step 2: Evaluate RHS $(g \circ f)(1)$:
$$f(1) = 2(1) + k = 2 + k$$ $$(g \circ f)(1) = g(2 + k) = 3(2 + k) - 4 = 6 + 3k - 4 = 3k + 2$$Step 3: Equate LHS and RHS:
$$k - 2 = 3k + 2 \implies 2k = -4 \implies k = -2$$Answer: $k = -2$.
Tier 3 Complex Applications & Synthesis (Problems 8–10)
Let $f(x) = \sqrt{2x + 3}$ for $x \ge -1.5$.
(a) Find an expression for $f^{-1}(x)$ and state its domain.
(b) Find the exact coordinates of the point of intersection of the graphs of $y = f(x)$ and $y = f^{-1}(x)$.
View Full Worked Solution
Part (a): Find $f^{-1}(x)$
$$x = \sqrt{2y + 3} \implies x^2 = 2y + 3 \implies y = \frac{x^2 - 3}{2}$$Since $\text{range } f = [0, \infty)$, the domain of $f^{-1}$ is $x \ge 0$.
$$f^{-1}(x) = \frac{x^2 - 3}{2}, \quad x \ge 0$$Part (b): Find intersection point
Since $f(x)$ is strictly increasing, intersection points between $f$ and $f^{-1}$ lie along $y = x$. We solve $f(x) = x$:
$$\sqrt{2x + 3} = x \quad (x \ge 0)$$ $$2x + 3 = x^2 \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0$$Since $x \ge 0$, we reject $x = -1$. Therefore $x = 3$.
Since the point lies on $y = x$, $y = 3$.
Answer: $(3, 3)$.
A rational function is given by $f(x) = \frac{ax + 5}{3x - 2}$ for $x \ne \frac{2}{3}$, where $a$ is a constant. Given that $f$ is a self-inverse function, find the value of $a$.
View Full Worked Solution
Method 1: By finding $f^{-1}(x)$
$$x = \frac{ay + 5}{3y - 2} \implies x(3y - 2) = ay + 5 \implies 3xy - 2x = ay + 5$$ $$3xy - ay = 2x + 5 \implies y(3x - a) = 2x + 5 \implies f^{-1}(x) = \frac{2x + 5}{3x - a}$$Since $f$ is self-inverse, $f(x) \equiv f^{-1}(x)$:
$$\frac{ax + 5}{3x - 2} = \frac{2x + 5}{3x - a}$$Comparing the numerators and denominators directly gives $a = 2$.
Answer: $a = 2$.
A manufacturing process converts raw material mass $m$ (in kg) into product volume $V(m)$ (in liters) according to:
$$V(m) = 2.5m + 4, \quad m \ge 0$$
The total production cost $C$ (in dollars) for a product volume $V$ is given by:
$$C(V) = 0.80V + 15$$
(a) Find an expression for $(C \circ V)(m)$ and interpret its practical meaning.
(b) Find the inverse function $(C \circ V)^{-1}(c)$.
(c) Calculate the mass of raw material required when the total production cost is $\$79.00$.
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Part (a): Composite function
$$(C \circ V)(m) = C(V(m)) = 0.80(2.5m + 4) + 15 = 2.00m + 3.20 + 15 = 2m + 18.20$$Practical meaning: $(C \circ V)(m)$ gives the total production cost in dollars directly as a function of the mass of raw material $m$ (in kg).
Part (b): Inverse function
Set $c = 2m + 18.20$ and solve for $m$:
$$2m = c - 18.20 \implies m = \frac{c - 18.20}{2} = 0.5c - 9.10$$ $$(C \circ V)^{-1}(c) = \frac{c - 18.20}{2}$$Part (c): Mass for cost $\$79.00$
$$m = \frac{79 - 18.20}{2} = \frac{60.80}{2} = 30.4\text{ kg}$$Answer: (a) $(C \circ V)(m) = 2m + 18.20$; (b) $(C \circ V)^{-1}(c) = \frac{c - 18.20}{2}$; (c) $30.4\text{ kg}$.
Composition & Inverse Functions
- Composite Function: $$(f \circ g)(x) = f(g(x))$$
- Identity Property: $$(f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$$
Domain, Range & Inversion Identities
- Domain & Range Exchange: $$\text{dom } f^{-1} = \text{range } f \qquad \text{and} \qquad \text{range } f^{-1} = \text{dom } f$$
- Geometric Symmetry: $$\text{Graph of } f^{-1} \text{ is the reflection of } f \text{ across } y = x$$ $$(a, b) \in f \iff (b, a) \in f^{-1}$$
- Self-Inverse Function Condition: $$f(x) = f^{-1}(x) \iff (f \circ f)(x) = x$$
- Solving $f(x) = f^{-1}(x)$: $$\text{Solve } f(x) = x \quad (\text{for strictly increasing functions})$$
Critical IB Examination Pitfalls
- Confusing $f^{-1}(x)$ with $[f(x)]^{-1}$: $f^{-1}(x)$ is the inverse function, while $[f(x)]^{-1} = \frac{1}{f(x)}$ is the reciprocal. Writing $\frac{1}{f(x)}$ for the inverse receives an automatic zero.
- Reversing Composition Order: Writing $g(f(x))$ when asked for $(f \circ g)(x)$ is one of the most frequent errors in Paper 1. Always evaluate the right-hand function first.
- Neglecting the Restricted Domain of $f^{-1}$: When inverting a quadratic $f(x) = (x-3)^2 + 2$ with $x \ge 3$, the inverse is $f^{-1}(x) = 3 + \sqrt{x - 2}$. It is only defined for $x \ge 2$ ($\text{range } f$). Forgetting to state this domain loses full credit.
- Choosing the Wrong Sign Branch: When taking square roots in quadratic inverses $y - h = \pm\sqrt{x - k}$, choose $+$ if the original domain was $x \ge h$, and choose $-$ if the original domain was $x \le h$.
IB AA SL Topic Check: Composite & Inverse Functions
Section A has diagnostic questions with feedback. Section B has Paper 1 and Paper 2 exam-style problems with markschemes.
Section A: Diagnostic Concept Check
Score: 0 / 4Question 1: What is the inverse function of $f(x) = 3x - 7$?
Question 2: Given $f(x) = x^2 - 1$ and $g(x) = 2x + 1$, what is the value of $(f \circ g)(2)$?
Question 3: What is the domain of $f^{-1}(x)$ if $f(x) = \sqrt{x - 4} + 2$ for $x \ge 4$?
Question 4: What value of $a$ makes the rational function $f(x) = \frac{ax + 2}{x - 3}$ a self-inverse function?
Section B: Exam-Style Free Response Questions
Question 5: Rational Function Inverse
Consider the rational function $f(x) = \frac{2x + 5}{x - 1}$, where $x \ne 1$.
(a) Find the inverse function $f^{-1}(x)$. [4 marks]
(b) State the domain and range of $f^{-1}$. [2 marks]
View Official Markscheme & Model Solution
Part (a) [4 Marks]:
Let $y = \frac{2x + 5}{x - 1}$. Swap variables $x$ and $y$:
$$x = \frac{2y + 5}{y - 1} \quad \text{\bf (M1)}$$ $$x(y - 1) = 2y + 5 \implies xy - x = 2y + 5 \quad \text{\bf (A1)}$$ $$xy - 2y = x + 5 \implies y(x - 2) = x + 5 \quad \text{\bf (M1)}$$ $$f^{-1}(x) = \frac{x + 5}{x - 2} \quad \text{\bf (A1)}$$Part (b) [2 Marks]:
$\text{Domain of } f^{-1}$: $\{x \in \mathbb{R} \mid x \ne 2\}$ (A1)
$\text{Range of } f^{-1}$: $\{y \in \mathbb{R} \mid y \ne 1\}$ (A1)
Question 6: Restricted Quadratic & Intersection
Let $f(x) = x^2 - 6x + 11$ for $x \ge 3$.
(a) Write $f(x)$ in the form $(x - h)^2 + k$. [2 marks]
(b) Find an expression for $f^{-1}(x)$. [3 marks]
(c) Find the coordinates of the point of intersection of the graphs of $f$ and $f^{-1}$. [2 marks]
View Official Markscheme & Model Solution
Part (a) [2 Marks]:
$$f(x) = (x^2 - 6x + 9) - 9 + 11 = (x - 3)^2 + 2 \quad \text{\bf (M1)(A1)}$$Part (b) [3 Marks]:
$$x = (y - 3)^2 + 2 \implies (y - 3)^2 = x - 2 \quad \text{\bf (M1)}$$Since $x \ge 3$, $y \ge 3$, choose positive root: (R1)
$$y - 3 = \sqrt{x - 2} \implies f^{-1}(x) = 3 + \sqrt{x - 2} \quad (\text{for } x \ge 2) \quad \text{\bf (A1)}$$Part (c) [2 Marks]:
Since $f(x)$ is increasing for $x \ge 3$, intersections with $f^{-1}$ occur on $y = x$:
$$(x - 3)^2 + 2 = x \implies x^2 - 6x + 9 + 2 = x \quad \text{\bf (M1)}$$ $$x^2 - 7x + 11 = 0$$ $$x = \frac{7 \pm \sqrt{49 - 44}}{2} = \frac{7 \pm \sqrt{5}}{2}$$Since $x \ge 3$, $\frac{7 - \sqrt{5}}{2} \approx 2.38 < 3$ is rejected. Thus $x = \frac{7 + \sqrt{5}}{2} \approx 4.62$.
Intersection coordinates: $\left(\frac{7 + \sqrt{5}}{2}, \frac{7 + \sqrt{5}}{2}\right)$ (or $(4.62, 4.62)$). (A1)
Topic 2 Curriculum
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About this guide
Independent SJMaths resource; not an official IB publication. Follows the AA syllabus first assessed in 2021 (final assessment: November 2028). Use the separate 2029 guide for the new course.
Check the IB Mathematics programme page and the official AA subject guide for the authoritative syllabus and examination-session requirements.