What are the key algebraic methods covered in Algebraic Methods 2 for Edexcel International A Levels?

In Algebraic Methods 2, part of the Pure Mathematics 3 syllabus for Edexcel International A Levels, students focus on advanced algebraic techniques. This includes solving polynomial equations, manipulating algebraic fractions, and using partial fractions. These methods are essential for simplifying complex expressions and solving higher-level mathematical problems.

How do you solve polynomial equations in Algebraic Methods 2?

To solve polynomial equations in Algebraic Methods 2, students typically use techniques such as factoring, the quadratic formula, and synthetic division. For higher-degree polynomials, the Rational Root Theorem and Descartes' Rule of Signs can be helpful. Understanding these methods is crucial for tackling polynomial equations effectively in the Edexcel International A Levels curriculum.

What is the process of decomposing a fraction into partial fractions?

Decomposing a fraction into partial fractions involves expressing a complex rational expression as a sum of simpler fractions. This is done by equating the original fraction to a sum of fractions with unknown coefficients and solving for these coefficients. This technique is particularly useful in integration and is a key component of Algebraic Methods 2 in Pure Mathematics 3.

Why is understanding algebraic fractions important in Algebraic Methods 2?

Algebraic fractions are crucial in Algebraic Methods 2 because they allow students to simplify and manipulate expressions that involve division of polynomials. Mastery of algebraic fractions is essential for solving equations and inequalities, and for performing operations such as addition, subtraction, multiplication, and division on rational expressions, which are common in the Edexcel International A Levels Mathematics curriculum.

How are algebraic methods applied in solving real-world problems?

Algebraic methods, as taught in Algebraic Methods 2, are applied in solving real-world problems by modeling situations with equations and inequalities. These methods help in optimizing solutions, analyzing trends, and making predictions. For instance, they are used in fields such as engineering, economics, and physics, where mathematical models are essential for decision-making and problem-solving.

Why is "Squaring $|f(x)| = g(x)$ and not checking for extraneous roots" a common mistake in Algebraic methods 2?

Why it happens: Squaring removes the modulus but introduces solutions where g(x) < 0 (which were never valid). How to avoid it: After squaring and solving, substitute each candidate root back into the original equation OR into g(x) ≥ 0. Reject any that fail. Source: WMA13/01 examiner reports — flagged annually

Why is "$|f(x)| > a$ written as $-a < f(x) < a$" a common mistake in Algebraic methods 2?

Why it happens: Memorising one form (|f(x)| a is a SPLIT (outside the bounds). Sketch y = |f(x)| and y = a if unsure.

Why is "Reading $fg(x)$ as 'apply $f$ first'" a common mistake in Algebraic methods 2?

Why it happens: Natural left-to-right reading. How to avoid it: fg(x) = f(g(x)) — the function CLOSEST to x is applied FIRST. Always evaluate from the inside out.

Why is "Quoting an inverse without checking that $f$ is one-to-one" a common mistake in Algebraic methods 2?

Why it happens: Treating the algebraic rearrangement as the whole story. How to avoid it: Before finding f^-1, confirm f is one-to-one over its given domain (look at the graph or compute f'(x) and check sign). For f(x) = x^2 on R, no inverse exists; restrict the domain (e.g. x ≥ 0) first. Source: WMA13/01 examiner reports — flagged annually

Why is "Stating $f^{-1}(x) = (x - 3)^{2} + 2$ without quoting its domain" a common mistake in Algebraic methods 2?

Why it happens: Focusing on the algebra and forgetting to transfer the range of f to the domain of f^-1. How to avoid it: ALWAYS write 'domain of f^-1 = range of f' explicitly. Without the domain restriction the expression is not a true function (e.g. (x - 3)^2 + 2 is a parabola with two branches — only one branch is the inverse).

Pure Mathematics 3Edexcel International A Levels Mathematics (XMA01-YMA01)

Algebraic methods 2

Q: Why is "$|f(x)| > a$ written as $-a < f(x) < a$" a common mistake in Algebraic methods 2?

Why it happens: Memorising one form (|f(x)| a is a SPLIT (outside the bounds). Sketch y = |f(x)| and y = a if unsure.

Q: Why is "Reading $fg(x)$ as 'apply $f$ first'" a common mistake in Algebraic methods 2?

Why it happens: Natural left-to-right reading. How to avoid it: fg(x) = f(g(x)) — the function CLOSEST to x is applied FIRST. Always evaluate from the inside out.

Q: Why is "Quoting an inverse without checking that $f$ is one-to-one" a common mistake in Algebraic methods 2?

Why it happens: Treating the algebraic rearrangement as the whole story. How to avoid it: Before finding f^-1, confirm f is one-to-one over its given domain (look at the graph or compute f'(x) and check sign). For f(x) = x^2 on R, no inverse exists; restrict the domain (e.g. x ≥ 0) first. Source: WMA13/01 examiner reports — flagged annually

Q: Why is "Stating $f^{-1}(x) = (x - 3)^{2} + 2$ without quoting its domain" a common mistake in Algebraic methods 2?

Why it happens: Focusing on the algebra and forgetting to transfer the range of f to the domain of f^-1. How to avoid it: ALWAYS write 'domain of f^-1 = range of f' explicitly. Without the domain restriction the expression is not a true function (e.g. (x - 3)^2 + 2 is a parabola with two branches — only one branch is the inverse).

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Algebraic Methods 2 Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

The modulus function and its graph $y = |f(x)|$ ; equations and inequalities involving $|f(x)|$ ; composite functions $fg(x)$ ; inverse functions $f^{-1}(x)$ and their existence (one-to-one), domain/range relationships, and the reflection in $y = x$ . The bedrock of P3 function theory.

At a glance

Modulus: $|x| = x$ if $x \geq 0$ ; $|x| = -x$ if $x < 0$ . Always non-negative.
$|f(x)| = a$ (a $\geq 0$ ): split into $f(x) = a$ or $f(x) = -a$ .
$|f(x)| < a$ : SANDWICH $-a < f(x) < a$ . $|f(x)| > a$ : SPLIT $f(x) > a$ or $f(x) < -a$ .
$|f(x)| = |g(x)|$ : square both sides OR split into $f(x) = \pm g(x)$ .
Composite $fg(x) = f(g(x))$ . Order matters; $fg \neq gf$ in general.
Inverse $f^{-1}$ exists iff $f$ is one-to-one. Domain of $f^{-1}$ = range of $f$ .
Graphical: $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in $y = x$ . Gradients are reciprocals at corresponding points.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P3 1.1 — Solve equations and inequalities involving the modulus function.
P3 1.2 — Form composite functions $fg(x)$ and find their domain and range.
P3 1.3 — Determine whether a function has an inverse; find the inverse $f^{-1}(x)$ and state its domain.
P3 1.4 — Use the geometric relationship between the graphs of $y = f(x)$ and $y = f^{-1}(x)$ (reflection in $y = x$ ).

The modulus function

Always non-negative. Defined by cases.

Definition. The modulus (or absolute value) of a real number $x$ is $|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}.$

Examples: $|3| = 3$ , $|-2| = 2$ , $|0| = 0$ .

Geometric interpretation. $|x|$ is the distance of $x$ from $0$ on the real line. So $|x - a|$ is the distance between $x$ and $a$ .

Properties.

$|x| \geq 0$ always.
$|xy| = |x| \cdot |y|$ .
$|x + y| \leq |x| + |y|$ (triangle inequality).
$\big| |x| - |y| \big| \leq |x - y|$ (reverse triangle inequality — mentioned briefly in P3).

Common P3 modulus-form expressions. $|2x - 3|$ , $|f(x) - g(x)|$ , $|\sin x|$ .

$|x| \geq 0$ always.
$|x| = x$ if $x \geq 0$ ; $-x$ if $x < 0$ .
Distance interpretation: $|x - a|$ = distance $x \to a$ .
Triangle inequality $|x + y| \leq |x| + |y|$ .

Modulus equations

Split into cases — or square — then check.

Case (a): $|f(x)| = a$ with $a \geq 0$ . Split into two equations and combine the solution sets: $f(x) = a \quad \text{or} \quad f(x) = -a.$

If $a < 0$ , NO solutions (modulus is non-negative).

Worked example. $|2x - 3| = 5 \Rightarrow 2x - 3 = 5$ or $2x - 3 = -5 \Rightarrow x = 4$ or $x = -1$ .

Case (b): $|f(x)| = g(x)$ . Two equations again, BUT we additionally need $g(x) \geq 0$ for any solution (since $|f(x)| \geq 0$ ).

Worked example. $|2x - 3| = x + 4$ . First require $x + 4 \geq 0$ , i.e. $x \geq -4$ .

$2x - 3 = x + 4 \Rightarrow x = 7$ (passes the condition).
$2x - 3 = -(x + 4) \Rightarrow 3x = -1 \Rightarrow x = -\tfrac{1}{3}$ (passes).

Both accepted.

Case (c): $|f(x)| = |g(x)|$ . Two moduli — both non-negative — so squaring is safe. $f(x)^{2} = g(x)^{2}.$

Equivalent: $f(x) = g(x)$ or $f(x) = -g(x)$ .

Worked example. $|2x + 1| = |x - 3|$ . Square: $(2x + 1)^{2} = (x - 3)^{2}$ . Expand and simplify: $3x^{2} + 10x - 8 = 0 \Rightarrow x = \tfrac{2}{3}$ or $x = -4$ .

$|f(x)| = a$ : two cases $f(x) = \pm a$ .
$|f(x)| = g(x)$ : two cases AND check $g(x) \geq 0$ .
$|f(x)| = |g(x)|$ : square both sides.
Always verify final solutions in the ORIGINAL equation.

Modulus inequalities

$<$ gives a sandwich; $>$ gives a split.

$|f(x)| < a$ ( $a > 0$ ): equivalent to $-a < f(x) < a.$

A SANDWICH (the values inside the bounds).

$|f(x)| > a$ ( $a > 0$ ): equivalent to $f(x) > a \quad \text{or} \quad f(x) < -a.$

A SPLIT (the values outside the bounds).

Worked example. Solve $|3x + 2| > 4$ .

$3x + 2 > 4 \Rightarrow x > \tfrac{2}{3}$ .
$3x + 2 < -4 \Rightarrow x < -2$ .

Union: $x < -2$ or $x > \tfrac{2}{3}$ .

Worked example. Solve $|x - 1| \leq 3$ . $-3 \leq x - 1 \leq 3 \Rightarrow -2 \leq x \leq 4.$

Strict vs non-strict. $\leq$ instead of $<$ just includes the endpoints. The structure is identical.

$|f(x)| < a \Leftrightarrow -a < f(x) < a$ (SANDWICH).
$|f(x)| > a \Leftrightarrow f(x) > a$ or $f(x) < -a$ (SPLIT).
Strict vs non-strict: just endpoint inclusion.
Always combine the two pieces into a single solution set.

Composite functions $fg(x)$

$fg(x) = f(g(x))$ . Inner first.

Definition. $fg(x) = f(g(x))$ — apply $g$ first, then $f$ .

Order matters. In general $fg(x) \neq gf(x)$ .

Worked example. $f(x) = 2x + 1$ , $g(x) = x^{2}$ .

$fg(x) = f(g(x)) = f(x^{2}) = 2x^{2} + 1$ .
$gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)^{2} = 4x^{2} + 4x + 1$ .

Very different!

Domain of $fg$ . $x$ must lie in domain of $g$ , AND $g(x)$ must lie in domain of $f$ : $\text{dom}(fg) = \{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}.$

Worked example. $f(x) = \ln x$ (dom $x > 0$ ), $g(x) = x - 2$ . Then $fg(x) = \ln(x - 2)$ .

$x \in \text{dom}(g)$ : always (any real).
$g(x) > 0$ : $x - 2 > 0 \Rightarrow x > 2$ .

So dom $(fg) = \{x : x > 2\}$ .

Range of $fg$ . First find the range of $g$ (restricted to dom $(fg)$ ), then apply $f$ to that set.

$fg(x) = f(g(x))$ — apply $g$ first.
$fg \neq gf$ in general.
Domain: $x \in$ dom $(g)$ AND $g(x) \in$ dom $(f)$ .
Range: apply $f$ to the range of $g$ over the domain of $fg$ .

Inverse functions $f^{-1}$

Exist iff $f$ is one-to-one. Graph: reflection in $y = x$ .

Existence. $f^{-1}$ exists if and only if $f$ is one-to-one (injective): $f(a) = f(b) \Rightarrow a = b.$

Equivalently: every horizontal line meets the graph of $y = f(x)$ at most once.

Algebraic method to find $f^{-1}$ .

Write $y = f(x)$ .
Rearrange to make $x$ the subject.
Swap $x$ and $y$ to write the inverse as a function of $x$ .
State the domain of $f^{-1}$ = range of $f$ .

Worked example. $f(x) = 3 + \sqrt{x - 2}$ , $x \geq 2$ .

Range of $f$ : $\sqrt{x - 2} \geq 0 \Rightarrow f(x) \geq 3$ . So range = $[3, \infty)$ .
$y = 3 + \sqrt{x - 2} \Rightarrow y - 3 = \sqrt{x - 2} \Rightarrow (y - 3)^{2} = x - 2 \Rightarrow x = (y - 3)^{2} + 2$ .
Swap: $f^{-1}(x) = (x - 3)^{2} + 2$ .
Domain of $f^{-1}$ : $x \geq 3$ .

Graphical relationship. $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ .

$(a, b)$ on $f$ ⇔ $(b, a)$ on $f^{-1}$ .
Gradient at $(a, b)$ on $f$ = $1/$ gradient at $(b, a)$ on $f^{-1}$ (provided non-zero).

Self-inverse functions. Some functions are their own inverse — their graph is symmetric in $y = x$ . Examples: $f(x) = 1/x$ , $f(x) = -x$ , $f(x) = a - x$ .

Restricting the domain. If $f$ is not one-to-one over its natural domain, you can RESTRICT the domain to make it one-to-one. E.g. $f(x) = x^{2}$ has no inverse over $\mathbb{R}$ ; restricted to $x \geq 0$ , $f^{-1}(x) = \sqrt{x}$ .

Inverse exists iff $f$ is one-to-one.
Algorithm: $y = f(x)$ → solve for $x$ → swap variables.
dom $(f^{-1})$ = range $(f)$ , range $(f^{-1})$ = dom $(f)$ .
Graph: reflect $y = f(x)$ in $y = x$ .
Gradients at corresponding points are reciprocals.

How it’s examined

Algebraic Methods 2 sits in the opening pages of every WMA13/01 paper as a 'warm-up' worth $4$ – $8$ marks. Typical question types: solve $|f(x)| = g(x)$ ( $4$ - $5$ marks); solve $|f(x)| < a$ or $> a$ ( $3$ - $4$ marks); state range / find inverse with domain ( $6$ - $8$ marks); identify the corresponding point on $f^{-1}$ given a point on $f$ and link the gradients. Mark schemes are STRICT on stating domains — students who quote $f^{-1}(x)$ without 'domain of $f^{-1}$ is …' lose B marks every time.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Algebraic methods 2

Step-by-step solutions to past-paper-style questions on algebraic methods 2, written exactly the way a tutor would explain them at the board.

1Solve a modulus equation $|2x - 3| = x + 4$ (5 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q3• modulus, equation
Question
Solve $|2x - 3| = x + 4$ .
Step-by-step solution
1. Step 1
  The right-hand side must be non-negative for any solution to exist, so we require $x + 4 \geq 0$ , i.e. $x \geq -4$ .
2. Step 2
  Case 1: $2x - 3 \geq 0$ , i.e. $x \geq \tfrac{3}{2}$ . Then $|2x - 3| = 2x - 3$ .
  $2x - 3 = x + 4 \;\Rightarrow\; x = 7$
3. Step 3
  Check $x = 7$ against the case condition $x \geq \tfrac{3}{2}$ . ✓ Accept.
4. Step 4
  Case 2: $2x - 3 < 0$ , i.e. $x < \tfrac{3}{2}$ . Then $|2x - 3| = -(2x - 3) = 3 - 2x$ .
  $3 - 2x = x + 4 \;\Rightarrow\; -3x = 1 \;\Rightarrow\; x = -\tfrac{1}{3}$
5. Step 5
  Check $x = -\tfrac{1}{3}$ against the case condition $x < \tfrac{3}{2}$ . ✓ Accept. Both solutions also satisfy $x \geq -4$ .
Answer
$x = 7$ or $x = -\tfrac{1}{3}$ .
Examiner tip
M1 for splitting into two cases (or squaring both sides). A1 for $x = 7$ . M1 for the second case. A1 for $x = -\tfrac{1}{3}$ . A1 for both correct and no extraneous roots. Common loss-of-mark: students who square both sides forget to check candidate roots against the original equation and quote spurious solutions.
2Solve a modulus inequality $|3x + 2| > 4$ (4 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q2• modulus, inequality
Question
Solve the inequality $|3x + 2| > 4$ .
Step-by-step solution
1. Step 1
  $|y| > 4 \Leftrightarrow y > 4$ or $y < -4$ . Apply with $y = 3x + 2$ .
2. Step 2
  Case A:
  $3x + 2 > 4 \;\Rightarrow\; 3x > 2 \;\Rightarrow\; x > \tfrac{2}{3}$
3. Step 3
  Case B:
  $3x + 2 < -4 \;\Rightarrow\; 3x < -6 \;\Rightarrow\; x < -2$
4. Step 4
  Take the union: $x < -2$ or $x > \tfrac{2}{3}$ .
Answer
$x < -2$ or $x > \tfrac{2}{3}$ .
Examiner tip
M1 for the disjunction $3x + 2 > 4$ or $3x + 2 < -4$ . A1 each branch. A1 for the correct combined solution set. Students who write the strict inequality the wrong way around ( $-4 < 3x + 2 < 4$ instead of the disjunction) lose all four marks — that case answers $|3x + 2| < 4$ .
3Composite function $fg(x)$ and domain reasoning (6 marks, P3)
Extended• Adapted from WMA13/01 January 2023 Q6• composite function, domain
Question
$f(x) = \dfrac{2}{x - 1}$ , $x \in \mathbb{R}, x \neq 1$ . $g(x) = x^{2} + 3$ , $x \in \mathbb{R}$ . (a) Find $fg(x)$ and state its domain. (b) Solve $fg(x) = 1$ .
Step-by-step solution
1. Step 1
  $fg(x) = f(g(x)) = f(x^{2} + 3) = \dfrac{2}{(x^{2} + 3) - 1} = \dfrac{2}{x^{2} + 2}$ .
2. Step 2
  Domain of $fg$ : any real $x$ for which $g(x)$ is defined AND $g(x)$ lies in the domain of $f$ . $g(x) = x^{2} + 3 \geq 3 > 1$ for all real $x$ , so $g(x) \neq 1$ automatically and the domain is $x \in \mathbb{R}$ .
3. Step 3
  (b) Solve $\dfrac{2}{x^{2} + 2} = 1 \Rightarrow 2 = x^{2} + 2 \Rightarrow x^{2} = 0$ .
4. Step 4
  Hence $x = 0$ .
Answer
(a) $fg(x) = \dfrac{2}{x^{2} + 2}$ , $x \in \mathbb{R}$ . (b) $x = 0$ .
Examiner tip
(a) M1 for substituting $g(x)$ into $f$ . A1 for the simplified expression. B1 for the correct domain (must justify that $x^{2} + 3 \neq 1$ holds automatically). (b) M1 for setting equal to $1$ and clearing the fraction. A1 for $x = 0$ . Common error: stating the domain as $x \neq 1$ — that is the domain of $f$ , not of $fg$ .
4Find an inverse function with domain/range (7 marks, P3)
Extended• Adapted from WMA13/01 June 2023 Q5• inverse function, domain, range
Question
$f(x) = 3 + \sqrt{x - 2}$ , $x \geq 2$ . (a) State the range of $f$ . (b) Find $f^{-1}(x)$ and state its domain.
Step-by-step solution
1. Step 1
  (a) For $x \geq 2$ , $\sqrt{x - 2} \geq 0$ , so $f(x) \geq 3$ . Range: $f(x) \geq 3$ .
2. Step 2
  (b) Set $y = 3 + \sqrt{x - 2}$ and solve for $x$ .
  $y - 3 = \sqrt{x - 2} \;\Rightarrow\; (y - 3)^{2} = x - 2 \;\Rightarrow\; x = (y - 3)^{2} + 2$
3. Step 3
  Swap variables to write the inverse in terms of $x$ :
  $f^{-1}(x) = (x - 3)^{2} + 2$
4. Step 4
  Domain of $f^{-1}$ = range of $f = \{x : x \geq 3\}$ . So domain of $f^{-1}$ is $x \geq 3$ .
Answer
(a) Range $f(x) \geq 3$ . (b) $f^{-1}(x) = (x - 3)^{2} + 2$ , $x \geq 3$ .
Examiner tip
(a) B1 range. (b) M1 swap $y \leftrightarrow x$ or rearrange. A1 squaring both sides. A1 final $f^{-1}(x)$ . B1 stating the domain of $f^{-1}$ equals the range of $f$ . The squaring step is reversible only because the original range fixes the branch $y - 3 \geq 0$ — without this, $f^{-1}$ would not be a function.
5Reflection $y = f^{-1}(x)$ in $y = x$ (4 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q4• inverse function, graph, reflection
Question
The point $(5, 2)$ lies on the graph of $y = f(x)$ and $f$ is one-to-one. (a) Write down the corresponding point on $y = f^{-1}(x)$ . (b) The line $y = f(x)$ has gradient $\tfrac{1}{4}$ at $(5, 2)$ . State the gradient of $y = f^{-1}(x)$ at the corresponding point.
Step-by-step solution
1. Step 1
  (a) Reflection in $y = x$ swaps coordinates: $(5, 2) \to (2, 5)$ .
2. Step 2
  (b) Reflection in $y = x$ inverts the gradient (corresponding to $\dfrac{dy}{dx}$ becoming $\dfrac{dx}{dy} = 1 / \dfrac{dy}{dx}$ ).
  $\text{Gradient of } f^{-1} \text{ at } (2, 5) = \dfrac{1}{1/4} = 4$
Answer
(a) $(2, 5)$ . (b) Gradient $= 4$ .
Examiner tip
(a) B1. (b) B1 stating gradient inverts; B1 for the value $4$ . The key fact: at corresponding points on $f$ and $f^{-1}$ , gradients are reciprocals. This anticipates the parametric / implicit differentiation theme of P3 Topic 6.

Key Formulae — Algebraic methods 2

The formulae you need to memorise for algebraic methods 2 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Modulus (absolute value)
$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$
$|x|$
non-negative distance of $x$ from $0$ on the number line
When to use
Whenever a modulus appears in an equation, inequality, or graph sketch — split into cases according to the sign of the expression inside. NOT in the IAL formula booklet — definition by case.
Example
$|2x - 3| = 2x - 3$ when $x \geq \tfrac{3}{2}$ ; $|2x - 3| = 3 - 2x$ when $x < \tfrac{3}{2}$ .
Modulus equation $|f(x)| = a$ and $|f(x)| = g(x)$
$|f(x)| = a \;\Leftrightarrow\; f(x) = a \text{ or } f(x) = -a \quad (a \geq 0)$
$f(x)$
any expression inside the modulus
$a$
non-negative constant (or $g(x)$ with check $g(x) \geq 0$ )
When to use
Solving modulus equations. If the RHS is a function $g(x)$ rather than a constant, after solving both branches check each candidate against $g(x) \geq 0$ AND against the original equation. NOT in the formula booklet.
Example
$|x - 1| = 5 \Rightarrow x = 6$ or $x = -4$ .
Modulus inequalities
$|f(x)| < a \Leftrightarrow -a < f(x) < a, \qquad |f(x)| > a \Leftrightarrow f(x) > a \text{ or } f(x) < -a$
$f(x)$
any expression inside the modulus
$a$
positive constant
When to use
Translating $<$ or $>$ inequalities involving a modulus into a conjunction (sandwich) or a disjunction. Memorise the direction: $<$ gives a sandwich, $>$ gives a 'two-pronged' OR. NOT in the formula booklet.
Example
$|3x + 2| > 4 \Rightarrow x < -2$ or $x > \tfrac{2}{3}$ .
Composite function
$(fg)(x) = f(g(x))$
$f, g$
two functions
$(fg)(x)$
apply $g$ first, then $f$
When to use
Forming a composite. Domain of $fg$ = $\{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}$ . Order matters: $fg(x) \neq gf(x)$ in general. NOT in the formula booklet.
Example
$f(x) = 2x + 1$ , $g(x) = x^{2}$ . Then $fg(x) = 2x^{2} + 1$ , $gf(x) = (2x + 1)^{2}$ .
Inverse function
$y = f(x) \;\Leftrightarrow\; x = f^{-1}(y), \qquad ff^{-1}(x) = f^{-1}f(x) = x$
$f^{-1}$
inverse function (exists iff $f$ is one-to-one)
$\text{dom}(f^{-1})$
= range $(f)$
$\text{range}(f^{-1})$
= dom $(f)$
When to use
Finding $f^{-1}$ : set $y = f(x)$ , rearrange for $x$ , then swap variables. The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ . NOT in the formula booklet.
Example
$f(x) = 3 + \sqrt{x - 2}$ , $x \geq 2 \Rightarrow f^{-1}(x) = (x - 3)^{2} + 2$ , $x \geq 3$ .

Key Definitions and Keywords — Algebraic methods 2

Definitions to memorise and the exact keywords mark schemes credit for algebraic methods 2 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Modulus function $|f(x)|$
Examiner keyword
The non-negative version of $f(x)$ : equals $f(x)$ where $f(x) \geq 0$ and equals $-f(x)$ where $f(x) < 0$ . Graph: reflect the parts of $y = f(x)$ that lie below the $x$ -axis in the $x$ -axis.
One-to-one (injective) function
Examiner keyword
A function $f$ is one-to-one if $f(a) = f(b) \Rightarrow a = b$ — equivalently, every horizontal line crosses the graph of $y = f(x)$ at most once. An inverse function $f^{-1}$ exists if and only if $f$ is one-to-one.
Domain and range
Examiner keyword
Domain: the set of allowable inputs $x$ . Range: the set of resulting outputs $f(x)$ . For a composite $fg$ : domain is $\{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}$ . For $f^{-1}$ : domain $=$ range of $f$ , range $=$ domain of $f$ .
Composite function $fg$
Examiner keyword
$fg(x) = f(g(x))$ — the function obtained by applying $g$ first and then $f$ . Composition is not commutative: in general $fg \neq gf$ .
Inverse function $f^{-1}$
Examiner keyword
The function that 'undoes' $f$ : $f^{-1}(f(x)) = x$ for all $x \in \text{dom}(f)$ and $f(f^{-1}(y)) = y$ for all $y \in \text{range}(f)$ . Its graph is the reflection of $y = f(x)$ in the line $y = x$ .
Related: one to one

Common Mistakes and Misconceptions — Algebraic methods 2

The traps other students keep falling into on algebraic methods 2 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Squaring $|f(x)| = g(x)$ and not checking for extraneous roots
WMA13/01 examiner reports — flagged annually
Why it happens
Squaring removes the modulus but introduces solutions where $g(x) < 0$ (which were never valid).
How to avoid it
After squaring and solving, substitute each candidate root back into the original equation OR into $g(x) \geq 0$ . Reject any that fail.
✕ $|f(x)| > a$ written as $-a < f(x) < a$
Why it happens
Memorising one form ( $|f(x)| < a$ becomes a sandwich) and reusing it for the opposite direction.
How to avoid it
Remember the rule: $|f(x)| < a$ is a SANDWICH (inside the bounds); $|f(x)| > a$ is a SPLIT (outside the bounds). Sketch $y = |f(x)|$ and $y = a$ if unsure.
✕Reading $fg(x)$ as 'apply $f$ first'
Why it happens
Natural left-to-right reading.
How to avoid it
$fg(x) = f(g(x))$ — the function CLOSEST to $x$ is applied FIRST. Always evaluate from the inside out.
✕Quoting an inverse without checking that $f$ is one-to-one
WMA13/01 examiner reports — flagged annually
Why it happens
Treating the algebraic rearrangement as the whole story.
How to avoid it
Before finding $f^{-1}$ , confirm $f$ is one-to-one over its given domain (look at the graph or compute $f'(x)$ and check sign). For $f(x) = x^{2}$ on $\mathbb{R}$ , no inverse exists; restrict the domain (e.g. $x \geq 0$ ) first.
✕Stating $f^{-1}(x) = (x - 3)^{2} + 2$ without quoting its domain
Why it happens
Focusing on the algebra and forgetting to transfer the range of $f$ to the domain of $f^{-1}$ .
How to avoid it
ALWAYS write 'domain of $f^{-1}$ = range of $f$ ' explicitly. Without the domain restriction the expression is not a true function (e.g. $(x - 3)^{2} + 2$ is a parabola with two branches — only one branch is the inverse).

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Algebraic methods 2 — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Algebraic methods 2

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Solve a modulus equation ∣2x−3∣=x+4|2x - 3| = x + 4∣2x−3∣=x+4 (5 marks, P3)

2Solve a modulus inequality ∣3x+2∣>4|3x + 2| > 4∣3x+2∣>4 (4 marks, P3)

3Composite function fg(x)fg(x)fg(x) and domain reasoning (6 marks, P3)

4Find an inverse function with domain/range (7 marks, P3)

5Reflection y=f−1(x)y = f^{-1}(x)y=f−1(x) in y=xy = xy=x (4 marks, P3)

Modulus (absolute value)

Modulus equation ∣f(x)∣=a|f(x)| = a∣f(x)∣=a and ∣f(x)∣=g(x)|f(x)| = g(x)∣f(x)∣=g(x)

Modulus inequalities

Composite function

Inverse function

Modulus function ∣f(x)∣|f(x)|∣f(x)∣

One-to-one (injective) function

Domain and range

Composite function fgfgfg

Inverse function f−1f^{-1}f−1

✕Squaring ∣f(x)∣=g(x)|f(x)| = g(x)∣f(x)∣=g(x) and not checking for extraneous roots

✕∣f(x)∣>a|f(x)| > a∣f(x)∣>a written as −a<f(x)<a-a < f(x) < a−a<f(x)<a

✕Reading fg(x)fg(x)fg(x) as 'apply fff first'

✕Quoting an inverse without checking that fff is one-to-one

✕Stating f−1(x)=(x−3)2+2f^{-1}(x) = (x - 3)^{2} + 2f−1(x)=(x−3)2+2 without quoting its domain

Practice questions

Past paper style quiz

Video lesson

Algebraic methods 2 — frequently asked questions

1Solve a modulus equation $|2x - 3| = x + 4$ (5 marks, P3)

2Solve a modulus inequality $|3x + 2| > 4$ (4 marks, P3)

3Composite function $fg(x)$ and domain reasoning (6 marks, P3)

5Reflection $y = f^{-1}(x)$ in $y = x$ (4 marks, P3)

Modulus equation $|f(x)| = a$ and $|f(x)| = g(x)$

Modulus function $|f(x)|$

Composite function $fg$

Inverse function $f^{-1}$

✕Squaring $|f(x)| = g(x)$ and not checking for extraneous roots

✕ $|f(x)| > a$ written as $-a < f(x) < a$

✕Reading $fg(x)$ as 'apply $f$ first'

✕Quoting an inverse without checking that $f$ is one-to-one

✕Stating $f^{-1}(x) = (x - 3)^{2} + 2$ without quoting its domain