AQA A-Level Maths: Pure Mathematics 2 — Year 2 Complete Revision Guide (7357)
AQA A-Level Maths: Pure Mathematics 2 — Year 2 Complete Revision Guide (7357)
Year 2 Pure is where AQA A-Level Maths (7357) stops being an extension of GCSE and starts becoming a genuinely synoptic subject. The Year 1 content gave you the building blocks — algebra, simple calculus, basic trigonometry, vectors. Year 2 takes those blocks and forces you to combine them in problems that look harder than they are because they pull on three or four topics at once. A typical Year 2 question might ask you to write a rate-of-change problem as a differential equation, separate variables, integrate using a partial-fraction decomposition, and then interpret the result. None of those steps is individually hard. The difficulty is in the chain.
This guide is a topic-by-topic walkthrough of the Year 2 Pure content in the AQA 7357 specification. It covers the ten major areas examined: algebraic and partial fractions, functions, modulus functions and transformations, further trigonometry, parametric equations, differentiation techniques, integration techniques, differential equations, numerical methods, and proof by contradiction. For each topic you will find the core skills, the typical pitfalls, a short worked example, and a link to the full lesson on the LearningBro course.
A note on emphasis. AQA tests Year 2 Pure heavily on Papers 2 and 3 — the harder synoptic content lives there. Integration techniques and parametric work in particular appear most commonly on the long synoptic 10-15 mark questions that decide the top grade boundaries. If you are aiming for an A or A*, the difference is almost always made on these long Year 2 questions rather than on the shorter Year 1 mechanics. Treat this section of the spec as the highest-leverage area in your revision plan.
What the AQA 7357 Specification Covers
The AQA A-Level Maths qualification (7357) is assessed through three two-hour papers. Paper 1 is pure mathematics only. Paper 2 is pure mathematics and mechanics. Paper 3 is pure mathematics and statistics. Each paper is worth 100 marks, and there is no choice of questions. Year 2 Pure content can appear on any of the three papers, but it is concentrated on Papers 2 and 3 because the synoptic style of those papers suits Year 2 techniques.
The table below shows the Year 2 Pure sub-topics, the part of the specification they sit under, and a realistic estimate of how many marks across a full sitting come from each. These are estimates based on the spread of typical 7357 papers, not guarantees for any single year.
| Topic | Spec Section | Typical paper marks weight |
|---|---|---|
| Algebraic and partial fractions | B (Algebra) | 5-8 marks |
| Functions and inverse functions | B (Algebra) | 5-8 marks |
| Modulus functions and transformations | B (Algebra) | 4-6 marks |
| Further trigonometry (compound, double, R-form) | E (Trigonometry) | 8-12 marks |
| Parametric equations | F (Coordinate geometry) | 8-12 marks |
| Differentiation techniques | G (Differentiation) | 10-15 marks |
| Integration techniques | H (Integration) | 12-18 marks |
| Differential equations | H (Integration) | 6-10 marks |
| Numerical methods | J (Numerical methods) | 6-10 marks |
| Proof by contradiction | A (Proof) | 2-4 marks |
What is reliable across years is that integration techniques and differentiation techniques together account for the largest share of Year 2 Pure marks. Trigonometry and parametric work follow closely. Numerical methods is small in headline marks but high-yield for accuracy candidates because the techniques are mechanical. Proof by contradiction is often a single 3-4 mark question, easy to drop if you have not seen the structure before.
Algebraic and Partial Fractions
Year 2 picks up algebraic-fraction work where Year 1 left off and pushes it into partial fractions. The reverse-direction skill — splitting a single fraction with a complicated denominator into a sum of simpler fractions — is high-value because it appears in three places: algebra (this section), the binomial expansion of $(1 + x)^n$ for non-positive-integer $n$, and integration.
There are three cases on the 7357 specification, distinguished by what is in the denominator after factorising.
| Case | Denominator factor | Partial fraction form |
|---|---|---|
| Distinct linear factors | $(ax + b)(cx + d)$ | $\frac{A}{ax+b} + \frac{B}{cx+d}$ |
| Repeated linear factor | $(ax + b)^2$ | $\frac{A}{ax+b} + \frac{B}{(ax+b)^2}$ |
| Linear and repeated linear | $(ax+b)(cx+d)^2$ | $\frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$ |
In each case, write the original fraction as a sum of unknown-coefficient fractions of the right form, multiply both sides by the original denominator, and solve. There are two standard methods: substituting strategic values of $x$ to make terms vanish (the cover-up method for distinct linear factors), and equating coefficients of like powers.
A short worked example. Decompose $\frac{3x + 4}{(x-1)(x+2)}$. Write it as $\frac{A}{x-1} + \frac{B}{x+2}$. Multiply through: $3x + 4 = A(x+2) + B(x-1)$. Substitute $x = 1$: $7 = 3A$, so $A = 7/3$. Substitute $x = -2$: $-2 = -3B$, so $B = 2/3$. The decomposition is $\frac{7/3}{x-1} + \frac{2/3}{x+2}$.
A common pitfall is using the wrong template for repeated factors — writing $\frac{Ax+B}{(x-1)^2}$ instead of $\frac{A}{x-1} + \frac{B}{(x-1)^2}$. Always check the case before assigning coefficients. Another is forgetting to convert improper algebraic fractions (degree of numerator at least the degree of denominator) into mixed form by polynomial division before applying partial fractions.
For worked examples in all three cases and the cover-up method, see the Algebraic Fractions and Partial Fractions lesson.
Functions
The functions topic at A-Level formalises ideas that Year 1 used informally. A function $f: A \to B$ is a rule that assigns to each element of the domain $A$ exactly one element of the codomain $B$. The set of values actually produced is the range.
You need to be fluent with composite functions, inverse functions, and the algebra that links them. The composite $fg(x)$ means apply $g$ first, then $f$. The order matters — $fg(x)$ and $gf(x)$ are usually different functions. The inverse $f^{-1}$ exists if and only if $f$ is one-to-one on its domain. Geometrically, $f^{-1}$ is the reflection of $f$ in the line $y = x$, and the domain and range swap roles.
To find an inverse algebraically, write $y = f(x)$, swap the roles of $x$ and $y$, and solve for $y$. For $f(x) = \frac{2x + 3}{x - 1}$, write $y = \frac{2x+3}{x-1}$, swap to get $x = \frac{2y+3}{y-1}$, multiply through by $y - 1$ and rearrange to find $y = \frac{x + 3}{x - 2}$. The inverse is therefore $f^{-1}(x) = \frac{x+3}{x-2}$, and you should state its domain (here $x \neq 2$).
A common pitfall is failing to restrict the domain so that the function is one-to-one. The function $f(x) = x^2$ on the whole real line has no inverse, but on $x \geq 0$ its inverse is $\sqrt{x}$. Another is forgetting that the range of $f$ becomes the domain of $f^{-1}$ — you must state both clearly.
For composite-function practice and inverse-function workflow, see the Functions lesson.
Modulus Functions and Transformations
The modulus function $|x|$ returns the magnitude of $x$, ignoring sign. So $|3| = 3$ and $|-3| = 3$. The piecewise definition is $|x| = x$ for $x \geq 0$ and $|x| = -x$ for $x < 0$, and this is the foundation of every modulus problem.
Two transformations come up constantly. $y = |f(x)|$ takes any part of $y = f(x)$ that lies below the x-axis and reflects it upward. $y = f(|x|)$ takes the part of $y = f(x)$ for $x \geq 0$ and reflects it across the y-axis. The two look similar at first but produce very different graphs.
To solve modulus equations like $|2x - 3| = 5$, split into cases. Either $2x - 3 = 5$ giving $x = 4$, or $2x - 3 = -5$ giving $x = -1$. Always check both candidates against the original equation. For modulus equations with a modulus on each side — like $|x + 2| = |2x - 1|$ — square both sides to remove the modulus signs and solve the resulting quadratic.
The standard graph transformations remain on the syllabus and are tested alongside modulus work. The table summarises them.
| Transformation | Effect on the graph |
|---|---|
| $y = f(x) + a$ | Translation by $a$ in the positive y-direction |
| $y = f(x + a)$ | Translation by $a$ in the negative x-direction |
| $y = a f(x)$ | Vertical stretch, scale factor $a$ |
| $y = f(ax)$ | Horizontal stretch, scale factor $1/a$ |
| $y = -f(x)$ | Reflection in the x-axis |
| $y = f(-x)$ | Reflection in the y-axis |
The single rule that catches most candidates is the inside-bracket-do-the-opposite rule. Anything happening to $x$ inside the bracket affects the graph in the opposite way. A common pitfall on modulus equations is treating them as automatic two-solution problems without checking validity — if the right-hand side of $|...| = c$ is negative, there are no solutions at all.
For transformation sketches, modulus equations and combined transformations, see the Modulus and Transformations lesson.
Further Trigonometry
Year 2 trigonometry extends Year 1 by introducing the reciprocal functions, the inverse functions, and a family of identities used heavily in proofs and integrals.
The reciprocal trig functions are $\sec x = \frac{1}{\cos x}$, $\csc x = \frac{1}{\sin x}$, and $\cot x = \frac{1}{\tan x}$. Their identities derive from the Pythagorean identity by dividing through. Dividing $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ gives $\tan^2 x + 1 = \sec^2 x$. Dividing by $\sin^2 x$ gives $1 + \cot^2 x = \csc^2 x$. These two identities are used constantly in proof and integration questions.
The compound angle formulae are the workhorse of Year 2 trig:
sin(A±B)=sinAcosB±cosAsinB
cos(A±B)=cosAcosB∓sinAsinB
tan(A±B)=1∓tanAtanBtanA±tanB
The double angle formulae are the special case $A = B$, giving $\sin 2A = 2 \sin A \cos A$, $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$, and $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$. The three forms of $\cos 2A$ are used in different contexts and you should know all three.
A particularly useful manipulation is the R-form or harmonic form: $a \sin x + b \cos x = R \sin(x + \alpha)$ where $R = \sqrt{a^2 + b^2}$ and $\tan \alpha = b/a$. This converts a sum of sine and cosine into a single sinusoid, which makes solving equations and finding maxima much simpler. To express $3\sin x + 4\cos x$ in R-form, $R = \sqrt{9 + 16} = 5$ and $\alpha = \arctan(4/3)$, so $3\sin x + 4\cos x = 5\sin(x + \arctan(4/3))$.
A common pitfall is the sign in the $\cos$ compound formula — the sign flips between the formula for $A + B$ and the trig functions inside it. Another is forgetting that small-angle approximations are valid only when the angle is in radians and is genuinely small.
For compound and double-angle proofs, R-form examples, and reciprocal-function sketches, see the Further Trigonometry lesson.
Parametric Equations
A curve given parametrically has its $x$ and $y$ coordinates each expressed as functions of a third variable, the parameter $t$. So instead of $y = f(x)$, you have $x = f(t)$ and $y = g(t)$. Many curves that are awkward in Cartesian form are clean parametrically — circles, ellipses, and trajectories in mechanics are the obvious examples.
The two core skills are converting between parametric and Cartesian forms, and differentiating parametrically. To convert, eliminate $t$ between the two parametric equations. For $x = 2t$, $y = t^2$, eliminate $t$ by writing $t = x/2$ and substituting: $y = (x/2)^2 = x^2/4$. For trigonometric parameters, the Pythagorean identity is usually the route — for $x = 3\cos t$, $y = 3\sin t$, square both and add to get $x^2 + y^2 = 9$, the equation of a circle.
To differentiate parametrically, use the chain rule:
dxdy=dx/dtdy/dt
This gives the gradient of the curve at any value of $t$ without needing to convert to Cartesian form. For $x = t^2$, $y = t^3 - 3t$, $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2 - 3$, so $\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$. To find the tangent at $t = 1$, evaluate at that point, find the corresponding $(x, y) = (1, -2)$, and write the tangent equation in the usual way.
A common pitfall is forgetting to give the answer in the form the question asked for — Cartesian or parametric. Another is dividing by $dx/dt$ when it equals zero, which corresponds to a vertical tangent and needs handling separately.
For parametric-Cartesian conversions and parametric calculus, see the Parametric Equations lesson.
Differentiation Techniques
Year 2 calculus extends differentiation from polynomials and trig functions to compositions, products, quotients, exponentials, logarithms, and implicit relations. The four named techniques are the chain rule, product rule, quotient rule, and implicit differentiation.
The chain rule for a composite $y = f(g(x))$ is $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. The product rule for $y = u(x) v(x)$ is $\frac{dy}{dx} = u'v + uv'$. The quotient rule for $y = u(x)/v(x)$ is $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$. Implicit differentiation treats $y$ as a function of $x$ and applies the chain rule whenever a $y$-term is differentiated, picking up an extra $\frac{dy}{dx}$ factor.
You also need the standard derivatives of the new Year 2 functions:
| Function | Derivative |
|---|---|
| $e^x$ | $e^x$ |
| $e^{kx}$ | $k e^{kx}$ |
| $\ln x$ | $1/x$ |
| $a^x$ | $a^x \ln a$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\csc x$ | $-\csc x \cot x$ |
A short worked example. Differentiate $y = x^2 e^{3x}$. Use the product rule with $u = x^2$ and $v = e^{3x}$: $u' = 2x$, $v' = 3e^{3x}$, so $\frac{dy}{dx} = 2x e^{3x} + 3x^2 e^{3x} = x e^{3x} (2 + 3x)$. Always factor at the end — examiners reward neat final form.
A common pitfall is forgetting the chain-rule factor inside a composite. $\frac{d}{dx}[\sin(2x)]$ is $2\cos(2x)$, not $\cos(2x)$. Another is mis-signing the quotient rule — the order of the numerator terms matters.
For full coverage of all four techniques and exam-style differentiation problems, see the Differentiation Techniques lesson.
Integration Techniques
Integration is the largest single topic in Year 2 Pure and the most heavily examined. AQA expects you to handle integration by substitution, by parts, using partial fractions, and to recognise standard forms by inspection. The volume and variety of practice required here is greater than for any other topic.
The standard integrals you must know include:
| Function | Integral |
|---|---|
| $x^n$, $n \neq -1$ | $\frac{x^{n+1}}{n+1} + c$ |
| $1/x$ | $\ln |x| + c$ |
| $e^{kx}$ | $\frac{1}{k} e^{kx} + c$ |
| $\sin(kx)$ | $-\frac{1}{k}\cos(kx) + c$ |
| $\cos(kx)$ | $\frac{1}{k}\sin(kx) + c$ |
| $\sec^2(kx)$ | $\frac{1}{k}\tan(kx) + c$ |
| $\frac{f'(x)}{f(x)}$ | $\ln |f(x)| + c$ |
Integration by substitution undoes the chain rule. Choose a substitution $u = g(x)$, compute $du = g'(x) dx$, rewrite the integral entirely in terms of $u$, integrate, and substitute back. The art is choosing the substitution well — usually pick the inner function of a composite, or a quantity whose derivative also appears in the integrand.
Integration by parts undoes the product rule:
∫udxdvdx=uv−∫vdxdudx
The choice of which factor is $u$ matters. The mnemonic LIATE (Logs, Inverse trig, Algebraic, Trigonometric, Exponentials) suggests which factor to call $u$ in priority order. For $\int x e^x dx$, choose $u = x$ (algebraic, higher than the exponential) and $dv/dx = e^x$. Then $v = e^x$, $du/dx = 1$, and $\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + c$.
Partial fractions turns an awkward rational integral into a sum of simple log-form integrals. $\int \frac{1}{(x-1)(x+2)} dx$ decomposes (using earlier methods) into $\int \frac{1/3}{x-1} - \frac{1/3}{x+2} dx = \frac{1}{3} \ln|x-1| - \frac{1}{3}\ln|x+2| + c = \frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + c$.
A common pitfall is forgetting the constant of integration on indefinite integrals — examiners deduct a mark every time. Another is mismanaging limits when changing variables: if you change to a new variable $u$, you must change the limits to match, or substitute back to $x$ before applying the original limits.
For all three techniques and the full standard-integral table, see the Integration Techniques lesson.
Differential Equations
A first-order differential equation is an equation involving $x$, $y$ and $\frac{dy}{dx}$. AQA Year 2 examines separable equations — those that can be written in the form $\frac{dy}{dx} = f(x) g(y)$, separated as $\frac{1}{g(y)} dy = f(x) dx$, and integrated termwise.
The workflow is: separate the variables (all $y$-terms with $dy$ on one side, all $x$-terms with $dx$ on the other), integrate both sides (one constant of integration is enough — combine constants from both sides), and apply any boundary condition to fix the constant. Then rearrange to find $y$ explicitly if the question asks for it.
A short worked example. Solve $\frac{dy}{dx} = xy$ given $y = 2$ when $x = 0$. Separate: $\frac{1}{y} dy = x , dx$. Integrate: $\ln|y| = \frac{x^2}{2} + c$. Apply the condition $y = 2$ at $x = 0$: $\ln 2 = c$. So $\ln|y| = \frac{x^2}{2} + \ln 2$, giving $y = 2 e^{x^2/2}$.
The other common style of question gives you a modelling context — population growth, cooling, drug concentration — and asks you to set up and solve the differential equation yourself. A typical setup is "the rate of change of a quantity is proportional to the quantity itself", which translates to $\frac{dN}{dt} = kN$ and produces exponential growth or decay depending on the sign of $k$.
A common pitfall is forgetting to add the constant before applying the boundary condition. Another is failing to interpret the modulus when exponentiating both sides — $|y| = e^{...}$ usually becomes $y = \pm e^{...}$, and you need the boundary condition to choose the sign.
For separable equations and modelling problems, see the Differential Equations lesson.
Numerical Methods
Numerical methods is small in headline weight but high in accessibility — the techniques are mechanical, and questions reward careful arithmetic over insight. AQA examines three methods: change of sign for locating roots, iteration of the form $x_{n+1} = g(x_n)$, and the Newton-Raphson method.
For change of sign, evaluate $f(x)$ at two points $a$ and $b$. If $f(a)$ and $f(b)$ have opposite signs and $f$ is continuous on the interval, there is at least one root between them. The method is simple and widely tested — but be aware of the failure cases: a discontinuity in $f$ between $a$ and $b$ can create a sign change without a root, and an even number of roots between $a$ and $b$ can hide behind a same-sign pair.
For iteration, rearrange $f(x) = 0$ into the form $x = g(x)$, choose a starting value $x_0$, and iterate $x_{n+1} = g(x_n)$. The iteration converges to a root if $|g'(x)| < 1$ near the root, and the rate of convergence is faster the smaller $|g'|$ is. The same equation can usually be rearranged into multiple iteration forms, some of which converge and some of which do not — choose the form with the smallest $|g'|$.
The Newton-Raphson formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. Geometrically, you draw the tangent at the current point and read off where it crosses the x-axis. Newton-Raphson typically converges much faster than fixed-point iteration but can fail spectacularly if $f'(x_n)$ is close to zero, and it requires you to compute the derivative.
A common pitfall is reporting answers to too few decimal places — AQA usually asks for a specific accuracy, and you must iterate until consecutive estimates agree to that accuracy and then round. Another is failing to check that the chosen iteration converges before producing answers — a divergent iteration produces nonsense.
For all three methods with worked numerical examples, see the Numerical Methods lesson.
Proof by Contradiction
Proof by contradiction is the one new proof technique introduced in Year 2. The structure is: assume the negation of what you want to prove, derive a logical contradiction, and conclude that the original statement must be true. The technique is used for results that are easier to disprove than to prove directly — for instance, results about irrationality and infinity.
The classic example is the proof that $\sqrt{2}$ is irrational. Assume the negation: $\sqrt{2}$ is rational, so $\sqrt{2} = p/q$ for integers $p$ and $q$ with no common factor. Square: $2 = p^2/q^2$, so $p^2 = 2q^2$, so $p^2$ is even, so $p$ is even, so $p = 2k$ for some integer $k$. Substituting: $4k^2 = 2q^2$, so $q^2 = 2k^2$, so $q$ is also even. But $p$ and $q$ both even contradicts the assumption that they had no common factor. Therefore $\sqrt{2}$ is irrational.
The other standard examples on the spec are the proof that there are infinitely many prime numbers, and the proof that if $n^2$ is even then $n$ is even (a useful lemma in the irrationality proof above).
A common pitfall is starting with a statement to prove and never properly setting up the contradiction. Always begin with "assume, for contradiction, that..." and state the negation clearly. Another is reaching a contradiction without articulating it — the final line of a proof by contradiction should explicitly say which assumption has been contradicted.
For the standard exam proofs and a clear contradiction template, see the Proof by Contradiction lesson.
Common Mark-Loss Patterns Across Year 2 Pure
Year 2 Pure rewards procedural fluency, and the same handful of habits accounts for a disproportionate share of lost marks across the section. None of these are about content you do not know.
- Forgetting the chain-rule factor when differentiating a composite. $\frac{d}{dx}[\sin(3x)]$ is $3\cos(3x)$, not $\cos(3x)$.
- Mis-signing the quotient rule or quoting it with the numerator terms in the wrong order.
- Forgetting the constant of integration on indefinite integrals. Examiners deduct a mark every time.
- Mismanaging limits during substitution. If you change variable, change the limits — or substitute back before applying them.
- Choosing the wrong factor as $u$ in integration by parts. LIATE is a guide, not a rule, but it is right far more often than wrong.
- Using the wrong partial-fraction template for repeated factors. Always check the case before assigning unknown coefficients.
- Treating modulus equations as automatically two-solution problems without checking which branches actually satisfy the original equation.
- Not stating the domain of an inverse function or failing to restrict the original domain so the function is one-to-one.
- Reporting numerical answers to the wrong accuracy. Iterate until consecutive estimates agree to the required precision, then round once.
- Skipping the contradiction setup in a proof by contradiction. Always begin "assume, for contradiction, that..." and end by naming which assumption has failed.
- Not showing enough working. AQA mark schemes award method marks generously when working is clear. A correct final answer with no working can score fewer marks than an incorrect one with clean method.
A revision plan that drills these habits — not just the content — moves your grade more than another pass through the textbook.
Recommended Six-Week Year 2 Pure Revision Plan
This plan assumes about 6-7 hours per week on Year 2 Pure. Adjust pace if you are starting earlier or later. The plan front-loads the highest-mark topics so you have time for second passes on integration and trigonometry near the exam.
| Week | Topics | Practice |
|---|---|---|
| 1 | Algebraic and partial fractions; functions and inverses | 15 partial-fraction decompositions across all three cases; 15 composite/inverse problems including domain statements |
| 2 | Modulus and transformations; further trigonometry (compound, double, R-form) | 10 modulus-equation problems; 15 trig identity proofs; 10 R-form rewrites and equation solves |
| 3 | Parametric equations; differentiation techniques | 10 parametric-Cartesian conversions; 20 differentiation problems mixing chain, product, quotient and implicit |
| 4 | Integration techniques (substitution, parts, partial fractions) | 25 integrals across the three techniques, with at least 5 definite integrals requiring careful limit handling |
| 5 | Differential equations; numerical methods | 8 separable-DE problems with boundary conditions; 6 modelling-context DEs; 12 numerical-method exercises across change-of-sign, iteration and Newton-Raphson |
| 6 | Proof by contradiction; mixed practice; targeted review | 4 contradiction proofs from the standard set; one full mixed Year 2 problem set per day; review marking-scheme working for any question scoring below 60% |
The point of the plan is to keep moving forward while maintaining contact with earlier topics. By the end of week 5, every Year 2 area should have had focused contact and a practice round. Week 6 is consolidation, weakness-targeting, and dry runs of full Paper 2 / Paper 3-style synoptic questions.
Treat any question you got wrong not as a mistake but as a diagnostic. Was it a content gap? A method error? A careless arithmetic slip? Logging the cause means your next review session targets the right thing.
How LearningBro's AQA Pure Mathematics 2 Course Helps
LearningBro's AQA A-Level Maths: Pure Mathematics 2 course is built around the structure of this guide. Each of the ten lessons covers one section of the 7357 Year 2 specification, in a sensible teaching order, with worked examples, practice questions, and full mark-scheme-style solutions. Lessons end with a short review and quick-recall questions designed for spaced revisits.
The course is designed to be used in two ways. As a first pass, you can work through the lessons in order, building each topic on the last. As a revision tool, you can drop into any lesson and work the practice independently — for example, drilling integration by parts for a week before mocks. The AI tutor is available throughout to give targeted hints when you get stuck, without giving away full solutions, and to mark your written working with structured feedback.
If you want one place to revise Year 2 Pure well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the AQA A-Level Maths: Pure Mathematics 2 course.