AQA A-Level Mathematics Revision
Explore AQA A-Level Mathematics revision topics. Check the official specification for complete requirements for your exam year and route.
Course details
Assessment The assessment includes Paper 1, Paper 2 and Paper 3; see the official specification for the full structure.
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Reviewed course guidance
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Topic group
Reviewed requirements
- Core
Open revision guidance
What to know
- For positive integer n, (a+bx)^n = Σ(r=0…n) nCr a^(n−r)(bx)^r, where nCr = n!/[r!(n−r)!]. For rational n the generalized coefficient is n(n−1)…(n−r+1)/r!; about a, the expansion is valid when |bx/a|<1. The same nCr notation appears in binomial probabilities.
What to do in an assessment
- Use the finite sum for positive integer n; for rational n, factor out a^n, substitute u=bx/a, and retain the requested powers only after checking |u|<1. For X~Bin(n,p), use P(X=r)=nCr p^r(1−p)^(n−r).
In plain language
The exponent tells you how the terms continue: an integer power eventually stops, but a non-integer power normally does not. Truncating a valid series gives an approximation, not an exact identity. For example, √(1 + 2x) = 1 + x − x²/2 + … when |2x| < 1. At x = 0.1 this gives 1.095, close to √1.2 ≈ 1.09545. The same choose-r coefficients count arrangements in a binomial probability model.
Self-check
Use the first three terms to approximate √1.2.
Show the answer
Set x=0.1 in √(1+2x) ≈ 1+x−x²/2: 1+0.1−0.005 = 1.095.
- Core
Scope: Core · §3.9 H2 — integrate x^n (n≠−1), e^(kx), 1/x, sin(kx), cos(kx), and related sums, differences and constant multiples
Open revision guidance
What to know
- For n≠−1, ∫x^n dx = x^(n+1)/(n+1)+C. Also ∫e^(kx)dx=e^(kx)/k+C, ∫(1/x)dx=ln|x|+C, ∫sin(kx)dx=−cos(kx)/k+C and ∫cos(kx)dx=sin(kx)/k+C, for non-zero k where applicable.
What to do in an assessment
- Apply the matching rule term by term, multiply by any outside constant, and preserve signs. Include C for an indefinite integral; differentiate the result to check every term.
In plain language
Integration reverses differentiation, but each function has its own pattern. The power rule cannot be used when n=−1; that case gives a logarithm. A factor k inside an exponential or trigonometric angle is balanced by dividing by k. For example, ∫(6x² + 2e^(2x) + 3/x)dx = 2x³ + e^(2x) + 3ln|x| + C. Differentiating this answer returns the original integrand.
Self-check
Evaluate ∫(6x² + 2e^(2x) + 3/x) dx.
Show the answer
2x³ + e^(2x) + 3ln|x| + C.
- Core
Scope: Core · §3.14 M2 — conditional probability using tree diagrams, Venn diagrams, two-way tables and P(A|B)=P(A∩B)/P(B)
Open revision guidance
What to know
- Conditional probability restricts attention to cases where B has happened: P(A|B)=P(A∩B)/P(B), provided P(B)>0. Tree diagrams, Venn diagrams and two-way tables can all represent the same overlap and condition.
What to do in an assessment
- Identify the condition after the vertical bar, use its total as the denominator, and count only outcomes that also satisfy A in the numerator. In a table, work within B's row or column; in a tree, use the relevant branch probabilities.
In plain language
The phrase “given B” changes the group being considered. If 24 of 40 students take route B and 12 of those also use the library, then P(library | route B)=12/24=1/2. It is not 12/40, because the question is no longer about all 40 students. A table makes the restricted total visible and helps prevent swapping P(A|B) with P(B|A).
Self-check
In a group of 30, 18 study French and 12 study both French and German. Find P(German | French).
Show the answer
12/18 = 2/3.
- Core
Scope: Core · §3.20 S1 — understand and use moments in simple static contexts
Open revision guidance
What to know
- The moment of a force about a pivot is force multiplied by its perpendicular distance from the pivot. In a simple body at rest, clockwise and anticlockwise moments balance.
What to do in an assessment
- Choose a pivot, use perpendicular distances in metres and forces in newtons, then set total clockwise moment equal to total anticlockwise moment. Solve the resulting equation and check the unit is newton metres for a moment.
In plain language
A longer lever arm makes the same force turn more strongly. If an 8 N force acts 1.5 m to one side of a pivot, its moment is 8 × 1.5 = 12 N m. A force F acting 2 m on the other side balances it when 2F=12, so F=6 N. This is a simple static balance; it does not assume the object is accelerating.
Self-check
What force 0.5 m from a pivot balances a 4 N force 1.5 m from the other side?
Show the answer
4×1.5=6 N m, so F×0.5=6 and F=12 N.
- Core
Scope: Core · §3.5 D2 — nth-term formulae, simple recurrences, and increasing, decreasing or periodic sequences
Open revision guidance
What to know
- A sequence may be specified by its nth-term formula or by a recurrence such as x_(n+1)=f(x_n), which tells you how to obtain the next term from the current one. A sequence can increase, decrease or repeat periodically.
What to do in an assessment
- Use a formula with the requested n, or start from the stated initial value and apply a recurrence one step at a time. Compare successive terms to classify the sequence; for periodicity, identify the repeating cycle.
In plain language
The rule for a sequence might give each term directly, or it might tell you how to move from one term to the next. For example, x_1=0 and x_(n+1)=1−x_n gives 0, 1, 0, 1, …, so the sequence is periodic with period 2. By contrast, x_n=5/n gives 5, 2.5, 1.67, … and decreases towards zero. A finite list alone does not prove a general pattern; use the given rule to generate later terms and classify the behaviour.
Self-check
Starting with x_1=2 and x_(n+1)=x_n+3, write the first four terms and classify the sequence.
Show the answer
2, 5, 8, 11; it is increasing and is generated by a recurrence.
- Core
Scope: Core · §3.5 D3 — understand and use sigma notation for sums of series
Open revision guidance
What to know
- The symbol Σ means to add the expression that follows for every integer value of the index from the lower limit to the upper limit. The index is a temporary counting variable and may be renamed if its limits are changed consistently.
What to do in an assessment
- Substitute each permitted index value, form the terms, and add them. For a compact expression, simplify or use a known series formula only after confirming that its index range matches.
In plain language
Sigma notation is a short way of writing a long addition. In Σ_(r=1)^4(2r+1), put r=1, 2, 3 and 4: the terms are 3, 5, 7 and 9, so the total is 24. The lower and upper limits matter: changing either one changes which terms are included. If a formula begins at r=0 but the question begins at r=1, adjust the range rather than copying the formula without checking.
Self-check
Evaluate Σ_(k=1)^3 k².
Show the answer
1²+2²+3²=1+4+9=14.
- Core
Scope: Core · §3.5 D4 — arithmetic sequences and series; nth-term and sum-to-n formulae
Open revision guidance
What to know
- In an arithmetic sequence, consecutive terms differ by the constant d. With first term a, the nth term is a+(n−1)d; the sum of the first n terms is n/2[2a+(n−1)d], equivalently n(a+l)/2 when l is the last term.
What to do in an assessment
- Find a and d from consecutive terms, check whether the question asks for a single term or a sum, then substitute the correct n. Verify the last term using the nth-term formula before using the endpoint sum formula.
In plain language
An arithmetic sequence adds the same amount each time. For 3, 5, 7, …, a=3 and d=2, so the tenth term is 3+9×2=21. The first ten terms sum to 10/2×(3+21)=120. Pairing the first and last terms, then the second and second-last, explains why the mean of the endpoints gives the average term. Do not use an arithmetic formula when the differences are not constant.
Self-check
Find the 12th term and the sum of the first 12 terms of 4, 7, 10, … .
Show the answer
a=4, d=3; u_12=4+11×3=37 and S_12=12(4+37)/2=246.
- Core
Scope: Core · §3.5 D5 — geometric nth term, finite sum, convergent infinite sum, |r|<1 and modulus notation
Open revision guidance
What to know
- A geometric sequence has common ratio r: its nth term is ar^(n−1). For r≠1, S_n=a(1−r^n)/(1−r); the infinite sum is a/(1−r) only when |r|<1. Modulus bars are essential because a negative ratio can also converge.
What to do in an assessment
- Divide consecutive terms to find r, identify a, then choose the nth-term, finite-sum or infinite-sum formula according to the question. For an infinite sum, explicitly check |r|<1 before using the limit.
In plain language
A geometric sequence multiplies by the same ratio each time. For 4, 2, 1, …, a=4 and r=1/2. The sum of the first n terms is 4[1−(1/2)^n]/(1−1/2), while the infinite sum tends to 4/(1−1/2)=8. A ratio such as −1/2 also satisfies the convergence condition because its modulus is less than one; a ratio of −2 does not. An infinite-series formula cannot be used merely because the terms look smaller at first.
Self-check
Does 6+3+1.5+… have an infinite sum, and if so what is it?
Show the answer
Yes. r=1/2, so |r|<1 and S∞=6/(1−1/2)=12.
- Core
Scope: Core · §3.5 D6 — use sequences and series in modelling
Open revision guidance
What to know
- A sequence models successive values and a series models their total. A constant additive change suggests an arithmetic model; a constant multiplicative change suggests a geometric model. The model is conditional on assumptions remaining suitable.
What to do in an assessment
- Define the first term and the step or ratio from the context, state what the index represents, select the matching sequence or sum formula, and interpret the numerical result with units and the model's assumptions.
In plain language
Suppose a regular savings plan deposits £10 in week 1 and increases the deposit by £2 each week. The weekly deposits form an arithmetic sequence with a=10 and d=2. Over six weeks the total is S_6=6/2[2(10)+5(2)]=£90. This answer assumes every deposit is made and the increase stays exactly £2; it does not include interest or fees. A model is useful because it turns a repeated real-world pattern into a sequence or sum, but the assumptions must be checked before interpreting it.
Self-check
A payment starts at £5 and rises by £3 each month. What is the total paid in four months?
Show the answer
5+8+11+14=£38 (equivalently S_4=4/2[2(5)+3(3)]).
- Core
Scope: Core · §3.9 H1 — know and use the Fundamental Theorem of Calculus
Open revision guidance
What to know
- If F is an antiderivative of a continuous function f on [a,b], then ∫_a^b f(x) dx=F(b)−F(a). The theorem connects differentiation, antiderivatives and definite integrals.
What to do in an assessment
- Find an antiderivative, substitute the upper and lower limits separately, and subtract lower from upper. Do not add an arbitrary constant to a definite integral; it cancels in the difference.
In plain language
The Fundamental Theorem turns the accumulated signed area under a curve into an endpoint calculation. For f(x)=3x², an antiderivative is F(x)=x³, so ∫_0^2 3x² dx=F(2)−F(0)=8. The upper limit is evaluated first, then the lower value is subtracted. A negative result can occur when the function is mostly below the axis; the theorem gives signed area, not automatically a positive geometric area. Continuity is a sufficient condition for using this standard result on the interval.
Self-check
Evaluate ∫_1^3 2x dx using an antiderivative.
Show the answer
F(x)=x², so F(3)−F(1)=9−1=8.
- Core
Scope: Core · §3.9 H3 — evaluate definite integrals and use them for area under a curve and between two curves
Open revision guidance
What to know
- A definite integral gives signed area. For geometric area, split at axis crossings or use upper curve minus lower curve on each interval; area between y=f(x) and y=g(x) is ∫(f−g)dx where f≥g.
What to do in an assessment
- Find intersection points or bounds, identify which graph is above, integrate the difference, and evaluate at the limits. If curves cross within the interval, split there so each area contribution is positive.
In plain language
Between y=4 and y=x² on 0≤x≤2, the line y=4 is above the curve. The enclosed area is ∫_0^2(4−x²)dx=[4x−x³/3]_0^2=16/3 square units. Integrating x² alone would give the area under that curve, not the area between the two graphs. If an integrand changes sign, its definite integral can cancel positive and negative parts even though geometric area cannot be negative; handle the intervals separately.
Self-check
Find the area under y=2x from x=0 to x=3.
Show the answer
∫_0^3 2x dx=[x²]_0^3=9 square units.
- Core
Scope: Core · §3.9 H4 — understand and use integration as the limit of a sum
Open revision guidance
What to know
- A definite integral can be defined as the limit of increasingly fine rectangular sums. On [a,b], divide the interval into widths Δx and add representative heights f(x_i)Δx as the width tends to zero.
What to do in an assessment
- Identify interval width and sample points, write the finite sum, then take the limit as the number of strips grows. Keep track of whether the sum estimates signed area and which endpoint convention is being used.
In plain language
For y=x on [0,1], divide the interval into n equal strips of width 1/n and use right endpoints i/n. The sum is Σ_(i=1)^n(i/n)(1/n)=(n+1)/(2n), which tends to 1/2 as n grows. Each finite sum is an approximation; the limit is the exact integral. The approximation becomes more faithful because the strips become narrower, not because one particular rectangle rule is magically exact for every curve.
Self-check
What is the limit of the right-endpoint sums (n+1)/(2n) as n→∞?
Show the answer
1/2, matching ∫_0^1 x dx.
- Core
Scope: Core · §3.9 H5 — simple substitution and parts; substitution scope and repeated parts allowed, reduction formulae excluded
Open revision guidance
What to know
- Substitution reverses the chain rule; integration by parts reverses the product rule: ∫u dv=uv−∫v du. The specification permits more than one parts application, but excludes reduction formulae; substitution is limited to a suitable single substitution leading to an integrable function.
What to do in an assessment
- Choose u so its derivative appears in the integrand for substitution. For parts, select u and dv, differentiate u and integrate dv, then repeat only if needed. Differentiate the final antiderivative to check it.
In plain language
For substitution, let u=x² in ∫x cos(x²)dx; then du=2x dx, giving ½∫cos u du=½sin(x²)+C. For parts, take u=x and dv=e^x dx in ∫xe^x dx: the result is xe^x−e^x+C=e^x(x−1)+C. Both methods undo a differentiation rule. Repeated parts may be needed for a polynomial times an exponential, but this card does not extend to reduction formulae, which the source explicitly excludes.
Self-check
Evaluate ∫x cos(x²)dx.
Show the answer
½sin(x²)+C, since its derivative is x cos(x²).
- Core
Scope: Core · §3.9 H6 — integrate using partial fractions with linear denominators
Open revision guidance
What to know
- A rational function with suitable distinct linear denominator factors can be decomposed into simpler fractions. Each term A/(x−a) integrates to A ln|x−a|+C; the modulus is needed for real x on either side of the root.
What to do in an assessment
- Resolve the fraction into linear partial fractions, integrate each logarithmic term, and include one constant of integration. Check by differentiating the logarithms and recombining the fractions.
In plain language
First decompose 2/[(x+1)(x+3)] as 1/(x+1)−1/(x+3). Integrating gives ln|x+1|−ln|x+3|+C. Differentiating returns 1/(x+1)−1/(x+3)=2/[(x+1)(x+3)]. The method applies to the source's linear-denominator partial fractions; it does not imply that every rational function can be integrated by splitting it into logarithms without first checking its algebraic form.
Self-check
Integrate 1/[(x+2)(x+5)] after partial fractions.
Show the answer
1/3 ln|x+2|−1/3 ln|x+5|+C.
- Core
Scope: Core · §3.9 H7 — analytically solve simple first-order separable differential equations and find particular solutions
Open revision guidance
What to know
- A separable first-order equation can be rearranged so the y terms multiply dy and the x terms multiply dx, then integrated on both sides. An initial condition determines the constant in a particular solution.
What to do in an assessment
- Separate variables, integrate both sides, solve for the dependent variable where possible, and use the given condition to find the constant. Check the resulting function in the original differential equation.
In plain language
For dy/dx=2xy, move y to the left and x to the right: dy/y=2x dx. Integration gives ln|y|=x²+C, so y=Ae^(x²). If y(0)=3, then A=3 and the particular solution is y=3e^(x²). Differentiating gives dy/dx=6xe^(x²)=2x·3e^(x²), so it satisfies the equation. The source allows factorising a common factor where needed, but the scope here is still simple separable equations rather than general differential equations.
Self-check
For dy/dx=2xy and y(0)=3, what is y(1)?
Show the answer
y(1)=3e.
- Core
Scope: Core · §3.9 H8 — interpret solutions and limitations in context, including kinematics links
Open revision guidance
What to know
- A solved differential equation describes how a quantity changes under stated assumptions. Interpret its values, units and long-term behaviour in the original setting, and identify where the model or solution may no longer be realistic.
What to do in an assessment
- Substitute the relevant time or position, interpret the result with units, and use the derivative to explain the rate or direction of change. State at least one limitation tied to the modelling assumptions.
In plain language
If a falling object's velocity is modelled by v=10e^(−0.2t) metres per second, then v(5)=10e^(−1)≈3.68 m/s and dv/dt=−2e^(−0.2t), so the initial acceleration is −2 m/s². The model predicts velocity approaching zero as time increases. It assumes the resistance law and other conditions remain unchanged; wind, changing drag or a ground impact can invalidate that description. A mathematically valid solution must still be interpreted in the kinematics context.
Self-check
For v=10e^(−0.2t), find v(0) and state the limiting velocity as t→∞.
Show the answer
v(0)=10 m/s; v tends to 0 m/s as t→∞ under this model.
- Core
Scope: Core · §3.14 M1 — mutually exclusive and independent events; links to discrete and continuous distributions
Open revision guidance
What to know
- Events A and B are independent when P(A∩B)=P(A)P(B), equivalently when learning one occurred does not change the probability of the other. Mutually exclusive events have P(A∩B)=0; unless one event has probability zero, they are not independent.
What to do in an assessment
- Find the intersection and compare it with the product of the separate probabilities. Do not treat “cannot happen together” as a synonym for “independent”; use the relevant rule when modelling probabilities, including distribution settings.
In plain language
On a fair die, let A be “even” and B be “greater than 4.” Then P(A)=1/2, P(B)=1/3, and P(A∩B)=P({6})=1/6, which equals the product, so these events are independent. They are not mutually exclusive because 6 satisfies both. By contrast, disjoint non-empty events cannot be independent: knowing one happened makes the other impossible. The distinction matters when multiplying probabilities and when deciding whether a distribution model's independence assumption is reasonable.
Self-check
If P(A)=0.4, P(B)=0.5 and P(A∩B)=0.2, are A and B independent?
Show the answer
Yes, because 0.4×0.5=0.2=P(A∩B).
- Core
Scope: Core · §3.14 M3 — model with probability, critique assumptions and likely effect of more realistic assumptions
Open revision guidance
What to know
- A probability model represents a real process using assumptions such as constant probabilities, independence or equally likely outcomes. A model's numerical result is conditional on those assumptions, not a guarantee about an individual outcome.
What to do in an assessment
- State the model and its assumptions, compare them with the context, then explain the likely direction of change if a realistic assumption is relaxed. Distinguish model limitations from random variation.
In plain language
A binomial model for ten coin tosses assumes two outcomes per trial, the same probability of heads on every toss, and independent trials. If the coin is biased or tosses are affected by a mechanical process, the binomial probabilities may not describe the experiment well. A larger head count can still occur by chance even when the model is appropriate; one unusual result alone does not prove the assumptions are false. To critique the model, identify which assumption is doubtful and explain how it could change the predicted probabilities.
Self-check
A coin is tossed repeatedly by a device that sometimes sticks after a head. Which binomial assumption is questionable?
Show the answer
Independence and possibly a constant probability of heads; the device's state can affect later tosses.
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