2. Compulsory and optional content
The compulsory pure content is organised into labelled sections (A to J in the specification). In rough order they cover:
- Proof — proof by induction, applied to sums, divisibility and matrices.
- Complex numbers — modulus-argument form, de Moivre's theorem, roots of unity, loci in the Argand diagram.
- Matrices — algebra, determinants, inverses, linear transformations and invariant points and lines.
- Further algebra and functions — roots of polynomials and symmetric functions, the method of differences, the Maclaurin series.
- Further calculus — improper integrals, volumes of revolution, arc length, mean values.
- Further vectors — equations of lines and planes, scalar and vector products, distances and angles.
- Polar coordinates — curve sketching and areas enclosed by polar curves.
- Hyperbolic functions — definitions, identities, inverses and their calculus.
- Differential equations — integrating factors, second-order linear equations, and simple harmonic and damped motion as applications.
For Paper 3 you study two of three optional applications. Mechanics extends momentum, work-energy, circular motion and elastic strings. Statistics develops discrete and continuous distributions, hypothesis testing, confidence intervals and chi-squared tests. Discrete covers graphs and networks, algorithms, linear programming and critical path analysis. Choose the pair that fits your intended degree: economics and many sciences value statistics; engineering favours mechanics; computer science and operational-research routes suit discrete. If you are unsure, my note on the hardest Further Maths topics compares the options on difficulty and overlap.
5. Four worked exam-style questions
These four questions are written by me in the AQA 7367 style — one each from proof, complex numbers, further calculus and matrices, the four areas that most reliably decide the top grades. Work each one yourself before reading the solution, and pay attention to the examiner note: it tells you which marks are genuinely at stake.
Question 1 — Proof by induction (compulsory)
Question · 6 marksProve by induction that for all positive integers \(n\),
$$\sum_{r=1}^{n} r(r+2) = \frac{n(n+1)(2n+7)}{6}.$$
Solution. Let \(P(n)\) be the statement that \(\sum_{r=1}^{n} r(r+2) = \tfrac16 n(n+1)(2n+7)\).
Base case. When \(n=1\), the left-hand side is \(1\times 3 = 3\), and the right-hand side is \(\tfrac16(1)(2)(9)=3\). So \(P(1)\) is true.
Inductive step. Assume \(P(k)\) holds, that is \(\sum_{r=1}^{k} r(r+2) = \tfrac16 k(k+1)(2k+7)\). Then
$$\sum_{r=1}^{k+1} r(r+2) = \frac{k(k+1)(2k+7)}{6} + (k+1)(k+3).$$
Taking out the common factor \(\tfrac{1}{6}(k+1)\),
$$= \frac{k+1}{6}\Big[\,k(2k+7) + 6(k+3)\,\Big] = \frac{k+1}{6}\big(2k^2 + 13k + 18\big).$$
Factorising the quadratic, \(2k^2+13k+18=(k+2)(2k+9)\), so
$$\sum_{r=1}^{k+1} r(r+2) = \frac{(k+1)(k+2)(2k+9)}{6}.$$
This is exactly \(P(k+1)\), since replacing \(n\) by \(k+1\) in the formula gives \(\tfrac16(k+1)(k+2)\big(2(k+1)+7\big)=\tfrac16(k+1)(k+2)(2k+9)\).
Conclusion. \(P(1)\) is true, and if \(P(k)\) is true then \(P(k+1)\) is true. Therefore, by the principle of mathematical induction, \(P(n)\) is true for all positive integers \(n\). \(\blacksquare\)
What the examiner is looking forTwo marks are pure communication: a clear base case and the explicit closing statement. Most lost marks here are not algebraic — they are the missing final sentence, or stating \(P(k+1)\) without showing it matches the target form. Always write out what \(P(k+1)\) should be and show your expression equals it; that "therefore true for all \(n\)" line is worth a mark on its own.
Question 2 — Complex numbers and de Moivre (compulsory)
Question · 7 marks(a) Use de Moivre's theorem to show that \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\).
(b) Hence solve \(8x^3 - 6x + 1 = 0\), giving your roots in exact trigonometric form.
Solution (a). By de Moivre, \((\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta\). Expanding the left side with the binomial theorem,
$$\cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta.$$
Equating real parts, \(\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta\). Replacing \(\sin^2\theta = 1-\cos^2\theta\),
$$\cos 3\theta = \cos^3\theta - 3\cos\theta(1-\cos^2\theta) = 4\cos^3\theta - 3\cos\theta.$$
Solution (b). Put \(x=\cos\theta\). Then \(8x^3 - 6x = 2(4\cos^3\theta - 3\cos\theta) = 2\cos 3\theta\), so the equation becomes \(2\cos 3\theta + 1 = 0\), i.e. \(\cos 3\theta = -\tfrac12\).
Hence \(3\theta = \tfrac{2\pi}{3},\ \tfrac{4\pi}{3},\ \tfrac{8\pi}{3}\) (taking three values that give distinct \(\cos\theta\)), so \(\theta = \tfrac{2\pi}{9},\ \tfrac{4\pi}{9},\ \tfrac{8\pi}{9}\). The three roots are
$$x = \cos\frac{2\pi}{9},\quad x = \cos\frac{4\pi}{9},\quad x = \cos\frac{8\pi}{9}.$$
These are three distinct values, and a cubic has at most three roots, so we have them all.
What the examiner is looking forPart (a) is AO1/AO2: the binomial expansion must be shown and the real part correctly extracted — quoting the identity earns nothing. In part (b) the marks turn on the substitution \(x=\cos\theta\) and on choosing three angles that yield different cosines; candidates who list \(3\theta\) values producing repeated roots lose the final mark. A closing remark that a cubic can have only three roots secures the "all solutions" mark.
Question 3 — Further calculus: improper integral and volume (compulsory)
Question · 7 marksThe region \(R\) is bounded by the curve \(y = \dfrac{1}{x}\), the \(x\)-axis, and the line \(x = 1\), for \(x \ge 1\).
(a) Show that the area of \(R\) is infinite.
(b) Show that the volume generated when \(R\) is rotated through \(2\pi\) about the \(x\)-axis is finite, and find it.
Solution (a). The area is the improper integral \(\int_1^{\infty} \tfrac{1}{x}\,dx\). Replace the upper limit by \(t\) and take a limit:
$$\int_1^{t} \frac{1}{x}\,dx = \big[\ln x\big]_1^{t} = \ln t.$$
As \(t\to\infty\), \(\ln t \to \infty\), so the area diverges and is infinite.
Solution (b). The volume of revolution is \(V = \pi\int_1^{\infty} y^2\,dx = \pi\int_1^{\infty} \tfrac{1}{x^2}\,dx\). Again use a finite limit \(t\):
$$\pi\int_1^{t}\frac{1}{x^2}\,dx = \pi\Big[-\frac{1}{x}\Big]_1^{t} = \pi\Big(1 - \frac{1}{t}\Big).$$
As \(t\to\infty\), \(\tfrac1t \to 0\), so
$$V = \pi\lim_{t\to\infty}\Big(1 - \frac{1}{t}\Big) = \pi.$$
The volume is finite and equal to \(\pi\) cubic units, even though the generating region has infinite area.
What the examiner is looking forThis is the classic test of whether you handle improper integrals rigorously. The marks are gated on replacing \(\infty\) with a variable limit and writing an explicit \(\lim_{t\to\infty}\); a candidate who writes \(\big[-\tfrac1x\big]_1^{\infty}\) and substitutes infinity directly is penalised even with the right number. Note the \(\pi\) in the volume formula and the units on the final answer — both are easy AO1 marks. The contrast between (a) and (b) is the point of the question, so a one-line comment that the area diverges while the volume converges is rewarded.
Question 4 — Matrices and invariant lines (compulsory)
Question · 7 marksThe transformation \(T\) is represented by the matrix \(\mathbf{M} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\).
(a) Show that \(\mathbf{M}\) has eigenvalues \(1\) and \(3\), and find an eigenvector for each.
(b) Hence describe the two lines through the origin that are invariant under \(T\).
Solution (a). Eigenvalues satisfy \(\det(\mathbf{M}-\lambda\mathbf{I})=0\):
$$\begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0.$$
So \((\lambda-1)(\lambda-3)=0\), giving \(\lambda = 1\) and \(\lambda = 3\). For \(\lambda=1\), solve \((\mathbf{M}-\mathbf{I})\mathbf{v}=\mathbf{0}\): the equation \(x+y=0\) gives the eigenvector \(\begin{pmatrix}1\\-1\end{pmatrix}\). For \(\lambda=3\), solve \(-x+y=0\), giving \(\begin{pmatrix}1\\1\end{pmatrix}\).
Solution (b). An eigenvector spans a line through the origin that maps onto itself, so the invariant lines are those in the eigenvector directions:
$$y = -x \quad(\lambda=1, \text{ points fixed}), \qquad y = x \quad(\lambda=3, \text{ points scaled by }3).$$
On \(y=-x\) every point is mapped to itself because the eigenvalue is \(1\); on \(y=x\) every point is mapped to three times its distance from the origin because the eigenvalue is \(3\). Both lines are invariant as a whole.
What the examiner is looking forPart (a) wants the characteristic equation set up and solved explicitly — stating the eigenvalues without \(\det(\mathbf M-\lambda\mathbf I)=0\) loses the method marks. The distinction in part (b) is the AO2 discriminator: an invariant line under \(\lambda=1\) is a line of fixed points, whereas under \(\lambda=3\) the line is invariant but its points move along it. Candidates who simply write down both equations without explaining the difference between "invariant" and "fixed" miss the final reasoning mark.