OCR A Level Further Mathematics A (H245) is examined by four papers, each 1 hour 30 minutes, 75 marks and 25% of the grade. Two are the mandatory Pure Core papers (Y540 and Y541); the other two are chosen from Statistics, Mechanics, Discrete Mathematics and Additional Pure Mathematics. Marks are assessed under AO1 (technique, about 50%), AO2 (reasoning and proof, about 25%) and AO3 (problem solving and modelling, about 25%). The hardest marks sit in complex numbers, matrices, proof by induction, polar curves, hyperbolic functions and differential equations.
Who this guide is for
This is for students sitting OCR A Level Further Mathematics A, specification H245 — the linear OCR A course, not OCR's MEI route (which is a separate qualification, H645). If your exam papers are coded Y540 to Y545, you are on H245 and this guide is for you. Further Maths is sat alongside, and assumes you have met, the content of A-Level Mathematics; everything below builds on that foundation.
1. How H245 is built
The thing that surprises students moving from single Maths is that Further Maths is modular in its options but not in its core. Every candidate sits the same two Pure Core papers, then chooses the rest of the course. There is no coursework and no non-exam assessment: the whole qualification is four written papers.
Y540 (Pure Core 1) and Y541 (Pure Core 2). Each is 1 hour 30 minutes, 75 marks and 25% of the A Level. Crucially, both papers can draw on the whole of the Pure Core content — the split into "1" and "2" is about timetabling, not about dividing the syllabus into two halves. You revise the entire core for both.
The other 50% comes from any two of four options, each again 1h 30m, 75 marks, 25%: Statistics (Y542), Mechanics (Y543), Discrete Mathematics (Y544) and Additional Pure Mathematics (Y545). You may sit more than two to broaden the course; OCR then awards the grade from your best-scoring combination that meets the rules.
This optionality is a strategic decision, not an afterthought. A student aiming at an economics or engineering degree will usually pair Statistics and Mechanics; a computer-science applicant often takes Discrete; and a candidate who simply loves pure mathematics, or is targeting admissions tests like the TMUA and STEP, frequently chooses Additional Pure for the extra depth in calculus, series and number theory. Pick options that reinforce each other and your degree, not just the ones that feel easiest in Year 12.
2. The assessment objectives — where the grade is actually decided
OCR marks every paper against three assessment objectives. Knowing their weighting tells you where the grade boundaries live and why "knowing the content" is not the same as scoring an A*.
Accurate recall of facts, terminology and methods, and the correct execution of routine procedures. Roughly half of every paper. These are the marks you secure by being fluent and careful: differentiating cleanly, manipulating a matrix without slips, quoting a standard result correctly.
Constructing rigorous mathematical arguments and proofs, interpreting results and communicating them precisely. This is where proof by induction, "show that" questions and the logical structure of an answer are tested. Marks here are lost not for wrong arithmetic but for missing steps and loose reasoning.
Translating an unfamiliar or modelling problem into mathematics, selecting a method when none is given, and evaluating the solution. The synoptic, multi-step problems that separate the top grades sit here.
The lesson is blunt: a candidate who can do every routine technique but cannot prove, justify or strategise tops out around the AO1 ceiling. Half the marks reward fluency; the other half reward thinking. The four questions below are deliberately weighted towards that second half, because that is where OCR Further Maths candidates leave marks on the table.
3. The topics where OCR candidates lose marks
Reading across OCR's examiners' reports for the Pure Core, the same theme recurs: candidates are competent at the mechanics of a topic but lose marks on the reasoning around it. The recurring trouble spots are these.
- Complex numbers — de Moivre's theorem applied to find multiple-angle identities, and the geometry of roots of unity. Errors cluster around losing roots, mishandling the argument, or failing to show the modulus-argument working that earns the method marks. See the deeper treatment in the hardest Further Maths topics.
- Matrices — invariant points and invariant lines (frequently confused), and the characteristic equation for eigenvalues. Candidates compute determinants reliably but stumble when asked to interpret a transformation or to find a line of invariant points versus an invariant line.
- Proof by induction — the algebra is usually fine; the marks are lost in the logic. OCR wants the assumption stated for \(n=k\), the inductive step shown to give the \(n=k+1\) case, and an explicit concluding statement. Omit the structure and AO2 marks vanish even with correct algebra.
- Polar coordinates — sketching curves with the correct symmetry, and the area integral \(\tfrac12\int r^2\,\mathrm d\theta\) with the right limits. Wrong or missing limits are the classic error.
- Hyperbolic functions — the logarithmic forms of the inverses, and integration using hyperbolic identities. Sign errors in \(\cosh^2 x - \sinh^2 x = 1\) and its relatives are endemic.
- Differential equations — first-order via the integrating factor, and second-order linear equations with the complementary function plus particular integral. The frequent slip is choosing a particular integral that clashes with the complementary function and forgetting to multiply by \(x\).
None of these are hard because they are long. They are hard because they reward precision and structure — the AO2 and AO3 skills. The worked questions that follow each target one of these danger zones.
4. Four worked exam-style questions
These four questions are my own, written in the OCR Pure Core style. Each has a full solution and a short note on what the examiner is rewarding — the marks that candidates of equal ability win or lose.
Use de Moivre's theorem to show that \(\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\). Hence find, in exact form, the value of \(\cos 36^\circ\).
Solution. By de Moivre's theorem, \((\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta\). Expand the left side with the binomial theorem and take the real part:
Now replace \(\sin^2\theta = 1 - \cos^2\theta\) throughout, writing \(c=\cos\theta\):
Expanding, \(-10c^3(1-c^2) = -10c^3 + 10c^5\) and \(5c(1-2c^2+c^4) = 5c - 10c^3 + 5c^5\). Collecting terms:
For \(\cos 36^\circ\), set \(\theta = 36^\circ\) so that \(5\theta = 180^\circ\) and \(\cos 5\theta = -1\). Writing \(c = \cos 36^\circ\):
Since \(c = -1\) (i.e. \(\theta=180^\circ\)) is a known root, factor it out. The polynomial \(16c^5 - 20c^3 + 5c + 1\) divides by \((c+1)\) to give \((c+1)(4c^2 - 2c - 1)^2 = 0\). The repeated quadratic gives the relevant roots:
As \(\cos 36^\circ\) is positive, we take the \(+\) root:
The expansion and the real-part step carry the early AO1 marks — drop the \(i^2=-1\) bookkeeping and you lose them. The decisive AO3 mark is recognising that \(\theta=36^\circ\) makes \(5\theta=180^\circ\), turning a trig identity into a solvable polynomial. Candidates who never make that connection cannot start the "hence". Finally, justify which root you keep: stating that \(\cos 36^\circ>0\) is a required mark, not a formality.
The transformation \(T\) is represented by the matrix \(\mathbf M = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\). Find the equations of the lines through the origin that are invariant under \(T\).
Solution. A line through the origin \(y = mx\) is invariant if every point on it maps to another point on the same line. Take a general point \((x, mx)\) and apply \(\mathbf M\):
For the image to lie on \(y = mx\), its coordinates must satisfy \(y' = m x'\):
Dividing by \(x\) (true for all points on the line) gives \(3m = m(2+m)\), so \(m^2 - m = 0\) and \(m(m-1) = 0\). Hence \(m = 0\) or \(m = 1\), giving the invariant lines:
We should also check whether a vertical line \(x = 0\) is invariant: a point \((0, t)\) maps to \((t, 3t)\), which is not on \(x=0\) unless \(t=0\), so the \(y\)-axis is not invariant.
This is a classic AO2 discriminator. The method mark is for setting up the image of a general point and imposing the condition that it stays on the line — not for guessing eigenvectors. Many candidates confuse invariant lines (the line as a whole maps to itself) with lines of invariant points (every point fixed); here the points move along the line, so these are invariant lines, not fixed points. The examiner also credits the explicit check of the vertical case, which the \(y=mx\) parametrisation silently excludes.
Prove by induction that \(7^{n} - 1\) is divisible by 6 for all positive integers \(n\).
Solution. Base case. When \(n = 1\), \(7^1 - 1 = 6\), which is divisible by 6. So the statement holds for \(n = 1\).
Inductive step. Assume the statement is true for \(n = k\), that is, assume \(7^{k} - 1 = 6m\) for some integer \(m\). We must show it follows for \(n = k+1\). Consider:
Since \(7m + 1\) is an integer, \(7^{k+1} - 1\) is divisible by 6. So if the statement holds for \(n=k\), it holds for \(n=k+1\).
Conclusion. The statement is true for \(n=1\), and its truth for \(n=k\) implies its truth for \(n=k+1\); therefore, by the principle of mathematical induction, \(7^{n}-1\) is divisible by 6 for all positive integers \(n\).
The algebra here is almost trivial; the marks are entirely about structure, and this is where OCR candidates haemorrhage AO2 marks. The examiner wants four things, each worth a mark: a verified base case, an explicit assumption written as an equation, the algebraic manipulation that extracts a factor of 6 in the \(k+1\) case, and a concluding sentence that names the inductive logic. A correct factorisation with no "assume", or no final statement, is capped below full marks. Writing \(7(7^k-1)+6\) — peeling off the assumed term — is the standard move that makes the divisibility visible.
Solve the differential equation \(\dfrac{\mathrm d^2 y}{\mathrm d x^2} - 3\dfrac{\mathrm d y}{\mathrm d x} + 2y = 4x\), given that \(y = 0\) and \(\dfrac{\mathrm d y}{\mathrm d x} = 1\) when \(x = 0\).
Solution. Complementary function. The auxiliary equation is \(\lambda^2 - 3\lambda + 2 = 0\), which factorises as \((\lambda-1)(\lambda-2)=0\), so \(\lambda = 1\) or \(\lambda = 2\). Hence:
Particular integral. The right-hand side is linear, so try \(y_p = ax + b\). Then \(y_p' = a\) and \(y_p'' = 0\). Substituting:
Comparing coefficients: \(2a = 4\) gives \(a = 2\), and \(2b - 3a = 0\) gives \(b = 3\). So \(y_p = 2x + 3\) and the general solution is:
Apply the conditions. At \(x=0\), \(y=0\): \(A + B + 3 = 0\). Differentiating, \(y' = Ae^{x} + 2Be^{2x} + 2\); at \(x=0\), \(y'=1\): \(A + 2B + 2 = 1\), so \(A + 2B = -1\). Subtracting the first relation \(A + B = -3\) gives \(B = 2\), hence \(A = -5\). The particular solution is:
The CF carries AO1 marks for a correctly solved auxiliary equation. The discriminating mark is the form of the particular integral: a linear right-hand side needs \(ax+b\), not just \(ax\) — forgetting the constant \(b\) is the most common error and it propagates through everything. The examiner also looks for the conditions applied to the full solution (CF + PI) and to its derivative, in that order; applying them to the CF alone is a frequent and costly slip. Had the forcing term been \(e^{x}\) or \(e^{2x}\), the PI would have clashed with the CF and required an \(x e^{x}\) trial — a subtlety OCR likes to test.
5. OCR-specific exam technique
Beyond knowing the content, a few habits reliably lift an OCR Further Maths script by a grade.
- Read the command word. "Show that" means the answer is given, so every step must be present and visible — you are being marked on the working, not the destination. "Hence" means you must use the previous part; a fresh method, even if correct, can forfeit the marks.
- Use the formula book, but know its gaps. OCR provides a formulae booklet, but standard results such as the derivatives of \(\sinh\) and \(\cosh\), de Moivre's theorem and the sum of an arithmetic series are not in it and must be recalled. Map out early what you are expected to know by heart.
- Structure proofs explicitly. For induction, always write the base case, the assumption, the inductive step and the conclusion as four labelled stages. For "show that", end with a line that states what you have shown. Examiners reward the scaffolding.
- Keep exact form unless told otherwise. Surds, fractions and logarithmic forms of hyperbolic inverses should be left exact; premature decimals lose accuracy marks across complex numbers, polar areas and roots.
- Manage the synoptic core. Because both Y540 and Y541 can examine the entire Pure Core, do not "finish" a topic after one paper. Revise the whole core for each sitting and expect questions that link, say, complex numbers to series, or matrices to vectors.
Further Maths & admissions tuition
OCR Further Maths rewards the half of the paper that tests reasoning, not just technique. For one-to-one help with the Pure Core, your chosen options, or the proof and problem-solving that decide A* boundaries, see A-Level Maths and Further Maths tuition, A-Level and admissions tutoring or TMUA preparation for university entrance.