Running the Gauntlet (Hotseat: OCR Mathematics C1-C4)


The GCE A Level is a school-leaving qualification that students in the UK take at the end of high school. Students usually take exams for 3-4 subjects. The exams are graded on a scale from A* to U (though not with all characters in between); typically an A* is awarded to the roughly top 8-9 percent of students.

This is a rather different type of challenge – previous installments of this series have featured especially difficult exams (or rather, competitions; only the MAT is realistically speaking an exam there). I’ve usually struggled to finish in the time limit (I didn’t finish the AIME and barely finished the MAT; I had some spare time on the BMO R1, but still not that much). I could of course do this in the same way as the other tests, though the score distribution would likely be close to the ceiling, with random variation simply down to careless mistakes.

Interestingly, the UK has multiple exam boards, so for this discussion we’ll be looking at OCR, which here stands not for Optical Character Recognition, but for Oxford, Cambridge and RSA (the Royal Society of Arts). The A level Maths curriculum is split into five strands: core (C), further pure (FP), mechanics (M), statistics (S) and decision (D); each strand features between two and four modules, which generally are part of a linear dependency chain – apart from FP, where FP3 is not dependent on FP2 (though it still is dependent on FP1). For the Mathematics A level, students need to take four modules from the core strand, and two additional “applied” modules; Further Mathematics involves two of the FP strand modules plus any four additional modules (but these cannot overlap with the mathematics A level ones). Thus, a student pursuing a Further Mathematics A level will take 12 distinct modules, including C1 – C4 and at least two FP modules, for example C1-4, FP{1,3}, S1-4, D1 and M1.

(In high school I took the IB diploma programme instead, which did have Further Mathematics (FM), though I didn’t take it as I picked Computer Science instead. That was before Computer Science became a group 4 subject; even then, I think I would still have wanted to do Physics, and thus would not have taken FM in any case.)


I attempted the June 2015 series of exams (C1 – C4). Each of these papers is set for 90 minutes, and is a problem set that features between about seven and ten multi-part questions. The overall maximum mark is 72 (a bit of a strange number; perhaps to give 1 minute and 15 seconds per mark?). To make things a little more interesting, we define a performance metric

P = \dfrac{M^2}{T}

where M is the proportion of marks scored, and T is the proportion of time used. For example, scoring 100 percent in half of the time allowed results in a metric of 2; scoring 50 percent of the marks using up all of the time yields a metric of 0.25. The penalty is deliberately harsher than proportional, to limit the benefit of gaming the system (i.e. finding the easiest marks and only attempting those questions).

Most of the errors were results of arithmetical or algebraic slips (there weren’t any questions which I didn’t know how to answer, though I did make a rather egregious error on C3, and stumbled a little on C4 with trying to do a complex substitution for an integral, rather than preprocessing the term). There are a few things I noted:

  • The scores for the AS-level modules (C1, C2) were considerably higher than that for the A-level modules (C3, C4). This is fairly expected, given that students only taking AS Mathematics would still need to do C1 and C2. Furthermore, from reading examiners’ reports the expectation in these exams is that students should have enough time to answer all of the questions.
  • The score for C1 was much higher than that for C2. I think there are two reasons for this – firstly, C1 is meant to be an introductory module; and secondly, no calculators are allowed in C1, meaning that examiners have to allocate time for students to perform calculations (which as far as I’m aware is something I’m relatively quick at).
  • The score for C4 was actually slightly higher than that for C3 (contrary to a possibly expected consistent decrease). While there is meant to be a linear progression, I certainly found the C3 paper notably tougher than that for C4 as well. That said, this may come from a perspective of someone aiming to secure all marks as opposed to the quantity required for a pass or an A.

We also see the penalty effect of the metric kicking in; it might be down to mental anchoring, but observe that perfect performances on C1 and C2 in the same amount of time would have yielded performance numbers just above 5 and 3, respectively.

Selected Problems in Depth

C3, Question 9

Given f(\theta) = \sin(\theta + 30^{\circ}) + \cos(\theta + 60^{\circ}), show that f(\theta) = \cos(\theta) and that f(4\theta) + 4f(2\theta) \equiv 8\cos^4\theta - 3. Then determine the greatest and least values of \frac{1}{f(4\theta) + 4f(2\theta) + 7} as \theta varies, and solve the equation, for 0^{\circ} \leq \alpha \leq 60^{\circ},

\sin(12\alpha + 30^{\circ}) + \cos(12\alpha + 60^{\circ}) + 4\sin(6\alpha + 30^{\circ}) + 4\cos(6\alpha + 30^{\circ}) = 1

This might have appeared a little intimidating, though it isn’t too bad if worked through carefully. The first expression is derived fairly quickly by using the addition formulas for sine and cosine. I then wasted a bit of time on the second part by trying to be cheeky and applying De Moivre’s theorem (so, for instance, \cos(4\theta) is the real part of e^{i(4\theta)} which is the binomial expansion of (\cos \theta + i \sin \theta)^4), subsequently using \sin^2 x = 1 - \cos^2 x where needed. This of course worked, but yielded a rather unpleasant algebra bash that could have been avoided by simply applying the double angle formulas multiple times.

The “range” part involved substitution and then reasoning on the range of \cos^4\theta (to be between 0 and 1). The final equation looked like a mouthful; using the result we had at the beginning yields

f (12 \alpha) + 4 f (6 \alpha) = 1

and then using a substitution like \beta = 3 \alpha, we can reduce the equation to 8 \cos^4 \beta - 3 = 1. We then get \cos \beta = \pm \left( \frac{1}{2} \right)^{(1/4)} and we can finish by dividing the values of \beta by 3 to recover \alpha.

C4, Question 6

Using the quotient rule, show that the derivative of \frac{\cos x}{\sin x} is \frac{-1}{\sin^2x}. Then show that

\displaystyle \int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \dfrac{\sqrt{1 + \cos 2x}}{\sin x \sin 2x} = \dfrac{1}{2}\left(\sqrt{6} - \sqrt{2}\right)

The first part is easy (you’re given the answer, and even told how to do it). The second was more interesting; my first instinct was to attempt to substitute t = \sqrt{1 + \cos 2x} which removed the square root, but it was extremely difficult to rewrite the resulting expression in terms of t as opposed to x. I then noticed that there was a nice way to eliminate the square root with \cos 2x = 2 \cos^2 x - 1. The integrand then simplifies down into a constant multiple of \frac{-1}{\sin^2x}; using the first result and simplifying the resultant expression should yield the result. That said, I wasted a fair bit of time here with the initial substitution attempt.


To some extent this is difficult, because students don’t generally do A levels in this way (for very good reasons), and I’m sure that there must be students out there who could similarly blast through the modules in less than half the time given or better (but there is no data about this). Nonetheless, the A level boards usually publish Examiners’ Reports, which can be fairly interesting to read through though generally lacking in data. The C3 report was fairly rich in detail, though; and the 68/72 score was actually not too great (notice that “8% of candidates scored 70 or higher”). Indeed the aforementioned question 9 caused difficulties, though the preceding question 8 on logarithms was hardest in terms of having the lowest proportion of candidates recording full marks.

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