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General remark for MX2 Q11d: Most students were unable to decipher and extract the information from the question as presented in order to get the correct roots. All of the information in the polynomial is written into the question (and the polynomial itself), but you have to pull it out. Quite...

like any programming, copy and paste code from stack exchange

just going to take this out of context and put it aside, and yep, describes the entire paper

anything is possible through Go- *shot*

I'm not sure you have enough information to do a cubic polynomial bash... but at the very least, it's straightforward to prove all three numbers have modulus 1.

Remarks for MX1, Q13 a) Most were able to get at least 2 marks, quite a few made miscellaneous errata in their working and got the wrong result at the end, so lost a mark. For some reason, a fairly large percentage of students think you need a triangle diagram to justify the conversion from θ...

At the top of page 3, for Question 14, you need to specify non-trivial involution. Also I see you are unfamiliar with some of the underneath stuff of LaTeX. To get proper 66 and 99 quotation marks, the first pair should be replaced with two backticks `` .

Well I guess it goes to show you were not intimidated by my question. Some people think that the unorthodoxy of my question is enough to scare students :awesome:

Remarks for Q16, MX2 a) The overwhelming majority were able to correctly execute the syllabus cloned, word for word, irrationality proof. No students used infinite descent, which I was hoping would sneak in from any students who looked at a specific previous BoS Trials paper. b) i) Some...

While I'm finishing off the pile of marking: Remarks for Q11, MX2 a) More than a handful of students made the incorrect border transformation of (3π/2, π/2), instead of (-π/2,π/2) Most students were able to correctly derive the relation between dx and dθ and the transformed border values but...

I would say Complex has more potential to be difficult, but usually isn't because there's not much they can do without plagiarising BoS Trials :p

The first part of 7 is a standard exercise which can easily be done using contradiction (assuming you have seen the algebraic bash proof that √2 is irrational) The second part: If x is irrational, then 2x is irrational (is what we want to show here, for a specific value of x) Taking the...

\begin{align*} w\bar{z} &= \cfrac{3+4i}{5} \times \cfrac{5 - 12i}{13} \\ &= \left(\cfrac{3}{5} \times \cfrac{5}{13} - \cfrac{4i}{5} \times \cfrac{12i}{13}\right) + \left(\cfrac{4i}{5} \times \cfrac{5}{13} - \cfrac{3}{5} \times \cfrac{12i}{13}\right) \\ &= \cfrac{15 - 48i^2}{65} +...

ah yes the engineer's method

same omission of detail as above in the locus question

they didn't put an open circle around z=-π, since it's not clear whether the inclusion of the curve from below takes precedence or not (it doesn't)

Well the vector field doesn't appear to be nice at the origin, so the uniqueness and existence theorem doesn't apply here.

this isn't right, since 1 1 1 1 1 0 1 0 0 has exactly 2 of each of the possible non-zero sums

Matrix didn't put a dashed line for the straight line segment joining z=0 to z=-π. -1 mark.