STEP 2016 Solutions

I did this whole 'attempt 6 but unable to complete many questions' thing for both I and II this year, which is strange because in basically all of my (25 ish?) mocks I get mostly fulls but not 6 Qs. Not sure if this is because of my psychology for the actual exams or that these ones are just harder to complete.
So yeah basically I think my mindset changed to the one you've mentioned here just for these exams lol, I never usually attempt 6 Qs.

STEP III however the absolute priority is to get a 1, so certainly not going to 'restrict' myself to 30 mins per Q, I managed to get a very good 1 in last years paper from spending 45 mins on the first Q and 45 mins on the second, but they were 39 marks in total which was worth it.
If I for some reason could attempt 6 that would be ideal, but not aiming for it.

Honestly I don't know how hard these questions are gonna be, so I would recommend just using your own judgement in the exam and decide whether or not it's worth trying to finish a question or move on (which is where experience from mocks comes in).

A bit late for this post,

But for STEP 1's question 2, if I differentiate the given giant equation correctly, and manage to solve part (ii) correctly using an FP3 substitution (let x=sinhu) is that acceptable?

I failed to solve part (i) and part (iii) completely. How many marks should I get out of 20 for that, you guys think, approximately?

Yeah, I used my '30 minutes per question' strategy for STEP I and II as well. I have about 7 minutes of looking through the questions at the start, ranking them and finding out which ones I am gonna do, so if it happens that STEP III has more of my favourite questions then I am probably gonna continue this strategy so I can get 6 potentially good solutions on the script for a potential S grade. However, if I do not like most of the questions, I would probably try to give more fully complete solutions so I can at least get a solid/high Grade 1. I feel like my usual strategy is a bit more risky, but if last parts are worth less or I just really like the questions, then it is more beneficial.


Posted from TSR Mobile

https://www.desmos.com/calculator/muiga0eqk4

Slightly better framing, more zoomed in:

https://www.desmos.com/calculator/r7adx3smel

This relates to STEP II Q1.
Press the play button next to the t slider.

In a simultaneous effort to learn a bit of LaTeX and to rub in my own face just how stupid I was to miss out some of the questions, I've typed up the questions and solutions for STEP II (credit is given, even if I've extended, abridged, corrected, or tweaked solutions).

I figured you guys may as well have it. Give me a shout if I've messed something up.

Were you at the STEP Easter School this year?


Posted from TSR Mobile

I was not. Was there another farry there?

I was not. Was there another farry there?

I think its marvelous - definitely should be attached in the OP. Thanks for sharing!

There is one very small thing that you have missed out of question (8i) - I hesitate to mention it, since I don't think its worth much bother to add it in, and you have done a great job otherwise on the LaTex - which is that, for our succinct notation to work, we just need to clarify what happens when m=1; that is, we should define
$\sum_{i=1}^{0} f(i) = 0$

But this is a very nice resource for future STEPpers (and makes my solutions look a lot neater than they were! ), so thank you again for sharing!

Give me a shout if I've messed something up.

What's your reasoning?

Damn, I didn't read that bit

I agree

There's a unique solution to the DE as long as the solution satisfies that y(1) = 1 thing, so if you want to claim that both z_n and y_(2n) satisfy the same DE and hence must be the same, then shouldn't you show they both satisfy the condition that z(1) = 1 for you to actually show that they are a solution?

z(1) = 1 follows because y_n(1) = 1 and z = 2y_n^2 - 1 etc but surely that's all you'd have to say? It's given that there is one solution satisfying y_k(1) = 1 for each k, and you've provided one for k = 2n, so z = y_2n.

Q3, the LHS is the sum of non-negative terms, not positive terms.

Edit to add: Q6, not sure about this, but I think you need to explicitly check z(1) = 1 or y_(2n)(1) = 1 or both.

Another edit: you seem to be missing the "write down for four events" in Q12.

Right you are; sorted.



Not sure how I managed that - whoops.

Thank you very much.