STEP Prep Thread 2016 (Mark. II)

Seriously though they'd better make the C3 exam harder this year. Otherwise I'm protesting lol, literally same Qs different numbers every year. Well different numbers if you're lucky.

Thought it might have been (see my edit).

Good luck with whatever exams though

nope. I believe the exams officer at my school is still the chief examiner for c3 (or c4 cant remember). I believe he is 50% potato 50% sick coloured wall paper and 100% chump so I wouldnt get ur hopes up.

Damn haha, guess I'll have to memorise an interesting STEP Q beforehand and do that when I'm bored lol.

Yea it actually kills time, I did that in c1 last year. probably spent more time on that q than on the actual paper

You were doing STEP Qs last year too? Lol when I first sat C1 I didn't understand half the topics, half of my answers were just application of algorithms. Didn't even understand how completing the square worked XD.

done two mocks today can siklos claim me as sacrifice yet?

Nah he only sacrifices goats.

That is if they've been tied to the side of a square barn of side length 2a.


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Is Grade 2 easy to achieve? I'm so nervous about the STEP II on the 16th ......panicking ...

Is S,S easy to achieve

yes got it last week compelted it

Slightly regretting not applying to Trinity, who apparently only require a 1 in STEP I for CompSci with Maths. My life would be so chill without the horrors of STEP III (OTOH I'd then die from the maths tripos if I got in).

Hello, could you please have a look since you are the only one i know who has done 2015 step 1, thank you

Hello, could you please have a look since you are the only one i know who has done 2015 step 1, thank you

STEP II 1992 Q1

(iii) Don't get this one. Solutions say it's 0, which yeah ok at first glance you'd think that, but I thought $\dfrac{\sin x}{x}$ becomes a $\dfrac{0}{0}$ indeterminate form every $2\pi$ and so it appears that we do not have:

'for every $\epsilon >0\ \exists\ N$ such that $|f(x)|<\epsilon\ \forall x>N$.

Now I don't know whether there is a subtle extension to this limit to infinity definition that takes into account the domain of the function but I would think that this shows that the limit doesn't exist, which is what I put.

Also I have to say the last limit (vi) was really tough especially after than many easy limits.

This line requires parity considerations I believe.
Reason being is you can't just arbitrarily pair off terms and use the proven result (ii). You need to consider how the terms pair off in the middle.
That's why the ms approach was a bit more complicated.