STEP Prep Thread 2016 (Mark. II)

I believe that one of the 4k+1 and 4k+3 cases was somewhat elementary to prove and the other one was a lot more difficult.

3 is easy, 1 is harder

do you know how to prove it for all n?
It's not a step level question I'm so sure about this .

I just don't remember about the 4l+1 proof . Elementary yes but to understand the proof not to think of it even the euclidean proof which is trivial to understand it's not that trivial to think of it .
Also did you understand the limits thing you said 0/0 it is 0/infinity ok?

Yeah no I only thought it was 0/0 because for some stupid reason I took n to be 0 when sinn was 0. It should be obvious now what I'm on about when I say I ruin my mocks with really stupid mistakes.

yah

ok tell us , it's beyond me

think intensively

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.

Part vi is not that hard for someone who has done practice in limits .
When you have to find limits with square roots you try to use ( srt(x) - sqrt (a) ) * ( sqrt(x) + sqrt (a)) = x-a so you have to multiply with conjugate so that you get rid of the sqrts I know that will leave other sqrts but they would have positive signs so no problem in finding them .
the expression inside the limit is equal to 1/2 * ( (sqrt(1+2/n) +1) / (sqrt (1+1/n)+1 ) but sqrts tend to 1 so the fraction tends to 2/2 thus the answer is 1/2

@IrrationalRoot

Does Q4 of that paper reminds you of something ............

Of course the first thing I tried was multiplying top and bottom by the conjugate of the bottom but that didn't work.
It took me a while but I did get the whole binomial expansion thing, it's just that it took more time than any of the other limits to see.

No idea how you got the bit in bold though.

Yep I noticed and it was extremely easy as well so definitely not doing that again.
I swear this is the third time I've seen blatant duplicates of Qs. Considering how far back 2006 is I would think a large number of candidates would have done 1992 as practice and will have done that Q, which is pretty bad.

On STEP Easter School this year, Siklos presented a Hitler video of STEP III 2013 which was infested with complex numbers. He went into detail about how to calm down during exam IF we get complex numbers on this year's paper. Coincidence? I think not. The devil is in the details. Complex numbers is 8 questions this year. Watch.


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What advice did he give?

"Relax"


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