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STEP Prep Thread 2016 (Mark. II)

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Original post by Farhan.Hanif93
Yeah, that's right. No idea why LaTeX has been struggling as of late.

Yeah thanks. This question looks nice at the start but at end it got really nasty especially the last part.:frown:

Yeah really unlucky, happening just few days before the exam!
@jjsnyder @physicsmaths Have u looked at q7 II 2015?

Spoiler

Original post by EnglishMuon
@jjsnyder @physicsmaths Have u looked at q7 II 2015?

Spoiler


I made a post a few pges back about th
Just did 2015 III and have to say that was one of the hardest III papers I've done. I definitely got a 1, but a few marks of an S I think. I've noticed I'm really bad at being concise in my solutions in general, so need to correct that before the exams.

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Original post by Zacken
I've just done III 2014 and I'm laughing my ass off. :rofl:

Remember what we were talking about yesterday? Look at Q15d of this second year Cambridge course exam paper and then look at Q4 of III 2014.


Lol, looks like these Qs aren't as original as I thought XD.
I know I was planning on prioritising newer questions I haven't done over doing 90s papers, but I really want to do one timed mock today so I can make sure I have my exam technique back up and running (after one week of not doing any STEP prep due to exam overload) for the I/II 2015 mocks this weekend (Saturday is I and Sunday is II). Also, I feel like during my STEP prep sessions, I am way more confident if I do one whole mock because I am then doing questions with no breaks in between for 3 hours, which equates to way more practice than if I did like 2-3 questions + marking them and completing them. Also, how much easier are newer papers compared to old ones? I am doing 1997 III today and want to understand how I should be doing for 1/S.


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Original post by Insight314
I know I was planning on prioritising newer questions I haven't done over doing 90s papers, but I really want to do one timed mock today so I can make sure I have my exam technique back up and running (after one week of not doing any STEP prep due to exam overload) for the I/II 2015 mocks this weekend (Saturday is I and Sunday is II). Also, I feel like during my STEP prep sessions, I am way more confident if I do one whole mock because I am then doing questions with no breaks in between for 3 hours, which equates to way more practice than if I did like 2-3 questions + marking them and completing them. Also, how much easier are newer papers compared to old ones? I am doing 1997 III today and want to understand how I should be doing for 1/S.


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You should be looking at 5 fulls I think for the high 1/S to equate to now. It is a nice paper though :smile:.


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Original post by physicsmaths
You should be looking at 5 fulls I think for the high 1/S to equate to now. It is a nice paper though :smile:.


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So 100 for high 1/S ?
Original post by EnglishMuon
@jjsnyder @physicsmaths Have u looked at q7 II 2015?

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Didn't do the question, let me know when you want me to mark stuff :smile:


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Original post by EnglishMuon
@jjsnyder @physicsmaths Have u looked at q7 II 2015?

Spoiler


I do like this question, I think there are a lot of good concepts in here for pre-undergrads to be comfortable with.

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(edited 7 years ago)
Original post by Insight314
So 100 for high 1/S ?


Yeah because generally speaking older papers are a lot easier than the newer ones (sure there are some tough Qs but there are enough easy-medium Qs in each paper).
I tried STEP II 2015 today too. Didn't go so well. Attempted first six questions.

Absolutely gutted when I saw the mark scheme for the last six points of question 2, I can't believe I didn't see that.
Original post by IrrationalRoot
Yeah because generally speaking older papers are a lot easier than the newer ones (sure there are some tough Qs but there are enough easy-medium Qs in each paper).


I am not surprised about it m8, I estimated about the same grade boundary as physicsmaths suggested. I was more like verifying what he said, and making sure that 100+ equates to S.


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Original post by Insight314
I am not surprised about it m8, I estimated about the same grade boundary as physicsmaths suggested. I was more like verifying what he said, and making sure that 100+ equates to S.


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I mean that's why such a high mark equates to a high 1/S.
Q9 of STEP I, 2002

Why the condition for the lorry not to tip over is N(2)>0?:smile: Should it not be the thing of maximum toppling angle which is tan(a)=d/h and then when toppling angle is exceeded that's when it topples?

Solution attached
Original post by IrrationalRoot
Q2 was lovely, but for the last part I couldn't resist whipping out L'Hopital's Rule etc.


wut? how?
Original post by StrangeBanana
wut? how?


It's funny I just came online to ask whether this would be accepted.
Anyway my solution was a bit convoluted but here is an outline, not sure if completely justified:
EDIT: Just realised I don't need L'Hopitals Rule or any asymptotic behaviour considerations XD, so put that in brackets.

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(edited 7 years ago)
Original post by Zacken
I've just done III 2014 and I'm laughing my ass off. :rofl:

Remember what we were talking about yesterday? Look at Q15d of this second year Cambridge course exam paper and then look at Q4 of III 2014.


But you weren't expected to use EL in STEP (which I think makes the STEP question slightly harder - and different). I never liked it as a STEP question as it's difficult to understand why the question was asking what it was.
(edited 7 years ago)
Reply 758
why does arctan(A)+arctan(B)=arctan(A+B/1-AB) hold iff AB<1? I understand theres issues with the range of the functions, but idk why AB<1 fixes it
Reply 759
Original post by KFazza
why does arctan(A)+arctan(B)=arctan(A+B/1-AB) hold iff AB<1? I understand theres issues with the range of the functions, but idk why AB<1 fixes it


Let arctanA + arctanB = arctanC. Then taking tangents implies that C = (A+B)/(1-AB). This is the forward implication.

Therefore C = tan(arctanA + arctanB). Taking inverse tangents, we want arctanC = arctan ( tan (arctanA + arctanB ) ) = arctanA + arctanB. In general, arctan (tan x) = x if and only if - pi/2 < x < pi/2. So this equation holds if and only if -pi/2 < arctanA + arctanB < pi/2. Now, it is a fact that for A =/= 0, arctanA + arctan1/A = pi/2. Since arctan is strictly increasing, we require B < 1/A to keep arctanA + arctanB within our range.

Indeed, if A is positive, the condition AB < 1 suffices, because arctanA + arctanB > -pi/2. Likewise if B is positive. We now consider A,B both negative. arctan is odd, so arctanA + arctanB = - (arctan(-A) + arctan(-B)). Note that -A and -B are positive numbers. Our range is again - pi/2 < arctan(-A) + arctan(-B) < pi/2, the same as before. The lower inequality holds true, the upper inequality is true iff -B < 1/(-A) giving us AB < 1 again.


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