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    (Original post by Farhan.Hanif93)
    I hadn't quite read your question correctly in my earlier reply.

    Observe that x=\sin (5\pi /8) > \sin (6 \pi / 8) = \dfrac{1}{\sqrt{2}}, and x>0 so x^2 \geq \dfrac{1}{2}.
    Thanks!

    The latex is not displaying correctly, but are you saying that as sin(5pi/ 8)>sin(6pi/8) and sin^2(5pi/8) must be greater than sin^2(6pi8) which is equal to 1/2, thus we take the root that is greater than 1/2?

    Sorry I don't know to read latex
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    (Original post by Geraer100)
    Thanks!

    The latex is not displaying correctly, but are you saying that as sin(5pi/ 8)>sin(6pi/8) and sin^2(5pi/8) must be greater than sin^2(6pi8) which is equal to 1/2, thus we take the root that is greater than 1/2?

    Sorry I don't know to read latex
    Yeah, that's right. No idea why LaTeX has been struggling as of late.
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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.

    Yeah really unlucky, happening just few days before the exam!
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    jjsnyder physicsmaths Have u looked at q7 II 2015?
    Spoiler:
    Show
    Still not sure how we can make the graph continuous:
    Name:  image1 (2).jpg
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    So my thinking was  h(x) = \arctan \dfrac{x}{1-x^{2}} + n \pi at x=0 it goes through pi so I that says that n=1? But then clearly from the sketch of x/(1-x^2) we still jump from + to - infinity so our h(x) jumps pi down/up? Idk Im sure ive missed something here but havent seen it yet.
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    (Original post by EnglishMuon)
    jjsnyder physicsmaths Have u looked at q7 II 2015?
    Spoiler:
    Show
    Still not sure how we can make the graph continuous:
    Name:  image1 (2).jpg
Views: 93
Size:  412.0 KB
    So my thinking was  h(x) = \arctan \dfrac{x}{1-x^{2}} + n \pi at x=0 it goes through pi so I that says that n=1? But then clearly from the sketch of x/(1-x^2) we still jump from + to - infinity so our h(x) jumps pi down/up? Idk Im sure ive missed something here but havent seen it yet.
    I made a post a few pges back about th
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    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.
    Spoiler:
    Show
    Took like 45 mins for Q1 lol. But that Q was way too long.

    Also took like 45 mins for Q8, and got one of the graphs wrong at the end but apart from that all correct which is good.

    Next I chose Q4 which was a stupid decision. Only got half of it, mainly because I realised that it was complex numbers halfway through and was like nah.

    Next Q5 which was very easy but due my my completely inability to be concise, I spent way longer than I should've on it and waffled too much with explanations etc.

    Q2 was lovely, but for the last part I couldn't resist whipping out L'Hopital's Rule etc. which I don't even know whether they'll accept. But I made sure my argument was completely correct so...
    I need to try and resist crazy methods and take at least a little bit of time to find simpler ones...

    My Q choice for this paper (order-wise and possible Q-wise) was bad. I might've got an S in this one had I chosen my Qs more carefully.

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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.
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    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 .


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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 .


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    So 100 for high 1/S ?
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    (Original post by EnglishMuon)
    jjsnyder physicsmaths Have u looked at q7 II 2015?
    Spoiler:
    Show
    Still not sure how we can make the graph continuous:
    Name:  image1 (2).jpg
Views: 93
Size:  412.0 KB
    So my thinking was  h(x) = \arctan \dfrac{x}{1-x^{2}} + n \pi at x=0 it goes through pi so I that says that n=1? But then clearly from the sketch of x/(1-x^2) we still jump from + to - infinity so our h(x) jumps pi down/up? Idk Im sure ive missed something here but havent seen it yet.
    Didn't do the question, let me know when you want me to mark stuff


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    (Original post by EnglishMuon)
    jjsnyder physicsmaths Have u looked at q7 II 2015?
    Spoiler:
    Show
    Still not sure how we can make the graph continuous:
    Name:  image1 (2).jpg
Views: 93
Size:  412.0 KB
    So my thinking was  h(x) = \arctan \dfrac{x}{1-x^{2}} + n \pi at x=0 it goes through pi so I that says that n=1? But then clearly from the sketch of x/(1-x^2) we still jump from + to - infinity so our h(x) jumps pi down/up? Idk Im sure ive missed something here but havent seen it yet.
    I do like this question, I think there are a lot of good concepts in here for pre-undergrads to be comfortable with.

    Spoiler:
    Show
    Consider the above analysis for |x|<1 and |x|>1 separately, and then enforce the continuity condition to make the whole thing line up at |x|=1.

    [Hint: In particular, note that you can't use the h(0)=\pi condition in the latter case to determine n]
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    (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).
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    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.
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    (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.


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

    Why the condition for the lorry not to tip over is N(2)>0? 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
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    (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?
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    (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.
    Spoiler:
    Show
    n^2 \leq 2^n \Leftrightarrow \dfrac{n}{\ln n} \geq \dfrac{2}{\ln 2}.
    (By L'Hopital's Rule, \displaystyle\lim_{n\to\infty} \dfrac{\ln n}{n}=\displaystlye\lim_{n\to \infty} \dfrac{1}{n}=0 so as n \rightarrow \infty, \dfrac{n}{\ln n} \rightarrow \infty.)

    Then I showed that \dfrac{n}{\ln n} is strictly increasing \forall x>e (by differentiation). So we can take m=5 since 5>e and 5^2 \leq 2^5 \Leftrightarrow \dfrac{5}{\ln 5} \geq \dfrac{2}{\ln 2}, so combining all the above results, n^2 \leq 2^n\ \forall n\geq 5.
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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.
    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.
 
 
 
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