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    (Original post by sweeneyrod)
    f(x) = x
    (possibly not what you want)
    Should've mentioned, anything except f(x)=x.
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    (Original post by ComputerMaths97)
    Even if you end up with less marks?
    I think the trick is to know when you're stuck whether you're ever gonna be able to think of the solution or if it its pointless trying and would be more beneficial to move on. I think once you've got that down you've basically mastered step.
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    (Original post by Farhan.Hanif93)
    The LHS of that statement is untrue - N=1 shows you that.
    Yeah that's actually wrong, but don't know why the mark scheme used this statement. See attached photo
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    (Original post by IrrationalRoot)
    Can anyone think of a function f such that f(f(f(x)))=x? I can't, not sure if I'm missing something obvious or what.
    EDIT: I just realised there was some IMO question about iterating polynomials in this way, interesting...
    f(x) = x is a solution. There might be more too. I do remember doing a step question at some point of this type. It was a bit more complex but the basic idea was the same
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    (Original post by IrrationalRoot)
    Should've mentioned, anything except f(x)=x.
    f(x) = 1 ?

    (sorry, I'll stop now)
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    (Original post by IrrationalRoot)
    Sorry, 'cyclic with period 3' is not the same thing as what I meant lol.
    I meant f(f(f(x)))=x.
    Ah okay yeah I was wondering why I came up with one straight away

    Does cheating it like f(x) = x(cos^2(x)+sin^2(x)) count?

    (Original post by gasfxekl)
    that's period 2
    See this is why I'm never going to pass a STEP exam
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    (Original post by sweeneyrod)
    f(x) = 1 ?

    (sorry, I'll stop now)
    ...stahp.
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    (Original post by ComputerMaths97)
    Ah okay yeah I was wondering why I came up with one straight away

    Does cheating it like f(x) = x(cos^2(x)+sin^2(x)) count?



    See this is why I'm never going to pass a STEP exam
    No because that is the function f(x)=x!
    Writing it differently doesn't change it's mathematical meaning.
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    The only idea I've got for f(f(f(x)))=x so far is taking f to be a degree 2 or more polynomial and solving a bunch of ugly simultaneous equations.
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    (Original post by Geraer100)
    Yeah that's actually wrong, but don't know why the mark scheme used this statement. See attached photo
    Having just opened the question, it specifies that N is an integer greater than 1 and it is true that the inequality holds for N≥2.

    To see this, note that N-√(N^2-1) = 1/[N+√(N^2-1)] < 1/N ≤ 1/2, which yields 2N-1/2 < N+√(N^2-1) after a little rearrangement.
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    (Original post by IrrationalRoot)
    Can anyone think of a function f such that f(f(f(x)))=x? I can't, not sure if I'm missing something obvious or what.
    EDIT: I just realised there was some IMO question about iterating polynomials in this way, interesting...
    EDIT 2: Except the trivial solutions.
    Sure.

    Text: f(x) = ( x + sqrt(3) ) / (1 - xsqrt(3))

    LaTeX: \displaystyle f(x) = \frac{x + \sqrt{3}}{1 - \sqrt{3}x}

    Image:

    Hopefully at least one format will work...
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    When do results come in for step, same as A levels?


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    (Original post by drandy76)
    When do results come in for step, same as A levels?
    As soon as the clock hits midnight on the 18th of August (i.e: the end of 17th of August) you rush onto the website, frantically refresh, swear at your screen in tandem with the thousands of other candidates refreshing as the site throws up "Site error" - some furious clicks later, your continued refreshing bears fruit and your result pops up on screen, usually followed by a triumphant yell or a dive out of the window.
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    (Original post by Zacken)
    Spoiler:
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    This is what I think:

    Q1 - not sure if I showed it didn't hold for n=4 convincingly and left my answer for (ii)(b) as ((7^7 +)^3 + 7's)((7^7 + 1)^3 - 7's) so probably dropped 3 or so marks here. 17/20

    Q2 - full. 19/20Q3 - didn't realise sin(-4) was negative, so probs cut a mark here. 18/20

    Q4 - did some weird stuff? I differentiate first bit correctly, managed to get up to v/sqrt(v^2 + 1) = kx + c where v = f ' (x) by using the first part and then re-arranged for sqrt(v^2 + 1), cubed it to get (v^2 +1)^(3/2) and then plugged it back into the original equation to get f '' (x) / v^3 = k / (kx+c)^3 and then integrated both sides.

    Not sure how much to give myself for that? 10 marks?

    Q10 - full except that for showing e < 1/3, I made a teeny slip, so cut a mark. 19/20

    Q11 - did the first bit about getting the equation, found the maximum value via differentiation, plugged it back into the formula but along the way miscopied from one line to the next and hence didn't get a correct quadratic. (solved the incorrect quadratic as well) but then moved on and got the distance at the last part to be h/cos 2alpha but couldn't plug in my cos 2alpha since I'd got the incorrect version.

    Probably 11 marks.

    That totals 94... a middling 1?
    Spoiler:
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    Yeah it sounds like you'll be in the 90s. Surprised you didn't go for 7 (I mean the one with sets S and T which was modular arithmetic in disguise), or 13 (which is very straightforward, especially so if you know the underlying theory about this type of process). Still not a bad result though.

    Make sure you don't do stupid things in II or III, some of those marks you lost in I you shouldn't have. (Why didn't you bother finishing Q1?). If I'm sounding harsh, it's because I know what you're capable of
    .
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    (Original post by Zacken)
    As soon as the clock hits midnight on the 18th of August (i.e: the end of 17th of August) you rush onto the website, frantically refresh, swear at your screen in tandem with the thousands of other candidates refreshing as the site throws up "Site error" - some furious clicks later, your continued refreshing bears fruit and your result pops up on screen, usually followed by a triumphant yell or a dive out of the window.
    As I dive out of the window I'll check UCAS too so I'll know it was the right call, need step so kings can ignore my woeful chemistry results


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    (Original post by Farhan.Hanif93)
    Having just opened the question, it specifies that N is an integer greater than 1 and it is true that the inequality holds for N≥2.

    To see this, note that N-√(N^2-1) = 1/[N+√(N^2-1)] < 1/N ≤ 1/2, which yields 2N-1/2 < N+√(N^2-1) after a little rearrangement.
    Oh I see! Thanks!
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    (Original post by shamika)
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    Yeah it sounds like you'll be in the 90s. Surprised you didn't go for 7 (I mean the one with sets S and T which was modular arithmetic in disguise), or 13 (which is very straightforward, especially so if you know the underlying theory about this type of process). Still not a bad result though.

    Make sure you don't do stupid things in II or III, some of those marks you lost in I you shouldn't have. (Why didn't you bother finishing Q1?). If I'm sounding harsh, it's because I know what you're capable of
    .
    Agreed. Hope I can redeem myself a little for STEP II and III. Thanks!

    --

    Good luck to all sitting this tomorrow!
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    (Original post by raff97)
    Seems all the pure questions have been done, was never a fan of applied so cant help there. sorry!

    (Yeah I have. Was alright I guess, will find out how I did on Saturday... arguable easier than STEP II and III if you put the work in throughout the year. certainly less interesting questions compared to STEP!)
    The Tripos has become properly boring in the last few years. There were a lot more interesting questions about 20 years ago (but most of the questions were still boring). I think that's just because you actually learn interesting stuff in your degree, so the end of year exams just seem lame in comparison to some of the problems you've tried on problem sheets which are harder and longer because you have more time.
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    To clarify one doubt that I have,

    When you have cos(6x)=cos(4x), the solutions are 6x=4x+/-2nipi or 6x=2npi+/-4x? and for tan6x=tan4x, the solutions will always be 6x=4x+/-npi or 6x=npi+/-4x?
    And how do you usually deduce for example the solutions of cosx=-(sqroot(3)/2), the solutions are sometimes, x=2npi+/-2pi/3 and sometimes it's x=npi+/-pi/3. Are there any tricks where writing the general form of these of questions. (I always get a bit confused at this).
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    (Original post by shamika)
    Spoiler:
    Show
    Yeah it sounds like you'll be in the 90s. Surprised you didn't go for 7 (I mean the one with sets S and T which was modular arithmetic in disguise), or 13 (which is very straightforward, especially so if you know the underlying theory about this type of process). Still not a bad result though.

    Make sure you don't do stupid things in II or III, some of those marks you lost in I you shouldn't have. (Why didn't you bother finishing Q1?). If I'm sounding harsh, it's because I know what you're capable of
    .
    Uhhhh I'm posting in spoilers because you did but
    Spoiler:
    Show

    Can you just assume properties of modular arithmetic like if a = b (mod m) and c = d (mod m) then ac = bd (mod m)?
 
 
 
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