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STEP II 2012 discussion thread

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    (Original post by SParm)
    Here's my solution for question 8:

    Part 3:

    Spoiler:
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    For this part; we can simply find the number p for which this equation works. If p is a real number; then clearly the equation will work. So:

    u_{n + 1} - pu_{n} + u_{n - 1} = 0

    When n = 1

    u_{n - 1} = u_{0} = \alpha\\



u_{n} = u_{1} = \beta\\



u_{n + 1} = u_{2}

    From the sequence equation:

    u_{2} = \dfrac{\beta ^{2} - q^{2}}{\alpha}

    Substitute back into the original equation:

    u_{n + 1} - pu_{n} + u_{n - 1} = \dfrac{\beta ^{2} - q^{2}}{\alpha} - p\beta + \alpha = 0

    Divide by \beta:

    \dfrac{\beta ^{2} - q^{2}}{\alpha\beta} - p + \frac{\alpha}{\beta} = 0

    Note: \dfrac{\alpha}{\beta} = \dfrac{\alpha ^{2}}{\alpha\beta}

    Re-arrange our original equation to get:

    \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

    \alpha,\: \beta \neq 0\\



\alpha,\: \beta,\: q are real, so p is real, and therefore our original equation works when p is the aforementioned expression.

    (I’ve assumed I can do this because of the wording of the question which asks you to prove that statement for some number “p”; and doesn’t put any limitations on what “p” can be; just asks for it in terms of \alpha,\: \beta,\: q, so I thought that you could go the other way and show that such a “p” exists and therefore the statement is true for that “p” which is basically the same as what they’re asking.)
    Spoiler:
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    Suppose u_{n - 1} + u_{n - 1} = p u_n, and from u_n(u_n + u_{n + 2}) = u_{n + 1}(u_{n-1} + u_{n + 1}), you get


    \displaystyle \frac{u_n(u_n + u_{n+2})}{u_{n+1}} = p u_n.
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    (Original post by jack.hadamard)
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    Suppose u_{n - 1} + u_{n - 1} = p u_n, and from u_n(u_n + u_{n + 2}) = u_{n + 1}(u_{n-1} + u_{n + 1}), you get


    \displaystyle \frac{u_n(u_n + u_{n+2})}{u_{n+1}} = p u_n.
    Oh yeah that's true :/...but is there anything inherently wrong with my proof?
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    Can't wait to bash out some solutions
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    (Original post by TheMagicMan)

    so 1 < e(1+1/n)^{-n}< (1+1/n)^{1/2}

    The RHS has a limit of one so want we want follows by the squeeze theorem...the real question is is there a simpler way (tbh I'm never quite sure how STEp wants you to approach limits)
    You could rewrite it as \left(1 + \frac{1}{n}\right)^n\sqrt{1 + \frac{1}{n}}, consider it as a product and then mention that it is decreasing.
    I don't see anything wrong with mentioning the Squeeze theorem, and it even sounds better.
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    (Original post by SParm)
    Oh yeah that's true :/...but is there anything inherently wrong with my proof?
    I just skimmed it by far, but I would generally avoid commenting on other people's work.
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    (Original post by Peter8837)
    Hmmm.. fair enough. Also, I did the very first part of Q8 just to grab a mark or two (I showed the first inequality) - do you reckon I'd get any marks for that?
    There was a lot going on in that question. Probably 1 mark for that inequality at the start (and even then, to get that mark I think they would expect the proof of it to give a short explanation of why squaring doesn't change anything since \beta - \alpha must be positive etc).
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    (Original post by jack.hadamard)
    I just skimmed it by far, but I would generally avoid commenting on other people's work.
    Fair. I assume you took the exam yesterday then?
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    (Original post by TheJ0ker)
    Not great but I don't think I have failed, 2 full solutions and 2 half solutions. I didn't have a Cambridge offer anyway so I think my grade on STEP 1 will be easily enough to get into Warwick.
    lucky you! some of us with Cam offers only got 3 almost out and a couple of partials :/ heres hoping for low boundaries anyway!
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    For question 4, you need to consider the expansion:

    e^x = 1 + x + (x^2)/2! + (x^3)/3! + ...

    Let x = 1/y

    e^(1/y) = 1 + 1/y + 1/(2*y^2) + 1/(6*y^3) + 1/(24*y^3) + ...
    < 1+ 1/y + 1/(2*y^2) + 1/(4*y^3) + 1/(8*y^) + ...
    Note that we have a geometric series with common ratio of 1/2y and first term 1/y. Since y > 1/2, 1/2y < 1. This allows us to use the formula for the sum to infinity
    Hence:
    e^(1/y) < 1+ 2/(2y-1)
    => e^(1/y) < (2y+1)/(2y-1)
    => 1/y < ln[(2y+1)/(2y-1)], as required

    Also, for the last part, just consider the ratio between the 2 boundaries for e. As n goes to infinity, the boundaries are closer together. e lies in between. We can deduce immediately that e will get closer to (1 + 1/n)^n
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    Just out of interest, I completed 3 questions successfully and about 2/3 of another question. Would you reckon that's enough for a grade 1. Worry about that since I am a Cambridge offer holder too :confused:
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    (Original post by Farhan.Hanif93)
    There was a lot going on in that question. Probably 1 mark for that inequality at the start (and even then, to get that mark I think they would expect the proof of it to give a short explanation of why squaring doesn't change anything since \beta - \alpha must be positive etc).
    Thanks. Also, for Q5 part (ii), I drew the graphs right in both cases (I checked with wolfram) but I didn't explicitly solve the cubic I got when trying to find stationary points... I just sort of wrote the stationary points down after drawing the graphs. How many marks would I lose do you think? This question is open to anyone by the way. Thanks.
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    (Original post by Peter8837)
    Thanks. Also, for Q5 part (ii), I drew the graphs right in both cases (I checked with wolfram) but I didn't explicitly solve the cubic I got when trying to find stationary points... I just sort of wrote the stationary points down after drawing the graphs. How many marks would I lose do you think?
    Would probably lose nothing more than 4 or 5; possibly less if you've got the correct S.P's too.
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    DONE 6 on paper, will post the solution soon, but will try some other qs first .
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    Can someone do a solution to Q5? I'd be interested to see how you would solve that cubic to find turning points!
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    (Original post by Zuzuzu)
    How'd your teacher find it?
    tough
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    (Original post by Peter8837)
    Can someone do a solution to Q5? I'd be interested to see how you would solve that cubic to find turning points!
    I think you can factorise out (x-a)(x-b) and then youre left with something that again factorises. I'll try and do it again after my tea.
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    (Original post by Peter8837)
    Can someone do a solution to Q5? I'd be interested to see how you would solve that cubic to find turning points!
    i guessed it was symmetrical about the midpoint of a,b meaning (a+b)/2 is either a turning point (1st case) or a discontinuity (2nd case)
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    (Original post by Peter8837)
    Thanks. Also, for Q5 part (ii), I drew the graphs right in both cases (I checked with wolfram) but I didn't explicitly solve the cubic I got when trying to find stationary points... I just sort of wrote the stationary points down after drawing the graphs. How many marks would I lose do you think? This question is open to anyone by the way. Thanks.
    I did exactly the same thing as I was hurrying to get out another question in the last half an hour, think one of my stationary points was right but the others weren't. Hoping the majority of the marks were for the graphs as mine were both right. Did you just guess the stationary points? What did you get?
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    (Original post by Nam Nguyen)
    Just out of interest, I completed 3 questions successfully and about 2/3 of another question. Would you reckon that's enough for a grade 1. Worry about that since I am a Cambridge offer holder too :confused:
    Perhaps; perhaps not. Now you have to perform at the top of your game for STEP III without knowing your result in II.

    Suppose that at interview, you had been asked 'can you perform under pressure?'. I think you would have found the answer 'yes'. They've thought of that. So they've designed a process that invites you to demonstate it.
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    (Original post by MrDD)
    i guessed it was symmetrical about the midpoint of a,b meaning (a+b)/2 is either a turning point (1st case) or a discontinuity (2nd case)
    Yes, this exactly. I checked my graphs with wolframalpha and they were correct... I just don't know how many marks I'd lose for not 'explicitly' finding the stationary points; but FarhanHanif93 who is quite experienced said 4-5 marks so that seems a good estimate.

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