STEP I 2012 discussion thread

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  1. fGDu's Avatar
    181 21-06-2012  21:39
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    Re: STEP I 2012 discussion thread
    Here is a pdf version of the paper for your convenience
    Attached Files
  2. File Type: pdf step 1 2012.pdf (308.7 KB, 707 views)
  3. mikelbird's Avatar
    182 22-06-2012  10:03
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    Re: STEP I 2012 discussion thread
    Q7
    (i)
    Using the relation, t_n = x \implies (p+q -1) x = 0 and since x\not=0, we have the solution \boxed{p+q = 1; x\in \mathbb{R}/\{0\}}, as required.


    (iii)
    Consider the relation again:
    t_{2n} = x, t_{2n+1} = y \implies \begin{Bmatrix} x=py+qx & (1) & \mathrm{if} & n &\mathrm{ even} \\ y=px+qy & (2) & \mathrm{if } & n & \mathrm{odd}

    Note (q-1) \times (1) - p \times 2: [(q-1)^2-p^2] x =0 and since x\not= 0, this can only hold if (q-1)^2 - p^2 =0 \iff \boxed{q\pm p = 1}, as was to be shown.

    Note x\times (1) - y\times (2): (q-1)(x^2-y^2) = 0 and we have x^2 - y^2 = 0 \iff x=\pm y if q\not 1 AND p\not=0.

    (ii)
    Again, from the relation:
    t_{3n} = x, t_{3n+1} = y, t_{3n+2} \implies \begin{Bmatrix} z=py+qx & (1) & \mathrm{if} & n\equiv 0\mod 3 \\ x=pz+qy & (2) & \mathrm{if } & n\equiv 1\mod 3 \\ y=px+qz & (3) & \mathrm{if } & n\equiv 2\mod 3

    Note p\times (2) + (3) = (4): (pq-1) y + (p^2 + q)z = 0

    Also q\times (2) + (1) = (5): (pq-1)z + (q^2+p)y = 0

    So (pq-1) \times (4) - (p^2 + q) \times (5): [(pq-1)^2 - (p^2+q)(q^2+ p)]y = 0 and since y\not=0, this can only hold if (pq-1)^2 - (p^2+q)(q^2+p) =0 \iff \boxed{p^3 + q^3 + 3pq -1 =0} (*), as required.

    Setting p=1-q in the LHS of (*) yields 0 so we can observe that:
    p^3 + q^3 + 3pq -1 =0 \iff (p+q-1)(p^2 +p - pq + q + q^2 + 1) = 0
    \iff \dfrac{1}{2}(p+q-1)[(p-q)^2 + (p+1)^2 + (q+1)^2] = 0
    \iff p+q =1 or (p-q)^2 + (p+1)^2 + (q+1)^2 =0, as was to be shown.

    By plugging p=1-q, into the equations (1), (2) and (3) and attempting to solve, you end up with \boxed{x=y=z}. Note also that (p-q)^2 + (p+1)^2 + (q+1)^2 =0 \iff p=q=-1 as squares are non-negative. By plugging these into each of (1), (2) and (3), they all reduce to \boxed{x+y+z=0}[/latex], as required.



    If you're looking for any other specific solutions, just mention it and I'm sure someone will be able to post one up.[/QUOTE]

    A little knowledge of matrices and determinants organises this nicely....
    Attached Files
  4. File Type: pdf Step2012Paper1Question7.pdf (74.8 KB, 91 views)
  5. Sdiff's Avatar
    183 22-06-2012  19:02
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    Re: STEP I 2012 discussion thread
    Terrible. Whenever I did practice, i would usually get 4-6 fully done. I only got 1 in this. A combination of exhaustion (4 hours sleep - i couldn't sleep) and the fact that I had been doing exams for the past 2 weeks and hadn't practiced.
  6. angelicsuccubus's Avatar
    184 23-06-2012  21:32
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    Re: STEP I 2012 discussion thread
    (Original post by gymrat)
    I got about 40/120 lol , gap year starts Thursday after FP2.
    i donno.. if im even gonna make it to a 40.. i have like 4 fragmented answers.. and I need a 2 for imperial =[
  7. mikelbird's Avatar
    185 25-06-2012  16:34
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    Re: STEP I 2012 discussion thread
    (Original post by shamika)
    I really liked how you did the last bit. I don't know how you came up with equating areas though... is that something you've seen before or did you just think 'this is going to give me something simpler to differentiate'?

    On a separate note, I'm certain I've seen Q4 before - but the only place I would've seen it is in Further Maths conics. Had anyone else seen the question before? There's no way I would've known the constant is 2 before doing the question otherwise...
    yes...the standard way look like horrible algebra.....so i look around for something simple....and there it was right under my nose!!!!
  8. HashBrowns's Avatar
    186 26-06-2012  17:49
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    Re: STEP I 2012 discussion thread
    I took the STEP 1 paper, and was wondering how many marks I would get for my partial solutions.
    I think I solved 2 questions, but came very close to solving two others. i.e I did the easy initial parts and made some headway for the concluding part at the end. Will I be able to get 10+ marks for these - or should I be feeling lucky to get a 2 grade? Thanks
  9. msmith2512's Avatar
    187 26-06-2012  18:43
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    Re: STEP I 2012 discussion thread
    (Original post by HashBrowns)
    I took the STEP 1 paper, and was wondering how many marks I would get for my partial solutions.
    I think I solved 2 questions, but came very close to solving two others. i.e I did the easy initial parts and made some headway for the concluding part at the end. Will I be able to get 10+ marks for these - or should I be feeling lucky to get a 2 grade? Thanks
    It depends on how much you have done of each question.

    Which questions did you come close to - which parts did you do?
  10. mathslive101's Avatar
    188 27-06-2012  23:32
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    Roughly how many marks do you think one need to get a grade 2..


    This was posted from The Student Room's iPhone/iPad App
  11. deejayy's Avatar
    189 28-06-2012  00:07
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    Re: STEP I 2012 discussion thread
    (Original post by mathslive101)
    Roughly how many marks do you think one need to get a grade 2..


    This was posted from The Student Room's iPhone/iPad App
    https://docs.google.com/spreadsheet/...kdMMHotczRZOUE
  12. cpdavis's Avatar
    190 28-06-2012  16:37
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    Re: STEP I 2012 discussion thread
  13. mikelbird's Avatar
    191 22-07-2012  13:48
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    Re: STEP I 2012 discussion thread
    (Original post by snow leopard)
    Q11(i), I'll add (ii) later.

    Q11
    (i)
    let \theta=\arctan \frac{7}{24} \Rightarrow \cos \theta = \frac{24}{25} & \sin \theta = \frac{7}{25} let \alpha=\arctan \frac{4}{3} \Rightarrow \cos \alpha = \frac{3}{5} & \sin \alpha = \frac{4}{5} R(A) (\mu R + 5mg\sin\theta)=T R(B) (\mu S + 3mg\sin\alpha)=T \mu R + 5mg\sin\theta=\mu S + 3mg\sin\alpha \mu \frac{24}{5}mg + \frac{7}{5}mg = \mu \frac{9}{5}mg + \frac{12}{5}mg \Rightarrow \mu = \frac{1}{3} \Rightarrow T=3mg Mg=2T=6mg \Rightarrow M=6m

    As snow leopard did not come back and complete this here is a full version.....
    Attached Files
  14. File Type: pdf Step2012Paper1Question11.pdf (68.7 KB, 48 views)
    15. Last edited by mikelbird; 22-07-2012 at 17:49. Reason: details on file wrong....
  15. Ree69's Avatar
    192 01-08-2012  23:13
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    Re: STEP I 2012 discussion thread
    Does anyone know when the STEP grade boundaries will be released?

    I've felt so restricted without internet access for so long! The paper was ok imo, q8 and q1 were the most approachable - and then from there i did about 4 more partial attempts of varying quality.

    Agreeing with others that this paper was definitely harder than past ones, I thought q5 was a lot harder than what meets the eye.
  16. tiny hobbit's Avatar
    193 08-01-2013  12:33
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    Re: STEP I 2012 discussion thread
    (Original post by mikelbird)
    As snow leopard did not come back and complete this here is a full version.....
    You lost the factor of g in your answers in part ii
  17. mikelbird's Avatar
    194 14-01-2013  09:21
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    Re: STEP I 2012 discussion thread
    (Original post by tiny hobbit)
    You lost the factor of g in your answers in part ii
    Yes I know....I corrected on my Computer but never put it up on TSR..sorry!!
  18. studyingmad's Avatar
    195 22-02-2013  16:18
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    Re: STEP I 2012 discussion thread
    (Original post by snow leopard)
    Q11(i), I'll add (ii) later.

    Q11
    (i)
    let \theta=\arctan \frac{7}{24} \Rightarrow \cos \theta = \frac{24}{25} & \sin \theta = \frac{7}{25} let \alpha=\arctan \frac{4}{3} \Rightarrow \cos \alpha = \frac{3}{5} & \sin \alpha = \frac{4}{5} R(A) (\mu R + 5mg\sin\theta)=T R(B) (\mu S + 3mg\sin\alpha)=T \mu R + 5mg\sin\theta=\mu S + 3mg\sin\alpha \mu \frac{24}{5}mg + \frac{7}{5}mg = \mu \frac{9}{5}mg + \frac{12}{5}mg \Rightarrow \mu = \frac{1}{3} \Rightarrow T=3mg Mg=2T=6mg \Rightarrow M=6m

    Can you post the second part please.
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