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    (Original post by jahovasfitness)
    I stupidly marked the coordinates of the points at the beginning and ends of the horizontal lines. Given that I still got all of the shapes right, how many marks would I have lost do you think?
    If you still got the coordinates right, I don't see why you should lose any marks?
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    (Original post by smartalan73)
    If you still got the coordinates right, I don't see why you should lose any marks?
    Agreed.
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    Courtesy of physicsmaths:

    Question 6, STEP I:

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    Q5
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    Draw in radii to tangents and join centres giving you the following;

    AC = a+c
    AB = a+b
    BC = b+c
    AP = a
    BQ = b
    CR = c

    WLOG assume a>c
    draw in perpendiculars from c to AP
    b to AP and CR

    Now use pythagoras giving you the following relations

    (a+b)^2 = (a-b)^2 + PQ^2
    (a+c)^2 = (a-c)^2 + PR^2
    (b+c)^2 = (c-b)^2 + QR^2

    solving these we find that
    2*sqrt(ab) = PQ and so forth

    Obviously PQ + QR = PR giving us;
    sqrt(ab) + sqrt(bc) = sqrt(ac)
    divide by sqrt(abc) to get the desired result

    For the second part simply square the result to find 1/b and square again to find 1/b^2
    Substitute in a result follows easily

    (ii)
    square the RHS and cancel to get the following
    1/a^2 + 1/b^2 + 1/c^2 = 2(1/ab + 1/bc + 1/ac)
    multiply by (abc)^2
    b^2(a^2 + c^2) + (ac)^2 = 2(a^2bc + ab^2c + abc^2)
    b^2(a-c)^2 -2(ac)(a+c)b + (ac)^2 = 0
    quadratic formula;
    b = [2(ac)(a+c) +- 4(ac)^{3/2} ]/2(a-c)^2
    b = [ac(sqrt(a) +- sqrt(c))^2]/(a-c)^2
    sqrt(b) = [sqrt(ac)(sqrt(a) - sqrt(c))]/(a-c)
    sqrt(b) = sqrt(ac)/(sqrt(a) + sqrt(c))
    1/sqrt(b) = 1/sqrt(a) + 1/sqrt(c)
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    STEP 1 2016 Q11 solution. This is one I did in the paper... Thought it was a very long one actually, perhaps there was a shortcut I missed (would appreciate people looking for one). Ended in a show that which is always comforting.
    Attached Images
  1. File Type: pdf STEP1 2016 Q11.compressed.pdf (324.1 KB, 169 views)
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    (Original post by rohanpritchard)
    STEP 1 2016 Q11 solution. This is one I did in the paper... Thought it was a very long one actually, perhaps there was a shortcut I missed (would appreciate people looking for one). Ended in a show that which is always comforting.
    Lol you just best me. I proceeded in a slightly different way, feel free to compare

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    550695[/ATTACH]ww.thestudentroom.co.uk/app]Posted from TSR Mobile
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    Q13
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    Disclaimer; my wording here is bad, please let me know how i can fix it. Also i use g = lamda ( don't know how to latex)
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    P(T_1=t) = ge^{-gt}
    P(T_1<t) = 1-e^{-gt}
    P(T_1>t) = e^{-gt}
    All independent
    P(T_1,T_2,...,T_n > t) = e^{-ngt}
    P(not all T > t) = 1-e^{-ngt}
    This means that the lowest time is < t
    we want P(lowest time =t) so differentiate
    P(lowest time = t) = nge^{-ngt}

    E(t) = integral of ngte^{-ngt} from 0 to infinity
    Use integration by parts and you get
    E(t) = 1/gn
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    (Original post by KingRS)
    Q13
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    Disclaimer; my wording here is bad, please let me know how i can fix it. Also i use g = lamda ( don't know how to latex)
    Spoiler:
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    P(T_1=t) = ge^{-gt}
    P(T_1<t) = 1-e^{-gt}
    P(T_1>t) = e^{-gt}
    All independent
    P(T_1,T_2,...,T_n > t) = e^{-ngt}
    P(not all T > t) = 1-e^{-ngt}
    This means that the lowest time is < t
    we want P(lowest time =t) so differentiate
    P(lowest time = t) = nge^{-ngt}

    E(t) = integral of ngte^{-ngt} from 0 to infinity
    Use integration by parts and you get
    E(t) = 1/gn
    This is only part 1?


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    Q13

    Will rotate at some point


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    (Original post by Krollo)
    ..
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    (Original post by Zacken)
    ..
    Ty bro

    Forgot all my matrices dos they're not on step so I forgot how to rotate things


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    (Original post by Zacken)
    Agreed.
    That's a relief:/
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    (Original post by student0042)
    Q1. I feel like there was another part as it seems a bit small, but then again I only proved it for n=1 and said, do the same thing.
    Attachment 550405 I'll let you blow it up again.
    Note that p4(1) = 2^8 - 9(1)(3^3)=256-243=13
    q4(1) = 2/2 = 1

    So the two polynomials cannot be equal.

    Much easier than wading through half a page of algebra!
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    Zacken

    I'm obviously not going to upload any solutions until you give the go-ahead, but could I perhaps do the solutions for STEP II questions 6, 13 and that [redacted] one?
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    (Original post by Mathemagicien)
    Zacken

    I'm obviously not going to upload any solutions until you give the go-ahead, but could I perhaps do the solutions for STEP II questions 6, 13 and that one?
    Believe it was question 7 or 8, but probably shouldn't mention what the topic is just in case


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    (Original post by drandy76)
    Believe it was question 7 or 8, but probably shouldn't mention what the topic is just in case


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    You're right, my apologies!
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    (Original post by Mathemagicien)
    You're right, my apologies!
    Ith cool, only 21 hours to go!


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    (Original post by MathHysteria)
    Note that p4(1) = 2^8 - 9(1)(3^3)=256-243=13
    q4(1) = 2/2 = 1

    So the two polynomials cannot be equal.

    Much easier than wading through half a page of algebra!
    That probably could have saved me some time in the exam. It looks like I went a very long way around it. I was thinking, "surely there's an easier way to do this", but didn't think to do that.
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    can I do q5
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    Section Leader
    Note the STEP Prep thread is now closed and will be re-opened at 9am (ish...) tomorrow.

 
 
 
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