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    (Original post by Lusus Naturae)
    Could have been worse but naturally my mind went blank on a couple of trivial parts, they were "show that...4k+sin4k+2=0" and . Should shoot myself for that Quite sure rest was correct though so around 90UMS I guess.
    I still don't get that... :confused:
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    Yeah. Exactly.

    And can anyone remember the ln question?
    was it f--> ln(x+k) and
    g--> |2x -k|?
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    (Original post by BovineBeast)
    I don't really think so. Negative Infinity isn't a real number.
    Guess that makes sense, thanks
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    Think so.
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    is anyone remember how many marks the Teperture lost in per minutes worth? thanks
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    I'm still trying to work out why fg(k/4) is 0.5k...
    I got 1.5k :x

    The temp one was...4 marks I think. Maybe 3.
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    (Original post by BovineBeast)
    Er, that's 1.64.
    Ok so -1.64 then :P MB but its def negitive...
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    (Original post by Malik)
    Definetly not 56, compared to the Jan06 and Jun05 I think larger majority would have found this paper easier. At best its just going to be the typical 60/75 80% boundary.
    Do you really think this paper was easier than the past 2 years?

    Besides, we have some foreigners amongst us.. who are probably taking the answers from us right now :rolleyes: UP GO THE UMS BOUNDARIES :tsr2: :tsr2: :tsr2: :tsr2: :tsr2: :tsr2: :tsr2: :tsr2:


    sigh..
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    (Original post by Rabite)
    I'm still trying to work out why fg(k/4) is 0.5k...
    I got 1.5k :x

    The temp one was...4 marks I think. Maybe 3.
    Its comes down to how you treated the modulus in the question. I not sure which way was right so I cant really comment.
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    fg(k/4) IS ln(1.5k) why are people saying its not.
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    i think it is ln(1.5).

    well how many marks was that one then? cause they're gone for me now.
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    Ill show you again:

    f(x) = ln(x + k)
    g(x) = |2x - k|

    fg(x) = ln(|2x - k| + k)

    Lets x = 0.25k
    Ln(|-0.5k| + k) = ln(1.5k)
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    You'll still get method marks for the earler part of the question.
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    (Original post by spacko101)
    so absolutely NO ONE remembers how many marks the Sin2A thing was?

    5 mark question...
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    (Original post by Sollytear)
    Ill show you again:

    f(x) = ln(x + k)
    g(x) = |2x - k|

    fg(x) = ln(|2x - k| + k)

    Lets x = 0.25k
    Ln(|-0.5k| + k) = ln(1.5k)
    Yea thats how I did it.
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    that was quite a challenging paper.. anyone else feel like they rushed through it? i think i finished it.. although i know i didnt get that Sin2A questsion right.. the proving stuff was quite difficult but i think it was managable is u let it sit for a while..
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    (Original post by ellen8899)
    is anyone remember how many marks the Teperture lost in per minutes worth? thanks
    3 or 2 i fink
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    There were two questions I didn't finish, which totalled 10 marks but I think I got method marks for what I did do

    I'll just have to hope there won't be too many silly mistakes in there, as they can cost me dear.
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    i got up to:

    f(x) = ln(x + k)
    g(x) = |2x - k|

    fg(x) = ln(|2x - k| + k)

    then i went dumb. and cancelled the k's. lol.
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    ok paper!

    does anyone remember how they showed k=0.277 after the iterative questions? i just left it out cos i cud not do it, simple as.

    i hope the c4 paper is really eas, it neeeds to be so i can get an A!

    and it is Ln(1.5k) definite.
 
 
 

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