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    I thought it was pretty damn good....

    Apart from the Vector questions...seeing as i need 59/100 to get an A, i think it doesnt matter that i got most of them wrong!
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    For the 2nd vector question I dot producted everything hoping that one would right.. lol.. but to honest I didn't have a clue.. the rest went pretty good I think. I was expecting worse after that horrible P2 paper!
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    not a bad paper


    did u get

    y = x -1/2 for the first question, or + 1/2 cant remember



    2ln12.25 for the partial fraction integration

    0.25 for the cone sand question

    0.444 for the gradient of the curve question


    (5,-3,2) for the point where the perpendicual is on the last question
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    (Original post by friday13th)
    not a bad paper


    did u get

    y = x -1/2 for the first question, or + 1/2 cant remember



    2ln12.25 for the partial fraction integration

    0.25 for the cone sand question

    0.444 for the gradient of the curve question


    (5,-3,2) for the point where the perpendicual is on the last question
    Yes I got that for the first question.. I got two values for the gradient of curve question and the rest I can't remember.. lol.. even though it was only a few hours ago!
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    yeah, + and - 0.444
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    hmm.. no I got two completley different gradients. Although I think I got it wrong because my working for the question was v. long-winded- and it was only 4 marks!? Did you use the chain rule to differentiate the y and then take the values to the other side of the equals sign leaving dy/dx?
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    what the...??

    You use implicit differentiation for that question
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    That's what I mean.. Implicit differentiation using the chain rule in a different form
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    paper was ok...only needed a D on it for an A so that was pretty swish!

    annoying cos I knew how to do vector last question it just wasn't working, never mind! I can't remember most of the questions but your answers look familiar!

    could do the integration...kept getting a random mixture of ln4's ln7s and all sorts...dammit! really bugging me!

    if anyone could help me i think the question was

    30/ (x-4)(7-2x)

    integrate between 3 and 0.


    I really wanna see where I went wrong!
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    Yes, you split it into partial fractions, like question A told you to do

    then you integrate both,

    you had like 2ln|x+4| - 2ln|7-2x| or something


    when u put the numbers in, you get

    when x = 3

    2ln|7| - 2ln|1| or something...

    do that with 0 to, like normal definate integrals

    but u use the log rule that says

    ln|7| - ln|4| = ln(4/7) using that identity twice or something, gives the answer 2ln12.25
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    (Original post by friday13th)
    Yes, you split it into partial fractions, like question A told you to do

    then you integrate both,

    you had like 2ln|x+4| - 2ln|7-2x| or something


    when u put the numbers in, you get

    when x = 3

    2ln|7| - 2ln|1| or something...

    do that with 0 to, like normal definate integrals

    but u use the log rule that says

    ln|7| - ln|4| = ln(4/7) using that identity twice or something, gives the answer 2ln12.25

    NOOOOOOOOO!!!!!! I did that and crossed it out! What a DUMBARSE! He/she might be nice and see that I got that answer originally!

    also on the rates of change question how can you explain why the volume was whatever it was?
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    (Original post by friday13th)
    Yes, you split it into partial fractions, like question A told you to do

    then you integrate both,

    you had like 2ln|x+4| - 2ln|7-2x| or something


    when u put the numbers in, you get

    when x = 3

    2ln|7| - 2ln|1| or something...

    do that with 0 to, like normal definate integrals

    but u use the log rule that says

    ln|7| - ln|4| = ln(4/7) using that identity twice or something, gives the answer 2ln12.25
    i thought the answer was just ln(7/4).

    You had ln |7| - 1/2ln |2| which is equal to ln (7/4) (Whatever i did, i checked on my calculator so to make sure i re-arranged the logs correctly)
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    (Original post by friday13th)
    not a bad paper


    did u get

    y = x -1/2 for the first question, or + 1/2 cant remember



    2ln12.25 for the partial fraction integration

    0.25 for the cone sand question

    0.444 for the gradient of the curve question


    (5,-3,2) for the point where the perpendicual is on the last question
    i got +0.444 and -0.444 for the gradient. 0.24 for the cone. (3/(4pi)). I got 2ln12.25 for the integration. I got y= 4x - (7/2) for the first question. I cant remember what i got for the last question.
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    Piece of cake, as was P2 but in a different way. In P2 I enjoyed the way the answers came together. This time I just did them and dispensed with them, like a machine gun...
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    (Original post by hohum)
    also on the rates of change question how can you explain why the volume was whatever it was?
    Volume of a cone is 1/3 x Pi x r^2 x h

    It said that the radius stayed equal to the height, thus you could substitue r for h, giving the equation of 1/3 x Pi x h^3

    Shame it was only one mark

    It wasn't a bad paper over all, although I found myself a bit short of time, and didn't get the final vector question right but that's just me and my tendency to be too slow.
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    Woah, how do you remember your answers?!

    I got everything to work out neatly, plus need a ridiculously low mark to get an A as I found P2 really easy for some reason!

    I wouldn't have a clue how my answers compare anyway as I never convert them to decimals, I always leave them as fractions in their simplest form, much neater that way and means you don't have to worry about rounding!

    The only thing i remember was that on the second to last question, K = 1/6! (at least I hope it does!)
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    (Original post by figgetyfig)

    The only thing i remember was that on the second to last question, K = 1/6! (at least I hope it does!)
    sure did
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    on the integration question

    3,0

    2ln|x+4| - 2ln|7-2x|

    then it said get the answer in the form qlnp where q and p where rational

    so they can be fractions


    then putting the limits in gets to
    2ln7 - 2ln1 -2ln4 + 2ln7

    cancels to
    4ln7 -2ln4

    now at this point you can use the log rule to say this
    ln((7^4)/(4^2))

    or to be precise as the question asks

    factorise
    4(ln7 - 0.5ln4)
    then

    4(ln7 - ln4^0.5)
    then 4(ln7 - ln2)
    then 4ln(7/2)


    thats it
    both are rational
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    (Original post by ZJuwelH)
    Piece of cake, as was P2 but in a different way. In P2 I enjoyed the way the answers came together. This time I just did them and dispensed with them, like a machine gun...
    It will be interesting to see what the marks are like then! I doubt I'll be getting any 100's as in the past papers I usually make one or two slips (most commonly getting plus and minus signs wrong in binomial expansion!), But i'm hoping for at least 90 in P2, P3, and S1, at least 85 in M2, and hopefully at least 80 in P5 (that one didn't work out quite so neatly for me!)
 
 
 
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