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    (Original post by Zaphod77)
    Yes, in that you have to divide it through before splitting into partial fractions or create 3 fractions, as justinawe showed me earlier - I tend to divide through! Doing partial fractions normally don't work on that kind of fraction
    Awesome thanks very much mate
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    What's the hardest C4 paper since 2000? (I can't remember when our spec started :P )
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    (Original post by Fortitude)
    That would still be complicated as you'd have to cosx in terms of u??:confused:
    Practise some questions in the book to get the hang of it.
    Yes will do now,


    so they got

    sin3xcos2x + cos3xsin2x + sin3xcos2x - cos3xsin2x

    agree so far?
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    (Original post by usycool1)
    Yup, I am :yep:
    All the best of luck Even though I dont think you need it
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    (Original post by Mallika)
    What's the hardest C4 paper since 2000? (I can't remember when our spec started :P )
    June 2005 and January 2008 have the lowest grade boundaries
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    (Original post by kuku2013)
    is there any specific soloman press paper worth doing that may help in exam or shell I just do all of them?
    All of them:eek: The exam's tomorrow are you Sonic??
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    (Original post by Story)
    Would it be -1/3(3x+1)^-4 ?
    it should be:

    -1/3(3x+2)^-4

    think you just misread the original equation lol, you've got the fundamentals correct
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    (Original post by Zaphod77)
    You could be, you can use a substitution to do that or do it by inspection. Your second question, if you have an improper fraction I tend to divide through (long division) so then you get a number and a remainder, which will be over the original denominator. The remainder can then be split into partial fractions, which you can do either through comparing coefficients when you've set up the A and B fraction or through eliminating them (e.g. if the fraction included A(x+2) you can let x=-2 to eliminate A and just leave B). Does that answer your question at all?
    Right, if instead of long division you can use the alternative comparing coefficient method first right? and then use it again to find the values or use the sub in method like you said.
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    (Original post by Frankster)
    it should be:

    -1/3(3x+2)^-4
    Thats what I meant, typo mistake! Thank you.
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    Question:
    What's the HARDEST c4 paper you've ever done? Or just a list of the most hardest ones, doesn't matter what exam board really. Thanks
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    (Original post by Fortitude)
    All of them:eek: The exam's tomorrow are you Sonic??
    I did 8 yesterday my head actually exploded, i had to have an head implant.
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    (Original post by Fortitude)
    All of them:eek: The exam's tomorrow are you Sonic??
    I can do them trust me just tell me which ones. its takes me an hour each paper but might leave all the easy questions
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    (Original post by otrivine)
    Yes will do now,


    so they got

    sin3xcos2x + cos3xsin2x + sin3xcos2x - cos3xsin2x

    agree so far?
    yep
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    (Original post by Chris-69)
    I mean if it's top heavy, would you do for example:

    A + B/(x+1) + C(x+2)

    ye
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    (Original post by otrivine)
    All the best of luck Even though I dont think you need it
    Thanks, good luck to you too! Ha, I'm sure I will need it.
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    found this on a solomon paper and i did it the normal way but i didnt get the right answer

    4. The points A and B have coordinates (3, 9, −7) and (13, −6, −2) respectively.
    (a) Find, in vector form, an equation for the line l which passes through A and B. (2)
    (b) Show that the point C with coordinates (9, 0, −4) lies on l. (2)
    The point D is the point on l closest to the origin, O.
    (c) Find the coordinates of D. (4)
    (d) Find the area of triangle OAB to 3 significant figures. (3)
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    Does anyone have the markscheme for the original c4 june 2012 paper? Not the replacement
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    (Original post by Fortitude)
    yep
    but then why in the book gives


    sin5x+sinx = 2sin3xcos2x

    where did sin5x come from
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    (Original post by justinawe)
    3^2^x = (3^2)^x = 9^x
    What about 3^(2x+1)
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    can someone explain how sin^2tcost integrates to (sin^3t)/3 please, im really confused (june 2008 Q8)
 
 
 
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