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    Thought everything was pretty alright apart from Question 9
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    (Original post by chemari1)
    the asymptote was 1
    There was one at x = 1. The horizontal asymptote was at y=3.
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    (Original post by Duskstar)
    You had something like

    \dfrac{1}{6}n(n+1)(3n^{2}+(3-2p)n-3) = \dfrac{1}{6}(n^{2}+n)(3n^{2}+(3-2p)n-3)

    as the formula I think. The coefficient on n^{4} is \dfrac{1}{2} if you multiply it out, and the coefficient on n^{3} is given by multiplying \dfrac{1}{6} \times (1 \times 3 + 1 \times (3 - 2p)) by matching up terms in either bracket to get n^{3} terms.

    Set \dfrac{1}{2} = \dfrac{1}{6} \times (1 \times 3 + 1 \times (3 - 2p)) and solve for p.
    Ahhhhhhh thanks, how many marks was just that part?
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    (Original post by hoakenfull)
    Anyone remember the mark-split for the individual parts of q7? Roots of the polynomial eq'n.

    I did alpha*beta*given root*complex conjugate = -b/a
    I got that p = 3/2 and I think another answer might have been -1 ??

    If someone could post the question+answers would be v grateful
    Don't quote me on it but I think it was something like 5/6 for show a+b and find ab, then 2/3 to find a and b, and 4/5 for the rest of the question. Can't remember that well haha
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    My handwriting was so messy in that exam I think i'll lose a few extra marks because they cba to look for the answers in all of my scribble :lol: Seriously don't know how I messed that up so much, it took a huge chunk out of my time because I drew C1 as Pi/4 instead of Pi/2 so I couldn't for the life of me figure out how to work out the part of the question where you have to say where they intersect,

    I'm also worried that I didn't end up writing out the inverse of B matrix, because I started the question then thought my Mu value was wrong so I told myself I'd go back to it later but I can't remember if I actually did I also got the 3sf series one wrong because I got two diferent values each time I calculated it and I ended up writing the wrong one down, and for the polynomial question I got A=6 and B=39 :mad: I was really aiming for 100 ums in all my AS further exams, fp1 has been such a downer after M2
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    (Original post by lfcjatt)
    Ahhhhhhh thanks, how many marks was just that part?
    I don't think it was more than 3 or 4 marks but I can't remember.
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    someone please explain the method for 9iv)????
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    (Original post by DoNoMicrowave)
    For Qs 7 i think, how did you calculate alpha x beta. I didn't get that and so couldn't really do the rest of the qs and find A and B and then the w/j bit. I've done all the fp1 past papers and there has been nothing like this qs before. Also the series method of differences qs (i think it was near the end, honestly can't remember anymore) was a toughie, i think I may have gotten partly there but still hard
    To get \alpha \beta I think the easiest would have been to use the formulae for roots of equations (eg the sum of the roots is -\dfrac{b}{a}

    The one to use in the question was the product of the roots is \dfrac{e}{a} (for a quartic), so you got \alpha \beta \gamma \delta = \dfrac{-26}{2} = -13 and you had \gamma and \delta so you could multiply them together (gave you 13) and you get \alpha \beta = -1.

    And I agree the method of differences question was tough, but I'm guessing that's why it was the last question on the paper :P
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    (Original post by Lisb)
    someone please explain the method for 9iv)????
    If that was the last part, then...

    They gave you the sum \dfrac{45}{stuff} + ... + \dfrac{105}{stuff} so you had to work out the limits of the sum (2n_{1} + 5 = 45 \Rightarrow n_{1} = 20, \, 2n_{2} + 5 = 105 \Rightarrow n_{2} = 50), then use these and the formula to get the answer...

    \displaystyle \sum_{r=20}^{50} u_{r} = \sum_{r=1}^{50} u_{r} - \sum_{r=1}^{19} u_{r}



\mathrm{plug numbers into your calculator...}



= 0.00813
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    (Original post by Duskstar)
    There was one at x = 1. The horizontal asymptote was at y=3.
    will I get all the marks except for that of naming the asymptote?
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    (Original post by chemari1)
    will I get all the marks except for that of naming the asymptote?
    I think it was a 3 mark question, which will be 1 mark per correct asymptote, so you'll just get a mark for each correct one you wrote down.
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    (Original post by Duskstar)
    If that was the last part, then...

    They gave you the sum \dfrac{45}{stuff} + ... + \dfrac{105}{stuff} so you had to work out the limits of the sum (2n_{1} + 5 = 45 \Rightarrow n_{1} = 20, \, 2n_{2} + 5 = 105 \Rightarrow n_{2} = 50), then use these and the formula to get the answer...

    \displaystyle \sum_{r=20}^{50} u_{r} = \sum_{r=1}^{50} u_{r} - \sum_{r=1}^{19} u_{r}



\mathrm{plug numbers into your calculator...}



= 0.00813
    did exactly the same method but ended up with -6.25 x 10^-4 as my answer. strange. should i only lose one mark since i did the right method?
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    (Original post by Duskstar)
    I think it was a 3 mark question, which will be 1 mark per correct asymptote, so you'll just get a mark for each correct one you wrote down.
    thanks, what do you think the boundary will be for a b?
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    (Original post by Lisb)
    did exactly the same method but ended up with -6.25 x 10^-4 as my answer. strange. should i only lose one mark since i did the right method?
    Yep if you wrote down the method and just put it into your calculator wrong you should only drop the last mark
    (Original post by chemari1)
    thanks, what do you think the boundary will be for a b?
    I have absolutely no idea sorry :P
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    (Original post by Duskstar)
    Yep if you wrote down the method and just put it into your calculator wrong you should only drop the last mark

    I have absolutely no idea sorry :P
    in the method of differences question I replaced 1, 2 , 3 and 4 in the formula, I wrote the equivalence of the sum of the three terms, but there wasn't a pattern of what would cancel out to be seen, so I stopped there, will I get any marks?
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    (Original post by Duskstar)
    To get \alpha \beta I think the easiest would have been to use the formulae for roots of equations (eg the sum of the roots is -\dfrac{b}{a}

    The one to use in the question was the product of the roots is \dfrac{e}{a} (for a quartic), so you got \alpha \beta \gamma \delta = \dfrac{-26}{2} = -13 and you had \gamma and \delta so you could multiply them together (gave you 13) and you get \alpha \beta = -1.

    And I agree the method of differences question was tough, but I'm guessing that's why it was the last question on the paper :P
    Thank you! I don't know why, but I only really revised cubic equation root formulas so I wans't sure how to work it out, but I got the first part for alpha + beta. So I coulnd't do the rest of the question. I'm not sure how many marks that whole question was but I must have only gotten 2 or so from working out alpha + beta Hopefully the grade boundaries will be lower.
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    (Original post by Lisb)
    did exactly the same method but ended up with -6.25 x 10^-4 as my answer. strange. should i only lose one mark since i did the right method?
    OMG I SWEAR , I GOT THE SAME! XD
    at least u might get 3/4
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    (Original post by chemari1)
    in the method of differences question I replaced 1, 2 , 3 and 4 in the formula, I wrote the equivalence of the sum of the three terms, but there wasn't a pattern of what would cancel out to be seen, so I stopped there, will I get any marks?
    I think you might get a mark for the method. I wasn't sure as well, but if I remember correctly, I think three canceled out each other (as in two added together was the negative of one of them) but then what I got in the end I didn't really have enough time to get it to be what they gave us, but with enough working you might be able to salvage some mark
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    (Original post by chemari1)
    in the method of differences question I replaced 1, 2 , 3 and 4 in the formula, I wrote the equivalence of the sum of the three terms, but there wasn't a pattern of what would cancel out to be seen, so I stopped there, will I get any marks?
    Maybe 1, 2 at most sorry.
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    (Original post by Duskstar)
    Maybe 1, 2 at most sorry.
    How do you think marks will be allocated as it was a 'show that' question? I wrote out the first few terms and then the terms for n-2 n-1 and n but didn't really get much further other than writing down that it was equal to what we were given
 
 
 
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