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    (Original post by Adam_AFC)
    Yes to both questions. They made use the cosine rule and 0.5ABsinC for only 2 marks.
    I just used bh/2
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    (Original post by hoakenfull)
    You could split it into two right-angled triangles and use 0.5bh
    Or just one triangle on its side. I did (6*4)/2 = 12
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    (Original post by Duskstar)
    My answers:

    1. M^{-1}=\dfrac{1}{8+2p} \begin{pmatrix} 1 & 2 \\ -p & 8 \end{pmatrix}

A = \dfrac{4 \times 6}{2} = 12, \, |M| = 8 + 2 \times 3 = 14, \, \therefore A' = 168

2. -\dfrac{21}{29} + \dfrac{20}{29}i

a = \dfrac{8}{29}, \, b = \dfrac{38}{29}

3. \lambda = 3, \, \mu = 25

B^{-1} = \dfrac{1}{25}A = \dfrac{1}{25}\begin{pmatrix} 3 & 6 & -4 \\ 2 & 5 & -1 \\ -1 & 4 & 3 \end{pmatrix}

4. \mathrm{proof:} \, \displaystyle\sum_{r = 1}^{n} r^{2}(2r - p)

\mathrm{p s.t. coefficient on} n^3 == n^4, \, p = \dfrac{3}{2}

5. \mathrm{Argand diagram:}

C_{1}: |z+3-4i|=5, \, C_{2}: arg(z+3-6i) = \dfrac{\pi}{2}

\mathrm{intersection:} -3+9i
    6: \mathrm{proof \, by \, induction:}

u_{1} = 8, \, u_{n+1} = 3u_{n} + 2n + 5

\mathrm{RTS:} u_{n} = 4(3^{n}) - n - 3

\mathrm{I can write this out if we really need...}

7. f(z) = 2z^{4} - 9z^{3} + Az^{2} + Bz - 26

\alpha > \beta, \, \gamma \, \mathrm{and} \, \delta \, \mathrm{are \, imaginary}, \, \gamma = 3+2i

\mathrm{show} \, \alpha + \beta = -\dfrac{3}{2}

\alpha \beta = -1

\alpha = \dfrac{1}{2}, \, \beta = -2 \,\,\, \mathrm{(I \, think \, order \, will \, matter)}

A = 6, \, B = 51

\mathrm{find roots of} \, f(\dfrac{w}{i}) = 0 \Rightarrow z = \dfrac{w}{i}, \, \therefore w = zi

w = \dfrac{1}{2}i, \, -2i, \, 2 + 3i, \, -2 + 3i

8. y = \dfrac{3x^{2} - 9}{x^{2} + 3x - 4}

\mathrm{asymptotes:} x = 1, \, x = -4, \, y = 3

\displaystyle \lim_{x \rightarrow \infty}(y) = 3^{-}

\displaystyle \lim_{x \rightarrow - \infty}(y) = 3^{+}

y \geq 0 \Leftrightarrow x < -4, \, - \sqrt{3} \leq x < 1, \, x \geq \sqrt{3}
    9. \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}=\dfrac{3}{4(2r-1)}-\dfrac{1}{2r+1}+\dfrac{1}{4(2r+3  )}

\mathrm{proof for} \, \displaystyle \sum_{1}^{n}

\displaystyle \lim_{n \rightarrow \infty} = \dfrac{2}{3}

\displaystyle \sum_{20}^{50} = 000813

    I forgot to halve the area in 1, and started writing 3/2 instead of 2/3 in 9 but apart from that it was alright
    Got all these answers. Didn't notice alpha>beta in q7 but I'm thinking I probably put them in the correct order anyway.
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    (Original post by Mr Moon Man)
    More like 52 imo
    Where are you getting these grade boundaries from? :/
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    (Original post by ♥Samantha♥)
    Got all these answers. Didn't notice alpha>beta in q7 but I'm thinking I probably put them in the correct order anyway.
    Cheers for that! I definitely wasn't quite on 100% but this is a confidence boost

    Edit: Haha sorry about that, I was supposed to reply to what you replied to
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    (Original post by JoshRyan74)
    Cheers for that! I definitely wasn't quite on 100% but this is a confidence boost
    Same here!
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    Does it matter if u don't label the angle in the half line loci thing in the argand diagram?
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    (Original post by TheSurrealTV)
    it was n=50-n=19 not 24
    i thought it was n=50 -n=20
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    (Original post by aeroline1999)
    i thought it was n=50 -n=20
    yeah same
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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
    Gutted took away 20 not 19! do you think i will lose all four marks?
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    (Original post by TheSurrealTV)
    Do you think 68 marks will be 100 UMS this year?
    4 I think
    I hope it's 68 for 100ums- that paper was so hard!
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    (Original post by aeroline1999)
    i thought it was n=50 -n=20
    Good news alway looking over past papers and on Jan 2010 there is the same type of question for 3 marks and you get M1 for splitting into parts then use of result from previous part so hopefully we will all just drop 1 or 2 out of 4 on that
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    (Original post by phoebeisgreat)
    I hope it's 68 for 100ums- that paper was so hard!
    I felt it was easier than last year and grade boundaries were 62 for A so I'm going to say a little higher maybe 69 or 70 for 100 UMS
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    (Original post by phoebeisgreat)
    I hope it's 68 for 100ums- that paper was so hard!
    If the world was just the A boundary would be like 50/72 for that paper. But the world isn't so no doubt it'll be like 60/72 for an A. By far the hardest FP1 Paper I've ever done, and I've song all of them from 2008 onwards! Also its so unfair, like just last night I got 69/72 in a past paper with like 96UMS and on this I'd be lucky to get a B let alone an A. I don't have confidence the boundaries will reflect how hard it was. Although you never know, the did get done to 48/72 in January 2013. If the A is between 48-52/72 I should be absolutely fine, but I reckon it'll be 58/72. I just have a hunch. Also I hate OCR MEI WHY IS MY SCHOOL JEOPARDISING MY FUTURE BY CHOOSING SUCH AN AWFUL EXAM BOARD?! I want EDEXCEL or AQA or even just normal OCR!!!!!!! I swear, if I don't end up getting an A in almost all my modules (C1, C2, C3, FP1, D1, M1 & S1) I'm gonna move school just so I can get edexcel.
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    I got the wrong roots for alpha beta, then carried it through the whole of question 7, how many mark is that gone. I know I dropped at least 5 marks elsewhere on the paper due to carelessness. Will I get many error carried forward marks?
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    (Original post by chemari1)
    the asymptote was 1
    no it was 3 mate, 3x^2
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    Guys for f (w/j) i got two out of the four roots correct, how much do you think i would lose
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    (Original post by rajvirnagi)
    no it was 3 mate, 3x^2
    true, what do you reckon the boundary will be for a b?
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    I think, worst case scenario, I got about 54/72 best case scenario, about 59/72, what sort of grade am I looking at? Thought it was one of the most difficult fp1 papers yet
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    (Original post by jcouchman)
    I think, worst case scenario, I got about 54/72 best case scenario, about 59/72, what sort of grade am I looking at? Thought it was one of the most difficult fp1 papers yet
    That's pretty much the same ball park I am in. Hoping it will be enough or close to 80UMS.
 
 
 
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