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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

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    (Original post by maxguywell69)
    Hopefully two or three at most, that was the one bit i got wrong- i got -3n but got -11 cos i took away rather than added.
    I might have gotten the same though, -11. Is that definitely wrong? I remember changing a sign on a +6n after going back. If that's wrong I'm a *******.
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    (Original post by Ewanclementson)
    Answers - let me know if they need to be edited (edit - typo on 5 and 3 fixed)
    1) -2<x<-root2 -1<x<root2
    2) shown x(x^2-3x-9)/2(x+2)
    3)e^ipi(1/24, 13/24, -11/24, -23/24) shown
    4) r/q(1-e^-(q/p)t) (2sinx-cosx+e^-2)/5
    5) Coefficients are 1/16, -5/16, 10/16 shown
    6) coefficients are 1 2 2 8/3 then shown
    7) Ax^2+Bx-1/12x^(-2)
    8) 21/4, (+-)arcos(3/4) 32.519 which I think was 3root7/4-11arcos3/4+49pi/4
    Hope this helps everyone
    I got exactly the same

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    (Original post by smartalan73)
    yeah that's still different to what I got so would still like to know how you did it? I'm awful at complex numbers.
    Oh I did it in the principle argument range -pi<=x<=pi but you could do it in the 0<=x<=2pi range as well and get different values.
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    (Original post by Louisb19)
    STEPers across England smiled when they saw  \displaystyle\int e^{2 \theta} \sin{\theta} \mathrm{d} \theta .
    I recognised it from my procatination on such threads. I too smiled.
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    (Original post by ninjass)
    Is it just me - or was 4c a lot easier than first glace - I got it but in a really complicated way

    I think i got 70/75 - UMS predictions??
    Probably a B or C
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    (Original post by sweeneyrod)
    In C3/C4? I don't think our class learnt it in any case.
    Surely your class learnt how to re-arrange equations of the form a = b + ka?
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    (Original post by Zacken)
    It's in your formula booklet.
    Where abouts? I only saw The sum for r^2 and r^3.
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    (Original post by daniella_n)
    I meant I messed up so bad in that question that I didn't even have e2x sinx anywhere fun times


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    Ah I see that's unfortunate I guess we got to hope for low grade boundaries.
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    (Original post by Mathemagicien)
    I did question 4 directly, not by parts, using exponential version of sin x.
    But its sinx not sinh x
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    (Original post by Mattematics)
    Where abouts? I only saw The sum for r^2 and r^3.
    Arithmetic series. C1 section. S_n = n/2(1 + n).
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    (Original post by Mattematics)
    Where abouts? I only saw The sum for r^2 and r^3.
    I don't think it's in there. I had a quick glance before reaching into the depths of my mind to retrieve it.

    EDIT: Never mind, Zacken's right haha
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    (Original post by Mattematics)
    Where abouts? I only saw The sum for r^2 and r^3.
    arithmetic series:

    sn = n/2 (a + l)

    here a = 1, l = n
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    Predict the grade boundaries:
    http://www.thestudentroom.co.uk/showthread.php?t=4149979
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    (Original post by Zacken)
    Arithmetic series. C1 section. S_n = n/2(1 + n).
    you f***ing ninja
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    (Original post by anhaanha)
    But its sinx not sinh x
    So? sin x also has a exponential definition, you have heard of DeMoivre's, surely...?

    (Original post by cjlh)
    I don't think it's in there. I had a quick glance before reaching into the depths of my mind to retrieve it.
    Arithmetic series. C1 (!) section.
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    (Original post by Zacken)
    Surely your class learnt how to re-arrange equations of the form a = b + ka?
    I think it's a bit of a stretch to say it's on the syllabus. There aren't any examples of IBP being used that way in the text book so it's a bit of a sneaky question. Yes if you think about it it makes sense but if you've never seen anything of that type then it's not exactly a c4 integral
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    (Original post by cjlh)
    I might have gotten the same though, -11. Is that definitely wrong? I remember changing a sign on a +6n after going back. If that's wrong I'm a *******.
    I am quite sure what I put was correct (-9) as I used the sum function on my calculator to check it for n=1,2 and 3 and it worked for all of them; this would not have been so for a different value in place of the -9.
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    (Original post by Student403)
    Nope the coefficients of sinx and cosx didnt have e-2theta in the final answer
    yeah because e^2x y = 1/5e^2x * (2sinx - cosx) + c

    therefore

    y = 1/5 (2sinx - cosx) + ce^2x
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    (Original post by AlexFrangos;[url="tel:65555783")
    65555783[/url]]What did you get in question 2 and in 2nd order differential equation?
    Cant recall wts Q2, inequality?

    -1-rt11 <x< -2 , -1+rt11 <x< 4, 4 critical values

    2nd order: -5e^-x + fraction (3/4?i forgot)e^-2x + sth + sth + 17/4 😂 sorry i cant recall the answer
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    (Original post by RThornton)
    yeah because e^2x y = 1/5e^2x * (2sinx - cosx) + c

    therefore

    y = 1/5 (2sinx - cosx) + ce^2x
    Correct except it's ce^-2x
 
 
 
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