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    Did question 4 give initial conditions for part a? I got x=r/q+ce^(-qt/p). Would I drop a mark for not putting rc/q? I merged it as one constant. In the unofficial mark scheme by Ewan he has a minus and not a plus
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    (Original post by Student403)
    I know it wasn't
    Doesn't specify in what form. Any value, exact or not, given its correct will be accepted.
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    (Original post by AlexFrangos)
    I found the same in the differential equation. Is that sth in polar -9/4rt(3) ?
    What radius and centre did you also get in the transformation question?
    umm centre (-5/8,0) radius = 7/8,

    i think its not -9/4 rt3, its more complicated , like some two digits numbers. , what I'm sure is that i have calculated the answer in calculator and it is value is 0.5xxx
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    (Original post by samb1234)
    That isn't the same thing. Yes it's ibp twice, but it's obviously integration by parts twice as you can easily see how by doing by parts you are going to get to a point where you get an integral independent of x - this isn't the case for e^2x sinx as most people get taught that you can't use IBP for things like sine that go round in circles, so you might do it once and then you get to a very similar point and probably stop. You can say all you like about it but when all but 2 people at a grammar school, many of whom will go to top unis and get very high grades, can't do the question it suggests that it isn't just a normal question that people are stupid for not getting right. We get it, you do step but not everyone does.
    Agreed. Don't think it was unfair to ask but it clearly wasn't easy.



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    vote please! So we can get an idea of what the boundaries will be like.
    http://www.thestudentroom.co.uk/show....php?t=4150121
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    Did anyone get this for 4(c):
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    (Original post by notrodash)
    Damn I practiced so many complex transformation questions and then it didn't come up! Really struggled for time towards the end but looking at these answers does inspire some confidence.

    I probably spent the longest doing the integral in question 4(b), I couldn't figure out a better way to do it so I used FP3…

    The integral seems right according to wolfram alpha, yet when I differentiate the whole thing I don't get dy/dx = 0 when y = 0 and theta = 0, so I probably did it wrong
    You haven't divided your constant by e^2theta
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    (Original post by Ewanclementson)
    Answers - let me know if they need to be edited (edit - typo on 5 fixed)
    1) -2<x<-root2 -1<x<root2
    2) shown x(x^2-3x-9)/2(x+2)
    3)e^ipi(1/24, 13/14, -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!
    That feeling when you get 3root7/4 -11arccos(3/4) + 49pi/2 = 71.0
    I forgot to times by a 1/2 some where i seems :'(
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    (Original post by jamie.mair)
    Did anyone get this for 4(c):
    Yes, this is right.
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    (Original post by somevirtualguy)
    You haven't divided your constant by e^2theta
    Oh crap. That's the A1 down the drain
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    For 4 (b) I integrated by parts twice and then went wrong , how many marks would I lose?

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    (Original post by hank_the_tank)
    umm centre (-5/8,0) radius = 7/8,

    i think its not -9/4 rt3, its more complicated , like some two digits numbers. , what I'm sure is that i have calculated the answer in calculator and it is value is 0.5xxx
    hmm. How did you find the paper in general and what grade boundaries do you expect?
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    Anyone know the anwer for 7 and 8?
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    Someone upload hitlers reaction- it's needed!
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    (Original post by angelitajin)
    Anyone know the anwer for 7 and 8?
    Ax^2+Bx-1/(12x^2)
    21/4, arcos(3/4)
    21/4, -arcos(3/4)
    32.5, or 49pi/4+3sqrt7/4-11arcos3/4
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    (Original post by notrodash)
    Oh crap. That's the A1 down the drain
    Don't dwell on it too much, the way everyone seems to be reacting the boundaries might be slightly lower than usual
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    (Original post by Student403)
    no part c

    there was part ai and aii

    and part b



    anything so long as you have FP1 and FP2 or FP3

    And in maths you must have S1+S2, D1 + D2, M1 + M2, or 2 of D1, M1 and S1. So you can't have M1, D1 and S1 in FM
    okay so based on what you're saying could you have the following as a possible arrangement?

    math: c1-4, d1, m2
    further: fp1-2, m1, s1-3

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    As the last question didn't specify to leave the answer in exact form, would you still get all the marks for being right to 3sf (32.5)?

    Edit: oh my god 2 reps I've made it
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    (Original post by Waffles15)
    Nah,
    5. De Moivres (5M) + (5M)
    6. Tan(x) (5M) + (5M)
    7. Second Order Diff (4M) +(7M) + (2M)
    8. Polar Coordinates (3M)+ (8M)
    Second order Diff first question was 6 Marks
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    (Original post by Katiee224)
    okay so based on what you're saying could you have the following as a possible arrangement?

    math: c1-4, d1, m2
    further: fp1-2, m1, s1-3

    Nah you couldn't have d1 and m2 together.

    It would have to be d1 + m1, d1 + s1, s1 + s2, or m1 + m2 in maths

    all the rest goes to FM
 
 
 
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