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    (Original post by qq1102141262)
    Question 1 was
    -2<x<-sqrt(2)
    -1<x<sqrt(2)

    any one agree??
    Yep that's what I got too. Thought I made a mistake when I saw sqrt(2) but my calculator confirmed it.
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    Answers - let me know if they need to be edited
    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, -5/15, 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!
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    (Original post by lukejoshjames)
    yuuup
    I think I botched the last question though (the last part). I got some arccos(3/4) terms and some pi terms and some root 3 terms.

    Did anyone else do the area of C2 in the region arccos(3/4) to -arccos(3/4) + two times the area of C1 in the region pi/2 to arccos(3/4)?
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    For differences I just used the summation formulae for the first two term (n and minus 3), and solved the other two terms normally.
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    (Original post by RThornton)
    Question 4 Integral Solution

    Question 4b

    let In=e^2x*sinx

    By parts twice gives

    In= 0.5e^2x*sinx - 0.25e^2x*cosx - Integral of 0.25e^2x*sin x
    Integral of 0.25e^2x*sin x = 0.25 InHence, 5/4 In = 0.5e^2x*sinx - 0.25e^2x*cosxTherefore In = 1/5e^2x * (2sinx - cosx)
    Nope the coefficients of sinx and cosx didnt have e-2theta in the final answer
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    (Original post by taichingkan)
    Yep that's what I got too. Thought I made a mistake when I saw sqrt(2) but my calculator confirmed it.
    Heh, the IAL paper had -1 +- sqrt(11). :lol:
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    (Original post by JakeTan)
    I got r/q or something, because as t tends to infinity e^pt/q tends to zero.
    I really hope I put r/q instead of t/q. My ts look like rs...
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    How many marks for the area question?


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    that **** was hard af
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    (Original post by Ewanclementson)
    Answers - let me know if they need to be edited
    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, -5/15, 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!
    All same apart from 8b, I got 15.something but must be a mistake on my part as most people seem to agree with 32.5. I did get the same terms, but the wrong coefficients.
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    (Original post by anhaanha)
    I do it like this
    Let t = intergrate e^2x sin x
    then you by part you got
    -cosx.e^2x + intergrate(2e^2x. cosx)
    then by part again you got
    -cosx.e^2x + 2 ( e^2x. sinx - intergrate(2e^2x.sinx) )
    Therefore
    t = -cosx.e^2x + 2e^2x.sinx - 2t
    t = 1/3 ( 2e^2x. sint - cosx.e^2x)
    Careful, the last term should be -4t so the 1/3 should be 1/5.
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    (Original post by Ewanclementson)
    Answers - let me know if they need to be edited
    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, -5/15, 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!
    Thanks but q5 those were not my coefficients. Can you check yours again
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    Did anyone get 23.9 something for the area of c2 ??
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    (Original post by Ewanclementson)
    Answers - let me know if they need to be edited
    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, -5/15, 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!
    Finally someone who can do integration by parts! Though i think i got a different value of b for question 2, everything else is the same, besides the typos. Thanks!
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    Horrible horrible stuff.. I revised so hard and got 73/75 for the last paper I did yesterday all to have it wasted today.. I think I got 53/75, I'm just hoping that is a B because otherwise ill need 90% on M2 and S2 to get an A
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    (Original post by anndz3007)
    Did anyone get 23.9 something for the area of c2 ??
    I DID !!
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    (Original post by Ewanclementson)
    Answers - let me know if they need to be edited
    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, -5/15, 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's what I got. Hope I didn't mess up on 5 though I don't think I did.
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    (Original post by Ewanclementson)
    ...
    2) shown x(x^2-3x-9)/2(x+2)
    ...
    Didn't get -3, got -9 though I think. If this is wrong how many marks do people think would be lost?
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    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!
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    (Original post by Zacken)
    Heh, the IAL paper had -1 +- sqrt(11). :lol:
    haha, i personally thought the numbers were ridiculous. I checked my work for so long when I got (arccos 3/4) for Q8a, only to realise that it was actually correct.
 
 
 
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