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    (Original post by Louisb19)
    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)?
    I don't think it required an exact answer, so if it came to about 32.5 (or whatever the answer was) then it should still be correct.
    Also yeah, that's what I did.
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    (Original post by target21859)
    That's what I got. Hope I didn't mess up on 5 though I don't think I did.
    Except for your coefficients of sintheta I think. I know I got it right I checked it.
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    I left my answer for 8b exact, will I be penalised (It's 32.5 when you calculate it which is supposedly correct). Didn't do it purposefully, am just a little special
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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!
    Yes! If these answers are legit, I got most of this right.
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    (Original post by taichingkan)
    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.
    For last q I got it but put it in the form 49/something pi - 5.(something) it didn't state what form to put it so would that gain absolute marks?
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    (People who know how the first order part b was done please help)

    I started integrating the normal way, with e2theta sin theta, then realised I couldn't do that so I looked and part a and realised I could use it. I didn't cross this part out - just tried another method.
    I got to the point where I got two equations equating with r, p and q.
    I then managed to get x in terms of r and t.

    How many marks might I have gained??
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    For Q8:
    Does 2 * [(region C1 from 0 to arrcos(3/4)) + (region C2 from arccos(3/4) to pi/2)] sound valid?
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    (Original post by Rkai01)
    For last q I got it but put it in the form 49/something pi - 5.(something) it didn't state what form to put it so would that gain absolute marks?
    I left it as that horrible mess cause they didn't say to put it to 3sf or anything and an exact answer is always better, you should get the marks.
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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!
    Couldn't do q2partb, q4partb or q8partb.

    Oh dear..
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    Interesting (and viable(!)) solution to integrating e^(2x) sin x:

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    Anyone else not finish in time?
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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!
    For 4b I got C as 1/5
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    I was tempted to do 8 b using decimals however I would have probably lost marks for that. In stead I chose to keep everything to exact values using  \sin{ 2 \theta } = \sqrt{ 1 - (2 \cos^2{\theta} -1)^2 } and probably made a mistake or two along the way. At least I got pi, arccos(3/4) and root3 in my final answer!
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    (Original post by themechguy)
    (People who know how the first order part b was done please help)

    I started integrating the normal way, with e2theta sin theta, then realised I couldn't do that so I looked and part a and realised I could use it. I didn't cross this part out - just tried another method.
    I got to the point where I got two equations equating with r, p and q.
    I then managed to get x in terms of r and t.

    How many marks might I have gained??
    Pretty sure you can't use part a -- it wasn't in the same form. r was constant whereas sin(theta) isn't.
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    (Original post by cjlh)
    For Q8:
    Does 2 * [(region C1 from 0 to arrcos(3/4)) + (region C2 from arccos(3/4) to pi/2)] sound valid?
    It's what I did (except I forgot to multiply by 2 ****)
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    Been getting As and A*s in past papers and this exam was a definite U

    Just so many little things completely messed with me and so I answered barely anything

    There goes my shot at Imperial out the window unless I can get the grades with my other 3 subjects...


    #[email protected]&%*!
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    (Original post by Zacken)
    Heh, the IAL paper had -1 +- sqrt(11). :lol:
    What did you get in the last part of the last question?
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    (Original post by maxguywell69)
    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!
    .
    Right, it turns out i took away (n/2(n+2)) so got -11 instead of -9 on question 2.
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    Anyone remember the question of Q4 a?
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    (Original post by Jordan97)
    For 4b I got C as 1/5
    Yes but on the LHS we have e^2theta*x and so this 1/5 becomes (e^-2theta)/5
 
 
 
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