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Todays AEA Paper watch

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  • View Poll Results: How did you find the AEA paper
    1 - Easy
    2
    11.76%
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    3
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    4
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    5 - Average
    1
    5.88%
    6
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    7
    2
    11.76%
    8
    8
    47.06%
    9
    2
    11.76%
    10 -Hard
    2
    11.76%

    • Thread Starter
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    How did you all find it? I thought it was really hard - will be lucky to even get a merit :mad: :nn: . I did Step 1 on Wednesday and thought that went better - I am really annoyed.
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    I found it very hard aswell- especially the last question!
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    i thought it was going really well, but i tried to add up my definite marks at the end and i'll be lucky to get a merit!
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    how much was it out of, and what do you need for a merit?:confused:
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    (Original post by Dharini1987)
    I found it very hard aswell- especially the last question!
    Indeed, that question probably was harder than the usual AEA Maths question - the start was exactly the same as the start of Question 4 in STEP III 2004, so it's to be expected that most found it hard.

    Personally, I didn't find the last question too hard, but nonetheless managed to make a stupid mistake so I messed up two parts of the question.

    Apart from the mistake I made in the last question, and the second part of the logs question (I moved onto the next question after I couldn't see quite how to do it, hoping to come back later), I think I did the whole paper well.
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    It was tough. Only managed to complete 3 questions fully- the trig one, the vectors one, and the integration one. The first Q seemed so simple- (1-x)^-2 with a geometric series on the bottom, but I couldn't progress from getting the sum to infinity of the GS. The log one wasn't too bad, although on the last part I spent ages wondering why I kept getting x=x or -1=-1 when "solving" the first equation, thinking the other was a seperate part of the Q. And I couldn't even start the last Q- so akk I did was find dS/da and left out the rest.
    I reckon I've got around 40-45 marks, so I HOPE I've scraped a merit. Won't be suprised if I fail though
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    (Original post by Kernel)
    It was tough. Only managed to complete 3 questions fully- the trig one, the vectors one, and the integration one. The first Q seemed so simple- (1-x)^-2 with a geometric series on the bottom, but I couldn't progress from getting the sum to infinity of the GS.
    You were, I believe, meant to note that their series was just the series for \large (1-y)^{-2} , but with \large y = \frac{x}{1+x} .
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    Ahh, indeed, well spotted. But was there any way you could work it out (y was the sum to infinity of the geometric series, is that significant?), or were you just supposed to spot it?
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    (1-x/(1+x))^-2

    = (1-1+1/(1+x))^-2

    = (1+x)^2
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    (Original post by Kernel)
    Ahh, indeed, well spotted. But was there any way you could work it out (y was the sum to infinity of the geometric series, is that significant?), or were you just supposed to spot it?
    I imagine the only feasible way of doing the question was to spot it; in AEA and STEP-type questions, later parts often involve the earlier parts, as I'm sure you've noted.

    violinist's done it correctly above.
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    I did really bad in it. Wish I'd done STEP instead, the questions seem easier somehow, and I prefer the choice. Deffo failed today, but on the bright side...NO MORE EXAMS
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    Overall i found the paper really tricky, but think i may just have scraped a merit.

    Did anyone get (e^pi/2,1) as the maximum point and then 1 as the shaded area (intergation was sin(lnx))?

    Oh, how many solutions did you get to the trig question?
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    Hi

    At least i was not the only one that found this paper amazingly tricky and it is nothing like the past papers which seem much more straight forward, i really need a merit for uni,

    Did anyone get (e^pi/2,1) as the maximum point and then 1 as the shaded area (intergation was sin(lnx))?

    Yes i did get the e^pi/2,1 as my max point not too sure about the area i can't remember,

    did anyone get the circle question where it said y=mx where m satify's the equation......
    that was a HARD question i have never seen one like that, i had to make a educated guess that we hopefully get me some marks!!!!!!!!!!

    also the logs solving question was "undoable" as well as the last question
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    How did you solve the trig question? I spent about an hour trying to spot any of the basic trig identities and failing
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    I factorised by bringing up the sinx +cosx from the bottom on the LHS on to the other side if that makes sense then bringing everything to one side and then takes out a factor i can't remember the question though,

    then work out each of the brackets separelty by putting them equal to one
    i think i had tan2x=root3
    where you need to find out like 4 solutions, for this one
    but i got stuck with the other bracket but i guessed and i wrote no solutions (which i hope is correct)

    PLease tell me i got this one right!!!!!!!!
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    (Original post by time2think)
    I factorised by bringing up the sinx +cosx from the bottom on the LHS on to the other side if that makes sense then bringing everything to one side and then takes out a factor i can't remember the question though,

    then work out each of the brackets separelty by putting them equal to one
    i think i had tan2x=root3
    where you need to find out like 4 solutions, for this one
    but i got stuck with the other bracket but i guessed and i wrote no solutions (which i hope is correct)

    PLease tell me i got this one right!!!!!!!!
    yes, i got the tan bit too, does anyone know how to verify the m in that quadratic for question 4?
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    after tan2x you use the Rcos(x+a) formulae
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    I think i got something like sin2x = -0.5 which is prob wrong.

    I divided through by the bit that was on both sides - the divided by the two, expanded out cos^2x - sin^2x and then expressed this as (cosx-sinx)(sinx+cosx) then divided through the bit one bottom.

    Ended up with (cos-sinx) = root6 over 2

    Squared it all and used sin^2x + cos^2x =1 to get:


    1 -2sinxcosx = 6/4

    2sinxcosx = -0.5

    sin2x = -0.5

    Think I prob made some silly error though :confused:
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    (Original post by Scottus_Mus)
    I think i got something like sin2x = -0.5 which is prob wrong.

    I divided through by the bit that was on both sides - the divided by the two, expanded out cos^2x - sin^2x and then expressed this as (cosx-sinx)(sinx+cosx) then divided through the bit one bottom.

    Ended up with (cos-sinx) = root6 over 2

    Squared it all 1 -2sinxcosx = 6/4

    2sinxcosx = -0.5

    sin2x = -0.5

    Think I prob made some silly error though :confused:
    for cos-sinx you use compound angle formula Rcos(x+a) like in last years paper
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    i seem to remember 30, 120, 210 and 300 as four of my solutions to the trig question.

    How did people tackle the last part of the vectors question? I did the first two bits but stuggled with part (c) (think i ended up with a position vector where lamda equalled 2, for point A)
 
 
 
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