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    (Original post by danguetta)
    Well, because of the following (this was my working, but you might have done it differently)

    At one point, you get that some of the solutions are given by:

    2sinθ + 2cosθ = √6
    Then, squaring both sides and re-arranging gives you:
    sin 2θ = 1/2
    Now, it looks like 17Pi/12 and 13Pi/12 should be a solution, but if you look carefully, they result in a NEGATIVE value of sinθ - this is because you squared on the way, so negative values are also produced... They didn't fulfil the original thing...
    I dunno if my reasoning is even right - I noticed it in the exam (that for these values 2sinθ + 2cosθ = √6 doesn't hold, and so I hastily scrubbed them out!)
    Daniel
    You could convert that equation into the form Rsin(θ + Φ) = √6. I got Φ = π/4 and I think I got R = 2√2 (I certainly got sin(θ + Φ) = √3/2 though, so it's all good). I can calrify that you don't get any value bigger than π when using this method - obviously sinθ takes nagive values for π < x < 2π
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    (Original post by Gaz031)
    Same pretty much. I had an idea that it could be written in the form r=(Coordinate of C)+lambda(ai+bj+k) and you could use the fact it made an angle 0.5θ with the two vectors to find a and b using the scalar product, but it didn't really work out nicely.


    v^3+12v+16 = (v-2)(v^2+2v-8) = (v-2)^2(v+4) but v>0 as the graph is translated to the right, so v=2.
    To solve it there were two steps:
    (i) At dy/dx=0, the smaller value of x gives y=0.
    (ii) At dy/dx=0, the greater value of x gives the same value of y as x=0 does.
    DAMNIT! I got that equation for v too! Didn't factorise it though - I stupidly assumed it couldn't be done easily! Because I thought v could only take one value, I tried to see if I could get 64 rather than "16" (Don't know why!)
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    (Original post by Gaz031)
    I also got four solutions but used a different method. I put it in the form cosθ[Acos(x+a)-B]=0


    No that was the simplest form, it just took me ages to sort out the square roots etc. (6 lines!)
    Haha. I was sure I did it wrong before I realised that √324 (or something) was 18. Then I knew I'd done it right.
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    My paper was a bit of a mess unfortunately [I put two vertical lines through pages when the working I did was wrong] and I had attached sheets. If the actual working is perfectly legible but my answer book is a little messy do you think i'm like to lose many or all of the style/presentation marks?
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    Wouldn't a line from O to the midpoint of AC be a bisector?

    I found the paper quite tricky, missed out most of the integration stuff at the end (wasnt sure where to split the integral into parts, realised afterwards and kicked myself), and the later parts of the cosine rectangle thing. The graph stuff in the middle was a godsend (took me about 5 minutes) after struggling through the earlier questions.

    I was aiming for a distinction but it looks unlikely now, I did STEP II in the morning as well so I might just have been a bit drained for the AEA, not that it wasn't a difficult paper in its own right.
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    (Original post by Robert602)
    Wouldn't a line from O to the midpoint of AC be a bisector?
    It bisected the angle, not AC. I might be wrong though as I couldn't do that part.
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    (Original post by Robert602)
    Wouldn't a line from O to the midpoint of AC be a bisector?

    I was aiming for a distinction but it looks unlikely now, I did STEP II in the morning as well so I might just have been a bit drained for the AEA, not that it wasn't a difficult paper in its own right.
    I failed that STEP paper miserably! I just couldn't get started with any question except 3, but I assume everyone would have got that (it didn't seem worthy of STEP to me).

    I've got to say, I didn't feel at all affected by doing STEP in the morning. I thought I'd just end up curling up and dying half way through the AEA, but it actually made me more active.
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    (Original post by Nima)
    omg, i got 0.5[1 - ln(2)], u make me so happy! I cant believe i got full/near full on that 9 marker!
    Ohh I didn't get √2. I think I got ln(1+√2)

    If I remember correctly, You substituted θ = π/4 into the stated result in (b), which would give the log in the answer (θ=0 gives ln1 = 0).

    This, in ln|secθ + tanθ| would give ln|√2 + 1|. I suspect that I may be right for that reason - I don't know though because I was sick of the sight of π/4 (It was in practically every question at some point!) so can't remember much of substituting in.

    Well done, anyway . We seem to have been the only two who got some kind of answer for (c) - that stands us in good stead!
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    I agree with the 'ln(1+√2)' term. I think mine was of the form (1/4)(a√2-ln(1+rt2)) where 'a' is a constant I can't remember.
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    (Original post by Gaz031)
    I agree with the 'ln(1+√2)' term. I think mine was of the form (1/4)(a√2-ln(1+rt2)) where 'a' is a constant I can't remember.
    I don't have a clue where my 0.5[1 - ...] but came from so I can't help there :p: . I kinda remember getting √2's cancelling in the first term.

    The first term was secθtanθ. θ=0 gives 0. θ=π/4 gives 1/(1/√2) x (1/√2) = √2 x (1) = √2. Maybe you were right - I may have put tan(π/4) as 1/√2 without thinking.
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    I did last year's paper and in my opinion this paper was harder. If 50% last year = merit, will it be a bit less this year?
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    (Original post by Nima)
    I did last year's paper and in my opinion this paper was harder. If 50% last year = merit, will it be a bit less this year?
    I'm not sure but I don't think they could justify putting the pass mark below 50%.
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    (Original post by JohnSPals)
    *runs around screaming* EVERYTHING you got I got practically!

    1) was quite simple. I got the minimum of 2 and the maximum of 12. You could work out the radius and centre of the circle. Then it was a matter of working out the distance of the centre to the origin and adding/subtracting this from 7.
    What do you mean? I tried a circle argument, but I ended up getting a raidus of -1, so that didn't really work!!
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    (Original post by Nima)
    I did last year's paper and in my opinion this paper was harder. If 50% last year = merit, will it be a bit less this year?
    For 2003 (the examiners' report is on the Edexcel website), you needed 50 for a merit and 70 for distinction. 1007 sat, of which 30% for a merit and just under 10% got a distinction.
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    (Original post by danguetta)
    What do you mean? I tried a circle argument, but I ended up getting a raidus of -1, so that didn't really work!!
    It wasnt particularly hard! You probably just made a sign error or something. Radius = √(g2 + f2 - c). This comes out at √49 = 7. Did you remember to take the integral to the other side to make the equation of the circle equal to 0, THEN usng it in the above equation? If c was negative (I cna't remember), did you remember that it was - -c i.e. +c?. Who knows, eh?
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    Mine went ok. I did roughly 80pc of the paper, with probably 0 for presentation.
    What did everyone get for 7c? (Please tell me it had a root 2 in it).
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    (Original post by mkxz)
    Mine went ok. I did roughly 80pc of the paper, with probably 0 for presentation.
    What did everyone get for 7c? (Please tell me it had a root 2 in it).
    It's already been almost agreed on. Something like 1/k[√2 + ln(1 + √2)] where k is some constant.

    The P/S/C marks don't go on actual presentation. You get 2 for each question if you apply a novel or efficient method to get to a completely correct solution, and 1 if you make some basic arithmetical error and thus get the final answer wrong; up to a limit of six marks on three distinct questions. A seventh T mark goes if you generally answer many questions well - these are often only awarded if you're getting above 75/99 for the rest of the paper.
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    I thought it was one of the stranger AEA Maths papers, having done some past ones. Some questions seemed very tricky (like the integration and, for me, the area of the rectangle under the cos graph, where I couldn't work out where that inequality came from for the life of me) and some relatively easy (the graph one and the vector one, for me at least). Looking at what people are saying though, I think I may have just got a Merit if I've picked up a few marks for partial answers on questions.
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    ...eek, I went into the exam blind having been too frantic about other subjects' revision. Soo annoying, I got the circle one and the vector and graph sketching, and I thought I had the integration but looking at everyone else's answers I'm not so sure...

    Anyway, don't spose you know if they put it on your certificate if you get a U?
 
 
 
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