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    As above really; what did everyone think of AEA?
    Could have been harder? Could have been easier?
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    errrrrrrrrrrrrr none of us even answered enough qus to pass!
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    Considering I didn't have time to prepare much I thought the paper was really friendly. I could complete all questions (hopefully correctly). I think it could have been much harder - at least I had expected it to be much harder.
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    Aaaagh it was horrible. Roll on FAAILLL
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    Hmmm, wasn't as hard as i was expecting. That said, its still going to be pretty hard to get a merit.
    Got stuck on the vector one strangely trying to find the shortest distance... which shouldnt have been hard, Q6 part c) f(x-v) + w , and Q7 part c integration. Something else i got stuck on though ive forgotten!
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    All over the place that exam was. The difficulty of the questions fluctuated quite a bit.
    I might have got a 'merit' if:
    a) All the things I think i got right, I did get right.
    b) I get method marks for the working I did on questions that I couldn't quite finish off.
    c) I get a reasonable number of marks for clarity, presentation.
    d) God descends from Heaven, breaks into Edexcel and makes a few alterations to my paper for me.
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    Failed it. Quite badly. Requested that my script not be submitted.

    I never took the exam. Ahem.
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    I didn't think it was too bad.
    The paper didn't seem to be as bad as previous ones and I thought that I have probably got a merit but with a very small chance of a distinction.
    There were some real gift marks as well as some you had to work for but weren't impossible to get. I think i've lost a few marks for presentation and I couldn't find the equation of the line that bisected the angle but apart from that it was okay.
    'Q6 part c) f(x-v) + w' was tricky, it took me around half an hour to get. What did people get for v/w? I think I had v as either 2 or 4 (I can't remember which now) and f(x-v)+w=0 gave two integral values.
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    16 for 'w' methinks.
    God knows whether it's right.
    As for the rest of the question........................ ..........................
    Maybe I'll get a method mark or two............................. ... :confused:
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    Hullo - I dunno - it was a rly wierd paper. Some rly easy questions, some really hard...

    Here are some of my answers - nyone got smthing similar?
    1. Couldn't do
    2. π/12, 5π/12, π/2, 3π/2 (NOT 13π/12 and 17π/12)
    3. u = 2 / √(1 - 2x)
    4. Area = 2p cos p
    5. cos(angle) = 29/30 couldn't do nything after that
    6. P was at -2√ 3 R was at 2√ 3 and Q was at (2,16). v was 4 and w was 16. The solutions to the equation given where x = 2, x = 2 and x = 8
    7. Couldn't do the integral for the life of me!
    Daniel
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    Isnt it 50% for merit? I think I got that....i got v as 2 or 4 but didnt know how to say which one was right :P...w=16 i think :P.
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    *]π/12, 5π/12, π/2, 3π/2 (NOT 13π/12 and 17π/12)
    Looks good.

    [*]u = 2 / √(1 - 2x)
    I can't remember the numerator but it was definitely of that form so that looks good.

    [*]Area = 2p cos p
    Agreed. The inequality part was tricky (I just wrote f(theta)=tan(theta-1/(theta) and explained why there was a sign change) [I think there should have been a more elegant way as we aren't allowed calculators] but the later parts followed on nicely from that result.

    [*]cos(angle) = 29/30 couldn't do nything after that
    I get the same angle as you but it took me bloody ages to simplify it down. I had an expression for the distance but couldn't tidy it up so i left it as rt[(a/b)+(c^2/d)] so I won't have all the marks there.

    [*]P was at -2√ 3 R was at 2√ 3 and Q was at (2,16). v was 4 and w was 16. The solutions to the equation given where x = 2, x = 2 and x = 8
    That also looks good You scared me when you listed three solutions but I remember now one of the factors was a perfect square
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    Pardon my ignorance...
    Why not 17/12, 13/12 pi?
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    (Original post by Gaz031)
    I didn't think it was too bad.
    The paper didn't seem to be as bad as previous ones and I thought that I have probably got a merit but with a very small chance of a distinction.
    There were some real gift marks as well as some you had to work for but weren't impossible to get. I think i've lost a few marks for presentation and I couldn't find the equation of the line that bisected the angle but apart from that it was okay.
    'Q6 part c) f(x-v) + w' was tricky, it took me around half an hour to get. What did people get for v/w? I think I had v as either 2 or 4 (I can't remember which now) and f(x-v)+w=0 gave two integral values.
    I didn't even know where to start with the bisector equation. It's the only one that I didn't have a very good stab at.

    How did you do 6(c)? I did a load of crap and got w=16, and a cubic in v2 (somethine like v^3 + 12v + 16 = 0) which I couldn't solve.

    Some of the questions were absolutely lovely though. 7 was brilliant, as was questions 1, 2, 4. Can't remember what 3 was, and five and six were a bit yucky.
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    (Original post by Gaz031)
    I get the same angle as you but it took me bloody ages to simplify it down. I had an expression for the distance but couldn't tidy it up so i left it as rt[(a/b)+(c^2/d)] so I won't have all the marks there.
    Simplify it down? You mean there was a simpler way than 29/30 ? Gosh, should have thought of that

    Now that I think of it, also the "shortest distance" questions was mindnumbingly easy once you had the angle! I can't BELIVE I didn't get it! Thank G-d it was only 3 marks!!

    Anyhow, exams are over now
    :party:

    Still can't make myself belive it :-p

    Daniel
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    (Original post by trunksss6)
    Pardon my ignorance...
    Why not 17/12, 13/12 pi?
    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
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    Ah, I see............chipping away at my total
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    (Original post by danguetta)
    Hullo - I dunno - it was a rly wierd paper. Some rly easy questions, some really hard...

    Here are some of my answers - nyone got smthing similar?
    1. Couldn't do
    2. π/12, 5π/12, π/2, 3π/2 (NOT 13π/12 and 17π/12)
    3. u = 2 / √(1 - 2x)
    4. Area = 2p cos p
    5. cos(angle) = 29/30 couldn't do nything after that
    6. P was at -2√ 3 R was at 2√ 3 and Q was at (2,16). v was 4 and w was 16. The solutions to the equation given where x = 2, x = 2 and x = 8
    7. Couldn't do the integral for the life of me!
    Daniel
    *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.

    2) Those exact four answers. cosθ = 0, or 2sinθ + 2cosθ - √6 = 0 (the latter being worked out by converting it to Rsin(θ + Ø) = √6

    3) Got that exact form for u = f(x)

    4) Got the area, and proved those three results. I ended up with tan(alpha) = 1/(alpha) when I set dS/d(alpha) to 0. I dind't like my argument for (b) - it worked but it was a bit sloppy - but I know I was along the right lines because (c) depnded on the tan(alpha) result. (d) was easier than (c) yet was worth more marks!

    5) cosθ = 29/30 I got. Proving the modulus of the two vectors were equal was easy enough - surely you got that after you came so far? I didn't know where to start with the bisector question

    6) I got w = 16 and suspected that v = 4, but couldn't prove the latter at all. I ended up with a cubic in v; I'm desperately hoping that I didn't mess up and miss that v=4 was a solution. Don't think it was though! The rest of teh question was great though, so I bet I didn't lose many marks.

    7) The first two proofs were lovely. Part (a) was self-explanatory. Part (b) used a nice trick where you end up with the integral asked for on both sides so you took it to th other side. It was a bit tricky, but just persevering with it world get you to the result. (c) was fantastic! You could set u = √2cosθ to get the integral into the form ∫√(x2 - 1) dx, then you changed the limits for a second time based on the substitution x = secθ that you did in part (b). I got an answer like 0.5[1 - ln(√something)] in the end.


    Haha. The ironic thing is is that I probably failed STEP II this morning, yet I could get a distinction (definitely a merit) on an exam I didn't even try for
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    (Original post by JohnSPals)
    I didn't even know where to start with the bisector equation. It's the only one that I didn't have a very good stab at.
    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.

    How did you do 6(c)? I did a load of crap and got w=16, and a cubic in v2 (somethine like v^3 + 12v + 16 = 0) which I couldn't solve.
    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.
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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
    I also got four solutions but used a different method. I put it in the form cosθ[Acos(x+a)-B]=0

    Simplify it down? You mean there was a simpler way than 29/30 ? Gosh, should have thought of that
    No that was the simplest form, it just took me ages to sort out the square roots etc. (6 lines!)
 
 
 
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