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The Big 2009 AEA Maths Thread! :)

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    Thanks for everyone's input already!...Looks like this shall be a successful thread It's already turning out to be very useful for me, since I haven't yet done C4 differential equations (I'm not doing full further maths, just new syllabus FM AS privately!) so saw it as an opportune moment to get my textbook out and learn it! I think I really should get through c4 myself asap to really get to grips with the AEA questions, right?

    Also, June 2005 Question 7 (attached)..parts a is pretty straight forward, part b is manageable, but I have no idea how to attempt part c...any ideas?


    It's hard to see it there!, here's the direct link : http://s5.tinypic.com/df7waw.jpg
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    ...

    show;

     \displaystyle \frac{d}{dx} x^n = nx^{n-1}
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    My question, what's AEA? Never heard of that o.O
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    (Original post by Andylol)
    My question, what's AEA? Never heard of that o.O
    http://en.wikipedia.org/wiki/Advanced_Extension_Award
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    subscribed*
    havent started solving papers for it though
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    ... Providing;  e^{inx} = cos(nx) + isin(nx) \; and \; i = \sqrt{-1} use the sub  y = e^{kx}

     \displaystyle \frac{d^2y}{dx^2} +4y = 0

    Hence obtain the function so;

     \displaystyle \frac{d^2y}{dx^2} +4y = sin(x)
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    (Original post by Economic Historian 1)
    Thanks for everyone's input already!...Looks like this shall be a successful thread It's already turning out to be very useful for me, since I haven't yet done C4 differential equations (I'm not doing full further maths, just new syllabus FM AS privately!) so saw it as an opportune moment to get my textbook out and learn it! I think I really should get through c4 myself asap to really get to grips with the AEA questions, right?

    Also, June 2005 Question 7 (attached)..parts a is pretty straight forward, part b is manageable, but I have no idea how to attempt part c...any ideas?


    It's hard to see it there!, here's the direct link : http://s5.tinypic.com/df7waw.jpg
    Expand out cos(2x) as 2\cos^2 x - 1; now, can you make a substitution which'll enable you to use the result in (i)?
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    (Original post by Glutamic Acid)
    Expand out cos(2x) as 2\cos^2 x - 1; now, can you make a substitution which'll enable you to use the result in (i)?
    Yeah I get it now, thanks!
    (I think I better quickly learn how to use latex though, to actually start posting my working on here! :redface: )
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    (Original post by Horizontal 8)
    Ok I'll start us off:

    Solve the D.E.:

     \left( \frac{dy}{dx} \right)^2+sinx(cos^2x)\left( \frac{dy}{dx} \right)-sin^4x = 0

    Spoiler:
    Show
    it's a factorisable quadratic in dy/dx
    Spoiler:
    Show
    sinx(cos^2x)= sinx(1-sin^2x)=sinx-sin^3x
    They can't ask diff eqs in the AEA. It's not C1-C4.
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    (Original post by sohanshah)
    They can't ask diff eqs in the AEA. It's not C1-C4.
    It is in the current C4 edexcel syllabus I quote from the specification:

    "Analytical solution of simple first order differential equations with separable variables."

    Furthermore, in the question not even this knowledge of separation of variable is needed all you need is the 'intuition' to try to factorise the expression and consider the expressions in the two brackets separately.
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    (Original post by Horizontal 8)
    It is in the current C4 edexcel syllabus I quote from the specification:

    "Analytical solution of simple first order differential equations with separable variables."

    Furthermore, in the question not even this knowledge of separation of variable is needed all you need is the 'intuition' to try to factorise the expression and consider the expressions in the two brackets separately.
    At the risk of sounding stupid, C4 differential equations are well...much much easier though :redface:
    And I haven't found any questions like the one you posted in past papers as yet...
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    (Original post by sohanshah)
    They can't ask diff eqs in the AEA. It's not C1-C4.
    I'm pretty sure DEs are on C4. And to be fair, all you need to solve this question is:

    Spoiler:
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    a simple trig identity, quadratic factorisation, and integration of polynomials in sinx
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    (Original post by Economic Historian 1)
    At the risk of sounding stupid, C4 differential equations are well...much much easier though :redface:
    But AEA questions are supposed to be much harder than C4 ones.

    And I haven't found any questions like the one you posted in past papers as yet...
    That, however, is a fair point. The people posting on this thread (and the 2009 STEP one) are not doing a great job of reproducing the examination style. Even in maths, real exam questions tend to be a little wordier than "solve {long complicated equation}".
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    2X = 4

    Find X...:ninja:
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    (Original post by *MJ*)
    2X = 4

    Find X...:ninja:
    It is there... between 2 and the equation symbol *points*. :yep:
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    (Original post by (2Clever*0.5)^2)
    It is there... between 2 and the equation symbol *points*. :yep:
    Go on...

    This is better than the answer I was expecting...:tongue:

    Solve 7 + 2 = 9...:ninja:

    ...:cool: :tongue:
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    Given ln(x) = ax, where a is a constant, determine the values of a for which there is one solution to the equation.

    EDIT: Maybe this is too hard for AEA? I'm not sure.
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    (Original post by Mark13)
    Given ln(x) = ax, where a is a constant, determine the values of a for which there is one solution to the equation.

    EDIT: Maybe this is too hard for AEA? I'm not sure.
    ... seems reasonable (Camb interview questions)

    edit; you can even guess the correct answers from the graph.
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    (Original post by Mark13)
    Given ln(x) = ax, where a is a constant, determine the values of a for which there is one solution to the equation.

    EDIT: Maybe this is too hard for AEA? I'm not sure.
    Do you have the answer?
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    (Original post by Mark13)
    Given ln(x) = ax, where a is a constant, determine the values of a for which there is one solution to the equation.

    EDIT: Maybe this is too hard for AEA? I'm not sure.
    considering the graph y= lnx and y=ax

    if there is one solution, the graph of y=ax will be tangent to y=lnx

    differentiating, and noting that their gradients will be the same, we get  a = \frac{1}{x} \implies ax = 1 \implies \ln x = 1  \implies x = e \implies \boxed{a = \frac{1}{e}}

    also sketching a graph of lnx, we see that for a \leq 0, there exists one solution only as well.

    Spoiler:
    Show
    for  a > \frac{1}{e} there exists no solutions to  \ln x = ax

    for  \frac{1}{e} > a > 0 there exists two solutions to  \ln x = ax
Updated: June 24, 2009
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