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    6 day 2 exam!!!
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    I'm fairly sure FP1 and P4 are the same exam and syllabus so we might as well discuss both in the same thread.
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    (Original post by arash_gourtani)
    6 day 2 exam!!!
    Enfin!

    See my post on this thread:

    http://www.thestudentroom.co.uk/t124737.html


    I'll see if I can find some FOD/SOD equation nasties in the next couple of days from another source for practice...

    Aitch
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    (Original post by Aitch)
    Enfin!

    See my post on this thread:

    http://www.thestudentroom.co.uk/t124737.html


    I'll see if I can find some FOD/SOD equation nasties in the next couple of days from another source for practice...

    Aitch

    Here's a previous nasty as a starter:

    http://www.thestudentroom.co.uk/t81039.html

    Aitch
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    The hardest part in FP1 is the series (telescoping) and proof by induction, unless you practise these. Just remember the formulae of the series, n, n², n³.

    Complex numbers are an important part of math and are pretty much easy if you become used imagnery numbers, your garanteed to get a question to find all real and imagenary roots of a quadratic, this should be by far the easiest.

    Good luck for then exam
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    Edexcel P4 Jan 04 Q 6(b)

    [You've probably all done this. since there are few practice papers]

    Answer from (a) is

    y= e-2x(Acosx + Bsinx) + sin2x - 8 cos2x

    (b) Show that for large values of x this general solution may be approximated

    by a sine function,and find this function.

    [=> used for approaches]


    As x=> ∞, e-2x =>0

    so y=>sin2x - 8 cos2x

    Ok so far.

    Mark scheme then expects conversion to Harmonic Form:

    y=> Rsin[2x + α]
    and eventual arrival at R = √65 and α= -82.9º

    But what about

    y=> sin2x - 8cos2x = sin2x -8(cos²x - sin²x)

    so y => sin2x -8(1 - 2sin²x)
    Is this a sine function? or not? If it is, why no mention in the mark scheme? What is the definition of a sine function?

    Aitch
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    But what about
    y=> sin2x - 8cos2x = sin2x -8(cos²x - sin²x)
    so y => sin2x -8(1 - 2sin²x)
    Is this a sine function? or not? If it is, why no mention in the mark scheme? What is the definition of a sine function?
    Aitch
    Isn't that made up of a number of sine functions? Perhaps 'a sine function' implies only one use of 'sin', rather than one function that uses sine multiple times.
    There's always the possibility both are correct though!
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    (Original post by Gaz031)
    There's always the possibility both are correct though!
    What makes me a bit doubtful is that the mark schemes are usually quite good at including alternative answers, and alternative routes to similar answers...

    Also, if a sine function contains only sine functions, then the constant in my answer is a problem!

    Aitch
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    (Original post by Aitch)
    What makes me a bit doubtful is that the mark schemes are usually quite good at including alternative answers, and alternative routes to similar answers...

    Also, if a sine function contains only sine functions, then the constant in my answer is a problem!

    Aitch

    ...unless I change the constant to sin(∏/2)

    Aitch

    [Yes, it's still a constant!]
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    (Original post by Aitch)
    ...unless I change the constant to sin(∏/2)

    Aitch

    [Yes, it's still a constant!]
    That's certainly one way of doing it
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    hey guys, have any of you got the Jan 2005 mark scheme?
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    (Original post by orangazhang)
    hey guys, have any of you got the Jan 2005 mark scheme?
    Edexcel P4 Jan05. Mark Scheme.

    Yes. In 3 bits.

    PM or Email me if you want them.

    Aitch
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    Nice P4/FP1 SOD question. There was a clue at the beginning, but I've omitted it to make it a bit harder.

    Find the solution of the following for which y=1 and (dy/dx)=0 at x=0

    y'' + 2*y' + y = e-x

    Aitch
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    (Original post by Aitch)
    why no mention in the mark scheme?
    If it's a provisional markscheme they only put what they're expecting down there. When they get the papers in they realise that there are more perfectly good answers than those that they've thought of already and so they work out how to allocate marks to anything they weren't expecting.
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    (Original post by Aitch)
    Nice P4/FP1 SOD question. There was a clue at the beginning, but I've omitted it to make it a bit harder.

    Find the solution of the following for which y=1 and (dy/dx)=0 at x=0

    y'' + 2y' + y = e-x

    Aitch
    I presume you've all done this so easily that you can't be bothered to admit it...


    Aitch

    [Surely someone wants the answer which I've prepared to copy & paste...]
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    (Original post by Aitch)
    I presume you've all done this so easily that you can't be bothered to admit it...

    Aitch
    Lol
    I have y=e^{-x}[1+x+\frac{1}{2}x^{2}]
    Is that about right?
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    (Original post by Gaz031)
    Lol
    I have y=e^{-x}[1+x+\frac{1}{2}x^{2}]
    Is that about right?
    You have to spot (unusually here) that both the RHS and the RHS*x appear in the CF. So you need to start the PI calc with y= n.x².RHS

    Partial working follows pasted in next post.

    Aitch
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    (Original post by Aitch)
    You have to spot (unusually here) that both the RHS and the RHS*x appear in the CF. So you need to start the PI calc with y= n.x².RHS

    Partial working follows pasted in next post.

    Aitch

    The CF is y = (A + Bx) e^-x

    for the PI, spot that e^-x and x.e^-x are both part of the CF, so start with

    let y = n.x².e^-x

    ...

    arrive eventually at n = ½

    assemble to

    y = (A + Bx) e^-x + ½.x².e^-x


    put in y=1, x=0 to get A=1

    differentiate and put in x=0 (dy/dx)=0 to get B=1

    Aitch
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    (Original post by Aitch)
    You have to spot (unusually here) that both the RHS and the RHS*x appear in the CF. So you need to start the PI calc with y= n.x².RHS

    Partial working follows pasted in next post.

    Aitch
    I saw that but only because I usually work out the complementary function first, otherwise I would have wasted time trying ke^{-x} and ke^{-x} as particular integrals
    Looks good.
    I'm probably going to go listen to bbc radio now while I do an AEA paper and then maybe revise P4 differential equations [there's no need to revise complex numbers due to p6]. I've missed most of the bond movie and most of the rest of tv is terrible.
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    (Original post by Gaz031)
    I saw that but only because I usually work out the complementary function first, otherwise I would have wasted time trying ke^{-x} and ke^{-x} as particular integrals
    Looks good.
    I'm probably going to go listen to bbc radio now while I do an AEA paper and then maybe revise P4 differential equations [there's no need to revise complex numbers due to p6]. I've missed most of the bond movie and most of the rest of tv is terrible.
    Cricket was rained off.

    Watched Gardener's World...

    All that's left is finding instances of having to multiply the PI starter expression by x²...

    ...more claret.

    Aitch
 
 
 
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