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AQA A2 MFP3 Further Pure 3 – 18th May 2016 [Exam Discussion Thread] Watch

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    (Original post by JPL98)
    Ah, I see so basically because of the combination I'm taking I have to get 270 ums across FP3, M3, (either S2 or M2). But this link here suggests the other way: http://filestore.aqa.org.uk/admin/sp...TAR-GRADES.PDF 'best 3 A2 units' interestingly this doesn't exclude C3,C4 which I assume it must do, but that suggests I can just take my best 3 of 4 A2 units. Rather than be forced to take FP3 and M3 cause they can't go with core.
    Ok for NORMAL maths you have C1-4 and 2 applied modules. For FURTHER maths you have FP1 and FP3 in your case plus 2 applied AS modules plus 2 applied A2 modules.
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    (Original post by zetamcfc)
    Ok for NORMAL maths you have C1-4 and 2 applied modules. For FURTHER maths you have FP1 and FP3 in your case plus 2 applied AS modules plus 2 applied A2 modules.
    Hopefully, its this one (M3 is my weakest module). I really wish AQA were clearer with how this works.
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    (Original post by JPL98)
    Hopefully, its this one (M3 is my weakest module). I really wish AQA were clearer with how this works.
    Ok I've read that page fully now. Are you taking maths AND further maths at sixth form/college? Or are you doing Advanced double award (Something I've never seen before)?
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    (Original post by JPL98)
    Hope revision is going well, do the papers get harder after 2010 because I've been getting pretty much full marks on the older papers (2006-2010)?

    Also can someone clear this up for me because I've heard conflicting responses. How exactly is the A* grade for further maths AQA calculated? I'll list my modules:
    AS: C1,C2,S1,FP1,M1,D1
    A2: C3,C4,S2,FP3,M2,M3
    Basically because only one applied module a year can go with Core maths, M3 must go with Further Maths in terms of calculating your UMS (as would have M1 with core maths AS and M2 with core maths A2) and because of this the further maths grade is calculated as FP3, M3 and the best of S2 and M2. However I've heard some people say that the further maths A* grade is calculated as at least 90% across your best 3 A2 modules (excluding C3,C4) as well as getting a 80% on the further maths bit only.

    Might seem like a pointless distinction but I'm not sure about hitting 90% in both FP3 and M3, the second method allows combinations like 90% across M2,M3,S2 and FP3,M2,S2 whereas the first doesn't. Also by any chance does FP1 count as an A2 unit even though I sat it in AS?
    Seems like this has already been explained to you, my only question is, where did you get the information that only 1 applied module can go with core maths a year? I am almost certain that is false.


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    (Original post by zetamcfc)
    Ok I've read that page fully now. Are you taking maths AND further maths at sixth form/college? Or are you doing Advanced double award (Something I've never seen before)?
    First option not Advanced double award.
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    (Original post by jjsnyder)
    Seems like this has already been explained to you, my only question is, where did you get the information that only 1 applied module can go with core maths a year? I am almost certain that is false.


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    AS Normal Maths (not further) is composed of C1,C2,+1 applied module, A2 Normal Maths is composed of C1,C2,+1 applied module. As the C1-4 are mandatory for core/normal maths only one applied module a year can go with it. Also note you can't put harder modules with core eg. put S2 in core and have S1 in further. Because of this M3 must go with further because the core slots at max have M1 at AS and M2 at A2. Any other combination doesn't work. Or at least thats what my teacher said.
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    (Original post by Kvothe the arcane)
    You go up in degree of x if present in the complimentary function.
    Btw does this apply to other things other than e...

    E.g. RHS of DE = 10

    CF = Ae^x + B

    in this case PI cannot be y = c (where a is c constant)
    PI is instead y = ct + d right?
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    (Original post by JME_CHG)
    Btw does this apply to other things other than e...

    E.g. RHS of DE = 10

    CF = Ae^x + B

    in this case PI cannot be y = c (where a is c constant)
    PI is instead y = ct + d right?
    Yes in that case PI trial function will be y=ct + d
    because you already have a constant in your CF
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    (Original post by JME_CHG)
    Btw does this apply to other things other than e...

    E.g. RHS of DE = 10

    CF = Ae^x + B

    in this case PI cannot be y = c (where a is c constant)
    PI is instead y = ct + d right?
    This seems like a strange DE you're dealing with...
    Is it missing the y term or has variable coefficients? Otherwise how is it that substituting y=Ae^x+B give 0??
    Also I have no idea how substituting anything will give 10.
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    (Original post by Qcomber)
    Yes in that case PI trial function will be y=ct + d
    because you already have a constant in your CF
    Well there wouldn't be a '+d' since this is part of the CF. And ct substituted probably isn't going to give a constant on the RHS since I'm assuming the DE has a y term.
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    (Original post by IrrationalRoot)
    Well there wouldn't be a '+d' since this is part of the CF. And ct substituted probably isn't going to give a constant on the RHS since I'm assuming the DE has a y term.
    I guess you're right, the +d would come out as zero
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    (Original post by IrrationalRoot)
    This seems like a strange DE you're dealing with...
    Is it missing the y term or has variable coefficients? Otherwise how is it that substituting y=Ae^x+B give 0??
    Also I have no idea how substituting anything will give 10.

    Yep missing y... So that one root solution =0...

    And I guess it's not as complicated as eˣ then...
    Since with the CF we learn, it's not as if we can have CF = Ax + ...
    Whereas with eˣ I guess the highest power we'll have to deal with is kx²eˣ since we can't have CF = eˣ(Ax²+B) or something...


    Anyway if CF = Aeˣ + Beˣ

    If we used PI = kx²eˣ would it work?
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    (Original post by JME_CHG)
    Yep missing y... So that one root solution =0...

    And I guess it's not as complicated as eˣ then...
    Since with the CF we learn, it's not as if we can have CF = Ax + ...
    Whereas with eˣ I guess the highest power we'll have to deal with is kx²eˣ since we can't have CF = eˣ(Ax²+B) or something...


    Anyway if CF = Aeˣ + Beˣ

    If we used PI = kx²eˣ would it work?
    Oh of course, it's missing the y term, it makes sense, just me being stupid.

    You could just let u=dy/dx and use IF on the resulting first order DE (not sure whether this is particularly quick) but the usual CF+PI method should work.

    In this case the PI will be of the form kt since that's the only way you'll get a constant on the right.
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    (Original post by JPL98)
    AS Normal Maths (not further) is composed of C1,C2,+1 applied module, A2 Normal Maths is composed of C1,C2,+1 applied module. As the C1-4 are mandatory for core/normal maths only one applied module a year can go with it. Also note you can't put harder modules with core eg. put S2 in core and have S1 in further. Because of this M3 must go with further because the core slots at max have M1 at AS and M2 at A2. Any other combination doesn't work. Or at least thats what my teacher said.
    Yes I know all of this, I myself have spent an extended period of time fully understanding which modules can and cannot be taken, what I was saying is that there is no restriction on the year in which modules are taken, nor on the order, only that the correct ones are achieved. (I.e for normal maths this is all four cores and 2 applied modules that can be two level one applied or one level one and its corresponding level 2 module)


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    (Original post by JME_CHG)
    Yep missing y... So that one root solution =0...

    And I guess it's not as complicated as eˣ then...
    Since with the CF we learn, it's not as if we can have CF = Ax + ...
    Whereas with eˣ I guess the highest power we'll have to deal with is kx²eˣ since we can't have CF = eˣ(Ax²+B) or something...


    Anyway if CF = Aeˣ + Beˣ

    If we used PI = kx²eˣ would it work?
    Why have you stated your complementary function like this? For one this could just be written as Ae^x, but also if the roots obtained from the auxiliary equation are equal then the CF = e^kx(Ax+B). I cannot think of any differential equation on syllabus that will give you that complementary function.


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    (Original post by jjsnyder)
    Why have you stated your complementary function like this? For one this could just be written as Ae^x, but also if the roots obtained from the auxiliary equation are equal then the CF = e^kx(Ax+B). I cannot think of any differential equation on syllabus that will give you that complementary function.


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    Oh yeah, it was random, what I meant was:

    CF = Aeⁿˣ + Beᵖˣ

    Would PI = kx²eⁿˣ work...
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    (Original post by JME_CHG)
    Oh yeah, it was random, what I meant was:

    CF = Aeⁿˣ + Beᵖˣ

    Would PI = kx²eⁿˣ work...
    Ahh okay I see


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    (Original post by JME_CHG)
    Oh yeah, it was random, what I meant was:

    CF = Aeⁿˣ + Beᵖˣ

    Would PI = kx²eⁿˣ work...
    Depends what the RHS of the DE is. If it's some function of e^{nx} then you'd try y_p=kxe^{nx}, and it should work.
    The only time you'd try y_p=kx^2e^{nx} is when your complementary function has both kxe^{nx} and ke^{nx} in it, which would occur if the auxiliary equation had equals roots n.
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    Can someone explain where the minus in the 1/3 comes from in this integral? Almost certain I'm just doing something stupid but can't quite get the answer myself.

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    (Original post by Adam_98)
    Can someone explain where the minus in the 1/3 comes from in this integral? Almost certain I'm just doing something stupid but can't quite get the answer myself.

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    The derivative of cosx is -sinx. So -1/3 needed to cancel the -3 when you differentiate your answer to obtain the original integrand.
 
 
 
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