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AQA A2 MFP2 Further Pure 2 – 24th June [Exam Discussion Thread] Watch

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    (Original post by C0balt)
    Thanks, I see it but how did you spot going from line 3 to 4 without seeing the ms
    You just have to try and force a similar expression to the expression you got when you expanded out what they gave you in (ii).
    You have to see how you can get a similar expression using values that you know, like  \sum \alpha and  \sum \alpha \beta and  \alpha \beta \gamma .
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    (Original post by MrMagic12345)
    No
    And for that question from June 2014 I would just try to memorise that Sum(a^2b)=sum(alpha) x sum(alphabeta)-3alphabetagamma
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    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    On question 4(c)(i), how are you supposed to know that alpha = -2???
    The mark-scheme just suddenly assumes you already know it. Is it just trial and error or something?
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    (Original post by 2014_GCSE)
    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    On question 4(c)(i), how are you supposed to know that alpha = -2???
    The mark-scheme just suddenly assumes you already know it. Is it just trial and error or something?
    Yeah just trial and error
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    (Original post by 2014_GCSE)
    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    On question 4(c)(i), how are you supposed to know that alpha = -2???
    The mark-scheme just suddenly assumes you already know it. Is it just trial and error or something?
    You've got a quadratic in alpha.. (You just need to put the p value you found in it)
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    (Original post by -jordan-)
    Yeah just trial and error
    Thanks man!

    Also, does multiplying sinht by sinht equal -sinh^2t?

    I'm confused why the equations have you switch the sign on sinh^2(t).
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    (Original post by MrMagic12345)
    You've got a quadratic in alpha.. (You just need to put the p value you found in it)
    No, you have a cubic with alpha.
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    (Original post by 2014_GCSE)
    Thanks man!

    Also, does multiplying sinht by sinht equal -sinh^2t?

    I'm confused why the equations have you switch the sign on sinh^2(t).
    No that's Osborne's rule, because of the way the hyperbolics in expoential form are, it allows you to convert any standard trigonometric identity into the hyperbolic equivalent. When you know sinht has been multiplied by sinht then you put a negative in front of that term in the expression. If you look at the normal identities and then their equivalent hyperbolic counterparts you'll see what I mean.
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    (Original post by 2014_GCSE)
    No, you have a cubic with alpha.
    Yeah my bad
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    (Original post by 2014_GCSE)
    Thanks man!

    Also, does multiplying sinht by sinht equal -sinh^2t?

    I'm confused why the equations have you switch the sign on sinh^2(t).
    You only use osborne's rule when trying to 'convert' a trig formula describing non-hyperbolic function's into a formula describing hyperbolics
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    Name:  Screen Shot 2016-06-23 at 16.15.50.png
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    can someone show me their working to this question please? i know its probably really simple but the mark scheme is vague
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    (Original post by IrrationalRoot)
    There's no method to explain, the whole method is shown in two lines to the right on the mark scheme.
    If there's something specific you don't understand within it, let me know and I'll explain it.
    I don't get where the -2 came from
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    (Original post by Jm098)
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    can someone show me their working to this question please? i know its probably really simple but the mark scheme is vague
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    Then u can do X coordinate + X coordinate /2 to get midpoint etc
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    (Original post by Hjyu1)
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    Then u can do X coordinate + X coordinate /2 to get midpoint etc
    thanks! where is the first line from though? have you just set the both mods equal?
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    (Original post by sam_97)
    This is how I did it:

    (\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)

= (\alpha + \beta)(\alpha\beta + \beta\gamma + \gamma\alpha +\gamma^2)

= \alpha^2\beta + \gamma\alpha^2 + \alpha\beta^2 + \beta^2\gamma + \beta\gamma^2 + \gamma^2\alpha + 2\alpha\beta\gamma 

= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - 3\alpha\beta\gamma + 2\alpha\beta\gamma

= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma

= \Sigma\alpha\Sigma\alpha\beta - \alpha\beta\gamma

    Expanding the brackets on line 4 gives us an extra 3\alpha\beta\gamma which we don't want, hence the -3\alpha\beta\gamma.
    I mean, yeah it's very easy to show the equality, but we were wondering how one would recognise the equality in the first place; the method shown to the right in the mark scheme is what I used and makes a lot more intuitive sense.
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    (Original post by Hjyu1)
    I don't get where the -2 came from
    -2 sum of roots.
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    (Original post by Jm098)
    thanks! where is the first line from though? have you just set the both mods equal?
    Yeah just converted to cerstian and which line
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    (Original post by IrrationalRoot)
    -2 sum of roots.
    Ohhhhhhhhhh **** I get it now thanks
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    (Original post by Hjyu1)
    Yeah just converted to cerstian and which line
    (x-4)^2 + (y-2)^2 = x^2 + y^2
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    (Original post by MrMagic12345)
    You only use osborne's rule when trying to 'convert' a trig formula describing non-hyperbolic function's into a formula describing hyperbolics
    Do you convert tanh^2(x) in a formula such as 1 + tan^2(x) = sec^2(x) because doesn't this technically contain a sinh^2(x)?

    Sorry, FP2 is an extra voluntarily module so I've kind of ignored it until now and I forget a bunch of small things like this.
 
 
 
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