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AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme] Watch

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    Sorry for all this guys but I've gone wrong again 😑😑Name:  ImageUploadedByStudent Room1432740492.313745.jpg
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    Sorry it's a bit messy :/
    On the markscheme it said to split it into 1/15 and xroot(2x-1) but surely there's a way to do it if I didn't?
    :/
    It's from June 2012 q8b


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    Can someone explain to me the method of getting the minimum/max values of x (2b) for everything as I don't know any of those bits as don't think my teacher explained it :/ just using question 2b from jan 09 to give you an idea of the questions I mean? Thanks for everyone's help today!

    I've done 2a if that helps and got root10sin(x-1.25)


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    (Original post by EmilyC96)
    Sorry for all this guys but I've gone wrong again 😑😑Name:  ImageUploadedByStudent Room1432740492.313745.jpg
Views: 133
Size:  131.6 KB

    Sorry it's a bit messy :/
    On the markscheme it said to split it into 1/15 and xroot(2x-1) but surely there's a way to do it if I didn't?
    :/
    It's from June 2012 q8b


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    (Original post by EmilyC96)
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    Can someone explain to me the method of getting the minimum/max values of x (2b) for everything as I don't know any of those bits as don't think my teacher explained it :/ just using question 2b from jan 09 to give you an idea of the questions I mean? Thanks for everyone's help today!

    I've done 2a if that helps and got root10sin(x-1.25)


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    Just as cos theta alternates between -1 and 1, your expression will alternate between root 10 and negative root 10.

    When looking for a minimum, set the expression you've just worked out to equal the negative value of R.

    When looking for a maximum, set the expression you've just worked out to equal the positive value of R.


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    How do you exactly know not to take the 15 up as well when separating variables?


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    (Original post by Jimmy20002012)
    How do you exactly know not to take the 15 up as well when separating variables?


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    You can take the 15 up too. It will just be slightly harder. Keep it outside the integral for the remainder of calculations


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    I somehow managed to get 2/4 mark on Q6 b ii of the June 2014 C4 paper by purely guessing. xD
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    (Original post by Tiwa)
    I somehow managed to get 2/4 mark on Q6 b ii of the June 2014 C4 paper by purely guessing. xD
    :L which question was that? Vectors?


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    (Original post by CD223)
    :L which question was that? Vectors?


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    Yep! I took me a while to understand the question. even then, I made an educational guess.
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    (Original post by Tiwa)
    Yep! I took me a while to understand the question. even then, I made an educational guess.
    Oh right! Do you get it now?


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    Where do they get a solution of 90 degrees from? I thought cotx couldn't equal zero?

    I got the other solutions, just didn't know 90 degrees was one?

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    EDIT: Oops, got my x and y the wrong way around. Never mind

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    (Original post by CD223)
    Where do they get a solution of 90 degrees from? I thought cotx couldn't equal zero?

    I got the other solutions, just didn't know 90 degrees was one?

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    EDIT: Oops, got my x and y the wrong way around. Never mind

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    Write Cotx as cosx/sinx
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    (Original post by a123a)
    Write Cotx as cosx/sinx
    In which case \cos x = 0 and x = 90^{o}?


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    (Original post by CD223)
    In which case \cos x = 0 and x = 90^{o}?


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    Yes.
    I remember doing this question last year.
    Edit: I could be wrong though.
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    (Original post by a123a)
    Yes.
    I remember doing this question last year.
    Edit: I could be wrong though.
    Oh right thanks!

    Is this an acceptable way of showing the two tangents meet on the x axis?

    (Jan 13 Q4 a ii)

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    (Original post by a123a)
    Yes.
    I remember doing this question last year.
    Edit: I could be wrong though.
    You're right

    It's like how y=(x)/(2x-3) has an asymptote at y=0, but y can still =0 when the numerator is 0, if that makes sense... Sort of like a special case
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    (Original post by CD223)
    Oh right thanks!

    Is this an acceptable way of showing the two tangents meet on the x axis?

    (Jan 13 Q4 a ii)

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    If this is the question I'm thinking of, you need to show that y=0 rather than substituting y=0 into the equation.

    i.e. you end up with two simultaneous equations where if you add them together you are given y=0.

    Hope this helps
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    (Original post by Stepidermis)
    You're right

    It's like how y=(x)/(2x-3) has an asymptote at y=0, but y can still =0 when the numerator is 0, if that makes sense... Sort of like a special case
    How is that even possible 😁

    Surely the equation breaks down?

    Maths blows my mind.


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    (Original post by Stepidermis)
    If this is the question I'm thinking of, you need to show that y=0 rather than substituting y=0 into the equation.

    i.e. you end up with two simultaneous equations where if you add them together you are given y=0.

    Hope this helps
    Thanks! Are there no alternative methods? My thinking was, if I set y=0 in both equations and that gives the same x co ordinate for both, surely that means they both intersect at the x axis?


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    (Original post by CD223)
    Thanks! Are there no alternative methods? My thinking was, if I set y=0 in both equations and that gives the same x co ordinate for both, surely that means they both intersect at the x axis?


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    There must be other ways of showing that y=0, however in this instance I don't think your way works because in the question they have told you that q cannot equal 0 but they haven't given you a restriction for p (correct me if I'm wrong).

    So the fact that you've found two equivalent expressions for x in this case is nullified by the fact that they both contain (q/p), as if p=0 then they don't intersect as this would instead be an asymptote, does that make sense?
 
 
 
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