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

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    (Original post by CD223)
    Ew. I do not look forward to doing that. Can't believe I've forgotten if it was that bad - must have selectively blanked it


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    Haha let me know how it goes. Only managed to get 62.


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    (Original post by Jimmy20002012)
    Haha let me know how it goes. Only managed to get 62.


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    To be fair that's 98 UMS. 63 is 100 UMS.


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    Could someone explain where I've gone wrong on June 2010 Q7?

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    I've set up a quadratic in terms of \alpha like so:

    5 \alpha^{2} - 12 \alpha + 7 = 0

    To which the solutions are:
    \alpha = 1 or \alpha = \frac{7}{5}

    This gives co ordinates of C as:
    (3, -1, 3) which is correct, and (\frac{19}{5}, -1, \frac{17}{5}) which is incorrect.

    How can I get to the other possibility of (-1, -1, 1)?


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    (Original post by CD223)
    Could someone explain where I've gone wrong on June 2010 Q7?

    Name:  ImageUploadedByStudent Room1432669986.797865.jpg
Views: 187
Size:  102.1 KB

    I've set up a quadratic in terms of \alpha like so:

    5 \alpha^{2} - 12 \alpha + 7 = 0

    To which the solutions are:
    \alpha = 1 or \alpha = \frac{7}{5}

    This gives co ordinates of C as:
    (3, -1, 3) which is correct, and (\frac{19}{5}, -1, \frac{17}{5}) which is incorrect.

    How can I get to the other possibility of (-1, -1, 1)?


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    Not sure why it does not work using your method as I would has thought BP and AC have the same distance due to it being a parallelogram. But when I tried it the other way round of saying vector BC=AP, you get the right answer. Not too sure why you cannot do it the other way round :s.


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    (Original post by Jimmy20002012)
    Not sure why it does not work using your method as I would has thought BP and AC have the same distance due to it being a parallelogram. But when I tried it the other way round of saying vector BC=AP, you get the right answer. Not too sure why you cannot do it the other way round :s.


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    That is odd. If anyone would be kind enough to explain why I'm sure we'd both appreciate it!


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    Does anyone have any tips on how to get better at vectors? they're my only weakness on C4 and I've been trying different strategies to try and get better but it's just not happening 😭😭 any help would be really appreciated!!


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    (Original post by Bueno786)
    Does anyone have any tips on how to get better at vectors? they're my only weakness on C4 and I've been trying different strategies to try and get better but it's just not happening 😭😭 any help would be really appreciated!!


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    What revision guide do you use? The CGP one has great tips


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    Name:  ImageUploadedByStudent Room1432735618.815212.jpg
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Size:  134.1 KB Could someone please explain the working out process for this please


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    (Original post by CD223)
    What revision guide do you use? The CGP one has great tips


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    I don't actually use a revision guide 😬 I'm thinking to get one, thank you!


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    (Original post by EmilyC96)
    Name:  ImageUploadedByStudent Room1432735618.815212.jpg
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Size:  134.1 KB Could someone please explain the working out process for this please


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    Does this help? Let me know if you need any clarification.


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    (Original post by Bueno786)
    I don't actually use a revision guide 😬 I'm thinking to get one, thank you!


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    No problem! It does help when you get stuck


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    Also how do you integrate

    1 / (2rooty) dy


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    (Original post by CD223)
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    Does this help? Let me know if you need any clarification.


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    Sorry I'm no good at those questions. How do you do a? I just don't understand how to make differential equations :/


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    (Original post by EmilyC96)
    Also how do you integrate

    1 / (2rooty) dy


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    This should help.

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    (Original post by Jimmy20002012)
    This should help.

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    Yes!!! Thank you!!


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    (Original post by EmilyC96)
    Sorry I'm no good at those questions. How do you do a? I just don't understand how to make differential equations :/
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    The question states that the rate of decrease of surface area A in time t is constant.

    As the rate of change of surface area is constant, this suggests it doesn't depend on A or t, so they don't belong in the differential equation.

    From this, we can say that, as the rate of decrease of surface area is constant,

    \dfrac{dA}{dt} = -k

    Think of it like a linear graph with a negative gradient. The gradient is constant and negative because it is a whole number and doesn't depend on the x or y variable.


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    (Original post by CD223)
    The question states that the rate of decrease of surface area A in time t is constant.

    As the rate of change of surface area is constant, this suggests it doesn't depend on A or t, so they don't belong in the differential equation.

    From this, we can say that, as the rate of decrease of surface area is constant,

    \dfrac{dA}{dt} = -k

    Think of it like a linear graph with a negative gradient. The gradient is constant and negative because it is a whole number and doesn't depend on the x or y variable.


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    I understand! Woo 🎉 Hahaa thanks ☺️


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    From June 12 q5ai
    Can someone tell me where I went wrong :/


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    (Original post by EmilyC96)
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    From June 12 q5ai
    Can someone tell me where I went wrong :/


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    Yeah of course thanks


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