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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)
    What I wrote on the mark scheme was what I put down in the paper so I'm hoping I got the majority right but probably missed some method marks - 71-72 maybe?

    Core 4 will be hard


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    thats great for you then! I'm just worried Bath won't let me in if I don't get that A* in Maths it's annoying as they've increased the grades I need from AAB to A*AA I know just hopefully not as hard as this exam! :/
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    (Original post by UOE15)
    Yeah I do understand what you've done, it is logical but I just don't think I'd remember how to do it in the exam haha
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    Haha just keep attempting past questions on differential equations until you're confident?
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    (Original post by HennersPD)
    thats great for you then! I'm just worried Bath won't let me in if I don't get that A* in Maths it's annoying as they've increased the grades I need from AAB to A*AA I know just hopefully not as hard as this exam! :/
    Seems harsh?!


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    (Original post by saad97)
    Haha just keep attempting past questions on differential equations until you're confident?
    Yeah I am going to do that as I did that for vectors and can vaguely do them now!
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    (Original post by CD223)
    Seems harsh?!


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    I know that's why I'm panicking about getting an A* now as I'll need 100 UMS in core 4 as i only think I've got about 64/75 in core 3 can you suggest any tough vectors questions as thats what I'm worst on!
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    5b Jan 11 explanation anyone?
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    (Original post by amyrah)
    5b Jan 11 explanation anyone?
    Yeah, so it's just saying that when t=d, m = m0/16.

    i.e. 10/16 = 10 x 2^(-1/8)d

    => 1/16 = 2^-d/8 (as the 10's cancel)
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    (Original post by datpr0)
    Yeah, so it-s just saying that when t=d, m = m0/16.

    i.e. 10/16 = 10 x 2^(-1/8)d

    => 1/16 = 2^-d/8 (as the 10's cancel)
    So basically I can put any value as m0?
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    Could someone explain q7 from June 11 for me please :/


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    (Original post by amyrah)
    So basically I can put any value as m0?
    Well yes, because the values you enter would just cancel each other. I think, actually a better way of showing it mathematically would be to just indicate the m0's cancelling.
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    Could someone explain q7 from June 11 for me please :/


    EDIT: also 8b of the same paper. Why is it + 1/(1-x) as I keep getting minus. Also how do you change to the form they want the answer in

    Thanks!
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    (Original post by EmilyC96)
    Could someone explain q7 from June 11 for me please :/


    EDIT: also 8b of the same paper. Why is it + 1/(1-x) as I keep getting minus. Also how do you change to the form they want the answer in

    Thanks!
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    This is 8b. I did it a few days ago too and was confused by it but apparently you multiply the -1 by (-x/1-x) to get (x/1-x). Well that's what my teacher said anyway.
    Hope that helps
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    (Original post by EmilyC96)
    Could someone explain q7 from June 11 for me please :/


    EDIT: also 8b of the same paper. Why is it + 1/(1-x) as I keep getting minus. Also how do you change to the form they want the answer in

    Thanks!
    7
    a) basically rewrite what they've told you in a differential equation.

    They said surface area (A) is decreasing at a constant (-k) rate (t).
    => dA/dt = -k

    ... dA/dt is the rate of change of surface area. It is DECREASING at a CONSTANT rate, so -k.

    b) i. Integrate your differential equation.

    => A = -kt +C

    Its INITIAL (t=0) radius is 60cm so C = 4pi x 60^2 (as they tell you to assume surface area = 4pi x r^2).

    With C calculated, input the values they give you and C into the integrated equation. This gives:

    4pi x 30^2 = -9k + 4pi x 60^2

    Solve to find k.

    Substitute k & C into integrated equation and just take 1200pi out as a factor to get it in the correct form.

    b) ii. A = 0, => -t + 12 = 0...
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    (Original post by EmilyC96)
    Could someone explain q7 from June 11 for me please :/


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    (Original post by saad97)
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    Thank you!!


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    (Original post by saad97)
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    This is 8b. I did it a few days ago too and was confused by it but apparently you multiply the -1 by (-x/1-x) to get (x/1-x). Well that's what my teacher said anyway.
    Hope that helps
    Thanks


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    Anyone fancy explaining the necessary and sufficient notations?

    According to the spec we need to be able to use and understand them.
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    Guys I will be really grateful , if someone helps me with a question, I attempted it like 10 times and don't know what to do.
    Its a question from JUNE 2010 QUESTION 3(b)
    Thanks !
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    Could someone help me with 3c from jan 12 please


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    (Original post by EmilyC96)
    Could someone help me with 3c from jan 12 please


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    It's pretty simple actually.

    Cubed-root of 100 = 10^2/3

    Therefore just consider what value of x is required for 8 + 6x = 10 to be true. Then substitute this value of x into your answer to b and simplify into the form a/b.
 
 
 
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