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

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    (Original post by JRN-95)
    For 8b)i) isnt it dr/dt= (pi*r^2)/k
    No because you were told that dr/dt is inversely proportional to the area (therefore 1/pi*r^2).

    Also, I am pretty sure it told you in the question the constant was a. So the actual answer should have been dr/dt = a/pi*r^2.
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    (Original post by RavinderKlair97)
    how much ums would 55 marks get?
    I think I got around 55 as well :// i hope very much for 80+
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    (Original post by ForgottenApple)
    Probably your score
    That's what a moron would say because it hasn't been marked yet
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    (Original post by amyrah)
    coordinates for C were (-9, -2, 18) ?
    I got this too! Does anyone know whether this is right/wrong with explanation?
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    Do you remember the actual diff equation they asked you to solve?
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    (Original post by datpr0)
    No because you were told that dr/dt is inversely proportional to the area (therefore 1/pi*r^2).

    Also, I am pretty sure it told you in the question the constant was a. So the actual answer should have been dr/dt = a/pi*r^2.
    ****!!! i did dr/dt= pi*r^2/k damnnn!!! could i still get some marks and follow through marks for the next part
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    Does anyone remember the question for the implicit differentiation? Just want to go over it again as I don't remember if I got 11/8 or not
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    is there an agreed upon mark scheme anywhere?
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    (Original post by HennersPD)
    Do you remember the actual diff equation they asked you to solve?
    Yes.

    It was: dx/dt = (4+5x)^1/2 / 5(1+t)^2
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    Anyone got a worked solution for the vectors question part b and c?
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    (Original post by HennersPD)
    Do you remember the actual diff equation they asked you to solve?
    if think it was dx/dt= (4+5x)^1/2 divided by 5(t+1)^2
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    (Original post by jchap9776)
    Potentially, did anybody else show x tending to infinity = 2x-3 tending to infinity = 1/(2x-3) tends to 0? Therefore is decreasing. (Idk if it was 2x-3 but was something like that)
    Thats what i literally done.
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    for 2)b)ii) I got

    -1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

    Therefore √29cos(x+1.19) = √29

    cos(x+1.19) = 1
    x + 1.19 = 0
    x = -1.19

    So why is it 5.09 where as it could be both...
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    For the last question, I forgot to divide 0.5 by 60. I wrote a quick comment, and some working-out to correct the value, but then I realise that I forgot to square root the 60. Now I'm pissed >

    Also I forgot the +C in the differential equation. How many marks do you think I'll lose
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    (Original post by Kennethm)
    for 2)b)ii) I got

    -1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

    Therefore √29cos(x+1.19) = √29

    cos(x+1.19) = 1
    x + 1.19 = 0
    x = -1.19

    So why is it 5.09 where as it could be both...
    It said between 0 < x <2pi I think
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    question 2 needs sorting and for vectors I swear it was E(11,0,0) and E(23,4,-8)
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    (Original post by Kennethm)
    for 2)b)ii) I got

    -1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

    Therefore √29cos(x+1.19) = √29

    cos(x+1.19) = 1
    x + 1.19 = 0
    x = -1.19

    So why is it 5.09 where as it could be both...
    because 0<x<2pi
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    (Original post by Kennethm)
    for 2)b)ii) I got

    -1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

    Therefore √29cos(x+1.19) = √29

    cos(x+1.19) = 1
    x + 1.19 = 0
    x = -1.19

    So why is it 5.09 where as it could be both...
    Because they told you to find the maximum value in the range 0 < x < 2pi !
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    (Original post by jamal_m96)
    Nah love im pretty sure my core 3 exam had a question 9
    There was deffo 8 questions, are you sure you're in the right thread?
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    (Original post by JAW-97)
    There was deffo 8 questions, are you sure you're in the right thread?
    He's trolling, ignore him.
 
 
 
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