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

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    (Original post by Alexandrite)
    For the vector of C did anyone get T=4 and then sub it into L. I can't remember the exact vector but definitely got T=4.

    As for the acute angle yeah I got 20something but it looks like most got 60 so whoops.
    I got T to be 4 aswell
    And bad luck :/ calculator errors can be a pain...


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    How many marks for a C and how many for a D? Thats all I needed in this exam Thanks
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    I got x=1/5(21/10 - 1/10t-10)^2 -4/5 for the write x in terms of t but im not sure... anyone?
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    (Original post by LordGaben)
    What u guys thinking? 67 for full ums?
    70
    Really easy paper from what everyone has said to me
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    (Original post by George1231)
    please try and add answers as I found the paper really hard and probably got a lot wrong
    1) A=3 B=-1b) 8/3ln5
    2) r=root 29 alpha=1.19max value root 29 x=5.09x=0.568 or 5..72
    3) d=1
    b) (2x+1)(2x-1)(2x-3)c) h(x)=(1=2x)^-1 differentiate shows its decreasing
    4a) 1+x-2x^2
    b) 1/4 -x/16+5x^2/256
    c)0.2313
    5a)gradient of normal -1/2
    b) equation of normal y=2x-1/2
    c) x cooridinate -1/8
    6a) 60 degrees
    B) (15,3,2)
    C) got wrong (11,0,0)
    7a) can't remember b)11/8
    8a) x=9/20-1/2(t+1)-1/20(1+t)^2
    b)got wrong
    c) 3
    7a) 1/4 - ln2

    8b) dr/dt = k/pi(r)^2
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    Predicted grade boundaries?
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    (Original post by HennersPD)
    Me too !! Did you get a horrid differential equation for x in terms of t? And did you get r=3 for last question? How did you show h(x) is decreasing? I just subbed in x vales>2 amd found it was decreasing. Differentiation doesnt work as its increasing in thar case.
    Dude I couldn't find E. I got a horrid differential equation and can't even remember it? I got r=3 and drew the graph for h.


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    (Original post by WillowXxx)
    I got x=1/5(21/10 - 1/10t-10)^2 -4/5 for the write x in terms of t but im not sure... anyone?
    Yh I got something like that
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    How much marks would I drop if I accidently put y in as cos2q and x and as sin(q) for the normal and curve question
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    (Original post by sam_97)
    Well said!
    Thank you


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    (Original post by Alexandrite)
    For the vector of C did anyone get T=4 and then sub it into L. I can't remember the exact vector but definitely got T=4.

    As for the acute angle yeah I got 20something but it looks like most got 60 so whoops.
    Yes i got t=4 and C as 15,8,-2 and then the acute angle as 69.1 as found obtuse as 110.58 and took away from 180
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    what does everyone think about grade boundaries this year? higher than 55 or lower?
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    (Original post by jamal_m96)
    There was one on the back too, i nearly missed it out!!!!!
    Are you drunk?
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    I changed the rates they gave us in the last question :/ The rates we were given were m / s, but the first part of the question was in minutes so converted them to m / min -_- This paper really wasn't my finest hour!
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    did anyone get 7/4 for 7b?
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    (Original post by lyddsssss)
    Predicted grade boundaries?
    A mark or two higher than last year's imo

    Though to be fair a lot of people said last year's was really hard but it seemed fine when I did it as a mock (different situations of course).
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    (Original post by CD223)
    Dude I couldn't find E. I got a horrid differential equation and can't even remember it? I got r=3 and drew the graph for h.


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    I didn't even get r for the last one. and did the h(x) bit COMPLETELY wrong so youre not the worst mate. Whats makes everything so much worse is people are saying it was easy so high grade boundaries
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    (Original post by Emmanuel12344)
    What were people's solution for the differential equation
    1/5[-2(1+t)^-1]^2-4+c


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    My answers:

    1a. A=3 B=-1
    1b. 8/3ln5

    2a. √29cos(x+1.19)
    2bi. x=5.09
    2bii. x=0.570, 5.72

    3a. d=1
    3b. g(x) = (2x+1)(2x-1)(2x+3)
    3c. h(x) = 1/(2x+3), h'(x) = -2(2x+3)^-2; always negative, therefore decreasing

    4a. 1 + x - 2x^2?
    4bi. 1/4 - 1/16x + 5/256x^2
    4bii. 0.2313

    5a. m=-1/2
    5b. y=2x-1/2
    5c. Qx=-1/8

    6a. ø=60 degrees
    6b. (15, y, z)
    6c. (11, 0, 0) and (23, y, z)

    7a. k=1/4 - ln2
    7b. m=11/8

    8a.
    8bi. (got this wrong)
    8bii. (will have got this wrong)
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    As CD223 cannot, I will attempt to make a Mark Scheme based on memory, will be online soon if I can!


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