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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by cjlh)
    My full solutions:

    1) 1/8 - 15/16 x + 75/16 x^2 - 625/32 x^3

    2) 0.6595, 1.083, 8/3 ln(2) - 7/9

    3) (-2-4xy)/(2x^2 + 4 + pisin(0.5pi)), (3pi+62)/(pi+22)

    4) x=e^(ln(60)-2.5t)), 633 mins

    5) (-5root(3)/16), (4,5root3)

    6) 3ln(3y+2)-2ln(y)+c, lambda=8, 4pi/3 - root3

    7) k=17/2, 1696pi/5

    8) {3,5,0}, r = {1,5,2} + lambda{-5,4,3}, k=2, cos(theta)=4/5, area=12/5, possible positions {3, 33/5, 16/5} and {-1, 17/5, 4/5}
    Fairly sure E is actually (3, 17/5, 4/5) or (-1, 33/5, 16/5) or otherwise it doesnt fit on the line L2.
    E = (1,5,2)+k(-5,4,3) so
    1-5k=3 --> k= -2/5
    5 + 4 x -2/5 = 17/5
    2 + 3 x -2/5 = 4/5

    You can do the same for the other value of E to work out if its on the line
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    (Original post by Vannzzoo)
    I GOT EXACTLY THE SAME IM SOO HAPPY, hope they're right tho
    Got all the same, except i got 634 minutes, or 10 hours 34 minutes
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    The exam went really well for except for the last 5 marker, could someone please explain to me how you were meant to do it as i literally spent about 25minutes on it and knew i had done one similar in the past. I was using the fact P was the midpoint and magnitudes were 2root2. But was proberly going off on a tangent
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    (Original post by Smithy1597)
    How was it not 1024pi/5
    thats volume for under graph, S would be volume of the rectangular part (becomes a cone) subtracted by the volume under the graph
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    (Original post by riffy123)
    For the volume question, how was it 544pi? It had to be 576pi as u had to work out the whole cylinder and take it away by the integral surely!
    It was 544π because radius is 8. 8.5 was the length
    π x 8 x 8 x 8.5 =544π

    Not rocket science

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    any unofficial mark scheme yet?
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    (Original post by Woodford_97)
    Fairly sure E is actually (3, 17/5, 4/5) or (-1, 33/5, 16/5) or otherwise it doesnt fit on the line L2.
    E = (1,5,2)+k(-5,4,3) so
    1-5k=3 --> k= -2/5
    5 + 4 x -2/5 = 17/5
    2 + 3 x -2/5 = 4/5

    You can do the same for the other value of E to work out if its on the line
    answers can't always be exactly right but you get the picture
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    Grade boundary predictions?
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    (Original post by Zuzuvela)
    It was 544π because radius is 8. 8.5 was the length
    π x 8 x 8 x 8.5 =544π

    Not rocket science

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    you could integrate it if you wanted to
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    (Original post by Yua)
    thats volume for under graph, S would be volume of the rectangular part (becomes a cone) subtracted by the volume under the graph
    It becomes a cylinder no?
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    Can someone explain how you were supposed to do the last vector question (8e)?
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    (Original post by ryanroks1)
    What did everyone get for the normal question where you had to find X? I remember 22 + pi being on the bottom?
    Is that the a,b,c and are integers, find their values question?
    I remember one of the bottom values was '1', I think it was the coefficient of pi.
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    Someone quote me when solutions are out please
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    (Original post by Zuzuvela)
    It was 544π because radius is 8. 8.5 was the length
    π x 8 x 8 x 8.5 =544π

    Not rocket science

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    Wasn't the height of cylinder 17/2 +1/2 = 9 so Volume = 576pi
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    (Original post by ombtom)
    I got that
    So did I 😊


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    E was definitely: (-1, 33/5 , 16/5 ) or (3, 17/5 , 4/5 )
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    (Original post by Smithy1597)
    How was it not 1024pi/5
    I got that too, but I think there was a volume between x=0 and x=1/2 which we didn't account for, this I would expect to lose 2 marks, 1 for accuracy one for method
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    (Original post by Yua)
    (becomes a cone)
    Wait do you mean a cylinder? Cos other than that I completely agree x
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    (Original post by Yua)
    thats volume for under graph, S would be volume of the rectangular part (becomes a cone) subtracted by the volume under the graph
    Rectangles become Cylinders, Triangles become Cones.
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    (Original post by ombtom)
    I got that
    (Original post by thehollowcrown)
    i think this is what i got
    Even better, hopefully should be 75/75 in this then maybe a few marks lost in some places because I have a tendency to occasionally be an idiot in exams
 
 
 
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