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    (Original post by decombatwombat)
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    Hmm but isn't it rotated 180 degrees instead of 360?
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    (Original post by prepdream)
    Hmm but isn't it rotated 180 degrees instead of 360?
    There was a question identical to this last year where it said 180, but you still only multiply by pi.
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    thought that was quite a tricky paper. Predictions for grade boundaries?
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    I found section B much easier than Section A

    The most difficult question for me was Q5 but I can't remember what the question was
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    (Original post by aelahi23)
    I found section B much easier than Section A

    The most difficult question for me was Q5 but I can't remember what the question was
    The stacked triangles i think.
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    (Original post by prepdream)
    Hmm but isn't it rotated 180 degrees instead of 360?
    Yes, but usually the curve is only in one quadrant.

    Today, since it was in 2 quadrants, rotating it 180 degrees is the same as rotating the graph of it in only 1 quadrant by 360 degrees
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    The vol of revolution was both 8pi/3 an 16pi/3 they accept both but pretty sure mainstream is multiply by pi/2
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    Paper seemed tough but very doable. Hoping I still got full ums. Should be lower grade boundaries in my opinion

    If anyone has any questions about the exam today I'll be sure to help - a lot of my friends asked more pre-exam questions to me than usual. Was a tough paper.
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    (Original post by betapro)
    The vol of revolution was both 8pi/3 an 16pi/3 they accept both but pretty sure mainstream is multiply by pi/2
    http://vle.woodhouse.ac.uk/topicdocs.../C42015May.pdf

    The MS for last years paper only allows pi, and the situation is pretty much the same.
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    didn't get the second part of question 5, other than that amazing paper
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    (Original post by decombatwombat)
    The stacked triangles i think.
    Oh right yeah, I did part 5i) easily but 5ii) I didn't right the ratio because i ran out of time

    I got something like for the length of the triangle on top 2xSin(theta)
    and the side length of the bottom triangle to be xTan(theta)

    Which would have the ratio of 2Cos(theta) I think
    tell me if i am wrong
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    exaxctly what i did ^
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    Can someone make an unofficial markscheme 😅
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    I want C4 to get done in the butt by something roughly godzilla sized.
    The same butt that was clearly responsible for producing the questions on that paper.
    Thanks, OCR. Goodbye, university.
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    (Original post by decombatwombat)
    http://vle.woodhouse.ac.uk/topicdocs.../C42015May.pdf

    The MS for last years paper only allows pi, and the situation is pretty much the same.
    Alright I see. They allow pi/2 but only accept pi for the final result, so reckon I would lose 1-2 marks prolly
    Overall, was a good paper!
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    (Original post by ComputerMaths97)
    Paper seemed tough but very doable. Hoping I still got full ums. Should be lower grade boundaries in my opinion

    If anyone has any questions about the exam today I'll be sure to help - a lot of my friends asked more pre-exam questions to me than usual. Was a tough paper.
    what do you think the grade boundary will be for an A this year?
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    Horrible paper. Somebody please explain question 6 the parametric equations one
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    (Original post by Crozzer24)
    Horrible paper. Somebody please explain question 6 the parametric equations one
    Find dy/dx

    Do y -y1 = m(x-x1) to find the general equation of a tangent using x =2t y =1/t

    Find the x and y intercepts by subbing in 0 for x and y

    Then the Area = 1/2 x OQ X OR = 8 which is independent of t
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    (Original post by Crozzer24)
    Horrible paper. Somebody please explain question 6 the parametric equations one
    Basically turn it into an equation and then differentiate it. Then find the equation of the tangent. Sub x=0 and y=0 to find the side of the triange. When you try to calculate the area afterwards, the result should be a number without t
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    i used chain rule but got the same, ended up with 8t/t so t's cancel
 
 
 
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