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    (Original post by SunDun111)
    What was the original equation of the curve can you remember?
    y= x^4 - 3x^2 -2

    I think?
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    I got -10 as well
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    (Original post by aoxa)
    You got -10 by using the dy/dx method.

    If you used the method of y-y/x-x to get the gradient, you got -7 or something. I don't think you were supposed to se that method though.
    You can't use y-y/x-x. That is for a straight line. The straight line may have intersected at A but that doesn't make the gradient of the curve the same at that very point. You are finding the tangent to the curve and therefore the gradient must be the same as the curve, not the line. That isn't the right method and you'd probably score only 1 mark for attempting to calculate the equation of a line.
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    Finished the markscheme. Thanks everyone!
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    I got -10 for that one but I know I dropped a definite 5 marks on the first 2 parts of question 6:'(
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    (Original post by Sekaib)
    Yeah I get 162/5 for the area behind the curve and the line.
    I got the curve as 108/5 and the trapezium at 54 but I split the trapezium into a triangle and a rectangle but I couldn't remember the formula lol.
    I got exactly the same! It came out as a negative number so i just flipped it and think i got the shaded region's area of 32.4
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    (Original post by -jordan-)
    You can't use y-y/x-x. That is for a straight line. The straight line may have intersected at A but that doesn't make the gradient of the curve the same. You are finding the tangent to the curve and therefore the gradient must be the same as the curve, not the line. That isn't the right method and you'd probably score only 1 mark for attempting to calculate the equation of a line.
    I did not use the y-y/x-x method - I was saying this is probably what the teacher used to get the answer of -7. To get this you used the y-y/x-x on the straight line AB, and the worked out the reciprocal of ABs gradient, this gives the perpendicular to AB. The problem is that this method does not necessarily give the tangent to the curve.
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    (Original post by brokenlevel23)
    Finished the markscheme. Thanks everyone!
    You just need to change bi to 5x ... I put 4x by accident in my original comment.
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    (Original post by student0042)
    You just need to change bi to 5x ... I put 4x by accident in my original comment.
    Oh yeah lol
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    if i couldn't factorise and got the critical values of the last inequality wrong, would i gain a mark for drawing a sign diagram and having p < x < q ? or do i lose all marks completely
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    (Original post by brokenlevel23)
    FINAL MARKSCHEME:

    Question 1
    (a) -3/5 [2]
    (b)(i)
    perp. gradient = 5/3
    y-?=5/3(x-?)
    5x-3y+1=0 [3]
    (b)(ii) A(9,-4) [3]

    Question 2
    7 + √15 [5]

    Question 3
    (a) y=-10x-4 [4]
    (b)(i) 108/5 [5]
    (b)(ii). 162/5 [3]

    Question 4
    (a) (x+1)^2 + (y-3)^2 = 50 [2]
    (b)(i). C(-1,3) [1]
    (b)(ii) 5√2 [2]
    (c)
    k^2-6x-16
    (k-8)(k+2)
    k=8, k=-2 [2]
    (d) min. distance = 7 [2]

    Question 5
    (a) p = 3/2, q = -¼ [3]
    (b)(i) (-3/2, -¼) [2]
    (b)(ii) line of symmetry is x=-3/2 [1]
    (c) y = x^2-x+4 [3]

    Question 6
    **IT WAS AN OPEN TOP CYLINDER DAMNIT**
    (a)(i) h = 24/r -r/2 [3]
    (a)(ii) Sub πr^2 into part (i) to get V = 24πr -π/2 r^3 [3]
    (b)(i) 24π -3π/2 r^2 [2]
    (b)(ii)
    r=4 when dy/dx=0
    r=4 => d^2y/dx^2(=-12π)<0, therefore maximum [4]

    Question 7
    (a) http://i.imgur.com/rVJ169A.png [3]
    (b)(i) R = 36 [2]
    (b)(ii) R= 0 therefore root [2]
    (b)(iii)(x-2)(x^2-5x+10) [2]
    (b)(iv) x=2 [3]

    Question 8
    (a) (Show) x^2 + 3(k-2)x -13-k=0 [1]
    (b) 9k^2-32k-16<0 [3]
    (c) -4/9< k<4 [4]
    Very nice, glad i could help - just put n = 7 for question 2 as this is what the question actually specified. It gave you + √15 in the question.

    Also, I just re did the question for 7biii - it is most definitely (x+2) and therefore 7biv is x=-2
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    (Original post by JackC3001)
    Very nice, glad i could help - just put n = 7 for question 2 as this is what the question actually specified. It gave you + √15 in the question.
    was it not x=-2 for the only root?
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    (Original post by brokenlevel23)
    FINAL MARKSCHEME:

    Question 7
    (a) http://i.imgur.com/rVJ169A.png [3]
    (b)(i) R = 36 [2]
    (b)(ii) R= 0 therefore root [2]
    (b)(iii)(x-2)(x^2-5x+10) [2]
    (b)(iv) x=2 [3]
    Pretty certain that the graph 7a went through the point (0,3)
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    (Original post by Excuse Me!)
    Pretty certain that the graph 7a went through the point (0,3)
    Edit: it does.
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    (Original post by Excuse Me!)
    Pretty certain that the graph 7a went through the point (0,3)
    It does...
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    Shouldn't the graph be going through the origin and dip down to 2 and up again without crossing the x axis again?
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    1 got 108/5 too then 162/5 or something for the last bit
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    would i have lost mark for not saying n=7
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    (Original post by Sabrina1996)
    would i have lost mark for not saying n=7
    As long as it was included in your answer and your answer was in the right format, you should be fine.
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    (Original post by Sabrina1996)
    would i have lost mark for not saying n=7
    Yes, it specifically said "stating the value for n" as a separate point.
 
 
 
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