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    (Original post by eggfriedrice)
    I agree. The paper wasn't particularly hard. );
    Do you know the highest grade boundary we ever had?
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    (Original post by Dugald)
    Dy/dx=-3x^2+6x+4-k
    k=-5 so dy/dx=3x^2+6x+9

    3x^2+6x+9=9
    3x^2+6x=0
    3x(x+2)=0
    x=0 or x=-2

    I did this wrong because instead of doing 4-(-5) while differentiating I did 4-5 so go -1 on the end instead of 9 which gave me a quadratic with no real routes. So I couldn't attempt to find x or y.

    How many marks will I lose?

    How
    First you say the dy/dx = -3x^2 ect.
    Then you say dy/dx = 3x^2 ect.

    I dont understand where the Negative has come from. Was the original equation; (x-1)(x^2+4x+k)?
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    (Original post by Rrobba)
    I put the values of x and y into the original equation to see if they satisfied it. Putting x = -2 into the equation gave y = -27. The other value for x did not satisfy the equation and gave a different value for y.
    How did you know it didn't satisfy the equation?


    (Original post by BankOfPigs)
    Not x-1.

    you divide by 1-x

    I think you get something like:

    y = 9x -9 = -9(1-x)
    y = (1-x)(x^2+4x+k)

    So -9 = x^2+4x+k
    Since k = -5

    0 = x^2+4x+4 or (x+2)^2, which shows the repeated root.

    At least thats how i did it.
    Aha ok, I did a different method xD
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    (Original post by eggfriedrice)
    Yup, I got 9/2 too.
    ****, left it as root 81/4. Will i lose the answer mark?

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    (Original post by Cowcat)
    Do you even get 'error carried forward' marks in A level?
    Yeah you do for some questions, I've seen it in the past paper mark schemes a few times.
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    (Original post by flyhigh99)
    Do you know the highest grade boundary we ever had?
    off the top of my head 61
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    (Original post by eggfriedrice)
    Ah I did not think of that. you're probably right.
    I did the same, and put down both we will lose 1-2 marks maximum. Surely wait for mr m for a mark scheme?


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    (Original post by hannah1258)
    i did something very similar. Instead of making it equal to -3 i did dy/dx and then put in -3 for x and got some wierd answer which then made the part 3 completly wrong
    will there be method marks for this?
    Someone please put my mind at ease out of 12 (have a guess) how many method marks would I get???
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    (Original post by gibjb)
    for the last part i know you had to equate the dy/dx of the curve to 9 and solve to get x. i did this but got 2 x coordinates so i didnt know which one to choose and stopped there. how many marks out of the 5 do you think i would have gotten?? and could anyone explain what you had to do after that step?
    You had to say x could not equal zero because the gradient of 9x-9 is positive and at x the gradient was negative. I got (-2,-27) or something similar...
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    (Original post by flyhigh99)
    Do you know the highest grade boundary we ever had?
    No :s last year was 61 for an A I think.


    (Original post by cheetahs56)
    ****, left it as root 81/4. Will i lose the answer mark?

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    Aha, I did that and realised root a square would just cancel out. Hmm I'm not sure, at most you'll lose one mark.
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    (Original post by princess_parisa_22)
    can we compile a question paper together on here if we're allowed? these answers will be pretty useless unless everyone remembers what they wrote, and i'm sure not everyone will =/ i have a bad memory
    If you wait until about 6:30-7 Mr.M will be posting up the answers
    However, if anyone has the paper could they scan it and post it?
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    (Original post by Ellie_Rosa)
    Which minimum point?
    the one where you had to complete the square and give the coordinates of the minimum point/vertex
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    (Original post by gibjb)
    for the last part i know you had to equate the dy/dx of the curve to 9 and solve to get x. i did this but got 2 x coordinates so i didnt know which one to choose and stopped there. how many marks out of the 5 do you think i would have gotten?? and could anyone explain what you had to do after that step?


    I got the x co-ordinates, then used them to find out the y co-ordinates. Then to find out with co-ordinates are the correct ones I put them back into the original curve equation, and the ones that equalled 0 showed it was a tangent at that point, I think that is right? So I would guess you would only drop 1 mark, 2 tops.
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    (Original post by Goods)
    I did the same, and put down both we will lose 1-2 marks maximum. Surely wait for mr m for a mark scheme?


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    Hopefully it's just one mark, I was hoping for 100% >.> darn.
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    (Original post by Pataiiii)
    For the circles question, was the centre - (0,-4) or (-4,0)???
    (0,-4)
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    (Original post by Pataiiii)
    For the circles question, was the centre - (0,-4) or (-4,0)???
    i got (0,-4)
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    (Original post by Whyllee)
    Someone please put my mind at ease out of 12 (have a guess) how many method marks would I get???
    I would guess 6-7 provided your method was correct bar the arithmetical error


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    (Original post by gibjb)
    for the last part i know you had to equate the dy/dx of the curve to 9 and solve to get x. i did this but got 2 x coordinates so i didnt know which one to choose and stopped there. how many marks out of the 5 do you think i would have gotten?? and could anyone explain what you had to do after that step?
    I got 2 x values and then substituted them into the equation of the curve to get two corresponding y values. Then I subbed these values into y=9x - 9 and only x=-2 and y=-27 worked. My other values were x=0 and y=-5 which don't work in y=9x-9
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    Guys, for the inequalities question
    3-8x>4
    Did you get x<-1/8 or x>-1/8
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    (Original post by Goods)
    I would guess 6-7 provided your method was correct bar the arithmetical error


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    Thank you I know it's just a guess but thanks *sigh of relief*
 
 
 
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