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    Does Mr M have the answers?
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    Yeah got 9/2 for distance of PQ
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    (Original post by Dugald)
    1i 12route5
    1iii 5route5
    (-3/2,17/4) was the minimum point
    got 13/4 as my y co-ordinate. can anyone confirm that the quadratic was 3x^2 + 9x + 10? my memory is terrible at the moment
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    (Original post by jt663)
    What did people get for the length PQ? on y=4 on the x^2 graph? I got 11/2, I heard it might be 9/2 though
    I got 9/2 for that question but could be wrong.

    I got k = -5
    Coordinates for last question (-2, -27)
    x < 1/4 for another question (can't remember which one)

    Anyone else get those?

    (Original post by Bohla)
    got 13/4 as my y co-ordinate. can anyone confirm that the quadratic was 3x^2 + 9x + 10? my memory is terrible at the moment
    Yea I think it was that. I got 13/4 for the y-coordinate too.
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    (Original post by Dugald)
    1i 12route5
    1iii 5route5
    (-3/2,17/4) was the minimum point
    Which minimum point?
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    (Original post by Bohla)
    got 13/4 as my y co-ordinate. can anyone confirm that the quadratic was 3x^2 + 9x + 10? my memory is terrible at the moment

    Got the same as you and that quadratic sounds right.
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    Oh ****ity ****.
    I drew a graph, so it shows the length is between -2 and 5/2. I then decided my length would be 15/2 because **** you thats why.
    ERGH.
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    Allot of people said it was minus -5 for the last, I would have got that but I made a stupid mistake where I forgot to differenitate 3x^2 and left it as 3x so I got k as -14 or something which cost me I hate silly mistakes and I found out a min before the end so I couldn't change the whole question :/// I understood all the other parts to the question and would have got it correct if my k value were correct, how many marks do you think I would loose?
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    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
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    (Original post by Rrobba)
    I got 9/2 for that question but could be wrong.

    I got k = -5
    Coordinates for last question (-2, -27)
    x < 1/4 for another question (can't remember which one)

    Anyone else get those?



    Yea I think it was that. I got 13/4 for the y-coordinate too.
    Got the same as you but I got two values of A. How did you figure out which value of A it was?
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    Ermm I got -45 for k as my last answer, I don't know where I went wrong... I expanded, differenciated and then put it equal to 0 and subbed in -3 for x.. Then because I got this wrong the last question wouldn't work but I tried simultaneous equations / dy/dx do you think I'll get many method marks?? I'm aiming for an A and I can't afford to lose 12 marks.... ((((
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    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?
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    (Original post by Bohla)
    got 13/4 as my y co-ordinate. can anyone confirm that the quadratic was 3x^2 + 9x + 10? my memory is terrible at the moment
    I got 13/4 as my y-coordinate too
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    For the circles question, was the centre - (0,-4) or (-4,0)???
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    (Original post by eggfriedrice)
    Got the same as you but I got two values of A. How did you figure out which value of A it was?
    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.
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    (Original post by eggfriedrice)
    Why would you divide by x-1? 0:

    I got 3x^2 +6x=0 so x(x+2)^2 =0
    So x=0 or x=-2
    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.
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    (Original post by olivers16)
    Does Mr M have the answers?
    He said he had a meeting until 6 but would post them afterwards
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    (Original post by squeakypop)
    He said he had a meeting until 6 but would post them afterwards
    I sent him a PM now saying if the answers aren't up by 6 then i'll hunt him down

    Posted from TSR Mobile
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    Do you even get 'error carried forward' marks in A level?
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    (Original post by Whyllee)
    Ermm I got -45 for k as my last answer, I don't know where I went wrong... I expanded, differenciated and then put it equal to 0 and subbed in -3 for x.. Then because I got this wrong the last question wouldn't work but I tried simultaneous equations / dy/dx do you think I'll get many method marks?? I'm aiming for an A and I can't afford to lose 12 marks.... ((((
    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?
 
 
 
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