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    (Original post by the_googly)
    Anybody got a copy of the paper?
    LOL! You aint gonna see one of those for a loooong time
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    If you get the correct answer, but don't show all the working, will you still get the method marks (full marks basically)?
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    (Original post by chriselle15)
    If you get the correct answer, but don't show all the working, will you still get the method marks (full marks basically)?
    Hopefully mate, fingers crossed, youll deffo get some though
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    guys look at my post, ive posted the full unofficial markscheme
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    (Original post by adnan12345)
    guys look at my post, ive posted the full unofficial markscheme
    were I cant see
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    http://www.thestudentroom.co.uk/show....php?t=2228423
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    complete markscheme :-) , this was by my knowledge, i may have made mistakes here and there but i am pretty confident in my answers lol :-)
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    1a) Sub in the x value of 7 from (7,k). This proves that k is indeed -4.
    1b) -10/6 or -5/3
    1c) Can't remember exactly, think I got -5x+3y-21=0
    1d) think I got either (12,-7) or (7,-12)

    2a) -1.5
    2b) 0>-1.5 so the height is decreasing
    2c) Ended up as 6-2, 4 for second derivative

    3a) root 18 = three root 2
    3b) 2/7 i.e. 2 root 2 over 7 root 2 (this is the quick method rather than the convoluted method to get the same answer)
    3c) I got 5-root6, but I don't believe it's right.

    4a) (x-3)^2+2
    4b) 2>0, always above x-axis - no solutions.
    4bii) Vertex = (3,2)

    Sketch crosses y at +11, vertex at (3,2)

    Transformation:

    [3]
    [2]


    5a) Remainder is 16 i.e. -5+21
    p(3)=0

    p(x)=(x-3)(x-3)(x+2)

    Curve crosses x at -2 and kisses x-axis without crossing it at +3

    6)a Gradient = 9
    y-4=9(x-1)
    y=9x-5
    6b) Integrate it - 2x^5-2x^3+5x+c
    Forgot to actually find c. Whoops.


    7) A circle with C(-3,2) has equation x^2+y^2+6x-4y=12
    a). x=0 and y=-2, y=6
    b). R=5.
    Pythagoras - root 34
    c) Got wrong answer for this, I am sure. Real answer was apparently root 9 i.e. 3 from root (34-25)=3

    8a(ii) show that 4k^2-20k+9>0
    b = (2k+1) a = 2 c = (3k+1)
    (2k+1)^2 - 4*2*(3k+1) simplifies to 4k^2-20k+9 which factorises to...
    (2k-1)(2k-9)>0

    1/2, 9/2 are critical values

    k<1/2 and k>9/2
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    (Original post by adnan12345)
    complete markscheme :-) , this was by my knowledge, i may have made mistakes here and there but i am pretty confident in my answers lol :-)
    Same, only one I appear to got wrong was the pythagoras bit on circle I did 34 - 5 instead of 34 - 25 and for 1A(iii)shouldn't it be -5x+3y-21=0 ? in the form of px+by+c did u get this as u recall
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    For 1a(iii) I did :
    y-2=5/3(x+3)
    3y-6=5x+15

    Carried over to the left:

    -5x+3y-21=0
    Not confident in this, but thats what i did.
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    (Original post by Fillly)
    For 1a(iii) I did :
    y-2=5/3(x+3)
    3y-6=5x+15

    Carried over to the left:

    -5x+3y-21=0
    Not confident in this, but thats what i did.
    You got it right, but I think they wanted it as 5x-3y=21. You should get some credit though, it's basically the same.
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    (Original post by Anon217)
    You got it right, but I think they wanted it as 5x-3y=21. You should get some credit though...
    Ye but when it says in the form of px+by+c=0 sometimes they can trick u and u can put a negative right ? just like in surds sometimes.
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    (Original post by dannyb2012)
    Ye but when it says in the form of px+by+c=0 sometimes they can trick u and u can put a negative right ? just like in surds sometimes.
    I guess so yeah. I though they wanted it as px+by=c ..? I might be wrong though or thinking of another question lol.
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    (Original post by Anon217)
    I guess so yeah. I though they wanted it as px+by=c ..? I might be wrong though or thinking of another question lol.
    Lol, it just didn't work out for me so I thought that was only solution, its annoying that we will never know until next year .
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    I thought it was px+by+c=0 Well, whatever it was, my version is basically x-1 of the other version. Nothing i can do now :
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    (Original post by MissGeek)
    I have a C1 exam on 14th January. I have been doing a bit of revision everyday of the Christmas holiday. I've been answering questions from my textbook and going over the past papers, and focusing on the bits where I tend to lose marks on.

    Is that enough or is there anything more which I should do if want to ensure that I get a comfortable high A for C1?
    I seriously don't understand why you've been negged for this? Maybe people have not done enough preparation like you. Anyway, good luck and I'm aiming for a solid A as well. XD
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    (Original post by Fillly)
    I thought it was px+by+c=0 Well, whatever it was, my version is basically x-1 of the other version. Nothing i can do now :
    Lol what if your username started with the letter W instead of the letter F.
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    (Original post by Fillly)
    I thought it was px+by+c=0 Well, whatever it was, my version is basically x-1 of the other version. Nothing i can do now :
    You might be right though, I think I'm confusing it with another question (since more than one person disagrees).
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    (Original post by Enginerd.)
    I seriously don't understand why you've been negged for this? Maybe people have not done enough preparation like you. Anyway, good luck and I'm aiming for a solid A as well. XD
    Look at post #238. I really want to know why I got negged for that post.
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    (Original post by Ferrari_1996)
    Look at post #238. I really want to know why I got negged for that post.
    Hi mate, I started it I missclicked and put negative even though its correct and when you do everyone follows lmao. yeah sorry about that. your right by the way it should be 4x^2
 
 
 
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