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    I couldn't find a thread for this so I figured I'd make you guys one.
    Good Luck with your exams (If you're reading this before the exam starts.)- I'm sure it went well.
    To the best of my knowledge there aren't any rules regarding discussion of OCR Exams so talk away
    If you want to make an “unofficial mark scheme” , just quote me and I’ll set up links in the OP for you all to have a look at.

    MrM's C1 13/05/13 OCR (Not MEI) Solutions

    Just some people's thoughts: - If you can remember a question, post it and I'll answer.
    Q1:
    Spoiler:
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    (Original post by Zakee)
    Just to say, for the first question you can just do:

     (4\sqrt{15})(\sqrt{3})



= 4\sqrt{45} = 4\sqrt{(9)(5)}



= (4x3)\sqrt{5}



= 12\sqrt{5}


    Q10:
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    (Original post by Big-Daddy)
    Cheers to joostan for making us this thread even though he didn't take the exam!

    I remember all sorts of little answers but not the questions.

    10.i.) k=-5
    ii.) minimum (d2y/dx2=12>0)
    iii) (-2,-27)

    To people who got 2 coordinates for the last question - unfortunately, it's wrong. One of those coordinates has x=0, right? Put x=0 into the curve, you get y=-5. Put x=0 into y=9x-9 and you get -9. I checked, because the question wrote "coordinate" not "coordinates". And realistically it will only be tangent at one point.

    The discriminant question was 9^2-4*3*10=-39. The coordinates on the complete-the-square question were (-3/2,13/4). The two transformations were: f(x+5) (sorry, can't remember the exact question) and stretch by Scale Factor 1/2 parallel to the y-axis.

    One of the quadratics was solved to x=-1, 1/2 (after rooting x^3=-1,(1/8) - I hope I didn't stupidly write down something different from what was in my head though!) The "decreasing function" question - set f'(x)<0 and solve the linear inequality. Forgotten the exact question.

    If anyone else remembers any of the questions I'd be happy to provide my answers.


    The whole lot . . .
    Spoiler:
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    An Unofficial Mark Scheme.
    Congratulations to majeue - it's a couple of posts up from here. . . - I'll see if I can improve the link


    Paper's part way down page 14.
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    (Original post by Pras123)
    How did you find it?
    Its not been taken yet . . .
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    What did people get for the last question? For the point A?
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    Community Assistant
    I am in a meeting until 6 p.m. but will post the answers once I get home.
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    (Original post by Han456)
    What did people get for the last question? For the point A?
    (-2, -27) here
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    (Original post by Bohla)
    (-2, -27) here
    SNAP
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    (Original post by Han456)
    What did people get for the last question? For the point A?
    (1,0) what about you ?
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    (Original post by Han456)
    What did people get for the last question? For the point A?
    I got two points, (-9,0) and (-2,-27).
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    Not seen it anywhere, how did you guys find it??
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    Got two points for A, was i supposed to pick one of the two or?
    Had k=-5
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    What was the question again I just recognise the (-2,-27) but don't remember the actual question
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    (Original post by eggfriedrice)
    I got two points, (-9,0) and (-2,-27).
    but (-9,0) did not fit into the first equation so i thought only (-2,-27) was the answer
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    ffs was the factored version equal to -4. I had plus 4
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    (Original post by nicked95)
    but (-9,0) did not fit into the first equation so i thought only (-2,-27) was the answer
    Thats exactly what my logic was because its a tangent to the curve at (-2,-27) yes at the other point the gradient was the same but it did not intersect the curve so my belief is its just the (-2,-27) point
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    What value did people get for k? And how?
    I expanded the brackets, then differentiated it and said when dy/dx = 0 its a stationary point, then subbed in x=-3 and found k to be 49. I know that's wrong but would I get any method marks? Thanks
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    (Original post by Z.M.1)
    What value did people get for k? And how?
    I expanded the brackets, then differentiated it and said when dy/dx = 0 its a stationary point, then subbed in x=-3 and found k to be 49. I know that's wrong but would I get any method marks? Thanks
    That sounds very similar indeed to my method but I got -5 I think that there was probably an issue with your expansion that just threw it off cant promise anything but pretty sure you will pick up some marks !
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    (Original post by Z.M.1)
    What value did people get for k? And how?
    I expanded the brackets, then differentiated it and said when dy/dx = 0 its a stationary point, then subbed in x=-3 and found k to be 49. I know that's wrong but would I get any method marks? Thanks
    Exactly the right method i think. just a wrong calculation somewhere so you should get method marks.
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    (Original post by nicked95)
    but (-9,0) did not fit into the first equation so i thought only (-2,-27) was the answer
    Ah I did not think of that. you're probably right.
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    (Original post by Z.M.1)
    What value did people get for k? And how?
    I expanded the brackets, then differentiated it and said when dy/dx = 0 its a stationary point, then subbed in x=-3 and found k to be 49. I know that's wrong but would I get any method marks? Thanks
    I did the same... But got k=+5
    I then got (1,0) as my coordinate.
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    If I recall, you got a nice quadratic in that question, by dividing by 1-x then rearranging into (x+2)^2 = 0

    I got -2 -27 as well, since it was meant to be tangent I'm pretty sure it was meant to only have one point.

    I think there are two methods for finding k, which i got as -5.
    You can either use the formula and get 18 = 6+-root84-12k, which gets you k as -5, or sub it into the dy / dx equation with x as -3.
 
 
 
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