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OCR (MEI) Mathematics C3 20/01/2010 watch

  • View Poll Results: How did you find the paper?
    Easy
    13
    19.12%
    OK-ish
    28
    41.18%
    Hard
    25
    36.76%
    I'm not sure/ i didnt take the exam/ ?
    2
    2.94%

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    O and P and the second half of 6 was beautiful!!!
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    (Original post by kennykwan)
    Was there a meant to be an insert for question 9 for the sketch the graph question?
    No just sketch it onto your answer paper.
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    (Original post by kennykwan)
    Was there a meant to be an insert for question 9 for the sketch the graph question?
    I don't think so, I just drew the graph out again with f(x) and the inverse on it, with labels etc :yep:
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    f(-x) = -f(x). so, g(-f(x)) = g(x). I think. ><
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    yeah it was a simple sketch graph drawing.
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    (Original post by JumpingJonny)
    I don't think so, I just drew the graph out again with f(x) and the inverse on it, with labels etc :yep:
    On graph paper?
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    (Original post by kennykwan)
    On graph paper?

    no just on the answer sheet paper. graph paper isnt needed for a sketch


    how did everyone else find the rest of q9?
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    (Original post by kennykwan)
    On graph paper?
    No I just did it in my answer booklet as a rough(ish) sketch, aslong as you have shown the curve and where it cuts the axis you should be fine :yep:
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    hmm nearly every1 found it SOLID, but i thought it was ok. i got k = 0.19 sumin. the P cord was pi/6 n Q was pi/2. any1 kno wat they got as the gradient
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    (Original post by theycallmesarah)
    can anyone tell me how they answered the f(x) g(x) question where you had to prove gf(x) was even?
    thankyou

    p.s don't worry about it too much marks2, i mean it wasn't like u were the only one who found it hard. there were loads of people who didnt like it
    It was definatly harder than any of the past papers but I managed it pretty well I think, had about 20-30 mins to check my answers at the end .

    For the f(x) and g(x) question I did the following:

    For f(x) to be odd, f(-x) = -f(x)
    For g(x) to be even, g(-x) = g(x)

    then for g[f(x)] to be even, g[-f(x)] = g[f(x)]

    Since g(-x) = g(x) , g[-f(x)] = g[f(x)] .

    I think that was it :p:
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    (Original post by waris_ahmed)
    hmm nearly every1 found it SOLID, but i thought it was ok. i got k = 0.19 sumin. the P cord was pi/6 n Q was pi/2. any1 kno wat they got as the gradient

    i think i got the gradient as - pi/2 at P. cant remember if that was the correct question though lol
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    (Original post by josh_a_y)
    Yeah! you equate the first line

    And you should end up with x^2= 1/3

    So x= 1/sqrt(1/3) or sqrt(3)/3 depending on what calculator you got


    Well done, soundsl ike you got it right!
    you then have to sub it into f'(x) dont ya?
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    For the gradient I ended up with f'(sqrt(1/3)) = 1.95... something I think
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    (Original post by MarkS2)
    For the gradient I ended up with f'(sqrt(1/3)) = 1.95... something I think
    Me too! :p:
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    Inverse functions a reflection in Y=X, all i did is use the questions as tracing paper to draw xD
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    I did the how f(x) is odd and g(x) is even. And explained why gf(x) is even but only in words, not algebraically. Any idea if that'll be included in the mark scheme?
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    (Original post by MarkS2)
    For the gradient I ended up with f'(sqrt(1/3)) = 1.95... something I think
    yeah thats what i got for the value of f'(x), something like 1.949.
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    (Original post by Rich-1991)
    i think i got the gradient as - pi/2 at P. cant remember if that was the correct question though lol
    i got this too
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    What did you guys get for P and Q.
    P( pie/6, 0)
    Q( pie/2, 0)
    anyone agree?
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    I thought it was oook I guess. I did crap on my c/w (11.5/18) and i think I have dropped around 11 marks, so taking into account mistakes made, I hope I havn't dropped more than 15 cos I want a solid +75% overall cos I'll still be on a high A.
    If everyone else has found it hard though I might be able to make an A cos of boundries. But AHHH WELLL now for Chem4 :woo:
 
 
 
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