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    (Original post by Garn197)
    isn't it reflection in y=x ?
    That is what I meant lol
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    (Original post by chizz1889)
    Yeah that's right 😊
    Think I got that. What was the radius equal to?
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    (Original post by chizz1889)
    Did everyone remember to subtract their answer to the integration from 3 for the area between y=1 and x axis? ... ☺️
    Completely forgot to do that, how many marks do you think that will have costed?
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    (Original post by dandav33)
    Rate one i got 0.023 or something like that


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    Oh that, i think it was 5/100pi = 0.0159
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    I've done some sort of a mark scheme here: http://www.thestudentroom.co.uk/show...332257&page=13

    I just found the question I missed, which was the chain rule one - my answer was 1/(20pi)
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    (Original post by chizz1889)
    That is what I meant lol
    XD i was worried there for a second haha
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    (Original post by bruce11)
    isnt it reflection in y=x?? andhow about inverse
    Yeah sorry I meant that, you had to make F(x)->x and the x->f-1(x) and then rearrange to give f-1(x) = ln((x+1)^0.5 +2) or something like that
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    (Original post by pierre225)
    yeah I put 0.016 to 3 s.f
    Thats 2 sig fig...
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    (Original post by Duskstar)
    All the functions that appeared in the paper, as follows:

    (I'll try and add questions as I remember them, though I have almost no idea of question numbers :P )

    \displaystyle f(x) = e^{2x} \cos x

    ~ differentiate and find the coordinate of the maximum (1.1, 4.1)

    \displaystyle \int \sqrt [3]{2x - 1} dx

    ~ Integrate for:
    \displaystyle \frac {3}{8} (2x - 1)^{\frac {4}{3}} + c


    \displaystyle \int^2 _1 x^{3}\ln x

    ~ Integration by parts for something I can't quite remember, but someone will bother to work out or ask about...


    \displaystyle y^2 + 2x \ln y = x^2

    ~ verify point (1, 1) which means sub it in and prove it still equates, then differentiate implicitly to get the gradient at that point, with was 1/2.


    \displaystyle f(x) = \frac {1 - x}{1 + x}

    ~ prove f(f(x)) = x, which involves subbing f(x) in in place of each x, then rearranging.
    ~ hence write down the inverse of f(x), which is f(x) (I personally don't understand why this was a 'hence' question - I haven't seen a rule that would describe this before, although it's obvious there is one)

    \displaystyle g(x) = \frac {1 - x^2}{1 + x^2}

    ~ show g(x) is even - sub (-x) for x, and show that is the same as g(x). This means that g(x) has a line of symmetry in the y-axis, or x = 0.


    \displaystyle f(x) = \frac {(x - 2)^2}{2}

    ~ you had to differentiate this twice. I found it easier to expand the top and divide through by x than use the quotient rule, but either should work and I think the answer was f'(x) = 1 - 4x^(-2), then differentiate again for f''(x) = 8x^-3 you should fin the point Q(-2, -8), then show it is a maximum because f''(x) < 0

    ~ I think this was the question where you had to integrate to find the area enclosed by the function and the line y = 1 (after proving the intersections were x = 1 and 4), which was easy enough - either use a rectangle, or take the integral of 1 - f(x). I think my answer was 15/2 - 4ln4

    translate (-1, -1) for:

    \displaystyle g(x) = \frac {x^2 - 3x}{x + 1}

    ~ you had to show this, which just meant showing f(x + 1) - 1 = g(x).
    ~ then, it asked you to write down the integral from 0 to 3 of g(x) with no further calculation, which is the integral from the second part (I don't know if they would want you to indicate that it was this value but negative), and then describe why this was the case. I guess you just talk about the translation and stuff ~

    \displaystyle f(x) = (e^x - 2)^2 - 1

    ~ I can't remember quite what the questions were here, but I think you had to differentiate it, find x-intercept at ln3, and minimum at (ln2, -1)

    I got for the inverse:
    \displaystyle f^{-1}(x) = \ln {( \sqrt {x + 1} + 2)}

    ~ then sketch the inverse by reflecting in y = x. The easy way to do this was plot the images of Q and P, and make sure you cross the line y=x in the same place, and continue the line without changing gradient much (it was pretty much a flat line past the intersection with y=x).


    and somewhere in the paper, two parts to a question:

    \displaystyle 6 \sin ^{-1} x - \pi = 0 , x = \frac {\pi}{6}
    \displaystyle  \sin ^{-1} x = \cos ^{-1} x ,x = \frac {1}{\sqrt 2}
    Domain was -1<= x and range was ln2 <= f-1(x)
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    (Original post by StudentH817)
    Thats 2 sig fig...
    no it isn't
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    (Original post by JamesExtra)
    Think I got that. What was the radius equal to?
    radius was equal to h
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    I got really flustered after the first question and reckon I dropped 20 marks at least despite getting A*s in past papers. I want to die.
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    How does your c3 coursework affect your grade.
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    (Original post by hukdealz)
    no it isn't
    .016 is three decimal places but 2 sig fig as the 0 after the decimal place is not significant. Just as the 160 would be 2 sig fig not three.

    Not that it really mattera that much.
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    (Original post by Garn197)
    radius was equal to h
    Ok now I'm feeling very confident I did well on this. Thanks
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    (Original post by StudentH817)
    .016 is three decimal places but 2 sig fig as the 0 after the decimal place is not significant. Just as the 160 would be 2 sig fig not three.

    Not that it really mattera that much.
    Hallelujah !!!
 
 
 
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