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OCR MEI C3 Jan 2012 Unofficial Mark Scheme

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    (Original post by just george)
    If you can remember the questions il do the workings/answers for you.. but i cant remember them myself :L only that for one of them the area was 30.67 (i think that was last part of q8) :L

    Yeah that was the area of PQR I think - don't remember the rest of the question though
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    I remember some questions: 8) y = x/(x-2) (I think the numerator is wrong but somebody can correct me on what it actually was) then using the substitution of u = x-2 show that the integral between x=3 and x=11 is 25 and a third.

    9) y=ln(2x/x-2) (think the denominator is wrong here but again somebody can correct me) Find the gradient at the point where x=3 [Hint using the rule of logs that ln(a/b) = ln(a) - ln(b)] which works out to be -0.5 I believe.

    Hence using this result (the gradient you just worked out when x=3) show that the curve is NOT symetrical about the line y=x

    The last question was to do with finding the integral of ln(2x/x-1) or whatever the denominator was using a substitution to show the area was ln(3/2) and using this result show that the area of the shaded region of the curve was something like ln(27/35)????? Hope this jolts someone's memory about some of the questions cause I'm sure people can correct me
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    thanks, for 8 the denominator was sqrt(x-2)

    8) show that the integral (x / sqrt(x-2)) between x=3 and x=11 equals 25 1/3

    let u = x-2
    du/dx = 1

    New limits: when x=11, u=9, when x=3, u=1

    I = integral(x / u^0.5) between 1 and 9

    = integral [(u+2) / (u^0.5)] between 1 and 9

    = integral(u^0.5) + 2*integral(u^-0.5)

    =(2u^(3/2))/3 + 4u^0.5 between u=1 and u=9

    =(18 + 12) - (2/3 + 4)

    =30 - 4 2/3

    =25 1/3 as required


    EDIT:
    And the next part of q8



    find the area enclosed by the curve, the line y=x and the line x=11

    For this, you find the area of the trapezium between x=3 and x=11 (the area under y=x)

    =(0.5*8*8)+(3*8)
    =56

    you then take away the area under the curve between x=3 and x=11 which you have just shown to be 25 1/3

    56 - 25 1/3 = 30 2/3
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    EDIT: This post is incorrect, the equation is wrong and the part stating the answer should be -0.5 is us getting mixed up between q8 and q9

    (Original post by Jack0234)

    9) y=ln(2x/x-2) (think the denominator is wrong here but again somebody can correct me) Find the gradient at the point where x=3 [Hint using the rule of logs that ln(a/b) = ln(a) - ln(b)] which works out to be -0.5 I believe.

    Hence using this result (the gradient you just worked out when x=3) show that the curve is NOT symetrical about the line y=x
    y = ln(2x/x-2)

    =ln|2x| - ln|x-2|

    dy/dx = 1/x - 1/(x-2)

    when x=3, dy/dx = 1/3 - 1

    not sure what the denominator was supposed to be here, but your right the answer should come to -0.5


    Edit: you then go on to say that to be symmetrical across y=x the gradient must be the negative reciprocal of the gradient of y=x, i.e. -1

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    (Original post by just george)
    x
    Done, thanks. That was some hard LaTeX'ing!
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    Ah there we go, searching back through the original C3 thread got the right equations for 9..

    y=ln|2x/x+1|

    Here we've mixed up question 8 and question 9. the one where dy/dx = -0.5 and is therefore not symmetrical across y=x is the question 8 curve.

    On this question we had to show that the 2nd line on the graph (there was a graph with 2 lines on) was the inverse of the line labelled y = ln|2x/x+1|. The line was labelled as y = (e^x)/(2 - e^x) so it was a show that question.

    y = ln|2x/x+1|

    inverse: x = ln|2y/y+1|

    e^x = 2y/(y+1)

    (y+1)e^x = 2y

    ye^x + e^x = 2y

    e^x = 2y - ye^x

    e^x = y(2 - e^x)

    (e^x)/(2 - e^x) = y as required




    Another part of the question was showing that the integral of (e^x)/(2 - e^x) between x=0 and x=ln|4/3| is equal to ln|3/2|

    let u = (2 - e^x)
    du/dx = -e^x

    new limits: when x=ln|4/3|, u = 2/3, when x=0, u=1

    I = - integral(1/u) du between u=1 and u=2/3

    = [-lnu] between 1 and 2/3

    = -ln(2/3) - -ln(1)

    =-ln|2/3|

    =ln|3/2| as required



    Then finally it asked you to find the shaded area on the graph (under y=ln|2x/x+1|) and show that it is equal to ln|32/27| (between the point the graph crosses the x axis and x=2) = between x=1 and x=2

    To do this the easiest way is that because the other line is the inverse of the first, you can find the area of the equivalent part of the curve y = (e^x)/(2-e^x), which is between x=0 and x=ln|4/3|.

    Therefore the area of the shaded region is the area of the rectangle enclosed by y=0, y=2, x=0 and x=ln|4/3| and then taking away the area that you have worked out in the previous question

    =2ln|4/3| - ln|3/2|

    =ln|16/9| - ln|3/2|

    =ln16 - ln9 - ln3 + ln2

    =ln|(16*2)/(9*3)|

    =ln|32/27| as required


    now time for some dinner :L
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    so i guess the first part of question 8 was find the gradient of the curve at P? (x=3). Hence show that the line y = x/(x-2)^0.5 is not symmetrical across the line y=x.

    y = x/(x-2)^0.5

    = x(x-2)^-0.5

    dy/dx = [-0.5x(x-2)^-3/2] + (x-2)^-0.5

    = [(x-2)^-3/2][-0.5x + (x-2)]

    = [(x-2)^-3/2][0.5x - 2]

    when x = 3

    = 1*(1.5-2)

    = -0.5

    and then the bit about negative reciprocals that i wrote above
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    I believe there was also a question on finding the gradient at P and R (which were reciprocals). Can't quite remember what is was, hope this jogs someones memory.
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    m I the only who remember differentiating y = ln|2x/x+1| though? Maybe at the start of question 9? and finding the gradient when x=1 or something....??
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    (Original post by godkid308)
    I believe there was also a question on finding the gradient at P and R (which were reciprocals). Can't quite remember what is was, hope this jogs someones memory.
    I remember it but cant remember the x coordinate of P or the answer it was supposed to make :L

    basically it was find the gradient of the curve y=ln|2x/x+1| at point P. (HINT use ln|a/b| = lna - lnb)
    Hence write down the gradient of the curve y=(e^x)/(2-e^x) at R.
    (or something like that)

    so y = ln|2x/x+1|

    = ln|2x| - ln|x+1|

    dy/dx = 1/x - 1/(x+1)

    substitute in the x coordinate of P (maybe x=1?)

    = 1 - 1/2
    =0.5

    Hence gradient of y=(e^x)/(2-e^x) at R is reciprocal which = 2

    i think thats the right workings, no idea if they are the right values though, so feel free to correct me on that
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    Anyone remember the show that question with x-4 on top of the fraction and 2(something)^3/2 on the bottom?
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    All up to date now, I think!
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    (Original post by joshgoldman)

    (ii) Another part of the question was showing that the integral of \displaystyle\int^{ln|\frac{4}{3  }|}_{0} \dfrac{e^x}{2-e^x} \ dx = ln|\frac{3}{2}|

    Let u = 2 - e^x
    \frac{du}{dx} = -e^x

    New limits: when x=ln|\frac{4}{3}| \rightarrow u = \frac{2}{3} when x=0 \rightarrow u=1
    - \displaystyle\int^{1}_{\frac{2}{  3}} \ \frac{1}{u} \dx
    =[-lnu] between 1 and 2/3.

    = (-ln(\frac{2}{3}) - (-ln1)
    =-ln|\frac{2}{3}|
    =ln|\frac{3}{2}| as required
    When you put the new limits on the integral, the top limit should be 2/3 and the bottom limit should be 1, because the ln|4/3| went to the 2/3 and the 0 went to 1.

    The way you have written it the next line would be (-ln1 - -ln|2/3) = ln|2/3| instead of -ln|2/3|

    also im assuming that the part of q9 answer in my previous post (gradient of y=ln|2x/x+1| at P) is correct, seems to match up with other people posts anyway, so if you could add that? perhaps point out that we are unsure if the values are correct though?

    and also the post starting:

    "so i guess the first part of question 8 was find the gradient of the curve at P? (x=3). Hence show that the line y = x/(x-2)^0.5 is not symmetrical across the line y=x."

    could be added to the mark scheme as part of q8?


    Thanks a lot for compiling it all anyway in all the nice writing much easier to read then my workings out!
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    (Original post by ILM16)
    Anyone remember the show that question with x-4 on top of the fraction and 2(something)^3/2 on the bottom?
    Sorry, i remember vaguely doing something with a ^3/2 like what your talking about, but not any where near well enough to write out how to do it without more information :L
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    Does anyone have the C3 Jan 2012 question paper?

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Updated: June 20, 2012
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