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AQA Core 3 - Thursday 6th June 2013 (AM) - Official Thread Watch

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    I only did this paper yesterday and got the right answer but I can't really rememeber my reasoning I think its something to do with the original domain being x<0
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    OK it's done.

    Some things I wanna check...

    The question where you show 'with a diagram' if the mid-ordinate rule was too high or low of an estimate - how?!

    Mid-ordinate question, answer was... 25.6..?

    Volume of vase... 160/3..?

    Final question... 2ln(2/3)..? (or something like that, can't quite remember).

    And another 'ln' question, I got an answer of 0.5ln8..?

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    Not too bad ey? What say peeps?

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    (Original post by stuart_aitken)
    OK it's done.

    Some things I wanna check...

    The question where you show 'with a diagram' if the mid-ordinate rule was too high or low of an estimate - how?!

    Mid-ordinate question, answer was... 25.6..?

    Volume of vase... 160/3..?

    Final question... 2ln(2/3)..? (or something like that, can't quite remember).

    And another 'ln' question, I got an answer of 0.5ln8..?

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    Too high, yes yes no yes is what I got, got 2ln2 squared

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    (Original post by stuart_aitken)
    OK it's done.

    Some things I wanna check...

    The question where you show 'with a diagram' if the mid-ordinate rule was too high or low of an estimate - how?!

    Mid-ordinate question, answer was... 25.6..?

    Volume of vase... 160/3..?

    Final question... 2ln(2/3)..? (or something like that, can't quite remember).

    And another 'ln' question, I got an answer of 0.5ln8..?

    Posted from TSR Mobile

    Erm underestimate me thinks.. 25.6 was right

    Volume of vase yeah 160/3 X PI

    last question I got 2(ln2+1)

    and yeah that was correct it was 0.5ln8 for the fg(x)=ln 5 question..

    Was the transformation one a reflection in the y axis??

    Also the PI- COS^-1X was annoying was it basically bring the graph of cos^-1x down by PI
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    Not too bad, messed up the last question (I got 14.4)
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    (Original post by Lay-Z)
    Too high, yes yes no yes is what I got, got 2ln2 squared

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    Pretty sure it was underestimate. Do the integration separately and you get 25.7 ish. The mid ordinate rule gave 25.6 so underestimate.

    And I got 2(ln2+1) for the final q, still unsure about it tbh though
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    Also was the integral of (lnx)^2 simply 2xlnx-2x+c?
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    (Original post by stuart_aitken)
    OK it's done.

    Some things I wanna check...

    The question where you show 'with a diagram' if the mid-ordinate rule was too high or low of an estimate - how?!

    Mid-ordinate question, answer was... 25.6..?

    Volume of vase... 160/3..?

    Final question... 2ln(2/3)..? (or something like that, can't quite remember).

    And another 'ln' question, I got an answer of 0.5ln8..?

    Posted from TSR Mobile
    -it was like trapezium rule in core 2, draw a trapezium
    the curve falls below the trapezium, so area using mid-ordinate rule
    is too high

    yea I got 25.6
    same volume of vase

    I got 2ln(3/2)

    yeah I got 0.5ln8

    the transformation of f(x) to f(-0.5x)
    I got reflection in y axis, horizontal stretch scale factor 2

    for the pi-inverse cos graph
    It was translation (0, -pi) then reflection in the x axis
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    Last q was ln(9/4), which I didnt get bit should get some marks for 2u/u^2 + u

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    grade boundries predictions?? i think it will be higher than jan13, this paper was alot more straight forward. A lot why so many graph AQA??
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    (Original post by MSI_10)
    Also was the integral of (lnx)^2 simply 2xlnx-2x+c?
    \int\ln(x)^2 dx \equiv \int 2\ln(x) dx \equiv 2[(ln(x))(x) - \int x\times \frac{1}{x}] dx = 2(xln(x) - x) = 2xlnx-2x+c

    Yeah, I guess it is!
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    (Original post by stuart_aitken)
    OK it's done.

    Some things I wanna check...

    The question where you show 'with a diagram' if the mid-ordinate rule was too high or low of an estimate - how?!

    Mid-ordinate question, answer was... 25.6..?

    Volume of vase... 160/3..?

    Final question... 2ln(2/3)..? (or something like that, can't quite remember).

    And another 'ln' question, I got an answer of 0.5ln8..?

    Posted from TSR Mobile
    I got similar things to you cant remember the other ln question though and I got too low cause I worked it out on my calculator and the actual figure was 27 something and seen as the answer I got was 25.6 it was lower then I showed it by doing a trapezium rule thingy (which I'm so happy I looked over last night)
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    how did you guys think the exam went??
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    (Original post by L'Evil Fish)
    \int\ln(x)^2 dx \equiv \int 2\ln(x) dx \equiv 2[(ln(x))(x) - \int x\times \frac{1}{x}] dx = 2(xln(x) - x) = 2xlnx-2x+c

    Yeah, I guess it is!
    but it was (lnx)^2 not just ln(x)^2

    so you'd do chain rule

    when differentiating and stil use byparts
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    (Original post by purpleshadows)
    how did you guys think the exam went??
    Some bits went okay others I've made stupid mistakes in but speaking to other people in my college we all think the grade boundaries will be low. How'd you do???
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    (Original post by Hdizzle)
    but it was (lnx)^2 not just ln(x)^2

    so you'd do chain rule

    when differentiating and stil use byparts
    Aah right, I didn't do the exam anyway, just answered above question
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    can someone please start an unofficial markscheme?:c fankyew x
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    I got ln3 for the last question :/
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    (Original post by Hdizzle)
    -it was like trapezium rule in core 2, draw a trapezium
    the curve falls below the trapezium, so area using mid-ordinate rule
    is too high

    yea I got 25.6
    same volume of vase

    I got 2ln(3/2)

    yeah I got 0.5ln8

    the transformation of f(x) to f(-0.5x)
    I got reflection in y axis, horizontal stretch scale factor 2

    for the pi-inverse cos graph
    It was translation (0, -pi) then reflection in the x axis
    This guy is on it

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