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AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread] Watch

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    (Original post by fpmaniac)
    Find the values for x for which
    I need to check this but I got x = -1, x = 4 so:

    - 1 < x < 4
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    Yep, thats right

    (Original post by Suits101)
    I need to check this but I got x = -1, x = 4 so:

    - 1 < x < 4
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    (Original post by fpmaniac)
    Yep, thats right
    Yup just checked my working and I'm happy with it and then checked it on graphical calc (life saver)!

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    Guys they is something dodgy with my graphical calcutor casio I type in Sin (inverse) (3X) so its been stretched by 1/3 but for some reason the Y end points changes for some reason any ideas why?
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    (Original post by Yo12345)
    Just use substitution. 3x^2du = dx. So you get : 3(x^3-10)^7/7 + C

    Posted from TSR Mobile
    it was meant to be done without a substitution, as it's in the form f'(x)(f(x))^n

    For C3 they hardly ever ask this type of question, I've seen it come up once for 2 marks. But for units like FP3 etc, this trick is more useful as it's quicker if you know how it works.
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    (Original post by SunDun111)
    Guys they is something dodgy with my graphical calcutor casio I type in Sin (inverse) (3X) so its been stretched by 1/3 but for some reason the Y end points changes for some reason any ideas why?
    That's strange... And what's even stranger I didn't know you could plot inverse trig function on a calculator!
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    (Original post by AlexM10)
    Any ideas for 5B june 2014??
    Complete the square on the quadratic and you'll find it'll make things a lot easier
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    [QUOTE=Suits101;65785575]These have never come up before, but I have written a solution to it:

    Page 160
    Exercise 9B
    Question 7

    I have never seen questions like these before!

    Could I see your solution? I can't seem to get the answer.
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    (Original post by MahuduElec)
    Integrate x^2(x^3-10)^6 .dx

    I made this up, but using that 'reverse chain rule', you can do this in two or so steps without parts/substitution.
    Could you/anyone write this in LaTex?
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    (Original post by SunDun111)
    Guys they is something dodgy with my graphical calcutor casio I type in Sin (inverse) (3X) so its been stretched by 1/3 but for some reason the Y end points changes for some reason any ideas why?
    My calc messes up inverse graphs too.
    You need to draw the graph of inverse sign then just apply the transformations to your endpoints in the exam.

    Th calc draws normal inverse graphs fine (without transformations) so its ok.
    I think there is a way to fix it though, don't know much about my calculator =(.
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    (Original post by Suits101)
    Could you/anyone write this in LaTex?
    The question or the solution using reverse chain rule?

    Also could you post the solution to that question please. the quotething messed up.
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    (Original post by Suits101)
    Could you/anyone write this in LaTex?
    Sorry, I've never used/ know how to use LaTex
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    How is everyone feeling then? I am getting 60's on some papers and high 50's on others, Not sure what to expect, Ngl i usually get 10 marks lower in the exam.
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    (Original post by whydoidothis?)
    .
    The bit after below:

    = 2sin^-1(2x) + C

    Therefore overall integral = -0.25(1-4x^2)^0.5 - 2sin^-1(2x)

    Then just put limits in

    (I didn't see limits originally hence all the + C stuff)
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    hey guys if i was asked to sketch a random inverse graph how would I do it
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    Can anyone help I'm not sure how to go about finding the area on Q8d
    Question paper -
    http://theallpapers.com/papers/AQA/A...W-QP-JUN05.pdf
    Mark scheme -
    http://theallpapers.com/papers/AQA/A...W-MS-JUN05.pdf
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    (Original post by FluorescentM)
    Can anyone help I'm not sure how to go about finding the area on Q8d
    Question paper -
    http://theallpapers.com/papers/AQA/A...W-QP-JUN05.pdf
    Mark scheme -
    http://theallpapers.com/papers/AQA/A...W-MS-JUN05.pdf
    Find the y value when x is 2 and 4. Then area B = Rectangle of where x=4*y(4) - area A - rectangle where x=2*y(2)


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    (Original post by FluorescentM)
    Can anyone help I'm not sure how to go about finding the area on Q8d
    Question paper -
    http://theallpapers.com/papers/AQA/A...W-QP-JUN05.pdf
    Mark scheme -
    http://theallpapers.com/papers/AQA/A...W-MS-JUN05.pdf
    Split everything up into shapes:

    Your integral finds the shaded area
    Overall area made by dotted lines is a rectangle
    There's a smaller rectangle in LHS

    Therefore B = shaded area - integral - smaller rectangle in LHS (with width of 2)
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    (Original post by SunDun111)
    hey guys if i was asked to sketch a random inverse graph how would I do it
    Use your graphical calculator
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    (Original post by aoxa)
    Find the y value when x is 2 and 4. Then area B = Rectangle of where x=4*y(4) - area A - rectangle where x=2*y(2)


    Posted from TSR Mobile
    (Original post by Suits101)
    Split everything up into shapes:

    Your integral finds the shaded area
    Overall area made by dotted lines is a rectangle
    There's a smaller rectangle in LHS

    Therefore B = shaded area - integral - smaller rectangle in LHS (with width of 2)

    Thank you both
 
 
 
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