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    (Original post by Wooty)
    How was the 6 mark question proof done? The one where it ended up 4k+sink4k-2=0

    And also the last proof where you had to differentiate something to get sin2A..?
    This is how i did it:

    y = (2x-1)(tan(2x))

    dy/dx = (2k-1).2Sec^2(2k) + tan(2k).2

    = (2k-1)(2sec^2(2k) + 2tan(2k)

    At S.P (Min point) dy/dx = 0

    Therefore (2k-1)(2sec^2(2k) + 2tan(2k) = 0

    So, (2k-1)(2/Cos^2(2k)) + 2(Sin(2k)/Cos(2k)) = 0

    Mutliply through by Cos2k

    (2k-1)(2/Cos(2k)) + 2(sin(2k)) = 0

    (4k-2)/(Cos(2k)) + 2(Sin(2k)) = 0

    Multiply through by Cos(2k)

    (4k-2) + 2Sin(2k)Cos(2k) = 0

    [Identity Sin2x = 2SinxCosx, so Sin4x = 2sin(2x)cos(2x)]

    4k+ sin4k -2 =0

    (Original post by puck_fella)
    The domain i seem to remember was restricted. This confused me, i put f(x) > ln 0, which, on reflection is not right. But oh well, only one mark.
    If i remember correctly, the domain was restricted to x> -k, because x=-k would be the asmyptote. This does not limit the range in any way. The graph of lnx has simply been shifted k units in the negative direction.
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Updated: June 12, 2006

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