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    y = (2x – 1) tan 2x, 0< x <pi/4 .

    The curve has a minimum at the point P. The x-coordinate of P is k.

    (a)Show that k satisfies the equation
    4k + sin 4k – 2 = 0

    So I got
    u= 2x-1
    du=2

    v=tan2x
    dv=sec^2 2x

    V x du = 2tan2x
    U x dv = (2x-1)(sec^2 2x)

    Giving 2tan2x + 2x-1 sec^2 2x

    This is bit correct?

    2(sin2x/cos2x) + (2x-1)(1/cos^2 2x)

    And I'm lost from there....(it said my latex formulas were dangerous?!)
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    yes your diff is correct....procede from here
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    thats on a solomon paper i think, and you need to insert K as the X co ordinate and proceed.
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    Its in the June 2006 paper and the mark scheme put a 2 infront of the (2x-1) when it's differentiated and I don't know why...:confused:
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    (Original post by jaz_jaz)
    Its in the June 2006 paper and the mark scheme put a 2 infront of the (2x-1) when it's differentiated and I don't know why...:confused:
    you get the factor of 2 when you differentiate the tan 2x
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    Can someone please tell me how to do it, I can't even do figure it out looking at the mark scheme!
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    tan 2x

    diffs to 2sec^2 2x, cus the 2 comes it being 2x. just like sin 2x would diff to 2cos 2x.
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    The derrivative of tan 2x is 2sec^2(2x)
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    ok thanks but i still have no idea how to get to want they want :o:
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    y = (2x – 1) tan 2x

    dy/dx = 2(2x - 1)sec^2 + 2tan 2x

    x=k

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

    (2k - 1)sec 2k + sin 2k = 0

    sin2k = -(2k - 1)sec 2k

    sin2kcos2k = -2k + 1

    2sin2kcos2k = -4k + 2

    [formula sheet states 2sin2kcos2k is equal to sin4k]

    sin4k = -4k + 2

    4k + sin4k - 2 = 0
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    Basically when you differentiate, you will get:

    2tan2x + 2(2x-1)sec2x

    Then you change everything into terms of sine and cosine:

    (2sin2x)/(2cos2x) + (4x-2)/cos2x

    to find the turning point, dy/dx = 0

    therefore:

    (2sin2x)/(2cos2x) + (4x-2)/cos2x=0

    now, you can multiply through by cosx to get rid of the fractions:

    2sin2xcos2x+4x-2=0

    2sin2xcos2x is the identity sin2A

    that should help.

    edit: ojpowermake got there before me!
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    (Original post by ojpowermake)
    y = (2x – 1) tan 2x

    dy/dx = 2(2x - 1)sec^2 + 2tan 2x

    x=k

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

    (2k - 1)sec 2k + sin 2k = 0

    sin2k = -(2k - 1)sec 2k

    sin2kcos2k = -2k + 1

    2sin2kcos2k = -4k + 2

    [formula sheet states 2sin2kcos2k is equal to sin4k]

    sin4k = -4k + 2

    4k + sin4k - 2 = 0
    Thanks
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    (Original post by unamed)
    Basically when you differentiate, you will get:

    2tan2x + 2(2x-1)sec2x

    Then you change everything into terms of sine and cosine:

    (2sin2x)/(2cos2x) + (4x-2)/cos2x

    to find the turning point, dy/dx = 0

    therefore:

    (2sin2x)/(2cos2x) + (4x-2)/cos2x=0

    now, you can multiply through by cosx to get rid of the fractions:

    2sin2xcos2x+4x-2=0

    2sin2xcos2x is the identity sin2A

    that should help.

    edit: ojpowermake got there before me!
    thanks i understand yours much better
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    for c4 paper people, i feel this shall definitely help..........
    generally... to differentiate.. sinx ---> cosx ---> - sinx
    to integrate .. -sinx ---> cosx ---> sinx ( so to integrate sinx you get -cos x ) + constant obvs...

    hope this helps you further on..
 
 
 
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