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Math Help - Edexcel C3 past paper question

  1. #1
    Junior Member Turple's Avatar
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    Edexcel C3 past paper question

    I'm using past papers to help with revision but found this question really difficult. Hope someone can help

    Equation of curve: y=(2x-1)tan2x, x is between 0 and pi/4 (including 0)

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

    a) Show that k satisfies the equation

    4k-sin4k-2=0

    I wasn't sure where to start but tried substituting k into the equation of the graph which hasn't helped :s

    Thanks
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  2. #2
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    Quote Originally Posted by Turple View Post
    I'm using past papers to help with revision but found this question really difficult. Hope someone can help

    Equation of curve: y=(2x-1)tan2x, x is between 0 and pi/4 (including 0)

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

    a) Show that k satisfies the equation

    4k-sin4k-2=0

    I wasn't sure where to start but tried substituting k into the equation of the graph which hasn't helped :s

    Thanks
    used a calculator and determined k \approx 0.276515...

    this value of k does not satisfy the equation 4k - \sin(4k) - 2 = 0
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  3. #3
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    Quote Originally Posted by Turple View Post
    I'm using past papers to help with revision but found this question really difficult. Hope someone can help

    Equation of curve: y=(2x-1)tan2x, x is between 0 and pi/4 (including 0)

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

    a) Show that k satisfies the equation

    4k-sin4k-2=0

    I wasn't sure where to start but tried substituting k into the equation of the graph which hasn't helped :s

    Thanks
    Differentiate the function using product rule:

    y'= 2 tan(2x)+ (2x-1) \cdot \dfrac2{(\cos(2x))^2}

    Set y ' = 0 and start to solve for x:

    2 tan(2x)+ (2x-1) \cdot \dfrac2{(\cos(2x))^2} = 0~\implies~2\sin(2x) \cdot \cos(2x) + 4x - 2=0 ~\implies~2 \cdot \frac12 \cdot \sin(4x) + 4x -2=0

    which yields:

    \sin(4x)+4x-2=0

    Could it be that there is a typo in your text?
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  4. #4
    Junior Member Turple's Avatar
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    Quote Originally Posted by earboth View Post
    Could it be that there is a typo in your text?
    Ah yeah sorry about that! Thanks for pointing it out though

    It is actually the same but with 4k+sin4k-2
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  5. #5
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    Quote Originally Posted by Turple View Post
    Ah yeah sorry about that! Thanks for pointing it out though

    It is actually the same but with 4k+sin4k-2
    I don't know a method to solve the equation algebraically. If you use an iterative method (for instance Newton's method) you'll get

    x = k \approx 0.276515039... which is the value skeeter posted yesterday.

    I used as initial value

    x_0 = 0.5
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