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    You're allowed to multiply top and bottom by the same number, leaving the RHS alone. It still works. Take a simpler example:

    1/2 = a

    If we multiply top and bottom of the LHS by 2, we get:

    2/4 = a

    Obviously, that's the same thing.

    If you only want to multiply the top by 2, let's say, then you have to do both sides:

    1/2 = a

    2/2 = 2a
    1 = 2a

    Once again, we'll arrive to the same equation.
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    hi

    how do u differenciate : cot²x ???

    HELP!!
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    Can anyone who's done the A1 paper help me please?

    Q5: Why can't the answer be simplified more to (x-3)/x(x-2) ?

    Q6a: I get 225/32 but the answer is 255/32... How?

    Q6c: How do you show it is an increasing function?

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    smatz: I think you would do:

    cot²x = (cotx)² and use chain rule

    2cotx*-cosec²x
    -2cotxcosec²x

    Anyone tell me if this is right? :/
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    (Original post by smatz01)
    hi

    how do u differenciate : cot²x ???

    HELP!!
    We use the chain rule.
    y=t^{2}, t=cotx\\

\frac{dy}{dt}=2t, \frac{dt}{dx}=-cosec^{2}x\\

\frac{dy}{dx}=\frac{dy}{dt}\frac  {dt}{dx}\\

\frac{dy}{dx}=-2tcosec^{2}x\\

\frac{dy}{dx}=-2cosec^{2}xcotx
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    Just to let people know there are 2 mistakes in the projectmaths c2 markscheme.
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    Wrong thread ?
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    (Original post by SunGod87)
    Wrong thread ?
    yep sorry.
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    (Original post by violet)
    Can anyone who's done the A1 paper help me please?

    Q5: Why can't the answer be simplified more to (x-3)/x(x-2) ?

    Q6a: I get 225/32 but the answer is 255/32... How?

    Q6c: How do you show it is an increasing function?

    5. It can.

    6. (a) You have done it wrong.

    f ' ' (x) = 2x - 2x-3

    so: f ' ' (4) = 8 - 2/64

    => 255/32

    6. (c) There are a few ways.

    Take f ' (x) and set it to less than zero and show there are no solutions.

    Sketch the graph of f ' (x) showing that it is a tangent to the x axis so no solutions less than zero ==> not a decreasing function ==> an increasing function.
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    (Original post by samdavyson)
    5. It can.

    6. (a) You have done it wrong.

    f ' ' (x) = 2x - 2x-3

    so: f ' ' (4) = 8 - 2/64

    => 255/32

    6. (c) There are a few ways.

    Take f ' (x) and set it to less than zero and show there are no solutions.

    Sketch the graph of f ' (x) showing that it is a tangent to the x axis so no solutions less than zero ==> not a decreasing function ==> an increasing function.
    Erm, this might be a stupid question, but what is an increasing function?
    I don't understand why when f ' (x) is less than 0 then it can't have no solutions.
    Thanks for the other 2 btw
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    (Original post by violet)
    Erm, this might be a stupid question, but what is an increasing function?
    I don't understand why when f ' (x) is less than 0 then it can't have no solutions.
    Thanks for the other 2 btw
    If f(x) is an increasing function, its gradient f'(x) will be greater than zero for all values of x.

    f'(x)
    = x² - 2 + 1/x²
    = (x-1/x)²

    Factorising it in this form shows us that f'(x) > 0 for all values of x, since any number squared is positive.
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    (Original post by violet)
    Erm, this might be a stupid question, but what is an increasing function?
    I don't understand why when f ' (x) is less than 0 then it can't have no solutions.
    Thanks for the other 2 btw
    An increasing function as a fuction for which dy/dx > 0. [ C2 chapter 9 for more info. ]

    (e.g. a line with a positive gradient).

    Another answer that you might prefer is:

    f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

    Square of anything ==> Positive.

    So f ' (x) > or = to 0.

    ==> Increasing Function.
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    Hi,

    Im stuck like nothing else with a c3 trig question...
    Find, in terms of pie the solutions of the equation
    sin5x+sinx=0
    0<X<Pie
    Any help would be much appreciated!
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    (Original post by Goody1)
    Hi,

    Im stuck like nothing else with a c3 trig question...
    Find, in terms of pie the solutions of the equation
    sin5x+sinx=0
    0<X<Pie
    Any help would be much appreciated!
    sin5x + sinx
    = sin(3x + 2x) + sin(3x-2x)
    = sin3xcos2x + sin2xcos3x + sin3xcos2x - sin2xcos3x
    = 2sin3xcos2x = 0

    2sin3x = 0 OR cos2x = 0
    sin3x = 0
    3x = arcsin0, 180°-arcsin0, 360° + arcsin0
    3x = 0°, 180°, 360°
    x = 0°, 60°, 120°

    cos2x = 0
    2x = arccos0, 360° - arccos0,
    2x = 90°, 270°
    x = 45°, 135°

    x = 45°, 60°, 120°, 135°

    Ignore the values 0° and 180° since they aren't in range.
    Unless the range is 0</=x</=pi
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    f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

    Is that completing the square?
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    (Original post by violet)
    f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

    Is that completing the square?
    factorising.

    (x-1/x)(x-1/x)
    = x² - x/x - x/x + 1/x²
    = x² - 1 - 1 + 1/x²
    = x² - 2 + 1/x²
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    (Original post by violet)
    f ' (x) = x2 - 2 + 1/x2 = (x - 1/x)2

    Is that completing the square?
    It would be if it wasn't a perfect square.

    But as it is there is nothing to "complete" (now addition or subtraction on the end) as it is a square.
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    (Original post by Widowmaker)
    factorising.

    (x-1/x)(x-1/x)
    = x² - x/x - x/x + 1/x²
    = x² - 1 - 1 + 1/x²
    = x² - 2 + 1/x²
    Oh I see!!!!
    Ok so increasing function is when dy/dx is greater than 0... And decreasing is when it's less than 0 right?
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    (Original post by violet)
    Oh I see!!!!
    Ok so increasing function is when dy/dx is greater than 0... And decreasing is when it's less than 0 right?
    yep, or in the range x = a and x = b
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    (Original post by violet)
    Oh I see!!!!
    Ok so increasing function is when dy/dx is greater than 0... And decreasing is when it's less than 0 right?
    Yeah.

    When it is equal to zero I think it is still classed as increasing. Someone confirm this??
 
 
 
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