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    (Original post by marioman)
    I've just updated my post with the link to the 2015 paper.
    Thanks
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    Is it me or is question 4 in the 2015 paper worth way too many marks for what it is?
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    Can anyone think of any small things which i might not know i should know for tomorrow?
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    (Original post by ughexams)
    the answer is 1/36 pi.... (for part iii)
    I know
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    Hi! How do you find the range and domain from the equation? I always forget!
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    (Original post by r34_786)
    Hi! How do you find the range and domain from the equation? I always forget!
    not sure if there is a perfect way of doing it, but look for important things. If you have a fraction the denominator can not be 0, a square root can not be negative etc, and for range maybe find the stationary point
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    How do I show that:

    1 – ln(2) = ln(1/2e)
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    (Original post by micycle)
    How do I show that:

    1 – ln(2) = ln(1/2e)
    1 = ln(e)

    so 1 - ln(2) = ln(e) - ln(2) = ln(e/2) = ln(0.5 e)
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    Thank you so much!
    (Original post by LukeWeatherstone)
    not sure if there is a perfect way of doing it, but look for important things. If you have a fraction the denominator can not be 0, a square root can not be negative etc, and for range maybe find the stationary point
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    I think proof in vectors will come up. Although I've never seen one in a paper, I've seen it in my textbook 😬
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    (Original post by ArbaazMalik)
    I think proof in vectors will come up. Although I've never seen one in a paper, I've seen it in my textbook 😬
    Vectors is C4.
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    Could anyone explain how to do question 8 ii on this specimen paper??
    the mark scheme makes the second derivative equal to 0 but I don't know why.
    Attached Images
  1. File Type: pdf Specimen QP - C3 OCR.pdf (42.6 KB, 94 views)
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    (Original post by 16characterlimit)
    Vectors is C4.
    Oh yeah. My mind is just mental atm.
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    can someone help me with this question from jan 2006

    3sin6Bcosec2B=4 where beta lies between zero and 90 degrees

    any help would be appreciated
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    (Original post by LukeWeatherstone)
    Can anyone think of any small things which i might not know i should know for tomorrow?
    perhaps the graphs of cosec sec cot etc?
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    (Original post by SI 1)
    Could anyone explain how to do question 8 ii on this specimen paper??
    the mark scheme makes the second derivative equal to 0 but I don't know why.
    the second derivative is the rate of change of gradient. so when the second rerivative = 0, the rate of change or gradient is at a turning point, thus it is maximum.
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    (Original post by duncant)
    the second derivative is the rate of change of gradient. so when the second rerivative = 0, the rate of change or gradient is at a turning point, thus it is maximum.
    Thank you so much!
    So at a turning point the gradient of a curve is the maximum. I always thought the gradient was 0 at turning points.
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    (Original post by SI 1)
    Thank you so much!
    So at a turning point the gradient of a curve is the maximum. I always thought the gradient was 0 at turning points.
    The gradient is always 0 at turning points, be it maximum or minimum

    The turning point could be either a minimum or a maximum, hence why you differentiate again to prove. If the value of x substituted into D^2y/Dx^2 is greater than 0, it's a minimum point and if it's less then it's a maximum point!
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    (Original post by AlfieH)
    The gradient is always 0 at turning points, be it maximum or minimum

    The turning point could be either a minimum or a maximum, hence why you differentiate again to prove. If the value of x substituted into D^2y/Dx^2 is greater than 0, it's a minimum point and if it's less then it's a maximum point!
    Sorry I probably sound really stupid but how would you know that the maximum value of the gradient is 0? Could the max gradient not be more than 0? I would think that there'll be a point on the curve where the gradient is more than 0.
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    (Original post by SI 1)
    Sorry I probably sound really stupid but how would you know that the maximum value of the gradient is 0? Could the max gradient not be more than 0? I would think that there'll be a point on the curve where the gradient is more than 0.
    It's not the point at which the gradient is the maximum - so you may be getting confused there.

    Nor is it the point where the gradient is the lowest.

    Instead the gradient is always 0 at the turning points/ min or max points. It is the maximum/ minimum as in the highest/lowest POINT of the curve not gradient.

    If that makes sense?
 
 
 
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