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    I don't get part 3, I don't see where I'm going wrong.



    That's the mark scheme, but I get theta + 70.9 = -22.2

    so theta = -93.1

    Can someone explain where I'm wrong!?
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    is this an edexcel past paper?
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    The range is the set of values of y you can get by putting in x. You know first of all that x\geq0, so from this it is clear that f(x)\geq0, since both numerator and denominator will be positive. So the lowest positive value we can get is 0. Now we know the minimum value of y, we need to find the maximum value of y to find the complete range of values y can take. How would you go about this?
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    I think you must find dy/dx to find the x-coordinate of the maximum point and then substitute this value in the main equation to find its y value. The range would be
    greater than or equal to zero and less than or equal to the y value you'll find.

    BTW, where'd you get this question from?
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    part a) to find the range of f(x) you have to find the highest point of the graph,which is the peak.
    to find the y value of the peak you are going to have to differentiate the function and then make f'(x)=0 -----> then find the x value, put it into the original equation to find the y value. the range would the be yvalue > f(x)>0
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    :lolwut:
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    (Original post by namedeprived)
    :lolwut:
    Argh I've confused everyone!! I switched the questions as I printscreened the wrong one! Sorry!!
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    (Original post by Josh-H)
    Argh I've confused everyone!! I switched the questions as I printscreened the wrong one! Sorry!!
    Thought I was becoming delusional. :p:

    The question states smallest positive value of \theta.
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    (Original post by namedeprived)
    Thought I was becoming delusional. :p:

    The question states smallest positive value of \theta.
    Yep, but I don't get how to go from my value to 131

    (sorry for the previous confusion btw)
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    (Original post by Josh-H)
    Yep, but I don't get how to go from my value to 131

    (sorry for the previous confusion btw)
    (Assuming the marking scheme is right, I haven't done the previous ones)

    \theta+70.9=-158 \ and \ -22 \ and \ 202 \ and \ 338 so \theta= -158 - 70.9 \ and \ -22 - 70.9 \ and \ ...
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    Where did you get both questions from ?
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    (Original post by FZka)
    Where did you get both questions from ?
    Core 3 June 2008 OCR
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    (Original post by namedeprived)
    (Assuming the marking scheme is right, I haven't done the previous ones)

    \theta+70.9=-158 \ and \ -22 \ and \ 202 \ and \ 338 so \theta= -158 - 70.9 \ and \ -22 - 70.9 \ and \ ...
    I'm sorry.. I still don't get it?
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    (Original post by Josh-H)
    I'm sorry.. I still don't get it?
    \theta + 70.9 = -158 so  \ \theta=-158-70.9
    \theta + 70.9 =  -22 so  \ \theta=-22-70.9
    \theta + 70.9 = 202 so  \ \theta=202-70.9
    \theta+70.9 = 338 so  \ \theta=338-70.9

    So the smallest positive value of \theta is 202-70.9 which is 131
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    The mark scheme gives the solutions for theta + 70.9 to be:

    -158
    -22
    202
    338

    The corresponding theta values would therefore be:

    -228.9
    -92.9
    131.1
    267.1

    The smallest is 131.1.
 
 
 
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