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    (Original post by Patrick2810)
    Hi

    Was there a june 2015 IAL FP1 paper?

    thanks
    There was a June 2015 IAL F1 paper, not a June 2015 IAL FP1 paper.
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    Can somebody explain this one? Damn, F2 is hard.

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    Attachment 530427530431
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    (Original post by Chirstos Ioannou)
    Can somebody explain this one? Damn, F2 is hard.

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    Attachment 530427530431
    That's FP2, by the way transformations of the complex plane is one of the hardest topics in fp2!!
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    (Original post by Patrick2810)
    That's FP2, by the way transformations of the complex plane is one of the hardest topics in fp2!!
    https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    Question 4.
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    Did you read my reply?
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    (Original post by Patrick2810)
    Did you read my reply?
    Yes, i meant that this is F2, not the GCE one but the IAL one.

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    (Original post by Chirstos Ioannou)
    Yes, i meant that this is F2, not the GCE one but the IAL one.

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    I know it's an F2 question, you said so in your original post! Transformations between the real and complex planes is on the fp2 spec and not fp1
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    (Original post by Patrick2810)
    I know it's an F2 question, you said so in your original post! Transformations between the real and complex planes is on the fp2 spec and not fp1
    True, i'm in the wrong place. my bad
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    This question is interesting - it seems that when enlarging shapes (matrices) you need to use |Det(T)| * Old area = New Area and not just Det(T) * Old area = New area
    I.e you need to consider the det(T) and the det(T) * (-1)
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    (Original post by Patrick2810)
    This question is interesting - it seems that when enlarging shapes (matrices) you need to use |Det(T)| * Old area = New Area and not just Det(T) * Old area = New area
    I.e you need to consider the det(T) and the det(T) * (-1)
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    Well yeah, area is a positive quantity.
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    (Original post by Patrick2810)
    This question is interesting - it seems that when enlarging shapes (matrices) you need to use |Det(T)| * Old area = New Area and not just Det(T) * Old area = New area
    I.e you need to consider the det(T) and the det(T) * (-1)
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    I guess it's just common sense because an area cannot be negative, much like lengths are modular et cetera


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    (Original post by Chirstos Ioannou)
    But it accepts the 150 degrees answer.

    Attachment 530221

    Which is basically the same thing but in the other direction.
    I find it much easier to just draw it and find the angle that it makes with the x-axis and then find the total rotation.
    It accepts 150 degrees clockwise, but not 150 degrees anticlockwise like the formula gives thats why you need to check both cos and sine
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    (Original post by iMacJack)
    I guess it's just common sense because an area cannot be negative, much like lengths are modular et cetera


    Posted from TSR Mobile
    Yeah but I just thought to do (when det(T) = -4k+30) -4k + 30 =2 and forgot to do the 4k-30=2

    easy mistake to make?
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    (Original post by Patrick2810)
    This question is interesting - it seems that when enlarging shapes (matrices) you need to use |Det(T)| * Old area = New Area and not just Det(T) * Old area = New area
    I.e you need to consider the det(T) and the det(T) * (-1)
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    (Original post by Zacken)
    Well yeah, area is a positive quantity.
    (Original post by iMacJack)
    I guess it's just common sense because an area cannot be negative, much like lengths are modular et cetera


    Posted from TSR Mobile
    If the determinant is negative, the shape has been turned over as well as enlarged. If you don't know whether it has been turned over or not, then yes, you need to consider the two options.
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    (Original post by tiny hobbit)
    If the determinant is negative, the shape has been turned over as well as enlarged. If you don't know whether it has been turned over or not, then yes, you need to consider the two options.
    Thanks very much for your reply. Could you explain what you mean by turned over?

    thanks
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    (Original post by Patrick2810)
    Yeah but I just thought to do (when det(T) = -4k+30) -4k + 30 =2 and forgot to do the 4k-30=2

    easy mistake to make?
    I think the way to avoid this is to just recall the formula as modulus det M not just det M because a lot of questions give you negative det M values. I thought that's how it is textbooks anyways; Area of A'= l DetM l x Area of A
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    (Original post by Patrick2810)
    Thanks very much for your reply. Could you explain what you mean by turned over?

    thanks
    The transformation included a reflection.
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    Please can anyone explain how to do question 14 on here http://crashmaths.com/wp-content/upl...orksheet-1.pdf as there's no mark scheme and the answer I get seems very ugly and not correct, thanks
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    (Original post by economicss)
    Please can anyone explain how to do question 14 on here http://crashmaths.com/wp-content/upl...orksheet-1.pdf as there's no mark scheme and the answer I get seems very ugly and not correct, thanks
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    Just had a go - not sure about my value for the constant of integration and therefore f(r) though. Values of a and b seem reasonable though.
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    (Original post by kingaaran)
    Just did it. Tbf, it does come out quite ugly: a = 8, b = 1225/204 and f(r) = 4r - (4742591/3468)

    Sorry I guess the numbers didn't quite go to plan -_-
    How would f(r) be 4r - k when the f'(r) = ar/2 + b/17?
 
 
 
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