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    (Original post by LelouchViRuge)
    There is no normal series, if you did change it to 13 you'd be adding  (13)^2 + 2^1^3 at the end, which would be incorrect.

    To calculate the amount of numbers in any summation, do whatever the n value is subtract the starting value plus 1
    Thanks!
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    This might be a stupid question, but on question 1c of exercise 4D please, https://644625398389466aee0063322305...0NRNU0/CH4.pdf why isn't x to xy linear, as I thought it would be since it's only to the power 1?! Thanks
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    (Original post by economicss)
    This might be a stupid question, but on question 1c of exercise 4D please, https://644625398389466aee0063322305...0NRNU0/CH4.pdf why isn't x to xy linear, as I thought it would be since it's only to the power 1?! Thanks
    A linear transformation is linear if it satisfied T(\lambda a + \mu b) = \lambda T(a) + \mu T(b) where a and b are column vectors.

    Obviously, y \to x + xy isn't.
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    Hi, please could anyone explain what is meant by 'orientation is reversed' on exercise 4h question 2c https://doc-00-0g-docs.googleusercon...kZmYTlpUmpNYXM Thanks
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    (Original post by economicss)
    Hi, please could anyone explain what is meant by 'orientation is reversed' on exercise 4h question 2c https://doc-00-0g-docs.googleusercon...kZmYTlpUmpNYXM Thanks
    Can't view the PDF for some reason.
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    Do the IAL F1 and IAL FP1 papers follow the same syllabus?
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    (Original post by NotNotBatman)
    Do the IAL F1 and IAL FP1 papers follow the same syllabus?
    More or less, I didn't really see any differences. Then again, I barely read the specification.
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    (Original post by NotNotBatman)
    Do the IAL F1 and IAL FP1 papers follow the same syllabus?
    Yes, minor differences such as knowing that a quadratic equation equation can be written as x^{2}-\left ( \alpha +\beta \right )x+\alpha \beta = 0 and matrix rotations about any angle. That's about it, I think.
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    (Original post by Zacken)
    More or less, I didn't really see any differences. Then again, I barely read the specification.

    Thank you.

    (Original post by aymanzayedmannan)
    Yes, minor differences such as knowing that a quadratic equation equation can be written as x^{2}-\left ( \alpha +\beta \right )x+\alpha \beta = 0 and matrix rotations about any angle. That's about it, I think.
    Isn't the quadratic equation thing in FP1? It's the only method I've been taught although I do know how to use other methods.
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    (Original post by NotNotBatman)

    Isn't the quadratic equation thing in FP1? It's the only method I've been taught although I do know how to use other methods.
    It isn't compulsory knowledge on FP1 - questions can be done without this formula. However, many questions on IAL are based on this.
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    (Original post by Zacken)
    Can't view the PDF for some reason.
    Hi, sorry about the links, here's a pic of the question, I'm just not too sure by what they mean by 'the orientation is reversed' on part c? Thank you Attachment 519045519047Name:  image.jpg
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    Hi, please could anyone explain 6c on this paper, how do we know that QP=I and in particular how do we know it's QP rather than PQ? Thanks Attachment 519053519055Name:  image.jpg
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    (Original post by economicss)
    ...
    Does reading this ( http://mathinsight.org/determinant_l...transformation ) help?
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    (Original post by economicss)
    ...
    1. Triangle T is transformed to T' by P. Triangle T' is transformed back to T by P^{-1}. The question has simply called Q = P^{-1}. Inverses satisfy PP^{-1} = PQ = I.

    2. Inverses are commutative so PP^{-1} = PQ = I = QP = P^{-1}P. That is, it doesn't matter which order you pick here, either will work.
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    Thank you, I think so, so is it saying that if one determinant is the negative of another determinant then the orientation of the second determinant is reversed compared to the original determinant?
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    Please could anyone explain how on this paper https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf question 2, you know that the focus is -3,0 rather than 3,0? Thanks
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    (Original post by economicss)
    Please could anyone explain how on this paper https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf question 2, you know that the focus is -3,0 rather than 3,0? Thanks
    Because the fixed point is (-3 , 0). The fixed line is x = 3 so is to the right of the focus and the whole curve is a mirror image of the usual one.
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    There is a question that says:
    Given  arg(\lambda +9i + w) =\frac{\pi}{4}

    where \lambda is a real constant

    find the value of \lambda

    w=10-5i from the first part.

    I've written:

    arg(\lambda + 4i +10) = \frac{\pi}{4}

    \arctan 1 = \frac{\pi}{4}\newline\newline \frac{10+\lambda}{4} =1\newline\newline \lambda = -6

    Would I receive full marks for doing it like this?
    Markscheme method (no ALT) https://gyazo.com/e1386625d2e22da34152ceb4df8b246a
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    Name:  Series.png
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    How do I do part ii? I did the summations for the 2n+1 I did (2n+1)(n) + 1 and added this to the summation, but I got the wrong answer.
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    (Original post by NotNotBatman)
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    How do I do part ii? I did the summations for the 2n+1 I did (2n+1)(n) + 1 and added this to the summation, but I got the wrong answer.
    It starts from r=0 where as the standard sums start from r=1. So you need to do:

    \displaystyle \sum_{r=0}^{n} \left(r^2 - 2r + 2n + 1\right) = 0^2 - 2(0) + 2n + 1 + \sum_{r=1}^{n} \left(r^2 - 2r + 2n + 1\right)

    Your error was doing (2n+1)(n) + 1 when you should have done (2n+1)(n+1).
 
 
 
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