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    (Original post by MintyMilk)
    Attachment 398155

    Our unofficial mark scheme, (2 of us).

    Pinch of salt.
    1 is the wrong way round I think and 4 is definitely wrong
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    (Original post by Aph)
    1 is the wrong way round I think and 4 is definitely wrong
    Yeah 1 is the other way around. The diagram on the right for 4 is correct.
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    (Original post by Aph)
    1 is the wrong way round I think and 4 is definitely wrong
    Nah, look closely at the one labeled most likely this one. It's correct - I too thought they had shaded inside the circle when I looked, but nope so it's correct
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    (Original post by Aph)
    \sum\limits_{r=1}^n (2r+1)=N^2
    \sum\limits_{r=n+1}^{2n} (2r+1)=\sum\limits_{r=1}^{2n} (2r+1) -\sum\limits_{r=1}^{n} (2r+1) = 4n^2-n^2
    Thanks
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    (Original post by Connorbwfc)
    Yeah 1 is the other way around. The diagram on the right for 4 is correct.
    But shouldn;t the shaded portion be enclosed by 2 lines, instead of 1? arg=0 and arg = pi/4
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    (Original post by Willtyler)
    Nah, look closely at the one labeled most likely this one. It's correct - I too thought they had shaded inside the circle when I looked, but nope so it's correct
    (Original post by Connorbwfc)
    Yeah 1 is the other way around. The diagram on the right for 4 is correct.
    no because iit was between 0 and pi/4 so it should've been a triangle with a point at -1-j not what they have drawn
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    (Original post by Buses)
    But shouldn;t the shaded portion be enclosed by 2 lines, instead of 1? arg=0 and arg = pi/4
    Oops, yeah it's wrong. There should be a horizontal line across at J=-1.
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    (Original post by Buses)
    But shouldn;t the shaded portion be enclosed by 2 lines, instead of 1? arg=0 and arg = pi/4
    Yeah you're right so question 4 is also wrong - I got that right in the exam thankfully
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    (Original post by Buses)
    But shouldn;t the shaded portion be enclosed by 2 lines, instead of 1? arg=0 and arg = pi/4
    exactly, and they haven't done that.
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    Did anyobe get something like this for the argand diagram? Name:  20150514_124924.jpg
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    Wrong thread
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    (Original post by Darcy1)
    Did anyobe get something like this for the argand diagram? Name:  20150514_124924.jpg
Views: 442
Size:  435.8 KB

    Yes, but rotated the other way and with all the boundaries included.
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    (Original post by Aph)
    exactly, and they haven't done that.
    cool. what was your transformation matrix that got A''B''C'' to ABC (q9)
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    (Original post by Connorbwfc)
    Yes, but rotated the other way and with all the boundaries included.
    Yeah came out wrong in the picture LOL
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    (Original post by Buses)
    cool. what was your transformation matrix that got A''B''C'' to ABC (q9)
    For that i did T-1 times M-1 (the inverses)
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    (Original post by Buses)
    cool. what was your transformation matrix that got A''B''C'' to ABC (q9)
    I don't really remember
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    Was it something like: 1/48(0,8 top row. -6 4 bottom row) ?
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    How do you guys feel the difficulty of this paper was compared to the previous years ?
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    (Original post by -Gifted-)
    How do you guys feel the difficulty of this paper was compared to the previous years ?
    slightly easier than last year. And less lengthy. Grade boundaries should remain the same as last year though
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    I think apart from the sigma question which is quite different from what people are used to, this years paper is one of the less difficult ones

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