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    Good paper I thought! The induction question in particular was great. 5ii caused some problems though.

    Took some notes out with me if anyone wants a look at any solutions.
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    For the transformation for matrix M i put a two way strech, sf 4 in the x direction and sf 2 in the y direction is this correct?
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    (Original post by Darcy1)
    For the transformation for matrix M i put a two way strech, sf 4 in the x direction and sf 2 in the y direction is this correct?
    Think so. That's what I put. Bit generous for 3 marks I thought, unless I've missed something.
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    (Original post by Darcy1)
    For the transformation for matrix M i put a two way strech, sf 4 in the x direction and sf 2 in the y direction is this correct?
    I did exactly this as well. But, I think we drop a mark for not saying two-way stretch.
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    (Original post by Aph)
    k=1/3

    area =4*6*8 = 192
    wouldn't the area be 48 since the determinants of T and M were 6 and 8 and the original shape had an area of 1?
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    This was the graph. Asymptotes at x = -1, 2 and y = 3
    Crosses at x = 3, -2/3 and y = 3
    Above at x = -100 Below at x = 100
    part 4 was where is y >= 3:
    x > -1 and 0 <= x < 2

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    (Original post by frackingmusic)
    wouldn't the area be 48 since the determinants of T and M were 6 and 8 and the original shape had an area of 1?
    The original shape had an area of 4...
    co-ordinates were (0,0), (0,2), (4,1) that's a isosceles triangle base 2 hight 4
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    (Original post by Aph)
    The original shape had an area of 4...
    co-ordinates were (0,0), (0,2), (4,1) that's a isosceles triangle base 2 hight 4
    I could've sworn the coordinates were (0,0) (2,0) and (4,1)
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    (Original post by frackingmusic)
    I could've sworn the coordinates were (0,0) (2,0) and (4,1)
    No Aph is right it was an area of 4 !
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    (Original post by frackingmusic)
    I could've sworn the coordinates were (0,0) (2,0) and (4,1)
    I got the same answer as you as well. I think we got the 2,0 and 0,2 co-ordinates other way round.
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    (Original post by frackingmusic)
    I could've sworn the coordinates were (0,0) (2,0) and (4,1)
    I did the same for a while but no they were definately what I stated.
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    I'm hoping around 70/72 for 100UMS. I definitely think it was a harder paper than last year!
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    (Original post by Aph)
    I did the same for a while but no they were definately what I stated.
    you're right those are the co-ordinates and the right answer
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    I was allowed to take this bad boy in. Did everything Name:  20150514_140820.jpg
Views: 319
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    Guys what do you think about grade boundaries? I would've got 100% but misread the (z-(-1-j))=pi/4 and instead did (z-(-1+j))=pi/4 whyyyyyyyy? ((((((((((((( So screwed ffs lol

    Easy paper as well, I expect 63 as A again :/
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    I cannot remember if I actually put p= r= in one of the questions, but I did right the equation. Will they still imply that I got the right answeR?
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    (Original post by MintyMilk)
    Attachment 398155

    Our unofficial mark scheme, (2 of us).

    Pinch of salt.

    Edit: X and Y on Q1 are the other way around (we both got them the right way in the actual exam)

    Edit 2: Q4 is meant to be from -1-j, and the area below it isn't meant to be shaded, I wasn't the one who actually wrote all the stuff out and we went through it all pretty quickly, sorry about the discrepancies
    Sorry about that.
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    (Original post by MintyMilk)
    Sorry about that.
    Was just about to say something :P

    Quick question, for question 4, for some stupid reason I did everything correct but with arg(z-(-1+j)) for some reason :/ Throughout the whole question, how many marks do you reckon I lost? Max would be 4 but probs 2/3 :/
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    (Original post by joe12345marc)
    Was just about to say something :P

    Quick question, for question 4, for some stupid reason I did everything correct but with arg(z-(-1+j)) for some reason :/ Throughout the whole question, how many marks do you reckon I lost? Max would be 4 but probs 2/3 :/
    Well, the part that required you to plot that particular inequality was two marks, then part 3 was another 2 marks.

    Not quite sure, but it could be anywhere from 1 to 4
    1 mark lost if they allow full follow-through marks on part 3 and the actual plotting of the point in part 2 was only worth 1 mark (assuming you got the inequality correct).

    4 marks lost if it's 0 for anything that isn't fully correct in part 2, then if that carries over to part 3. I think 2 to 3 is the most likely range, sorry I can't give an exact figure though, maybe someone more experienced can.
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    (Original post by MintyMilk)
    Well, the part that required you to plot that particular inequality was two marks, then part 3 was another 2 marks.

    Not quite sure, but it could be anywhere from 1 to 4
    1 mark lost if they allow full follow-through marks on part 3 and the actual plotting of the point in part 2 was only worth 1 mark (assuming you got the inequality correct).

    4 marks lost if it's 0 for anything that isn't fully correct in part 2, then if that carries over to part 3. I think 2 to 3 is the most likely range, sorry I can't give an exact figure though, maybe someone more experienced can.
    Nah it's cool I just was checking and making sure I'm being realistic with my score! Was expecting 100 UMS today but a stupid mistake like that can cost me like 10 UMS :/
 
 
 
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