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    (Original post by Pyslocke)
    Wait, for the QR and OP I don't remember them being parallel. Wasn't R = -5 on the real axis, P = -1+2i. I don't remember what they were exactly
    I can't rember exactly however if you look at OP as being the vector z (as that is what point P was)

    Then travelling from the point z^2 two lots of the vector z means they have both travelled in the same direction, just one QR was two lots of z hence travelling twice as far
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    If anybody gets a hold of the paper, I don't mind doing model solutions.
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    Here Are some of the answers I managed to write onto my name sheet:

    1) k = +/- sqrt(2)/2
    2b) 12.468
    3b) a = 38; b = 15; c = 1;
    4c) p = -4; q = 8;
    6a) 135 anticlockwise
    6b) p= -3; q = -9;
    7b) a = -1;
    7e) parallel
    9c) 15*sqrt(2) / 2

    please tell me if I am correct on the last two
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    (Original post by anndz3007)
    I was trying to be sure so i literally put all the digits in the calculation in the equation ._. Idk why everyone got .007
    yeah i made sure I didn't have rounding errors, my calculator could only just do it if I had the powers as decimals instead of fractions. glad someone else got the same as me, just looking unlucky that we're right haha
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    Yeah you're right don't worry

    (Original post by ICantFindMyName)
    Here Are some of the answers I managed to write onto my name sheet:

    1) k = +/- sqrt(2)/2
    2b) 12.468
    3b) a = 38; b = 15; c = 1;
    4c) p = -4; q = 8;
    6a) 135 anticlockwise
    6b) p= -3; q = -9;
    7b) a = -1;
    7e) parallel
    9c) 15*sqrt(2) / 2

    please tell me if I am correct on the last two
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    (Original post by ICantFindMyName)
    Here Are some of the answers I managed to write onto my name sheet:

    1) k = +/- sqrt(2)/2
    2b) 12.468
    3b) a = 38; b = 15; c = 1;
    4c) p = -4; q = 8;
    6a) 135 anticlockwise
    6b) p= -3; q = -9;
    7b) a = -1;
    7e) parallel
    9c) 15*sqrt(2) / 2

    please tell me if I am correct on the last two
    I got all of those mate
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    (Original post by veldt127)
    agreed on 12.468.

    Think I wrote mod z as root 5 but arg to 3sf. Looks like I lost one there
    I'm pretty certain it didn't specify the degree of accuracy for the modulus but did for the argument - I got the same, think it should be fine
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    (Original post by ICantFindMyName)
    Here Are some of the answers I managed to write onto my name sheet:

    1) k = +/- sqrt(2)/2
    2b) 12.468
    3b) a = 38; b = 15; c = 1;
    4c) p = -4; q = 8;
    6a) 135 anticlockwise
    6b) p= -3; q = -9;
    7b) a = -1;
    7e) parallel
    9c) 15*sqrt(2) / 2

    please tell me if I am correct on the last two
    7e for the second mark I think you needed to say QR was twice the length of OP and for 9c I got 15sqr(2)/4
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    Is the markscheme out yet?
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    (Original post by veldt127)
    agreed on 12.468.

    Think I wrote mod z as root 5 but arg to 3sf. Looks like I lost one there
    The wording in that questuon was terrible, i put root 5 and think its reasonable, and then it said put your answer in 3sf, so i put root5=2.something and be like is this what you want edexcel, i'll do whatever you want lol
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    (Original post by jonahz123)
    Nah the very last two questions were 3 each.
    Yes tha last two questions were , but he said the proof by induction questions, and they were 5 marks each
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    arsey where are ya
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    (Original post by ICantFindMyName)
    Here Are some of the answers I managed to write onto my name sheet:

    1) k = +/- sqrt(2)/2
    2b) 12.468
    3b) a = 38; b = 15; c = 1;
    4c) p = -4; q = 8;
    6a) 135 anticlockwise
    6b) p= -3; q = -9;
    7b) a = -1;
    7e) parallel
    9c) 15*sqrt(2) / 2

    please tell me if I am correct on the last two
    think the last one was 15*sqrt(2)/4
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    (Original post by ws12758)
    nein
    3 each
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    (Original post by LewisClothier)
    Got the inductions with a bit of disgusting work, wasn't nice though, and I realised that the imaginary part=0 in the last 5 minutes so had Five minutes to finish off the question, just about got it all apart from the one marker in part e). I was so annoyed about not knowing how to do that question until I realised though!

    Posted from TSR Mobile
    exact same thing happened to me i also managed to get the first induction but i only managed to do half of the 2nd induction as i wasn't thinking straight
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    i m confused is both f1 n fp1 discussions goin on in here
    n plz link a f1 unofficial markscheme if possible
    i think f1 was pretty hard dan fp1 paper
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    This exam made me want to kms.

    Wtf was with question 8 and 9.
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    I got 15root2/4 , its the midpoint so you have to devide the whole thing by 2
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    (Original post by anndz3007)
    12.468
    N/6(38n^2 +15n +1)
    X^2 -4x +8
    P=-3 q=-9
    M = (15root2/4,0)
    Only being pedantic, but they explicitly asked for the x-coordinate of 'M', given your value of p. Not the full coordinate, you would still get full marks I'm sure
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    It was a lot harder than last years paper, in my opinion
 
 
 
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