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    Dropped at least 2 marks, not too fussed though since the grade boundaries will probably be a bit lower than standard.
    Don't think it was too tough though
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    (Original post by NotNotBatman)
    For the last question, did it say the midpoint lied on the positive x axis or just that it was positive?
    Positive x-axis
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    How many marks was 9b worth?
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    Couldn't you say that T^2= T therefore its self inverse???
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    (Original post by oShahpo)
    Lied on the positive x axis so y=0
    (Original post by midgemeister7)
    Positive x-axis
    ah, I messed that question up.
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    (Original post by chewitt77)
    +-sqr2/2 (+-1/sqr2)
    9/2*x^1/2 + 25/2*x^-3/2
    12.467 (I double checked this however may still be wrong)
    a=38 b=15 c=1
    2-2i, -8i, p=-4 q=8
    1/p
    anti clockwise rotation 45 degree about (0,0) (this may be wrong)
    p=-3 q=-9, matrix (0 1 (bottom line 1 0))
    a^2 +2a -4 + 4i(a +1), a=-1, sqr5 and 2.03 rad, QR twice the length of OP and QR is parallel to OP
    x=15sqr2/4
    I put rotation 135 degree anticlockwise, cus when i put 135 back into the cos and sin formula i got the same matrix as the question
    And 12.468 3dp
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    (Original post by midgemeister7)
    those are all right except the anti-clockwise rotation which was 135 degrees I believe
    Yeah I got 135 anti, if you don't put (0,0), will I lose marks, I swear all the matrices are in FP1
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    (Original post by TheRandomGenius)
    Couldn't you say that T^2= T therefore its self inverse???
    No, T^2 = I, therefore T=T^-1, therefore its selfinverse
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    (Original post by ombtom)
    Yes, it was -1. You just had to find the inverse matrix and say that it was the same, i.e. R^-1 = R .'. self-inverse.
    My signs for T^-1 was flipped though. Did you get exactly the same signs?
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    I think I've forgot to switch the elements to find inverse R, that's why I couldn't prove it. How many marks was that question?
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    (Original post by NotNotBatman)
    I think I've forgot to switch the elements to find inverse R, that's why I couldn't prove it. How many marks was that question?
    Just 1 I think, may be2
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    (Original post by Chirstos Ioannou)
    How? It's the first time i've seen such a question.
    I also had difficulty with this question. Not sure if my ans is correct.
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    (Original post by Sallekmo)
    MY friend found one question soo hard he put his paypal email address as the answer. Lol I saw him do it.
    sounds like your friend should be taking BTEC media studies not Further maths then...
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    (Original post by AhnafR)
    My signs for T^-1 was flipped though. Did you get exactly the same signs?
    Did you remember to divide by the determinant (which was -1)?
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    I think I could get 73/75. Lol
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    (Original post by Person18)
    Yeah I got 135 anti, if you don't put (0,0), will I lose marks, I swear all the matrices are in FP1
    I think might have lost 2 marks there then , because I put pi/4 anti

    Posted from TSR Mobile
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    Sorry yeah thats what I meant- T^2 = I. That would get the mark?
    (Original post by anndz3007)
    No, T^2 = I, therefore T=T^-1, therefore its selfinverse
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    (Original post by NotNotBatman)
    For the last question, did it say the midpoint lied on the positive x axis or just that it was positive?
    No it said that the midpoint was on y=-x lol dumbdumb
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    (Original post by Skygliderjack)
    It wanted the modulus and argument of z, not z^2+2z
    FML I've got that wrong then.
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    (Original post by anonymousGangsta)
    sounds like your friend should be taking BTEC media studies not Further maths then...
    Lol, you know finding 1 question hard isn't that big of a deal,
 
 
 
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