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Edexcel Maths FP1 UNOFFICIAL Mark Scheme 19th May 2017

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    (Original post by KingsAlpaca)
    My Answer for FP1 2017,
    1. ST 6.45
    2. (3/10, 1/10
    -2/5, 1/5)
    (2,1
    1,-4)
    3. y=x/2 xy=16 (4sqrt(2), 2sqrt(2)) and (-4sqrt(2),-2sqrt(2))
    4. (2p+12)/13 + (3p-8)i/13 p=20 lambda=+-4sqrt(5)
    5. (12-5p, 4p-10
    6p-15, 12-5p)
    p=3/2 k=15/2 a=9/2 or -27/2
    6. -3-2i a=2 b=-11
    7. ST (-a, -3a/2) s=15a^2/16
    8. ST k=2/9
    9. ST ST
    If I did not get number 7 and number 6 totally right, would it be A still? I am sure I got the other right.
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    I got the same answer but youre supposed to calculate q and simplify the answers with it which i didnt.
    How many marks do u think i will lose?
    (I left the y coordinate and area in terms of a and q)

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    how many marks for the -27/2 one? i missed out this!
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    for the last induction question i changed f(k+1) into the same format as f(k) and then when i added 11f(k) it was divisible by 19. does this sound vaguely right to anyone?? had no idea what i was doing lol
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    (Original post by X_IDE_sidf)
    How did you solve this one, tbh I just added it 12 times
    So you know the summation formula which was n/2(n+3)(2n+5) if i recall correctly, put 12 in and you got 2610.
    so 2610 + k(2^(r-1)) = 3520
    k(2^(r-1)) = 910
    now as r is increasing you need to do K(2^0 + 2^1 + 2^2... 2^11) = 910
    it came to 910/4095 = 2/9
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    (Original post by glad-he-ate-her)
    Am i the only one who forgot the sum for 2 to the power of r-1 from c2 so manually wrote out all 12 terms :clap2:such mathematical elegance
    I love that xD
    Did the same thing.
    Worried I may loose marks on it?( Method marks)
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    (Original post by 20XX Fox)
    Misread the equation for newton raphson to have be +4/x^2 rather than -. How many marks will I lose
    Bump
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    (Original post by erawein)
    So you know the summation formula which was n/2(n+3)(2n+5) if i recall correctly, put 12 in and you got 2610.
    so 2610 + k(2^(r-1)) = 3520
    k(2^(r-1)) = 910
    now as r is increasing you need to do K(2^0 + 2^1 + 2^2... 2^11) = 910
    it came to 910/4095 = 2/9
    This is how I did it, but the comment was made that the geometric series formula in C2 would have been more elegant ( https://www.varsitytutors.com/hotmat...ometric-series )
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    (Original post by Charles Omer)
    I believe this is correct too.
    Attachment 648996
    I tHought we were meant to find k+2 or am I wrong?
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    Any ideas what 73/75 would be in UMS?
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    in question 7 dont you have 2 coordinates, -a;3a/2 and -a;-3a/2? because parabola so its symmetric
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    me too
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    How many marks was question 7
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    (Original post by TrueDAN)
    Any ideas what 73/75 would be in UMS?
    My guess, above 90, probably 95/96
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    (Original post by Edmar Hadad)
    what question was the p=20 one?
    Was the complex numbers second term 1/13i(3p-8) - I got p =20 though
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    Was it k(2^r-1) or k(2^r -1), i'm pretty sure it was the latter?
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    (Original post by Prodanski)
    in question 7 dont you have 2 coordinates, -a;3a/2 and -a;-3a/2? because parabola so its symmetric
    No because q>0
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    (Original post by Nik298)
    a would be ±4.5 surely as 27/2 wouldn't lead to the correct area
    You needed to realise the area could have been negative by going under the x-axis, hence area is -270 as well as 270
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    do you think its going to be ignore further working, or will i lose marks?
    (Original post by Faznaz55)
    No because q>0
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    I'm an idiot :banghead: somehow simplified the k thing wrong so I ended up with 909/4094 instead of 910/4095 :grumble:
 
 
 
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