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Mr M's OCR (not OCR MEI) Further Pure 1 Answers May 2016 Watch

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    Mr M do you have a rough opinion of what the grade for an A may look like?
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    (Original post by Mr M)
    I expect so. I think this will be a very common mistake.
    how many would i drop in part 8.ii for missing the 2n on top?
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    (Original post by anon326589)
    why root5 for 6i)?

    Also how do you 10iii)?

    Thanks in advance
    Because the diameter is 2 \sqrt 5 !

    You do 10iii) by understanding the connection with 10ii) and 10i).
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    (Original post by emilyp2206)
    Just wondering how many marks i would lose for:

    7ii) somehow fluffed up my matrix and got

    -3 8
    0 3

    or something, but correctly described what it was and showed methods and everything of how i got there

    10ii) didnt do or really even imply over 41 (not sure how i missed that), i identified conjugate root and then the roots of those
    7ii) I can't see any marks there but the working on your script might get you something.

    10iii) You've probably got 2 marks there.
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    (Original post by proximity)
    Mr M do you have a rough opinion of what the grade for an A may look like?
    I don't like to guess boundaries.
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    (Original post by ComputeiT)
    Thanks again for your answers - would you get ECF for the sum to infinity if you forgot to half the method of differences from 8 part ii?
    Wait, why would you half?
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    (Original post by Mr M)
    Because the diameter is 2 sqrt 5 !

    You do 10iii) by understanding the connection with 10ii) and 10i).
    oh so you don't include i in the calculation

    only just spotted the connection

    thanks
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    (Original post by ComputeiT)
    how many would i drop in part 8.ii for missing the 2n on top?
    Probably just 1.
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    (Original post by SGHD26716)
    Wait, why would you half?
    You were summing half of the expression you derived in i).
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    (Original post by SGHD26716)
    Wait, why would you half?
    in 8i) you found 2/(2r+1)(2r+3)
    in 8ii) you are asked to use method of differences for 1/(2r+1)(2r+3) which is half of part i)

    Hope this helps
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    (Original post by Kyrus)
    For question 8 part ii if you did everything correct and just forgot to halve how many marks do you think you would lose
    Just 1 mark I expect.
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    (Original post by Mr M)
    You were summing half of the expression you derived in i).
    Damn it I didn't half
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    (Original post by SGHD26716)
    Damn it I didn't half
    I expect about half of the candidates will be in the same position. Don't worry about it.
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    8ii) how many marks would you lose if your answer was 1/6 - 1/(4n+6) and how many marks would you get in 8iii) because i must have carried on the error

    many thanks for your help
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    At this rate I've got around 63 marks
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    (Original post by anon326589)
    8ii) how many marks would you lose if your answer was 1/6 - 1/(4n+6) and how many marks would you get in 8iii) because i must have carried on the error

    many thanks for your help
    Drop 1 mark in ii) and another in iii). Both parts asked for a single fraction.
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    Hard paper for OCR?
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    (Original post by Mr M)
    Drop 1 mark in ii) and another in iii). Both parts asked for a single fraction.
    thanks
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    (Original post by TeeEm)
    Hard paper for OCR?
    No - normal difficulty.
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    How many marks do you think I will lose if I put root 3 +3i for question 2i and therefore got an incorrect answer for part ii as I rationalised the wrong thing?
 
 
 
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