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OCR MEI FP1 20th January 2010 pm watch

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    Those two questions were standard questions and hopefully everybody got full marks on those. What did you guys get for CA and AC, in the first question?
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    (Original post by Tallon)

    for the last part, just sub n = 50 into that. It ends up 1.49 I think.


    Was okay. I walked out and realised I got the last part of the second to last question wrong though because I forgot to take away 2-1j from 43+47j. Should be okay though.

    Don't you sub n = 49 and take that away from n = 100?

    And what do you mean you forgot to take away 2 - j, why would u do that? :confused:
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    (Original post by Straightpath)
    Don't you sub n = 49 and take that away from n = 100?

    And what do you mean you forgot to take away 2 - j, why would u do that? :confused:

    I don't think so. What was the question? the sum of (4+r)/r(r+1)(r+2) from r= 1 to 50, wasn't it?

    If you did n = 50 then n = 100 and took them away you'd just get the the sum from 50 to 100, wouldn't you?

    and you take away 2-j because the question was pi/4< arg(z-(2-j) < 3pi/4
    I didn't do that. I just figured out what arctan of 47/43 was then comapred it with pi/4 and 3pi/4. Maybe 1 mark there, don't know.
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    (Original post by Summerdays)
    Those two questions were standard questions and hopefully everybody got full marks on those. What did you guys get for CA and AC, in the first question?
    CA was a 1 x 1 matrix, I got 50

    AC was 3 x 3, but I don't know how to use da maths format thing here on tsr to write it
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    (Original post by Tallon)
    I don't think so. What was the question? the sum of (4+r)/r(r+1)(r+2) from r= 1 to 50, wasn't it?

    If you did n = 50 then n = 100 and took them away you'd just get the the sum from 50 to 100, wouldn't you?

    and you take away 2-j because the question was pi/4< arg(z-(2-j) < 3pi/4
    I didn't do that. I just figured out what arctan of 47/43 was then comapred it with pi/4 and 3pi/4. Maybe 1 mark there, don't know.
    Noo! The question asked you to find the sum from 50 to 100. and besides you don't used this formula anyway do u ( (4+r)/r(r+1)(r+2) ), don't u use the 3/2... one that you proved in part ii?

    I done the same as u for the angle one, but still can't see wat u mean by taking away 2 - j
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    That is what I got. For the last question, would you get ANY marks for attempting. My method was wrong, I think, because I didn't get quite the same value as the value on my calculator.
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    (Original post by Straightpath)
    Noo! The question asked you to find the sum from 50 to 100. and besides you don't used this formula anyway do u ( (4+r)/r(r+1)(r+2) ), don't u use the 3/2... one that you proved in part ii?

    I done the same as u for the angle one, but still can't see wat u mean by taking away 2 - j

    yeah you use the one involing 3/2 and n.
    didn't realise it was n=5- to 100. annoying.
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    (Original post by Straightpath)
    Noo! The question asked you to find the sum from 50 to 100. and besides you don't used this formula anyway do u ( (4+r)/r(r+1)(r+2) ), don't u use the 3/2... one that you proved in part ii?

    I done the same as u for the angle one, but still can't see wat u mean by taking away 2 - j
    The start of the line is at (z-(2+j)) and the end is (z-(43+47j)) so if your working out the arg of (z-(2+j)) you consider a right angled triangle from the start of the line to the end.

    The height is 47-1 and width is 43-2 and then you work out the arctan of 46/41 or whatever it was
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    (Original post by Donor)
    I found it quite straight forward up to the method of differences which I worked out in the end. I definitely made some silly mistakes though. I was lucky actually as all the revision I'd done was a couple of exam papers yesterday (resitting) and everything that came up was similar to that.

    For the last question did anyone just sub values of r = 50 and r=100 in and then take one from the other?
    i subbed iin 100 and 49 and took one away from the other
    0.01094...ish?
    x
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    No, I got the same answer as my calculator subbing 49 and 100. Damn (I hope I didn't lose all three marks )
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    (Original post by aaizaali)
    i subbed iin 100 and 49 and took one away from the other
    0.01094...ish?
    x
    Yep yepp. Only the question asked to give the answer to 3 sig fig
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    I thought the paper went really well. I had a graphical calculator so I checked most of my answers and can be sure I got them right.

    The only question that caused me slight problems was question 9 (method of differences). For part (i) I couldn't see a simpler way to arrive at the result so I used partial fractions (that I learned from C4). For the sum to n, I couldn't entirely spot the pattern and arrive at the result they wanted, but I probably only lost 1 or 2 marks for that.
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    I'm probably going to have a look at the paper tommorow afternoon, I'll do the paper and post answers here if noone else does
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    (Original post by Summerdays)
    No, I got the same answer as my calculator subbing 49 and 100. Damn (I hope I didn't lose all three marks )
    That was right! If you put in 50 and not 49, you would have found the sum from n = 51 to n = 100, and not from 50 to 100.

    The limit of the series was obvious, was it 3/2?

    Just checking :P
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    (Original post by OL1V3R)
    That was right! If you put in 50 and not 49, you would have found the sum from n = 51 to n = 100, and not from 50 to 100.
    Which, sadly, is exactly what I did :ashamed2: .
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    Will I have lost 3 marks for subbing in 50 and 100? :/

    I really wasn't sure what to do so stuck them in and hoped
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    Fingers crossed!

    Everybody, why not take a guess at the grade boundaries? Was it a harder or easier paper than expected? I can't really judge because I thought FP1 was easy so my opinion might be skewed.

    Thank god there were no long matrix transformation questions
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    Seen the paper now. Not a bad paper really, here's the answers:
    1) ab = -13 + 11j
    a/b = -17/29 + 1/29j

    2)i) AB impossible
    CA = 50
    B + D =
    ( 3 1 )
    ( 6 -2)
    AC =
    ( 20 4 32 )
    ( -10 -2 -16)
    (20 4 32)

    ii) need DB=
    ( -10 -2)
    ( 22 1)

    3) a = 1, d = 12, so roots 1/2, 1, 3/2. k = 11

    4) k =5, x = -2, y = 3 , z = 17

    5) show that

    6) induction

    7) i) x = 3/2 x = -7/2 y = 0
    ii) x large + ve, y -> 0 from above
    x large -ve, y -> 0 from below
    iii) graph.
    iv) x<-7/2 or 3/2<x<= 9/5

    8) a) i) |z - (2 + 6j)| = 4
    ii) | z - (2 + 6j)| < 4
    and |z - (3+7j)| > 1
    b) sketch
    c) set z = that number in the arg(z - something), show between pi/4 and 3pi/4. (so is in locus)

    9) i) show that
    ii) show that
    iii) 3/2
    iv) 0.104
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    For 7iv, Isn't it x<-7/2, 3/2<x<=9/5 (?)
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    (Original post by Summerdays)
    For 7iv, Isn't it x<-7/2, 3/2<x<=9/5 (?)
    I got this for 7iv aswell :rolleyes:
 
 
 
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