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P1 Jan 2005 watch

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    were u supposed to integrate between 27 and 0 for the last question and then take away the area of the triangle?


    i have a bad fealing it was i did summin silly like 27 and 9 for the boundaries and then took away the area of the triangle :confused:


    will i get any marks?
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    (Original post by ThugzMansion7)
    were u supposed to integrate between 27 and 0 for the last question and then take away the area of the triangle?


    i have a bad fealing it was i did summin silly like 27 and 9 for the boundaries and then took away the area of the triangle :confused:


    will i get any marks?
    Yes, probably most of them - provided your integration was correct.

    Ben
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    (Original post by Ben.S.)
    Yes, probably most of them - provided your integration was correct.

    Ben
    what did u get as ur answer?
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    (Original post by ThugzMansion7)
    what did u get as ur answer?
    I didn't sit the exam - sorry (2nd year university). But it's generally the method for these integration questions that count. Maybe you were right (I don't know what the question was).

    Ben
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    (Original post by ThugzMansion7)
    what did u get as ur answer?
    Weren't you supposed to integrate between 0 and 3?

    the y value was 18

    for the question i got area under graph as 36, and the triangle area as 27, giving the answer as 9cm2.

    May be completely wrong
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    (Original post by Lottelo)
    Weren't you supposed to integrate between 0 and 3?

    the y value was 18

    for the question i got area under graph as 36, and the triangle area as 27, giving the answer as 9cm2.

    May be completely wrong
    the value of y was 27 i think..pretty sure...then again i mite be wrong
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    it was between 3 and 0 wher you needed to integrate, im so stupid i integrated between 3 and 1 losing marks

    Can you remember the questions? dnt think it went the best
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    (Original post by Lottelo)
    Weren't you supposed to integrate between 0 and 3?

    the y value was 18

    for the question i got area under graph as 36, and the triangle area as 27, giving the answer as 9cm2.

    May be completely wrong
    That's what I did and it looked correct to me. Minimum point on the curve was 3 (x co-ordinate) so you would integrate between 3 and 0. The curve was y = x^2 - 6x + 18 so the y intercept would be 18. I think I got 9 as the area of R.
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    I also got 9 for the area of R
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    i take it this is the edexcel paper?
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    (Original post by PrOdiGy_)
    i take it this is the edexcel paper?
    Yes, its Edexcel.
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    ok......just i couldnt remember getting that question but thats cos i did the OCR one heh
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    Yeh I intergrated between 0 and 3 then took away area of triangle.
    How did you guys find it? I thought it was alright, but I was hoping for a solid A. Methinks i prob managed about 55 to 60/75, probably a mid B? that area of a sector of a circle was annoying as well, when i finally figured out how to do it, the time was up :hmpf: .

    Ah well, I'll just retake it in June (i think its the same day as P3)
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    i got the area under the graph to be 9.

    how did you verify that the line was parallel to the x axis? i got the same x - values of 3 but didnt know how to explain.
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    ok this may have just been me. But did anyone else notice the paper was longer than all the previous P1 papers. Im sure ive never done a 9 Q paper. N i seem to remember somin like y = 18, then when intergrate got 36 then minus 27 for triangle.
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    1. Yep edexcel
    2. and yes i got the area of R to be 9, however not cm2, just 9 , or 9 units to be precise.
    3. i do afree with u syncman, it was a longer paper than all the others. when i saw the table on the front page going up to question 9, i was schocked, coz it's ually only 7 or 8 Qs, but it was a good test.
    4. for the proving it's a parallel, well since u've got two x-values being the same which is 3, meaning u'd have to divide by the 0, which is not possible thus it's parallel, thts wht i wrote, although i might be wrong
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    yeah because there is no change in x they it must be parallel. thats wot i put
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    (Original post by k0rrupter)
    . and yes i got the area of R to be 9, however not cm2, just 9 , or 9 units to be precise.
    9 units squared, to be precise-er!

    Ben
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    (Original post by Ben.S.)
    9 units squared, to be precise-er!

    Ben
    crap, I put cm squared.

    Oh well
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    What did everyone think of the paper compared to other P1 papers? I thought it was easy but a bit longer than the others.
 
 
 
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