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    (Original post by TheNicholas)
    Thanks! My answers are the same except I got a different alpha value, think I just mistyped it into my calculator when typing in all my answers, do you happen to remember the questions? Something -e^x =0?
    (2x-1)^2-e^x=0
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    (Original post by Sid1234)
    did something similar to this, but didnt use a/b just said is rational. did I(1) as it was easier
    I did that at the start and it worked, but then saw that n>1 so did 2 instead
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    (Original post by Sid1234)
    1.23 is right, c was 0.15 +2x but you can solve for x and x=0.1, so C was 0.35
    Ah cool, I'm pretty confident with the rest of the paper so hopefully not too many dropped there
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    (Original post by Sid1234)
    (2x-1)^2-e^x=0
    Thank you
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    It asked for n>1. You may have lost a mark

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    (Original post by Mr.Raskolnikov)
    It asked for n>1. You may have lost a mark

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    ****.
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    (Original post by Sid1234)
    1: +-ln(1.5+0.5sqrt(5))
    2. ln2 - pi/4
    3. sketch - one of those ones with a circle ish shape intersecting x axis at 90, then nothing between 2 and 5, then another bit. (0,root2) (0, -root2) A, B, C were coordinates.
    4. Beta = 3.733, Alpha = 1.6291
    5. Maclaurin : 2x - (8/3)x^3
    Pi approx 2/3 which rounds to 3
    6. Theta = pi/10
    (0.951,0.309)
    area pi/20
    7. ln(infinity) for first part, 2 is upper limit of second part
    8. bunch of show thats, messed up reduction.
    I agree with all of them.
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    Think the main thing you had to show was the square root of 2 became normal 2 as k=multiples of 2. Since k is rational the whole thing must be rational. True for n=2. True for 2k, etc.

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    (Original post by Mr.Raskolnikov)
    Think the main thing you had to show was the square root of 2 became normal 2 as k=multiples of 2. Since k is rational the whole thing must be rational. True for n=2. True for 2k, etc.

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    yeah i got to (root2)^2k and said that is always rational
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    (Original post by Sid1234)
    did something similar to this, but didnt use a/b just said is rational. did I(1) as it was easier
    I believe, n>1 was stated in the question. I think that we had to prove for even n, though not sure about it.
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    (Original post by tangotangopapa2)
    I believe, n>1 was stated in the question. I think that we had to prove for even n, though not sure about it.
    oh s h i t !
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    Yes, as if it's odd you'll have root 2 in the expression and the integral is no longer rational. You didn't have to say it was even, but that's why the condition was given

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    The only thing you had to show was there are no square roots and you get an expression for the integral. As k is an integer and becomes the only variable down to n=2, the integral must be rational

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    Does anyone remember exact question?
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    Is the integral from 0 or 1 to infinity. I spent ages trying to figure it out. Can't see how it's from 1 because the rectangle become above the curve

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    (Original post by Mr.Raskolnikov)
    Is the integral from 0 or 1 to infinity. I spent ages trying to figure it out. Can't see how it's from 1 because the rectangle become above the curve

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    You have to integrate from one to infinity and add the rectangle at one because you can't integrate from 0 as you would get 1/0
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    My answer became ln(infinity)<sum<ln(infinity)+1/0
    I assumed 1/0 is equal to infinity so my answer became
    Infinty<sum<infinity+infinity

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    ffs, i self-taught this module with no help during my gap year and wanted to do well , think i've just scraped an A -.-
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    (Original post by tobybes)
    I agree with all of them.
    For 1 I had ln(1.5 +/- 0.5rt(5)) is that equivalent?
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    Come in guys its great if it went well for you personally but please stop scaring peopl
 
 
 
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