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# OCR (non mei) FP2 Monday 27th June 2016

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1. (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
2. (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
3. (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
4. (Original post by Sid1234)
(2x-1)^2-e^x=0
Thank you
5. It asked for n>1. You may have lost a mark

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It asked for n>1. You may have lost a mark

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****.
7. (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.
8. 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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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
10. (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.
11. (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 !
12. 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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13. 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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14. Does anyone remember exact question?
15. 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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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
I assumed 1/0 is equal to infinity so my answer became
Infinty<sum<infinity+infinity

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18. 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 -.-
19. (Original post by tobybes)
I agree with all of them.
For 1 I had ln(1.5 +/- 0.5rt(5)) is that equivalent?
20. Come in guys its great if it went well for you personally but please stop scaring peopl

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