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# OCR MEI AS Mathematics Core 1 13/05/2013 Watch

1. (Original post by Dan_JR_12)
It could have 2 tangents, depending on the value of k, because it was a reciprocal graph. It had to different curves
Ah yes! May not be as wrong as I thought then Hopefully I got one of them right.
2. (Original post by rach100)
yhh same, but i ended up with surds for most of them :/
Same - really annoying but I'm sure we did it the right way

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3. Yeah I got that for the 125 root 5 question
4. (Original post by jo7777)
I used the quadratic formula three times I think
For the 125 root 5 question did you get 5^7/2?
Spot on

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5. What did anyone get for question 9? and for question 10 what was people intersecting circles?
6. what did you guys get for the minimum value of y?
7. Can I have a prediction of the grade boundaries for an A btw ?

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8. (Original post by Sun_Bear)
what did you guys get for the minimum value of y?
-7

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9. question nine anyone?
10. (Original post by hayjohn30)
What did anyone get for question 9? and for question 10 what was people intersecting circles?
Which one was question 9?
11. Someone create an unofficial mark scheme to with predicted grade boundaries?
12. (Original post by Dan_JR_12)
Which one was question 9?
The divisible by 3? I ended up from the sum as 3x^2+2
13. What did people get for the coefficient of x^3 in the binomial expansion question?
14. I got that but did you divide it by 3 as means of explaining you couldn't fully divide it by 3 ?

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15. I reckon it will be about 55-58 for an A i thought the paper was quite easy apart from the last question
Q 12 is where most candidates will loose loads of marks
16. I think I got everything right apart from retardedly writing 8a^8 instead of 8a^9...

last question was k=0 or 4, by discriminant = 0

distance between chord and centre was root10

on one of them the y intercepts (I think) were 2+root11 and 2-root11

cant remember much else
17. (Original post by Dan_JR_12)
What did people get for the coefficient of x^3 in the binomial expansion question?
-2560 but i think I've put positive 2560

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18. I got k as 0 and 12. The tried verifying the values k by subbing it back in the quadratic. Was 9 the proof if so I got 3n for the first one and said that 3 can be taken out. And the 9 part 2 I got 3n^2+2 and said 3 couldn't be removed. Not sure about part 2 though. I got (2,-7) as min point
19. (Original post by krishkmistry)
I got that but did you divide it by 3 as means of explaining you couldn't fully divide it by 3 ?

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Yeh eventually and for the coefficient i got it wrong but it was 10 x 4 x -64x^3
20. Much nicer than the January paper.

For the f(x-2) root question, did everyone get x=0, x=(7/2) and one other which I've forgot (-1?)

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Updated: October 25, 2013
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