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1. (Original post by NotNotBatman)
I did Worse than I thought, 67/75.
same
2. (Original post by BinaryJava)
Anybody have any ideas of grade boundaries, I think I might of just scraped an A which is good for me coming from a mid C from last years paper.
60 for an A imo
3. (Original post by iMacJack)
How did you get to that point?
Cuz qp=-1 then I rearrange into q=-1/p and then sub to the m1=1/q so I got -q as m1
4. I think this paper was glorious. Confident on 90+ UMS.
5. For the question about the self inverse, would it be reasonable to suggest it is a self inverse because the determinant was -1 because I remember reading somewhere that any matrix with a determinant of 1 or -1 is a self inverse.
6. (Original post by iMacJack)
Well here's what I did;

pq = -1 from the previous part

1/p = grad of tangent at p
therefore grad of tangent at q = 1/q

So for them to be perpendicular the product of the gradients = -1

1/p * 1/q = 1/pq = -1 as pq = -1 from the previous part
I did it like that. Used the pq=-1 from 5(b)
7. For the second induction question, I didn't prove for n=2. Will this definitely lose one mark? Sorry if this has already been asked.
8. Looks standard to me.
9. (Original post by Jam3898)
73/75 like I thought. I put too little effort in this module. Glad it's done.
leaving it as root 5 will be fine
10. (Original post by ombtom)
For the second induction question, I didn't prove for n=2. Will this definitely lose one mark? Sorry if this has already been asked.
My maths teacher said if they only give you n=1, you don't need to prove for n=2. Only for n=1.
11. Reasonable [standard], yes.

Thanks.
12. (Original post by oinkk)
My maths teacher said if they only give you n=1, you don't need to prove for n=2. Only for n=1.
you don't have to prove for n=2, it can be useful on recurrence ones though
13. (Original post by ombtom)
For the second induction question, I didn't prove for n=2. Will this definitely lose one mark? Sorry if this has already been asked.
Nah I've checked past papers and it doesnt lose a mark
14. Looks like a pretty reasonable paper to me. I sat last years, didn't resit. Difficult parts (to me at least) are 7e, potentially 8i and 9. Honestly looks about the same difficulty as last year's - the odd dodgy 1 or 2 marker with a difficult last question.

If I had to guess I'd say the same boundaries as last year.
15. (Original post by Arsey)
leaving it as root 5 will be fine
would i lose a mark for only mentioning gradients and proving them and not the lengths of or and qr?
surely you'd need to prove they're parallel, rather than relying purely on observation?
16. Reasonable [typical FP1], yes.

Thanks.
17. (Original post by Mattematics)
Looks like a pretty reasonable paper to me. I sat last years, didn't resit. Difficult parts (to me at least) are 7e, potentially 8i and 9. Honestly looks about the same difficulty as last year's - the odd dodgy 1 or 2 marker with a difficult last question.

If I had to guess I'd say the same boundaries as last year.
i'd say a mark lower than last years
18. (Original post by ombtom)
For the second induction question, I didn't prove for n=2. Will this definitely lose one mark? Sorry if this has already been asked.
I think you have to show that the formula is true using n=2 alongside your prove for n=1.
Because it was U(n+1), when n is one you work out what U2 is, and then show that is true for n=2. BUT you have only proven its true for n=1, because you'd have to work out the 3rd term to prove its true for n=2.

So if you did that you are correct. If you only wrote in your conclusion "True for n=1" then you will not lose any marks.
After all, if true for n=k, you have also proven its true for n=2, n=3, n=4 and so on.
19. Where do you think the 100 UMS boundary will be?
20. For everyone asking about the use of n=2 since we were only given U1, it's fine to just sub in and prove the base case for n=1, I've looked through the recurrence questions by topic and the marks schemes only look for n=1 providing only U1 is given and so in this case just using the base case on n=1 should be satisfactory.

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