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# AEA Prep Thread - 25th June 2015 (Edexcel) watch

1. (Original post by physicsmaths)
May you check this please? 2014 Q6 first part. Because the ms goes the other way. I'm sure my way is fine but not 100%. It is essentially the same thing except suggesting u is a dummy in replacement of x. Instead of using the sub in the expression.
Attachment 434457

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that is the correct answer. Isn.t it?
2. (Original post by TeeEm)
..

Should it not be -a^2 = a on the y-axis? So a = -1 is the only value that'll work as a = 0 gives P and Q as the same point?
3. (Original post by imusti96)
Thats a different question
was it this one?
Attached Images
4. solutionn.pdf (116.2 KB, 42 views)
5. (Original post by TeeEm)
that is the correct answer. Isn.t it?
Yes but I they just said let u=x-h into the expression needed. I used a dummy variable in the expressions given then subbed it into the expression.

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6. (Original post by TeeEm)
was it this one?
Yes thank you
7. (Original post by fatart123)
Should it not be -a^2 = a on the y-axis? So a = -1 is the only value that'll work as a = 0 gives P and Q as the same point?
there is no actual numbers ... the configuration could be slid anywhere on the xy plane. You just have to prove that the two curves intersect at the stationary points of one another
8. (Original post by physicsmaths)
Yes but I they just said let u=x-h into the expression needed. I used a dummy variable in the expressions given then subbed it into the expression.

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Are you asking is the method ok?
9. (Original post by TeeEm)
Are you asking is the method ok?
Yes.

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10. (Original post by physicsmaths)
Yes.

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I cannot see why not ... in most high end problems there are several clever ways of manipulating and no marking scheme could cater for all.

these paper are marked by mathematicians
11. (Original post by TeeEm)
I cannot see why not ... in most high end problems there are several clever ways of manipulating and no marking scheme could cater for all.

these paper are marked by mathematicians
Cheers. I hope its not like some of my teachers who mark for edexcel and couldnt see why the + square root in Q1 C3 was wrong for some x .........

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12. (Original post by TeeEm)
there is no actual numbers ... the configuration could be slid anywhere on the xy plane. You just have to prove that the two curves intersect at the stationary points of one another
Could you tell me where I've gone wrong? Don't really understand how that can work

13. (Original post by physicsmaths)
Cheers. I hope its not like some of my teachers who mark for edexcel and couldnt see why the + square root in Q1 C3 was wrong for some x .........

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I said mathematicians, not maths teachers.
very few are both!
14. (Original post by fatart123)
Could you tell me where I've gone wrong? Don't really understand how that can work

is this part b?
15. (Original post by TeeEm)
is this part b?
Mhm
16. (Original post by fatart123)
Mhm
if that is a yes then that is fine

(-1,0) and (0,-1)
17. (Original post by TeeEm)
if that is a yes then that is fine

(-1,0) and (0,-1)
Thanks, was a bit worried since they love asking graph transformation questions.
18. (Original post by fatart123)
Thanks, was a bit worried since they love asking graph transformation questions.
are you happy with the proof in part a?
19. (Original post by TeeEm)
are you happy with the proof in part a?
Yeah. What I wrote was a bit rough (not in as much detail as I'd do in exam) but the question was fine. Do you have any more of these graph ones? Kinda fun
20. (Original post by fatart123)
Could you tell me where I've gone wrong? Don't really understand how that can work

Lol are you working in a cave or something
21. (Original post by Gome44)
Lol are you working in a cave or something
I did not want to say ...
I copied the file
Pasted the picture into powerpoint...
Increased the brightness and the contrast so I could read it ...

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