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# FP2 (Not MEI) - Thursday June 14 2012, AM Tweet

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1. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
ahh, could you possibly upload it? would really be appreciated
2. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
Here it is
Attached Files
3. 4726-01Jan12.pdf (301.2 KB, 68 views)
4. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
Here's a brainteaser:

say a graph, has an oblique asymptope. How would this asymptope change when sketching ?
Last edited by Ree69; 06-06-2012 at 18:05.
5. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by Ree69)
Here's a brainteaser:

say a graph, has an oblique asymptope. How would this asymptope change when sketching ?
Would the asymptote no longer exist because of the propety of not existing for ?
6. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
By oblique you mean when a graph tends to some function of x?
7. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by Satsui no hado)
Would the asymptote no longer exist because of the propety of not existing for ?
...no

(Original post by SecondHand)
By oblique you mean when a graph tends to some function of x?
Yeah, well some linear function of x - in the form y = mx + c.
8. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by Ree69)
...no

Yeah, well some linear function of x - in the form y = mx + c.
Well it was worth a guess this topic is the one i need to work on as well as errors
9. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
So I guess that it would tend to ?
10. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
Has anyone done the June 2010 paper? If anyone has, can you help me with 9iii? Everything I try doesn't work out and is really long-winded.
11. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by hannnaaaaah)
Has anyone done the June 2010 paper? If anyone has, can you help me with 9iii? Everything I try doesn't work out and is really long-winded.
Okay it's quite clever and took me a while to spot. What you need to do is spot that the integral is some multiple of the region OA OB which you worked out earlier in the question. The next leap you need to make is working out how to rearrange into . You do this using the double angle formula, . You should be able to manage from there. The conclusion is that the area which you worked out in the first part is 1/8 of the area of the integral we are trying to work out. Hence it equals 24.
12. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
June 2010, 9(ii)

rearrange to polar

How? I know and but I can't rearrange to get the right answer. June 2010 was a really tough paper I thought.
Last edited by SecondHand; 07-06-2012 at 17:19.
13. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by SecondHand)
June 2010, 9(ii)

rearrange to polar

How? I know and but I can't rearrange to get the right answer. June 2010 was a really tough paper I thought.

and

Solve as a quadratic from here on in.

You can either do it in terms of cos or r... either way works
14. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
Sorry for the double post - just did Jan 11 which was tough in parts! I think the examiner must have been on the rag when they wrote that 56/72 for an A too which I thought was pretty high.
15. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by wibletg)

and

Solve as a quadratic from here on in.

You can either do it in terms of cos or r... either way works
Don't you think you made this difficult than it needed to be?

y^2= 2x+1

x^2+y^2= x^2+2x+1

r^2 = (x+1)(x+1)

r^2= (rcostheta+1)^2

r= rcostheta+1

r(1-costheta)=1

r= 1/(1-costheta)
16. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by f1mad)
Don't you think you made this difficult than it needed to be?

y^2= 2x+1

x^2+y^2= x^2+2x+1

r^2 = (x+1)(x+1)

r^2= (rcostheta+1)^2

r= rcostheta+1

r(1-costheta)=1

r= 1/(1-costheta)
Ooops, maybe I did

I did mine the same way as the markscheme though, suppose there's many ways of doing it.
17. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by wibletg)
Ooops, maybe I did

I did mine the same way as the markscheme though, suppose there's many ways of doing it.
Yeah there are, I'd like to think my way is one of the quickest .
Last edited by f1mad; 07-06-2012 at 23:04.
18. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by f1mad)
Yeah there are, I'd like to think my way is one of the quickest .
Yeah, does seem that way

One of the joys of further pure.
19. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
(Original post by f1mad)
Don't you think you made this difficult than it needed to be?

y^2= 2x+1

x^2+y^2= x^2+2x+1

r^2 = (x+1)(x+1)

r^2= (rcostheta+1)^2

r= rcostheta+1

r(1-costheta)=1

r= 1/(1-costheta)
Sexy.
20. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
Jan 10 9iib.

I keep getting as my minimum value of f(x). Also, how would you show that it's a minimum value without calculating the second derivative?
21. Re: FP2 (Not MEI) - Thursday June 14 2012, AM
^ Haven't looked at the paper, but try showing that a point to the left has a negative gradient and a point to the right has a positive gradient?
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