FP2 (Not MEI) - Thursday June 14 2012, AM
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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Re: FP2 (Not MEI) - Thursday June 14 2012, AMWould the asymptote no longer exist because of the propety of(Original post by Ree69)
Here's a brainteaser:
say a graph,
has an oblique asymptope. How would this asymptope change when sketching 
?
not existing for 
?
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Re: FP2 (Not MEI) - Thursday June 14 2012, AM...no(Original post by Satsui no hado)
Would the asymptote no longer exist because of the propety of not existing for ?
Yeah, well some linear function of x - in the form y = mx + c.(Original post by SecondHand)
By oblique you mean when a graph tends to some function of x? -
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Re: FP2 (Not MEI) - Thursday June 14 2012, AMOkay 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(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.
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.
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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
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Re: FP2 (Not MEI) - Thursday June 14 2012, AMDon't you think you made this difficult than it needed to be?(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
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) -
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Re: FP2 (Not MEI) - Thursday June 14 2012, AMOoops, maybe I did(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)
I did mine the same way as the markscheme though, suppose there's many ways of doing it. -
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Re: FP2 (Not MEI) - Thursday June 14 2012, AMYeah there are, I'd like to think my way is one of the quickest(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..
Last edited by f1mad; 07-06-2012 at 23:04. -
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Re: FP2 (Not MEI) - Thursday June 14 2012, AMSexy.(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)
this topic is the one i need to work on as well as errors
56/72 for an A too which I thought was pretty high.
of further pure.
