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Original post by Qwob
stuart, I also 256-16/pi^2 for c. Hoping this is the exact form of c, as that =254.4...

Can anyone remember the differential equation?

Was it dx/dt = (tcos(pi/4 t)/2x

?

Trying to redo my answer, but not sure if I've remembered the equation correctly.


Thank god someone else got the 16/pi^2 I'm not alone :biggrin:
Reply 521
Original post by georgeherby
i think you had 256 but you had to subtract 16 as the sin when to zero but the cos was 16 so C=240


Thinking about it when you were intergrating by parts you intergrated sin to give you (-)(-) cos... which would give you a positive.
Reply 522
Original post by MSI_10
Cardesian Q

K was 32?

And point Q was (-50,0)??


Wait what? There was a point Q we had to find the coordinates of? :s-smilie:

Could you explain the question more in detail like it was in the exam?
Original post by beccac94


I did and had to wait for ages for them to give me some :/ What did everyone get as the answer to the last question on the first page??? Out of everyone I've spoken to it seems like I was the only one who did that question...


What was the question :/
well, I've failed!


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Original post by stuart_aitken

For the final question....... First part, I did integration by parts to get something like tcos(tpi/4) + 16/pi^2 sin(tpi.4)
And for the next part I used that to do the various integrations. Only problem is my '+c' came out as 256-16/pi^2, which is different to what other people got.
And then when I plugged it all back into the original equation for when t=45... I forgot to add my +c into it!!! So my final answer was negative. I then realised I'd done it wrong but only had about 15 seconds left, so instead just wrote a note saying, "argh, forgot the +c, boohoo." Haha.



You got the same C as me :biggrin:

Original post by Mr Clingfilm
What was the question :/


Erm it was something like find the area beneath the curve of the answer to the one above it between x=0 and x=-1 or something.
(edited 10 years ago)
did anyone else run out of time? I never usually do but on this one, I looked up at the clock and it said I had half an hour left and I hadn't even attempted the last 2/3 questions :frown:


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Reply 527
Original post by Qwob
Wait what? There was a point Q we had to find the coordinates of? :s-smilie:

Could you explain the question more in detail like it was in the exam?


Normal at point P intersects x axis at point Q it was a 3 marker. Something like that
Reply 528
Was anyone able to combine logs at Q 1? It was annoying but I got 11/3 ln 2 at the end..

What was C in q1 tho? It was a 1 mark question but stumped me :s-smilie:
Reply 529
I found it disgusting. Yuk. Rather the January paper over this. SOOO hard ;/

I can't even remember half of my answers, and I don't actually want to. Not going to discuss it because what's the point

Hopefully my c3 resit will boost it up so I can still get an A overall :/

SO angry its unreal
Original post by MSI_10
Was anyone able to combine logs at Q 1? It was annoying but I got 11/3 ln 2 at the end..

What was C in q1 tho? It was a 1 mark question but stumped me :s-smilie:


I got the 11/3 ln 2 as well :smile:
I got 2 for C
Reply 531
Original post by Onoush
I think I dropped at least 20 marks. Hardest paper I've ever done.


is that it 20? i've dropped 30 -.-
Original post by MSI_10
Was anyone able to combine logs at Q 1? It was annoying but I got 11/3 ln 2 at the end..

What was C in q1 tho? It was a 1 mark question but stumped me :s-smilie:

I got 11/3 ln 2 ! :smile:
Reply 533
well... for what its worth I found the paper rather challenging. I thought (wrongly) that after Jan13 they would give us an easier paper. However, it turns out that this was certainly not the case, and although the paper was not as difficult as jan13 imo, the grade boundaries will not be too dissimilar (probably about 60 for an A*). Did anyone else get 365cm for the last question?
Reply 534
Original post by georgeherby
You used the identity (1-2sin^2) for cos2A. Then multiplied out the bracket and then simplified to give you -4sin^3+11sin+3=0 Then take it to the other side. to give 4sin^3-11sin-3=0 and then sub sin=x to give 4x^3+11x+3=0


I meant the part after that

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Reply 535
Original post by MSI_10
Normal at point P intersects x axis at point Q it was a 3 marker. Something like that


Oh you are ****ting me. I think I completely missed that question out.
I wasn't expecting it to be so hard, it was a horrible paper! :frown: I got C as 240 on the last question, but people seem to have gotten things involving pi^2. On the vectors question I couldn't do the last bit and didn't get anything over 7 for point D :/ I thought I'd be able to get an A* quite easily but I've completely screwed it up, unless the boundaries are about 58/75 I don't stand a chance.
Reply 537
1.3cos((pi/6)t) for q7???
Reply 538
Original post by EmilyC8057
I got 11/3 ln 2 ! :smile:


Same here.
What did people get for the bionomial expansion question? I normally get this right but I really struggled with this in this paper. :frown:

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