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# P3 - MEI OCR 16th June watch

1. Anybody take this paper today?

I thought it wasn't too bad and the comprehensive paper was quite self-explanatory.
How did everyone else find it? I did really badly in my M2 paper so hopefully this P3 paper can still scrape me an A
Hope everyone did well,
love rmyl xxx
2. I thought it was pretty hard. Question 3 was really hard and I didn't understand any of the double angle stuff, but the rest of it was alrite. Question 4 was wierd but I think it went quite well.
3. All of the questions were fine, i'll post up my answers just to see if i got them correct
4. yay! its over
i thought it was ok as far as a levels go
5. Well question one seemed abit too easy at times,
Question two didnt see too bad, for hte implicite differentiation did people do this method

y+ ln y = sinx
dy/dx + 1/y dy/dx = cosx
dy/dx (1 + 1/y) = cos x
dy/dx ((y+1)/y) = cos x
dy/dx = (ycosx)/(y+1)

Also the integration question

dy/dx = 5y/(2+x)(1-2x)

∫ 1/y dy = ∫ 5/(2+x)(1-2x) dx
ln y = ∫(1/(2+x) + 2/(1-2x)) dx
ln y = ln (2+x) - ln(1-2x) + c
ln y = ln ((2+x)/(1-2x) +c
y = (2+x)/(1-2x) + ec where ec = k
y = (2+x)/(1-2x) + k
when y = 2 x = 0
2 = 2/1 + k
k= 0
y=(2+x)/(1-2x)
6. (Original post by geekypoo)
Well question one seemed abit too easy at times,
Question two didnt see too bad, for hte implicite differentiation did people do this method

y+ ln y = sinx
dy/dx + 1/y dy/dx = cosx
dy/dx (1 + 1/y) = cos x
dy/dx ((y+1)/y) = cos x
dy/dx = (ycosx)/(y+1)

Also the integration question

dy/dx = 5y/(2+x)(1-2x)

∫ 1/y dy = ∫ 5/(2+x)(1-2x) dx
ln y = ∫(1/(2+x) + 2/(1-2x)) dx
ln y = ln (2+x) - ln(1-2x) + c
ln y = ln ((2+x)/(1-2x) +c
y = (2+x)/(1-2x) + ec where ec = k
y = (2+x)/(1-2x) + k
when y = 2 x = 0
2 = 2/1 + k
k= 0
y=(2+x)/(1-2x)
yeah i did the top one like that. But for some strange reason my brain froze at that next one, the only question i didn't do i think. It looks pretty right to me the way you've done it though.
7. How did people do the double angle part of question three,
x = 1 - cos2x and y=2sin2x

so if y= x
1-cos2x = 2sin2x
1- (cos²x - sin²x) = 2(2sinxcosx)
if 1 = cos²x + sin²x
cos²x + sin²x - cos²x + sin²x = 2(2sinxcosx)
2sin²x = 2(2sinxcosx)
2sin²x/2sinxcosx = 2
sinx/cosx = 2
tanx = 2

and the part where u do the equation thingy

cos2x = 1-x and sin2x = y/2

cos²2x = (1-x)² and sin²2x = y²/4

sin²2x+cos²2x = (1-x)² + y²/4
1 = 1 - 2x + x² + y²/4
y²= 8x - 4x²= 4x(2-x)
y= ±√(4x(2-x))

Well thats if i remember the quesion correctly
8. I didn't think the comprehension was bad either, section a was ok, couldn't do the integration bit of question 2, no one in my class did! Yay maths is over!!
9. Isnt the integration in question 2 what i just did? It's ok for some who have finished, i still have D1, M3, M4 and P5 to do
10. does anyone know what the grade boundaries were for the june 2004 exams (MEI)?
11. Well the grade boundaries are normally 80% for an A, 70% for a B, 60% for a C, 50% for a D, 40% for an E and lower for a U
A: 48
B: 42
C: 36
D: 30
E: 24
U < 24
12. I failed this paper so badly, I might as well have not turned up. Question 3 was awful, like, no marks awful/ , I screwed up the partial fractions integration question and many others. And then the comprehension paper was shockingly bad...
13. Well what were all the questions? can someone post them if they remember them all?
14. section a went really really well for me but i did horrible on the comprehension couldnt get anything to come out. I have to get 138/200 from this exam and mech 2 to get my A so i better keep my fingers crossed, wats done is done i suppose

good luck everyone for results day
15. Well thats a B and a C really isnt it? i need to get an extra 31marks in my retakes to get A, but i did really badly in this back in january, and on p4

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