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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Answers - let me know if they need to be edited
1) -2<x<-root2 -1<x<root2
2) shown x(x^2-3x-9)/2(x+2)
3)e^ipi(1/24, 13/14, -11/24, -23/24) shown
4) r/q(1-e^-(q/p)t) (2sinx-cosx+e^-2)/5
5) Coefficients are 1/16, -5/15, 10/16 shown
6) coefficients are 1 2 2 8/3 then shown
7) Ax^2+Bx-1/12x^(-2)
8) 21/4, (+-)arcos(3/4) 32.519 which I think was 3root7/4-11arcos3/4+49pi/4
Hope this helps everyone!
Original post by lukejoshjames
yuuup :biggrin:


I think I botched the last question though (the last part). I got some arccos(3/4) terms and some pi terms and some root 3 terms.

Did anyone else do the area of C2 in the region arccos(3/4) to -arccos(3/4) + two times the area of C1 in the region pi/2 to arccos(3/4)?
For differences I just used the summation formulae for the first two term (n and minus 3), and solved the other two terms normally.
Original post by RThornton
Question 4 Integral Solution

Question 4b

let In=e^2x*sinx

By parts twice gives

In= 0.5e^2x*sinx - 0.25e^2x*cosx - Integral of 0.25e^2x*sin x
Integral of 0.25e^2x*sin x = 0.25 InHence, 5/4 In = 0.5e^2x*sinx - 0.25e^2x*cosxTherefore In = 1/5e^2x * (2sinx - cosx)

Nope the coefficients of sinx and cosx didnt have e-2theta in the final answer
Original post by taichingkan
Yep that's what I got too. Thought I made a mistake when I saw sqrt(2) but my calculator confirmed it.


Heh, the IAL paper had -1 +- sqrt(11). :lol:
Original post by JakeTan
I got r/q or something, because as t tends to infinity e^pt/q tends to zero.


I really hope I put r/q instead of t/q. My ts look like rs...
How many marks for the area question?


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that **** was hard af
Original post by Ewanclementson
Answers - let me know if they need to be edited
1) -2<x<-root2 -1<x<root2
2) shown x(x^2-3x-9)/2(x+2)
3)e^ipi(1/24, 13/14, -11/24, -23/24) shown
4) r/q(1-e^-(q/p)t) (2sinx-cosx+e^-2)/5
5) Coefficients are 1/16, -5/15, 10/16 shown
6) coefficients are 1 2 2 8/3 then shown
7) Ax^2+Bx-1/12x^(-2)
8) 21/4, (+-)arcos(3/4) 32.519 which I think was 3root7/4-11arcos3/4+49pi/4
Hope this helps everyone!


All same apart from 8b, I got 15.something but must be a mistake on my part as most people seem to agree with 32.5. I did get the same terms, but the wrong coefficients.
Original post by anhaanha
I do it like this
Let t = intergrate e^2x sin x
then you by part you got
-cosx.e^2x + intergrate(2e^2x. cosx)
then by part again you got
-cosx.e^2x + 2 ( e^2x. sinx - intergrate(2e^2x.sinx) )
Therefore
t = -cosx.e^2x + 2e^2x.sinx - 2t
t = 1/3 ( 2e^2x. sint - cosx.e^2x)


Careful, the last term should be -4t so the 1/3 should be 1/5.
Original post by Ewanclementson
Answers - let me know if they need to be edited
1) -2<x<-root2 -1<x<root2
2) shown x(x^2-3x-9)/2(x+2)
3)e^ipi(1/24, 13/14, -11/24, -23/24) shown
4) r/q(1-e^-(q/p)t) (2sinx-cosx+e^-2)/5
5) Coefficients are 1/16, -5/15, 10/16 shown
6) coefficients are 1 2 2 8/3 then shown
7) Ax^2+Bx-1/12x^(-2)
8) 21/4, (+-)arcos(3/4) 32.519 which I think was 3root7/4-11arcos3/4+49pi/4
Hope this helps everyone!


Thanks but q5 those were not my coefficients. Can you check yours again
Did anyone get 23.9 something for the area of c2 ??
Original post by Ewanclementson
Answers - let me know if they need to be edited
1) -2<x<-root2 -1<x<root2
2) shown x(x^2-3x-9)/2(x+2)
3)e^ipi(1/24, 13/14, -11/24, -23/24) shown
4) r/q(1-e^-(q/p)t) (2sinx-cosx+e^-2)/5
5) Coefficients are 1/16, -5/15, 10/16 shown
6) coefficients are 1 2 2 8/3 then shown
7) Ax^2+Bx-1/12x^(-2)
8) 21/4, (+-)arcos(3/4) 32.519 which I think was 3root7/4-11arcos3/4+49pi/4
Hope this helps everyone!


Finally someone who can do integration by parts! Though i think i got a different value of b for question 2, everything else is the same, besides the typos. Thanks!
Reply 1413
Horrible horrible stuff.. I revised so hard and got 73/75 for the last paper I did yesterday all to have it wasted today.. I think I got 53/75, I'm just hoping that is a B because otherwise ill need 90% on M2 and S2 to get an A
Reply 1414
Original post by anndz3007
Did anyone get 23.9 something for the area of c2 ??


I DID !!
Original post by Ewanclementson
Answers - let me know if they need to be edited
1) -2<x<-root2 -1<x<root2
2) shown x(x^2-3x-9)/2(x+2)
3)e^ipi(1/24, 13/14, -11/24, -23/24) shown
4) r/q(1-e^-(q/p)t) (2sinx-cosx+e^-2)/5
5) Coefficients are 1/16, -5/15, 10/16 shown
6) coefficients are 1 2 2 8/3 then shown
7) Ax^2+Bx-1/12x^(-2)
8) 21/4, (+-)arcos(3/4) 32.519 which I think was 3root7/4-11arcos3/4+49pi/4
Hope this helps everyone!

That's what I got. Hope I didn't mess up on 5 though I don't think I did.
Original post by Ewanclementson
...
2) shown x(x^2-3x-9)/2(x+2)
...


Didn't get -3, got -9 though I think. If this is wrong how many marks do people think would be lost?
Answers - let me know if they need to be edited (edit - typo on 5 fixed)
1) -2<x<-root2 -1<x<root2
2) shown x(x^2-3x-9)/2(x+2)
3)e^ipi(1/24, 13/14, -11/24, -23/24) shown
4) r/q(1-e^-(q/p)t) (2sinx-cosx+e^-2)/5
5) Coefficients are 1/16, -5/16, 10/16 shown
6) coefficients are 1 2 2 8/3 then shown
7) Ax^2+Bx-1/12x^(-2)
8) 21/4, (+-)arcos(3/4) 32.519 which I think was 3root7/4-11arcos3/4+49pi/4
Hope this helps everyone!
Original post by Zacken
Heh, the IAL paper had -1 +- sqrt(11). :lol:


haha, i personally thought the numbers were ridiculous. I checked my work for so long when I got (arccos 3/4) for Q8a, only to realise that it was actually correct.
Original post by Louisb19
I think I botched the last question though (the last part). I got some arccos(3/4) terms and some pi terms and some root 3 terms.

Did anyone else do the area of C2 in the region arccos(3/4) to -arccos(3/4) + two times the area of C1 in the region pi/2 to arccos(3/4)?


I don't think it required an exact answer, so if it came to about 32.5 (or whatever the answer was) then it should still be correct.
Also yeah, that's what I did.

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