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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Original post by target21859
That's what I got. Hope I didn't mess up on 5 though I don't think I did.


Except for your coefficients of sintheta I think. I know I got it right I checked it.
Reply 1421
I left my answer for 8b exact, will I be penalised (It's 32.5 when you calculate it which is supposedly correct). Didn't do it purposefully, am just a little special
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!


Yes! If these answers are legit, I got most of this right.
Original post by taichingkan
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.


For last q I got it but put it in the form 49/something pi - 5.(something) it didn't state what form to put it so would that gain absolute marks?
(People who know how the first order part b was done please help)

I started integrating the normal way, with e2theta sin theta, then realised I couldn't do that so I looked and part a and realised I could use it. I didn't cross this part out - just tried another method.
I got to the point where I got two equations equating with r, p and q.
I then managed to get x in terms of r and t.

How many marks might I have gained??
For Q8:
Does 2 * [(region C1 from 0 to arrcos(3/4)) + (region C2 from arccos(3/4) to pi/2)] sound valid?
Original post by Rkai01
For last q I got it but put it in the form 49/something pi - 5.(something) it didn't state what form to put it so would that gain absolute marks?


I left it as that horrible mess cause they didn't say to put it to 3sf or anything and an exact answer is always better, you should get the marks.
Original post by Ewanclementson
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!


Couldn't do q2partb, q4partb or q8partb.

Oh dear..
Interesting (and viable(!)) solution to integrating e^(2x) sin x:

Anyone else not finish in time? :frown:
Original post by Ewanclementson
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!


For 4b I got C as 1/5
I was tempted to do 8 b using decimals however I would have probably lost marks for that. In stead I chose to keep everything to exact values using sin2θ=1(2cos2θ1)2 \sin{ 2 \theta } = \sqrt{ 1 - (2 \cos^2{\theta} -1)^2 } and probably made a mistake or two along the way. At least I got pi, arccos(3/4) and root3 in my final answer!
(edited 7 years ago)
Original post by themechguy
(People who know how the first order part b was done please help)

I started integrating the normal way, with e2theta sin theta, then realised I couldn't do that so I looked and part a and realised I could use it. I didn't cross this part out - just tried another method.
I got to the point where I got two equations equating with r, p and q.
I then managed to get x in terms of r and t.

How many marks might I have gained??


Pretty sure you can't use part a -- it wasn't in the same form. r was constant whereas sin(theta) isn't.
Original post by cjlh
For Q8:
Does 2 * [(region C1 from 0 to arrcos(3/4)) + (region C2 from arccos(3/4) to pi/2)] sound valid?


It's what I did (except I forgot to multiply by 2 ****)
Been getting As and A*s in past papers and this exam was a definite U

Just so many little things completely messed with me and so I answered barely anything

There goes my shot at Imperial out the window unless I can get the grades with my other 3 subjects...


#$@&%*!
Original post by Zacken
Heh, the IAL paper had -1 +- sqrt(11). :lol:


What did you get in the last part of the last question?
Original post by maxguywell69
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!

.
Right, it turns out i took away (n/2(n+2)) so got -11 instead of -9 on question 2.
Anyone remember the question of Q4 a?
Original post by Jordan97
For 4b I got C as 1/5


Yes but on the LHS we have e^2theta*x and so this 1/5 becomes (e^-2theta)/5
What is question 3 and question 6?

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