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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Reply 880
June 14 FP2 paper
Question: https://gyazo.com/7465ec0794cc7e5591ff6c1cf5c9c738
Mark scheme: https://gyazo.com/05340a73e15953bf33df8feb2fd276b5
My question: e^-x is part of the complementary function, so why did they not use Axe^-x as the particular integral instead? Because from what i've learnt if the particular integral occurs in the complementary function then you'll have to times by x
[QUOTE="Patrick2810;65461003"]but what is the equation of the circle C? (after the transformation)[/QUOTE

its (u- 9/8)^2 + v^2 = 9/64
Original post by Nerrad
June 14 FP2 paper
Question: https://gyazo.com/7465ec0794cc7e5591ff6c1cf5c9c738
Mark scheme: https://gyazo.com/05340a73e15953bf33df8feb2fd276b5
My question: e^-x is part of the complementary function, so why did they not use Axe^-x as the particular integral instead? Because from what i've learnt if the particular integral occurs in the complementary function then you'll have to times by x


It's because in the complimentary function it doesn't have a constant in front of it. It has Acos3x or Bsin3x in front of it. If the complimentary function was of the form Ae^-x +Be^-3x then yes you test xe^-x because both the CF's and PI's e^-x have a constant in front of it and nothing else.
Reply 883
Original post by target21859
It's because in the complimentary function it doesn't have a constant in front of it. It has Acos3x or Bsin3x in front of it. If the complimentary function was of the form Ae^-x +Be^-3x then yes you test xe^-x because both the CF's and PI's e^-x have a constant in front of it and nothing else.


But why?
Reply 884
Original post by target21859
It's because in the complimentary function it doesn't have a constant in front of it. It has Acos3x or Bsin3x in front of it. If the complimentary function was of the form Ae^-x +Be^-3x then yes you test xe^-x because both the CF's and PI's e^-x have a constant in front of it and nothing else.


If you expand the bracket into A(e^-x)sin(3x) + B(e^-x)cos(3x) doesn't that makes it "have a constant in front of it" ?
Original post by Nerrad
June 14 FP2 paper
Question: https://gyazo.com/7465ec0794cc7e5591ff6c1cf5c9c738
Mark scheme: https://gyazo.com/05340a73e15953bf33df8feb2fd276b5
My question: e^-x is part of the complementary function, so why did they not use Axe^-x as the particular integral instead? Because from what i've learnt if the particular integral occurs in the complementary function then you'll have to times by x


You can't use e^-x for the PI because it is an invalid solution. You would get 9 = 27 essentially. If you make the the PI y = ke^x you get a valid solution. ke^x wouldn't work IF it gave LHS = 0. This would be invalid because you can't have a solution that satisfies LHS = 0 and LHS = 27e^x at the same time. Only in this situation would you use
y = kxe^-x

(I used k as a constant)
(edited 7 years ago)
Reply 886
Original post by Pyslocke
You can't use e^-x for the PI because it is an invalid solution. You would get 9 = 27 essentially. If you make the the PI y = ke^x you get a valid solution. ke^x wouldn't work IF it gave LHS = 0. This would be invalid because you can't have a solution that satisfies LHS = 0 and LHS = 27e^x at the same time. Only in this situation would you use
y = kxe^-x

(I used k as a constant)


You might wanna rephrase your answer because I don't think i understand what you're saying
Original post by Nerrad
If you expand the bracket into A(e^-x)sin(3x) + B(e^-x)cos(3x) doesn't that makes it "have a constant in front of it" ?


The point is that it's still being multiplied by other non-constant functions
Original post by Nerrad
If you expand the bracket into A(e^-x)sin(3x) + B(e^-x)cos(3x) doesn't that makes it "have a constant in front of it" ?


But it also has sin3x or cos3x attached to it. It's not Ae^-x on its own.
Reply 889
Original post by target21859
But it also has sin3x or cos3x attached to it. It's not Ae^-x on its own.


Ok thanks
Original post by Nerrad
Ok thanks


No problem
This is a bit unrelated to the thread but is there a way to remove the UCAS clearing pop-up notification? It comes up every time I refresh or click through a few pages. It's quite annoying ^^
Reply 892
Original post by target21859
But it also has sin3x or cos3x attached to it. It's not Ae^-x on its own.


Just had a thought because no matter how much you differentiate A(e^-x)sin(3x) + B(e^-x)cos(3x) it'd always be a product of 2 functions, never a function of e by itself, which was the explanation i needed...
Original post by target21859
This is a bit unrelated to the thread but is there a way to remove the UCAS clearing pop-up notification? It comes up every time I refresh or click through a few pages. It's quite annoying ^^


Ad block probs
Original post by 1 8 13 20 42
Ad block probs


Can't have it on my phone.
Original post by Nerrad
Just had a thought because no matter how much you differentiate A(e^-x)sin(3x) + B(e^-x)cos(3x) it'd always be a product of 2 functions, never a function of e by itself, which was the explanation i needed...


True that is a much better way of explaining it
stuck on fp2 june 09 question 6 where I can't get the correct circle equation, can someone have a look please!!
Reply 897
Original post by thatapanydude
stuck on fp2 june 09 question 6 where I can't get the correct circle equation, can someone have a look please!!


iw=w|-iw| = |w|

To further explain:

Spoiler

(edited 7 years ago)
Original post by edothero
iw=w|-iw| = |w|

To further explain:

Spoiler



Thanks mate, I could originally see where the -1 went but not the i, but I understand it know. Thanks again!!

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