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FP3 june 2013 OCR (not MEI)

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http://www.examsolutions.net/maths-revision/syllabuses/Index/period-1/Further-Pure/module.php If you scroll down you can find a section on De-Moivres Theorem and First/Second Order Differential equations. Been very useful for my revision so far. :smile:
Reply 21
Does anyone have the Jan 2013 paper?
Reply 22
FP3 Jan 2013 attached,

I dont have the Mark Scheme unfortunately.
Original post by sj_1995
FP3 Jan 2013 attached,

I dont have the Mark Scheme unfortunately.


Luckily, I DO have the mark scheme :smile:

I found this to be one of the harder papers when I did it, some of the stuff really threw me off for a second :/
Reply 24
The grade boundary does suggest it was a bit harder (53 for an A compared to 64 for FP2 but that was ridiculous)

I thought it was alright, the last part of the first groups question confused, as well as the geometric progression in question 7.
Reply 25
Just did the jan 2013 paper found it okay :smile:. Seemed a bit harder than usual? I'm really worried for FP2 though!
Can someone send me the jan 13 paper please


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Reply 27
Already posted Coolina133! On the jan 13 paper I didn't get question 7ii, I would have thought that sin(1/2theta) if theta is 2npi you'd get 0 and that therefore S would be undefined? But apparently it makes it = 10!???
Reply 28
Anyone have stuff they're stuck on ?
Reply 29
Original post by kabutsu12
Anyone have stuff they're stuck on ?


Everythings going good so far, but sometimes get the integrals wrong when asked to do it using complex numbers.
Original post by kabutsu12
Already posted Coolina133! On the jan 13 paper I didn't get question 7ii, I would have thought that sin(1/2theta) if theta is 2npi you'd get 0 and that therefore S would be undefined? But apparently it makes it = 10!???


That's what I thought, but then it actually says for where theta isn't 2npi :')

Stick it into the original formula with the exponentials and you get 1+1+1+...=10, took me a few minutes to realise haha.

And the exam should be fine, so long as I don't do 2-2=5 when I'm getting vector products, they're my biggest downfall
(edited 10 years ago)
Reply 31
Original post by jamesisenglish
That's what I thought, but then it actually says for where theta isn't 2npi :')

Stick it into the original formula with the exponentials and you get 1+1+1+...=10, took me a few minutes to realise haha.

And the exam should be fine, so long as I don't do 2-2=5 when I'm getting vector products, they're my biggest downfall


Ive done that :P. also I don't like spiral enlargements.
Original post by Music99
Ive done that :P. also I don't like spiral enlargements.


Sometimes on the complex numbers questions I just can't see what they want me to do after using de Moivre to get some identity, think it's more a confidence issue than anything haha
Reply 33
Original post by jamesisenglish
Sometimes on the complex numbers questions I just can't see what they want me to do after using de Moivre to get some identity, think it's more a confidence issue than anything haha


Oh I'm fine of those. It's usually just trig identities, although some of the polynomial things after are difficult!
Original post by Music99
Oh I'm fine of those. It's usually just trig identities, although some of the polynomial things after are difficult!


That's what I mean, like I can get a bit confused on how they want me to get values from a polynomial, I was terrible at it at first but I'm usually fine now :')
Reply 35
Original post by jamesisenglish
That's what I mean, like I can get a bit confused on how they want me to get values from a polynomial, I was terrible at it at first but I'm usually fine now :')


Yeah just went through the jan 2013 paper again need to do some more practice on complex number applications.
Reply 36
Original post by jamesisenglish
That's what I thought, but then it actually says for where theta isn't 2npi :')

Stick it into the original formula with the exponentials and you get 1+1+1+...=10, took me a few minutes to realise haha.

And the exam should be fine, so long as I don't do 2-2=5 when I'm getting vector products, they're my biggest downfall


Hmm, I swear it says find S when theta is 2npi! although I see how the original method works, I reckon expanding it and then getting two fractions would sort it out.
Reply 37
With the polynomials the thing that got me at first was that even though you are putting a value of theta in it doesn't mean the values of theta on the otherside need to be subbed in. But that they're not in because you need to find the other values that would give that value, kind off... :/
Reply 38
Original post by kabutsu12
Hmm, I swear it says find S when theta is 2npi! although I see how the original method works, I reckon expanding it and then getting two fractions would sort it out.


Nah it's when it doesn't :smile:.
could anyone explain something about the roots of the auxiliary equation in 2nd order diff eqns being complex?
for example if the roots of the eqn are lambda = 3 +- 2i, this means complementary function is Aexp((3+2i)x) + Bexp((3-2i)x)
and this eventually boils down to y=exp(3x)((A+B)cos2x + (A-B)isin2x). Now you can continue to write A+B and (A-B)i as separate constants C and D. Can anyone explain why (A-B)i is not complex?

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