Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

Hi, Please could someone do a worked solution to question 12a please june 2004 paper https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202004%20QP%20-%20FP2%20Edexcel.pdf as I don't really understand the method on the mark scheme, thank you :smile:

Reply 322

Original post by economicss

Is there a specific but that you don't understand about the markscheme? Have you looked at examsolutions.net videos on complex transformations? :smile:

Reply 323

Original post by Zacken

Is there a specific but that you don't understand about the markscheme? Have you looked at examsolutions.net videos on complex transformations? :smile:

Yeah, I understand how you get to the fraction representing w but I don't really understand how you'd know to go from here to finding the modulus of the numerator and denominator as I've not seen this method used before? :smile:

Yeah I have, but I'm still struggling with bits of them quite a lot :/ Thanks

Reply 324

Can someone please help me with this step? I don't understand where the (pie/3 - theta ) goes

btw it is from Jan05 p4 Q7
_Thanks_{btw it}

(edited 7 years ago)

Jan05Q7.PNG8.8KB

Reply 325

Original post by Cpj16

Can someone please help me with this step? I don't understand where the (pie/3 - theta ) goes

btw it is from Jan05 p4 Q7
_Thanks_{btw it}

Not sure if this helps, but this is the working I did for the question :smile:

Reply 326

Original post by economicss

Not sure if this helps, but this is the working I did for the question :smile:

thanks

, I do understand this (this was part a of the question )

if possible could you explain the step I attached earlier

edit:the image is part of the mark scheme and is not a question I just don't understand how they got there

(edited 7 years ago)

Reply 327

Original post by Cpj16

thanks

Sure, is it part c you'd like a worked solution to? :smile:

Reply 328

Original post by economicss

Sure, is it part c you'd like a worked solution to? :smile:

yes please that would be really helpful

Reply 329

Original post by Cpj16

Can someone please help me with this step? I don't understand where the (pie/3 - theta ) goes

btw it is from Jan05 p4 Q7
_Thanks_{btw it}

\displaystyle \cos \left(\frac{\pi}{3} - \theta\right) = \frac{1}{2}\cos \theta + \frac{\sqrt{3}}{2}\sin \theta

\displaystyle 6\cos \theta \left(\frac{1}{2}(\cos \theta + \sqrt{3}\sin \theta\right) = 3\cos^2 \theta + 3\sqrt{3}\sin \theta \cos \theta

Therefore:

\cos^2 \theta + \sqrt{3}\sin \theta \cos \theta = 1 \iff \cdots

Reply 330

Original post by Zacken

\displaystyle \cos \left(\frac{\pi}{3} - \theta\right) = \frac{1}{2}\cos \theta + \frac{\sqrt{3}}{2}\sin \theta

\displaystyle 6\cos \theta \left(\frac{1}{2}(\cos \theta + \sqrt{3}\sin \theta\right) = 3\cos^2 \theta + 3\sqrt{3}\sin \theta \cos \theta

Therefore:

\cos^2 \theta + \sqrt{3}\sin \theta \cos \theta = 1 \iff \cdots

thank you!! I would rate you up but tsr is telling to not rate the same member again

Reply 331

Original post by Cpj16

thanks

Reply 332

Original post by economicss

Thank you!!!

Reply 333

Original post by Cpj16

thank you!! I would rate you up but tsr is telling to not rate the same member again

Nobody seems to be able to rep me, conspiracy. :erm:

jk, no problem! :smile:

Reply 334

DORI

Original post by edothero

I'm half-way through TeeEm's booklet, some really really tough questions at the end..

where can I find this booklet!! need some extra revision material.
thanks
and good luck with your studies :biggrin:

Reply 335

Original post by Zacken

Is there a specific but that you don't understand about the markscheme? Have you looked at examsolutions.net videos on complex transformations? :smile:

It's probably obvious but on the MS how do you know that the square root of lambda+1 squared plus lambda squared is 1? Thanks :smile:

Reply 336

Original post by economicss

It's probably obvious but on the MS how do you know that the square root of lambda+1 squared plus lambda squared is 1? Thanks :smile:

It's not saying the square root of the thingies squared is 1. It's saying that

w = \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}

So that

\displaystyle |w| = \left| \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}\right| = \frac{|\lambda + 1 + \lambda i|}{|\lambda + (\lambda +1)i|} = \frac{\sqrt{(\lambda+1)^2 + \lambda^2}}{\sqrt{\lambda^2 + (\lambda + 1)^2}} = 1

i.e: the modulus of the numerator is the same as the modulus of the denominator. Hence the modulus of the entire thing is just 1.

Reply 337

Original post by Zacken

It's not saying the square root of the thingies squared is 1. It's saying that

w = \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}

So that

\displaystyle |w| = \left| \frac{\lambda + 1 + \lambda i}{\lambda + (\lambda +1)i}\right| = \frac{|\lambda + 1 + \lambda i|}{|\lambda + (\lambda +1)i|} = \frac{\sqrt{(\lambda+1)^2 + \lambda^2}}{\sqrt{\lambda^2 + (\lambda + 1)^2}} = 1

i.e: the modulus of the numerator is the same as the modulus of the denominator. Hence the modulus of the entire thing is just 1.

Ah yes ofc, I'm being stupid! Thanks for your help :smile:

Reply 338