The Student Room Group

Edexcel FP2 June 2015 - Official Thread

Scroll to see replies

Does anyone know why the S13 FP2 paper was withdrawn?
Original post by 1 8 13 20 42
No; they may seem to enjoy challenging people lately but I don't think they're allowed to ask that


Assuming true for positive integers. Let n=-m where m<0
The proof follows.


Posted from TSR Mobile
Original post by 1 8 13 20 42
No; they may seem to enjoy challenging people lately but I don't think they're allowed to ask that


I know, S3 didn't go at all as planned....but I think asking for a proof by induction for negative integers goes beyond what they could reasonably ask. Let's hope...!
Original post by physicsmaths
Assuming true for positive integers. Let n=-m where m<0
The proof follows.


Posted from TSR Mobile


But how do you do it in terms of proof by induction? With K, K+1 etc, as how would that work with a negative integer proof for de moire?
Original post by physicsmaths
Assuming true for positive integers. Let n=-m where m<0
The proof follows.


Posted from TSR Mobile


I know a proof follows from that; I was saying that they wouldn't ask us to do it with induction
can somebody help me out as to why in june 09 q 6b they shade the outside region of the circle rather than inside esp as they stated its less than 3 :s
Original post by Maths degree
can somebody help me out as to why in june 09 q 6b they shade the outside region of the circle rather than inside esp as they stated its less than 3 :s


The locus of z was z < |3|, you're sketching the locus of w
If you go back through your workings to find the locus of w, imagine a less than sign instead of an equals sign (look carefully to see if it will have swapped at any point) and you'll find the inequality for w
I'm just hoping we won't need to sketch a polar curve..
Reply 148
I just have a couple questions:1. Do we need to know how to prove things like the integrating factor etc. or is it just De Moivres that we'll be expected to prove in the exam?2. In Exercise 3E Q7a When they are solving it they add the 2k(pi) AFTER they multiply the equation by 4 using De Moivres, and I tried to solve it by adding 2kpi before the multiplication but got different answers. So when we get a question like this do we only add 2kpi after multiplying and before ''dividing (by 3 in this case)'' and why does it make a difference?3.In Example 31 in Complex Numbers Pg 48 , For finding Min/Max values does the line ALWAYS have to pass through the centre of the circle regardless of where it starts ie max from z-1 4. And finally how do you draw arg(1-w)=pi/2 ?Thanks in advance
Original post by Mjmuk
I just have a couple questions:1. Do we need to know how to prove things like the integrating factor etc. or is it just De Moivres that we'll be expected to prove in the exam?2. In Exercise 3E Q7a When they are solving it they add the 2k(pi) AFTER they multiply the equation by 4 using De Moivres, and I tried to solve it by adding 2kpi before the multiplication but got different answers. So when we get a question like this do we only add 2kpi after multiplying and before ''dividing (by 3 in this case)'' and why does it make a difference?3.In Example 31 in Complex Numbers Pg 48 , For finding Min/Max values does the line ALWAYS have to pass through the centre of the circle regardless of where it starts ie max from z-1 4. And finally how do you draw arg(1-w)=pi/2 ?Thanks in advance


arg(1-w) is a half line going down. Hint tan is odd fucntion so arg(w-1)=-pi/2


Posted from TSR Mobile
Original post by physicsmaths
arg(1-w) is a half line going down. Hint tan is odd fucntion so arg(w-1)=-pi/2


Posted from TSR Mobile


Can you explain why arg(1-w)=pi/2 is the same as arg(w-1)=-pi/2?
Thanks
Reply 151
Original post by Anshul6974
Can you explain why arg(1-w)=pi/2 is the same as arg(w-1)=-pi/2?
Thanks


arg(1-w) = pi/2
=> arg[-1(w-1)] = pi/2
arg(-1) + arg(w-1) = pi/2
pi + arg(w-1) = pi/2
therefore: arg(w-1) =pi/2 - pi
arg(w-1) = -pi/2
(edited 8 years ago)
Reply 152
Original post by Boop.
arg(1-w) = pi/2
=> arg[-1(w-1)] = pi/2
arg(-1) + arg(w-1) = pi/2
pi + arg(w-1) = pi/2
therefore: arg(w-1) =pi/2 - pi
arg(w-1) = -pi/2


Thank you, this makes a lot more sense!
For the proof by induction of De Moivre, is it acceptable to prove it generally for k? What I mean is can we not just use the letter k then say somewhere at the top that (k<0)? Or do they want us to use the negative sign i.e. when n=-k, then when n = (-k-1) etc?
Original post by Anshul6974
Can you explain why arg(1-w)=pi/2 is the same as arg(w-1)=-pi/2?
Thanks


Tan is odd function.


Posted from TSR Mobile
Original post by Boop.
arg(1-w) = pi/2
=> arg[-1(w-1)] = pi/2
arg(-1) + arg(w-1) = pi/2
pi + arg(w-1) = pi/2
therefore: arg(w-1) =pi/2 - pi
arg(w-1) = -pi/2


I guess i should have explained it more.


Posted from TSR Mobile
Original post by physicsmaths
I guess i should have explained it more.


Posted from TSR Mobile


Bro, I need your help :frown:

http://www.thestudentroom.co.uk/showthread.php?t=3373953&p=56463361
Original post by Boop.
arg(1-w) = pi/2
=> arg[-1(w-1)] = pi/2
arg(-1) + arg(w-1) = pi/2
pi + arg(w-1) = pi/2
therefore: arg(w-1) =pi/2 - pi
arg(w-1) = -pi/2


Thanks a lot!
Out of interest would we be expected to sketch something like arg((z-z1)(z-z2)=pi/3 and if so what would it look like?

Quick Reply

Latest

Trending

Trending