# Edexcel FP2 June 2015 - Official ThreadWatch

3 years ago
#141
Does anyone know why the S13 FP2 paper was withdrawn?
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3 years ago
#142
(Original post by 13 1 20 8 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.

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3 years ago
#143
(Original post by 13 1 20 8 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...!
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3 years ago
#144
(Original post by physicsmaths)
Assuming true for positive integers. Let n=-m where m<0
The proof follows.

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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?
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3 years ago
#145
(Original post by physicsmaths)
Assuming true for positive integers. Let n=-m where m<0
The proof follows.

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I know a proof follows from that; I was saying that they wouldn't ask us to do it with induction
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3 years ago
#146
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
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3 years ago
#147
(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
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3 years ago
#148
I'm just hoping we won't need to sketch a polar curve..
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3 years ago
#149
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
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3 years ago
#150
(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

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3 years ago
#151
(Original post by physicsmaths)
arg(1-w) is a half line going down. Hint tan is odd fucntion so arg(w-1)=-pi/2

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Can you explain why arg(1-w)=pi/2 is the same as arg(w-1)=-pi/2?
Thanks
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3 years ago
#152
(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
1
3 years ago
#153
(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!
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3 years ago
#154
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?
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3 years ago
#155
(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.

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3 years ago
#156
(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.

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0
3 years ago
#157
(Original post by physicsmaths)
I guess i should have explained it more.

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http://www.thestudentroom.co.uk/show...953&p=56463361
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3 years ago
#158
I shall have a look now...

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3 years ago
#159
(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!
0
3 years ago
#160
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?
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