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Edexcel FP2 - June 3rd, 2015 [Exam discussion thread]

A thread for the FP2 exam this year.

UNOFFICIAL MARK SCHEME POSTED BY ARSEY:

http://www.thestudentroom.co.uk/showthread.php?p=56530957&highlight=arsey

ANSWERS:

1) −6<x<−3-6<x<-3 or x>1x>1, introducing mod gives x>1x>1

2) mod  =4arg⁡=2π3\mod=4 \arg=\frac{2\pi}{3}, z6=4096z^6=4096, w=±22w=\pm 2\sqrt2 and ±22i\pm 2\sqrt{2}i

3) y=3cos⁡x−2cos⁡3x+csin⁡xy=\dfrac{3\cos x-2\cos^3x+c}{ \sin x} but your answer may look very different and still be correct

4) Series proofs

5) Centre (−45,0),r=65(-\frac{4}{5}, 0), r=\frac{6}{5}, region inside of circle shaded

6) A=3,B=−1A=3, B=-1 Series coeffs: 3,83,40,176333, 8\sqrt3, 40, \frac{176}{3}\sqrt3

7) Polar turning point (9a2,π3)(\frac{9a}{2}, \frac{\pi}{3}), p=94,q=8116p=\frac{9}{4}, q=\frac{81}{16}

8)y=ln⁡x8+116+x4(Aln⁡x+B)y=\frac{ \ln x}{8}+\frac{1}{16}+x^4(A \ln x+B)
(edited 8 years ago)

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Reply 1
Avatar for Elcor
Elcor
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14
Show that ∣2z+1+i∣=2∣z+1+i∣|2z+1+i|=2|z+1+i| under the transformation w=z+1iz−1w=\dfrac{z+1}{iz-1} makes the circle ∣w−i∣=2∣w−1∣|w-i|=\sqrt{2}|w-1|

Can anyone post their solution to this? I went about half-way before realising that I can't see myself getting the right answer with what I had. I'm wondering if there's a sign error or something in the question (from the purple Oxford FP2 exercise book, Q8 Ex. 3.7).
Original post by Elcor
Show that ∣2z+1+i∣=2∣z+1+i∣|2z+1+i|=2|z+1+i| under the transformation w=z+1iz−1w=\dfrac{z+1}{iz-1} makes the circle ∣w−i∣=2∣w−1∣|w-i|=\sqrt{2}|w-1|

Can anyone post their solution to this? I went about half-way before realising that I can't see myself getting the right answer with what I had. I'm wondering if there's a sign error or something in the question (from the purple Oxford FP2 exercise book, Q8 Ex. 3.7).

Isn't there already a thread for FP2 already?
Ah well.
First I expanded the first equation, then I changed the subject in the second equation to z and equate real parts and imaginary parts, so I would get x and y in terms of u and v.
Donno why the image is sideways though.
IMG_1423.jpg523.1KB
Reply 3
Avatar for Elcor
Elcor
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Original post by simonli2575
Isn't there already a thread for FP2 already?
Ah well.
First I expanded the first equation, then I changed the subject in the second equation to z and equate real parts and imaginary parts, so I would get x and y in terms of u and v.
Donno why the image is sideways though.


I thought there was no thread which is why I made one

Thanks :smile:
Reply 4
Avatar for Elcor
Elcor
OP
14
For inequality questions, how are you meant to deduce the new inequality when they apply a modulus function to it? Is it just that any closed inequality disappears (this is just based on one example I've done)?
Reply 5
Avatar for Navo D.
Navo D.
2
Original post by Elcor
For inequality questions, how are you meant to deduce the new inequality when they apply a modulus function to it? Is it just that any closed inequality disappears (this is just based on one example I've done)?


I don't fully understand your question, could you post a problem where you're unsure about what to do?
Reply 6
Avatar for Elcor
Elcor
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Original post by Navo D.
I don't fully understand your question, could you post a problem where you're unsure about what to do?


This is from June 2010: http://gyazo.com/0d59e3efa592c83184dc168f5b821d6c

For part (b), is there some simple method like any closed inequalities disappear, or would you just test the sign of x+4−2∣x+3∣x+4-\dfrac{2}{|x+3|} in-between the critical values you found in part (a)?
Reply 7
Avatar for Navo D.
Navo D.
2
Original post by Elcor
This is from June 2010: http://gyazo.com/0d59e3efa592c83184dc168f5b821d6c

For part (b), is there some simple method like any closed inequalities disappear, or would you just test the sign of x+4−2∣x+3∣x+4-\dfrac{2}{|x+3|} in-between the critical values you found in part (a)?


Not sure if this would work but I'd sketch a graph of the two functions and see what range of values from my answer to part (A) would satisfy the new inequality
Reply 8
Avatar for Elcor
Elcor
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Original post by Navo D.
Not sure if this would work but I'd sketch a graph of the two functions and see what range of values from my answer to part (A) would satisfy the new inequality


That's a lot of time for 1 mark
Reply 9
Avatar for Navo D.
Navo D.
2
Original post by Elcor
That's a lot of time for 1 mark


Yeah you're right... :s-smilie: not sure then, does Arsey have worked solns for this paper?
Reply 10
Avatar for Elcor
Elcor
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14
Original post by Navo D.
Yeah you're right... :s-smilie: not sure then, does Arsey have worked solns for this paper?


Don't think so. I think you just plug in values like I suggested and check the sign.
Could they ask us to do a proof by induction on something other than the proof for de moivre (like the attached q) ?

Posted from TSR Mobile
1429348256974.jpg21.7KB
Also, how do you do this q?:
The transformation T from the complex z-plane to the complex w-plane is given by w=(z+1)/(z+i)

(a) Show that T maps points on the half-line arg(z) =pi/4 in the z-plane into points on the circle |w| =1 (4 marks)
Original post by BP_Tranquility
Also, how do you do this q?:
The transformation T from the complex z-plane to the complex w-plane is given by w=(z+1)/(z+i)

(a) Show that T maps points on the half-line arg(z) =pi/4 in the z-plane into points on the circle |w| =1 (4 marks)

That's just a half-line with equation y=x.
Reply 14
Avatar for Elcor
Elcor
OP
14
Original post by BP_Tranquility
Also, how do you do this q?:
The transformation T from the complex z-plane to the complex w-plane is given by w=(z+1)/(z+i)

(a) Show that T maps points on the half-line arg(z) =pi/4 in the z-plane into points on the circle |w| =1 (4 marks)


Find z in terms of w, sub that in the arg then you should be able to use arg rules to get the answer

Posted from TSR Mobile
If it helps at all I have written some free study materials for Edexcel FP2 (with some MEI sprinkled in there). I have also posted in the FP1 preparation thread if anyone is resitting that.

Good luck!
is the conventional range of theta for both complex nos and polar coordinates -pi to pi, and is that usually used in exams?
Original post by BP_Tranquility
Also, how do you do this q?:
The transformation T from the complex z-plane to the complex w-plane is given by w=(z+1)/(z+i)

(a) Show that T maps points on the half-line arg(z) =pi/4 in the z-plane into points on the circle |w| =1 (4 marks)


i think you equate rwal and imaginary parts
Reply 18
Avatar for Navo D.
Navo D.
2
Original post by Ilovemaths96
is the conventional range of theta for both complex nos and polar coordinates -pi to pi, and is that usually used in exams?


yep as far as i know this is the convention for edexcel
Original post by Navo D.
yep as far as i know this is the convention for edexcel


thanks

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