The Student Room Group
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>
<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

Scroll to see replies

Reply 480
Avatar for oinkk
oinkk
3
Original post by ChuckNorriss
guys has anyone come across a past paper question similar to example 13 from chapter 7?


Nah.

I think that example is there to help sketch polar curves. That is to say, remember the conditions it proves to assist you in drawing the curves.
Reply 481
Avatar for edothero
edothero
OP
15
11 days left.

Get revising! :biggrin:
Reply 482
Avatar for oinkk
oinkk
3
Original post by edothero
11 days left.

Get revising! :biggrin:


AAAH! I still feel like it's months away... it's scary!
Reply 483
Avatar for edothero
edothero
OP
15
Original post by oinkk
AAAH! I still feel like it's months away... it's scary!


FP3 is still a month away :laugh:
Reply 484
Avatar for oinkk
oinkk
3
Original post by edothero
FP3 is still a month away :laugh:


That is true.

But FP2 was a month away almost a month ago, and look where we are now :cry2:
Does anyone have a Casio fx-CG20? I'm having a problem with the graph sketcher as when I input mod(any big quadratic) it doesn't show it crossing the x-axis when there are supposed to be two roots. I tried 3x^2-19x+20 which has two roots but on the calculator it shows it translates upwards with no roots. Does anyone know how to fix this?
(edited 7 years ago)
Please could anyone explain how to do question 3c in here? Thanks :smile:image.jpg
Reply 487
Avatar for oinkk
oinkk
3
Original post by economicss
Please could anyone explain how to do question 3c in here? Thanks :smile:image.jpg


Hey,

Your solutions are plotted in the Argand diagram from part b). You can then find the magnitude of the lengths between each of the three points on your diagram to find the perimeter (think C1 distance formula).

Jack
Original post by oinkk
Hey,

Your solutions are plotted in the Argand diagram from part b). You can then find the magnitude of the lengths between each of the three points on your diagram to find the perimeter (think C1 distance formula).

Jack

Thanks for your help, I just gave it another go and these are the distances I got but they didn't give the right length, please could you have a look and see where I've gone wrong? Thanks :smile:image.jpg
Reply 489
Avatar for oinkk
oinkk
3
Original post by economicss
Thanks for your help, I just gave it another go and these are the distances I got but they didn't give the right length, please could you have a look and see where I've gone wrong? Thanks :smile:image.jpg


Allow me a few moments to do the question... I've had a bottle of prosecco so I hope my mathematical judgement is not harmed as a result :smile:
Original post by oinkk
Allow me a few moments to do the question... I've had a bottle of prosecco so I hope my mathematical judgement is not harmed as a result :smile:


Haha ofc, thank you :smile: wouldn't mind some prosecoo rn!
Reply 491
Avatar for oinkk
oinkk
3
Original post by economicss
Haha ofc, thank you :smile: wouldn't mind some prosecoo rn!


Best way to revise :wink:

I've just done the question from scratch and I agree with your solutions to the equation. But I am simply not getting the required perimeter either.

Can you show your working for each of the lengths?
Original post by economicss
Haha ofc, thank you :smile: wouldn't mind some prosecoo rn!


Original post by oinkk
Best way to revise :wink:

I've just done the question from scratch and I agree with your solutions to the equation. But I am simply not getting the required perimeter either.

Can you show your working for each of the lengths?


You get:

z=2eiπ/12,2ei3π/4,2−i7π/12\displaystyle z = 2e^{i\pi/12}, 2e^{i3\pi/4}, 2^{-i7\pi/12}

Which give you cartesian coordinates (2cos⁡π/12,2sin⁡π/12),(2cos⁡3π/4,2sin⁡3π/4)(2\cos \pi/12, 2\sin \pi/12), \quad (2\cos 3\pi/4, 2\sin 3\pi/4) and (2cos⁡(−7π/12),2sin⁡(−7π/12))(2\cos (-7\pi/12), 2\sin (-7\pi/12)).

The lengths between each one are 4(cos⁡π/12−cos⁡(−7π/4))2+4(sin⁡π/12−sin⁡(−7π/4))2=6+6=23\sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.

Multiply by 3 to get the required perimeter. Not quite sure what the problem is here?
Reply 493
Avatar for oinkk
oinkk
3
Original post by Zacken
You get:

z=2eiπ/12,2ei3π/4,2−i7π/12\displaystyle z = 2e^{i\pi/12}, 2e^{i3\pi/4}, 2^{-i7\pi/12}

Which give you cartesian coordinates (2cos⁡π/12,2sin⁡π/12),(2cos⁡3π/4,2sin⁡3π/4)(2\cos \pi/12, 2\sin \pi/12), \quad (2\cos 3\pi/4, 2\sin 3\pi/4) and (2cos⁡(−7π/12),2sin⁡(−7π/12))(2\cos (-7\pi/12), 2\sin (-7\pi/12)).

The lengths between each one are 4(cos⁡π/12−cos⁡(−7π/4))2+4(sin⁡π/12−sin⁡(−7π/4))2=6+6=23\sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.

Multiply by 3 to get the required perimeter. Not quite sure what the problem is here?


I totally had a sign issue... thanks for saving the day :-)
Original post by oinkk
I totally had a sign issue... thanks for saving the day :-)


Cheers, LaTeXing that was not pretty. :colondollar:
Reply 495
Avatar for oinkk
oinkk
3
Original post by Zacken
Cheers, LaTeXing that was not pretty. :colondollar:


Haha, I can imagine! :tongue:
Original post by Zacken
You get:

z=2eiπ/12,2ei3π/4,2−i7π/12\displaystyle z = 2e^{i\pi/12}, 2e^{i3\pi/4}, 2^{-i7\pi/12}

Which give you cartesian coordinates (2cos⁡π/12,2sin⁡π/12),(2cos⁡3π/4,2sin⁡3π/4)(2\cos \pi/12, 2\sin \pi/12), \quad (2\cos 3\pi/4, 2\sin 3\pi/4) and (2cos⁡(−7π/12),2sin⁡(−7π/12))(2\cos (-7\pi/12), 2\sin (-7\pi/12)).

The lengths between each one are 4(cos⁡π/12−cos⁡(−7π/4))2+4(sin⁡π/12−sin⁡(−7π/4))2=6+6=23\sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.

Multiply by 3 to get the required perimeter. Not quite sure what the problem is here?

Ah finally got it, thanks so much! :smile:
for this question (i have attached the mark scheme) can someone explain where the very first line of working came from? thanks :smile:
Screen Shot 2016-05-28 at 21.44.28.png26.2KB
Screen Shot 2016-05-28 at 21.44.33.png85.2KB
Reply 498
Avatar for edothero
edothero
OP
15
Original post by Zacken

The lengths between each one are 4(cos⁡π/12−cos⁡(−7π/4))2+4(sin⁡π/12−sin⁡(−7π/4))2=6+6=23\sqrt{4(\cos \pi/12 - \cos (-7\pi/4))^2 + 4(\sin \pi/12 - \sin (-7\pi/4))^2} = \sqrt{6 + 6} = 2\sqrt{3}.


What's going on here? can't quite make out what you did
Original post by edothero
What's going on here? can't quite make out what you did


sqrt((x1 - x2)^2 + (y1-y2)^2) give the distance between (x1, y1) and (x2, y2).

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you