I ain't Zain lol

Reason being we do a joint FP2-FP3 (I think the maths department wanted to kill us) and it's impossible to revise for both of them the night before :cry2:

Reply 191

Kevin De Bruyne

21

Original post by Student403

Flopped my FP2 mock today :lol:

It'll be alright on the night :wink:

Reply 192

Student403

19

Original post by SeanFM

It'll be alright on the night :wink:

Hope so mate - thanks :biggrin:

How was your test today?

Reply 193

Kevin De Bruyne

21

Original post by Student403

Hope so mate - thanks :biggrin:

How was your test today?

good memory, PRSOM (I see we are both really close to 11 gems...)

It was alright, not sure if I've met my GYG goal but we shall see :borat:

Reply 194

Student403

19

Original post by SeanFM

good memory, PRSOM (I see we are both really close to 11 gems...)

It was alright, not sure if I've met my GYG goal but we shall see :borat:

I'm sure it was alright!

And holy cow.. Yeah :eek:

Reply 195

economicss

3

Please can anyone explain why the limits aren't between 0 and pi/6 on question 4b of this paper https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/January%202006%20QP%20-%20FP2%20Edexcel.pdf and how you know to subtract a shape from the area? Thanks :smile:

Reply 196

EricPiphany

17

Original post by economicss

Please can anyone explain why the limits aren't between 0 and pi/6 on question 4b of this paper https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/January%202006%20QP%20-%20FP2%20Edexcel.pdf and how you know to subtract a shape from the area? Thanks :smile:

Reply 197

Cpj16

15

Original post by Zacken

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{z-3i}{z-5} = \frac{x + (y-3)i}{(x-5) + iy} = \frac{(x +(y-3)i)((x-5) -iy) }{(x-5)^2 + y^2}

Doing this all on LaTeX, forgive the excessive working:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{x(x-5) + y(y+3) - iyx + (y-3)(x-5)i}{(x-5)^2 + y^2} \end{equation*}

Imaginary and real parts need to be identical:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}x^2 - 5x + y^2 + 3y = -5y - 3x + 15\end{equation*}

Completing the square:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}(x-1)^2 + (y+1)^2 = 15 + 2 = 17\end{equation*}

As required.

Blah blah required conditions, pish posh. We require both the real and imaginary part to be positive, which gives us

\displaystyle x(x-5) + y(y+3) > 0

and

-5y - 3x + 15> 0 \iff 5y + 3x < 15

.

I don't really see how this is above the syllabus. :dontknow:

Whaaat is this in FP2? tbh we have done all the chapters except end of 3 :frown:

this looks horrible :frown:

edit I just went through it again and its not that bad :biggrin:

Reply 198

Zacken

22

Original post by Cpj16

Whaaat is this in FP2? tbh we have done all the chapters except end of 3 :frown:

this looks horrible :frown:

edit I just went through it again and its not that bad :biggrin:

Glad to hear that, it seems fairly standard to me but mileage varies; don't be worried if you weren't able to do it. :smile:

Reply 199

Cpj16

15

Hi can someone pls explain this integration, I don't understand it

thanks

Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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