Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

Do you have a specific question?

What happens when we're finding a Maclaurin expansion of a function and when it comes to evaluating a particular part of it, we end up doing 1/0?

Nah. I was just considering that scenario. Take y = cosec x for instance.

if its a "show that" question, then you can use the answer whilst solving it (to show that it is true). Generally using u+iv will work always, doing it with modulus' is usually less algebra/shorter but isn't always possible (and takes some thinking).

For example in the question we were talking about before, arg(z) = pi/4 is the half line y=x, so you can rewrite z = x + iy into z = x + ix (or z = y + iy if you prefer, it doesnt matter).

Then subbing into W = (Z+1)/(z+i) we get W = (x + ix + 1)/(x + ix + i)
grouping real and imaginary gives W = ( (x+1) + ix ) / (x + i(x+1) )

Looking at the answer, they want you to show that mod(W) = 1 so mod(our W) = 1

mod( (x+1) +ix) / (x + i(x+1)) ) = 1

mod((x+1) +ix) / mod(x+ i(x+1)) = 1

Remembering that mod(a + ib) = sqrt(a^2 + b^2) gives us

sqrt( (x+1)^2 + x^2) / sqrt( x^2 + (x+1)^2 ) = 1

which is clearly true because they're the same thing.

Not possible. For a MacLaurin expansion to be valid (i.e: for it to exist) you require all derivatives to be defined. i.e: if you get get 1/0, it means you're working out the Maclaurin series of a function that doesn't have a MacLaurin series. (like ln x)

Thanks. What do you think I should use in the exam, u + vi or x+yi. With u+vi you have to take long steps but its more straightfoward I guess but x+yi is easier but I might not get it . I also take long time to finish other questions so i dont know...

http://imgur.com/xKBlDyE

Hope this helps, it really brain teasing I know!

Product rule with $u = \dfrac{1}{x}$ and $v = \dfrac{dy}{dt}$

$\dfrac{d^{2}y}{dx^{2}} = uv' + u'v$

guys, could someone please explain the questions where they ask you to shade the area after a transformation in the complex plane? June 2009 6b or June 2015 5b for example

Please could anyone do a worked solution for question 12a https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202004%20QP%20-%20FP2%20Edexcel.pdf I've tried it by substituting in for lambda plus lambda i when x is greater than 0 but for some reason I can't get it to work but I thought the method was correct? Thanks

Hey have you guys done the review 1 ex in the FP2 book.
I m struggling with q47 pg70- could that come up tomorrow

Nah. I was just considering that scenario. Take y = cosec x for instance.

bump

bump

I'm almost certain that won't come up.

Pick a point (z) that satisfies whatever you have, then use the transformation to turn it into W and look at where that point is on the graph. For example in June 2015 5B:

mod(z) <= 2

pick a point that satisfies this. Easiest point is z = 0 + 0i

w = z/(z+3i) = 0/(0+3i) = 0

w = 0 + 0i is inside of the circle from part a, so you shade inside of the circle.