Edexcel FP2 June 2015 - Official Thread

omg this is me! i liked fp2 throughout the whole year (self-studying) and i understood everything, I only did the questions in the book though, and then i started past papers and i at first i was like ???!! why am i finding this difficult?? but then i kept doing a lot, and now i got the hang of it the more you'll do the better you'll get! also for the complex numbers, the transformations can be a bit tricky but once you get the hang of it it's fine!

if you need any help, don't be afraid to ask! good luck, it'll get better trust me!

Please could someone help me with this? Thanks

Is FP2 generally harder than M2, D2 and S2? I can get A*'s in C3 and C4 but lucky to get a C in FP2...

For differentiating (dy/dx)^2dx:

Letting t = dy/dx, t^2dx

So 2dy/dx * d^2y/dx^2

Is this correct?

Yes


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Thanks

What about if I want to differentiate dy/dx with respect to z?

Do we essentially just treat dy/dx type terms as any other term when using the chain and product rule?

Is there a method to follow that works for all complex numbers transformation questions? I really struggle with them

Depends on what you're trying to achieve. If you are given two moduli, then you should sub x + iy in and apply Pythagoras, because it's well faster, but if not, then you can either use the component method or the perpendicular bisector method, depending on which you prefer.


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What's the perpendicular bisector method?

If you have something lying on the real axis, then you can either use the fact that y=0 or that the real axis is the perpendicular bisector of | z - i | = | z + i |.

So, when you have $z = \frac{2}{w}$, you can either sub w = u + iv into there, make y=0 and so on, or you can sub z into | z - i | = | z + i | and rearrange.


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Can anyone explain this to me? Part B, when it asks you to find the solutions, I equate the RHS of the equation obtained in part (a) to 5 times the RHS of the equation given at the start of part (b). Subtracted RHS, setting equal to zero, pulled sin theta out as a common factor, to get sin theta = 0 or sin theta = +/- the 4th root of 5/8. I discounted sin theta = - the 4th root of 5/8 because theta turns out to be negative and theta must be between 0 and 2pi, but the mark scheme says I shouldn't have done that.

In the end I got theta = 0, 1.10, 2.05 and pi, which were all correct, but I was missing 4.237 and 5.188

Can anyone explain this to me? Part B, when it asks you to find the solutions, I equate the RHS of the equation obtained in part (a) to 5 times the RHS of the equation given at the start of part (b). Subtracted RHS, setting equal to zero, pulled sin theta out as a common factor, to get sin theta = 0 or sin theta = +/- the 4th root of 5/8. I discounted sin theta = - the 4th root of 5/8 because theta turns out to be negative and theta must be between 0 and 2pi, but the mark scheme says I shouldn't have done that.

In the end I got theta = 0, 1.10, 2.05 and pi, which were all correct, but I was missing 4.237 and 5.188