AQA Further Pure 2 - Wednesday 18th June 2014

I got $8 \cos^2 \theta \sin^2 \theta = 1 - \cos 4 \theta$

I got $8 \cos^2 \theta \sin^2 \theta = 1 - \cos 4 \theta$

I got 1 - (1/2)cos4x somehow

That was z^4=-9i

I got z=9^(1/4)e^i(kPi/2-Pi/8) for k=0,1,-1 and 2 I think

I concur.

I'm pretty sure it was -Pi/4 as the arg. which means instead of Pi/8 it would be Pi/16?

I had the arg as -Pi/2

I'm pretty sure it was -Pi/4 as the arg. which means instead of Pi/8 it would be Pi/16?

Yuk - what is AQA's problem with their maths papers this year?! I also got 1 - cos4theta - can anyone remember what the integral was? I think I had pi/3 - (root3)/8??

Also got 5rt2 for last integral, though completely botched up the working.

Differentials of arctans: I integrated then set x=0 so your c comes out as pi/4. Think kind of fudged this though because would only work if you do -c, not +c - is this still ok?

horrible paper

Hmm, I got -Pi/2 as the arg in part a), I may well have overlooked something though :/

Yuk - what is AQA's problem with their maths papers this year?! I also got 1 - cos4theta - can anyone remember what the integral was? I think I had pi/3 - (root3)/8??

Also got 5rt2 for last integral, though completely botched up the working.

Differentials of arctans: I integrated then set x=0 so your c comes out as pi/4. Think kind of fudged this though because would only work if you do -c, not +c - is this still ok?

horrible paper

-9i. If you think about it, the arg is definitely (-pi/2). Plot -9i on the graph.

Hmm, I got -Pi/2 as the arg in part a), I may well have overlooked something though :/