AQA FP3 June 2015 Unofficial Mark Scheme

What sid you get for AB? Off the top of my head it was $r = \dfrac{4}{3\sin \theta}$.You set this equal to $r = 1 + \cos 2\theta$ to get a cubic in $\sin \theta$, which you solve for $\sin \theta$. One value is less than -1, so can be discarded, one is associated with A and the other gives you the given result.

In 6b I think you're missing the end of the first bracket (should be after Bsin(lnx/2) maybe?), and for question 7 part (a) was 5 marks and (b)(i) 4 marks from what I remember. Can't believe how fast you put this together and how much you can remember, thanks!

Yeah did that just couldn't solve the cubic

What sid you get for AB? Off the top of my head it was $r = \dfrac{4}{3\sin \theta}$.
You set this equal to $r = 1 + \cos 2\theta$ to get a cubic in $\sin \theta$, which you solve for $\sin \theta$. One value is less than -1, so can be discarded, one is associated with A and the other gives you the given result.

It was in the Ln[(1+2x)(1-2x)] question, had to state the range of validity for values of x

Oh Damn! How many marks was it?

The validity will probably just be worth 1 mark, can't imagine it would be worth any more than that.

I would suggest a slight refinement - set sin theta = 3/4r

You then can change 1 + cos 2 theta into 2 - 2 (3/4r)^2

Rearranging you get 8r^3 -16r^2 + 9 = 0

The key point then is that you know r = 3/2 is a solution, so 2r - 3 is a factor. Use that to find the quadratic factor and then solve with the formula or cts.

It should be r = 3/(4sin theta).
Carrying on give me sin theta = 1/2 and (-1 +/- √13)/4. I substituted them back into the AB equation but its not the right answer.

I did say it was off the top of my head.

So you got the right answer to II?
I used the calculator to find the value of theta at B(the question said not exact value). I then used cosine rule to find the length of AB. There are probably a few ways to do it.