OCR MEI AS Further Mathematics FP1 13/05/2013 PM

Yep I agree with your answer for z2.

The diagram was really cluttered with tons of surds. I think I had like 3 complex numbers in the positive x quadrant and 1 somewhere else.

-3+j/-3-j

Yeah, in terms of them all I got:

$z_{1} = 3-2j[br]z_{2} = \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}j[br][br]z_{1} + z_{2} = \frac{6+5\sqrt{2}}{2} + \frac{-4+5\sqrt{2}}{2}j[br][br]z_{1} - z_{2} = \frac{6-5\sqrt{2}}{2} - \frac{4+5\sqrt{2}}{2}j[br]$

What was the original p coordinate can you remember?

The induction question was quite alright this time, I've done it here:


$\sum_{r=1}^{n}[r(r-1)-1] = \frac{1}{3}n(n+2)(n-2)$

The inductive step:

$\frac{1}{3}k(k+2)(k-2) + (k+1)((k+1)-1)-1$

$= \frac{1}{3}[k(k+2)(k-2) + 3k(k+1) -3]$

$= \frac{1}{3}[k(k^{2}-4) + 3k^{2} + 3k -3]$

$= \frac{1}{3}[k^{3}-4k + 3k^{2} + 3k -3]$

$= \frac{1}{3}[k^{3} + 3k^{2} - k -3]$

$= \frac{1}{3}(k+1)(k+3)(k-1)$

$= \frac{1}{3}(k+1)((k+1)+2)((k+1)-2)$

I got that for part iii or whatever it was before, not for the last part!

Do you think some kind of explicit working was required to go from the polynomial to the factorised form? :/ Seems kinda weak seeing as the answer was given, but it was awkward to do that on top of everything else. I was a bit thrown by this because I couldn't believe they'd give us something which didn't factorise neatly without needing complete expansion first :/ I would have much preferred the so called 'horrible' proofs from previous papers, I quite enjoy the algebra for those.

Yeah, in terms of them all I got:

$z_{1} = 3-2j[br]z_{2} = \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}j[br][br]z_{1} + z_{2} = \frac{6+5\sqrt{2}}{2} + \frac{-4+5\sqrt{2}}{2}j[br][br]z_{1} - z_{2} = \frac{6-5\sqrt{2}}{2} - \frac{4+5\sqrt{2}}{2}j[br]$


z₂ and z₁+z₂ were both in the top right quadrant, and z₁ was in the bottom right quadrant, and z₁-z₂ was in the bottom left quadrant.

I'm more than likely entirely wrong though..! Need to have a look back through the paper.

That's exactly what I was thinking. This induction question just ruined my day. I got to the cubic and though, there must be a way I can make this work without getting a cubic. spend so much time trying to find a solution that I had to do question 9 in 5 minutes. Luckily I found it Q9 really easy as I remember a very similar question from the 2005/2006 past papers.

all in all, the exam was fairly straightforward, but all the questions were long and was weird to explain.

For question 7, When it says determine, could you just give the right answers? It was 4 marks and 3 answers so I guess they wanted working out, but I didn't know what to put up lol.

I did all the induction (and got it right), apart from the last bit '((k+1)+2)' I tried but ended up doing something like '((k)+3)' how many marks would I lose for this, 0 or 1?

Do you think some kind of explicit working was required to go from the polynomial to the factorised form? :/ Seems kinda weak seeing as the answer was given, but it was awkward to do that on top of everything else. I was a bit thrown by this because I couldn't believe they'd give us something which didn't factorise neatly without needing complete expansion first :/ I would have much preferred the so called 'horrible' proofs from previous papers, I quite enjoy the algebra for those.

Guys what did you get in the matrix questions because I was kind of baffled except the first and the last parts. I think the (ii) I got p'(2,-2), (iii) y=-x, (iv) y =6

i) Rotation clockwise 90 degrees about the origin
ii) (2,-2) or (-2,2), I forgot which way around it was
iii) y=-x
iv) y=6
v) Showed that the determinant=0 then said it was singular and baffled on about what it implies for the transformation
vi) I found R, which was:
(01)
(01)
Then showed that it was y=x

I just wrote what the asymptotes were. Then said that the fractiony asymptote must be the one with a coeffecient in front of an x, the one without a fraction must be the other vertical asymptote. And then found the horizontal asymptote which was 3/2, then said that c must be 3 and b must be 2.

Hold tite 11/12 marks, did the EXACT same thing as yourself.

Part v, and the argand diagram were my only errors of 3 marks..

I just gave the right answers and was planning to come back to it, but because I spend half an hour trying to work out how to solve the cubic in the induction question properly (and I still didn't manage) I didn't have time to show what I did.

Hopefully I'll lose a maximum of 1 mark, What I'm fearing however is that the mark scheme will be like this:

A1 mark for a =2
A1 mark for b = 2
M1 mark for doing 3x^2/2x^2
A1 mark for c = 3 dependant on M1 being shown

Which means I'll lose 2 marks. If they wanted working out for "a" and "b" then I'll be way pissed as that will mean 4 marks lost LOL

The only bit I am worried about is the last question, and maybe the equation with the 1/3x+1 roots in it, as I got a load of big numbers as my coefficients. The Argand diagram I am reasonably happy about, although I only labelled my points as z1, z2 ect because I didn't want to write out a giant surd.

Does anyone have a past paper infront of them. If so could you tell me how many marks Q's: 6, and each part of 9, were worth please, thanks