Edexcel FP2 23rd June 2011 SOLUTIONS IN FIRST POST

OK here is my solution for the difference of series Q

any ideas what 89 ums will be as a raw score?

OK here is my solution for the difference of series Q:

$r^2 = \frac{(2r+1)^3}{24}-\frac{(2r-1)^3}{24} - \frac{1}{12}$

then series it from r =1 to r=n:

you get:

9/8 - 1/24 - 1/12

125/24 - 9/8 - 1/12

343/24 - 125/24 - 1/12
....

$\frac{(2n+1)^3}{24} - \frac{(2n-1)^3}{24} -\frac{1}{12}$

so overall when you add them all together you get:

$\frac{(2n+1)^3}{24} - \frac{1}{24} - \frac {n}{12}$

Multiply through by 24:

$(2n+1)^3 - 1 - 2n \rightarrow 8n^3 + 12n^2 +6n + 1 - 1 - 2n[br][br]\rightarrow 8n^3 + 12n^2 +4n[br][br]\rightarrow \frac{n}{6} (48n^2 + 72n +24n)[br][br]\rightarrow \frac{n}{6}(4n^2 + 6n + 2)[br][br]\rightarrow \frac{n}{6}(2n+1)(n+1)$

Phew. lol.

I also used this method I think it should be fine, the question itself is actually in the FP2 book, I remembered doing it with the same method, I will take a look at the model solution on the computer to see what they suggest

How many raw marks will you need for 100 UMS?

About 66/75

Hey Gibbo, thanks for the solutions. I just have a couple of queries regarding some marks if you would not mind answering

Question 2 - I got the wrong value for the third differential, everything else correct including the series apart from where I had used my wrong value, (2 marks lost?)

Question 6 - Integrated within the limits correctly but when working out the area of the triangle I did 0.5(5/2)x base rather than 0.5(5/2sin60) x base I then took this value away from the value of the integrated area. (2/3 lost?)

Question 7b) - I incorrectly identified 2pi to be valid and I must have forgotten to use the negative root has I only had 5 solutions, 4 of which were valid.

Question 8c) - Differentiated, set to 0 but ran out of time really.

Much appreciated

ahhh thanks! I think best case scenario i got 69, but worst case 64.... very very borderline for an a* then, as i need 89 for one! i think my most likely mark will be 65/66. kicking myself about silly one mark errors ive made..

This method is fine.

You've just subtracted 2n and divided by 24 same as me but I did it at the end while you did it at the beginning.

Thanks! Phew I thought with 3 terms being taken away at once it would not have been suitable lol.

Fingers crossed.
What other modules have you taken for Maths and F Maths??

Nah that's fine. You've only got a -1/12 hanging on the end of your method of differences.
It's a perfectly good method, mine's just a bit more economical on ink!!

for a2 modules ive taken s2 (got 93) and m2 (88), and this summer ive done fp3 and fp2.. think 89 is more likely in fp2 though, so counting on it!

Hmmm.

I would have gone for a re-take in M2. The grade boundaries are always higher for these A2 modules in Jan (I assume you did it in Jan).

Oh well. Good luck!!

a retake was tempting - but i didnt know about the grade boundaries (did take it in jan)! also, i was never that great at m2, so i thought that my 88 was unlikely to go up. s2 wish id got 100 though, especially as i thought id nailed the exam! thank you

If you did S2 in Jan 93ums was 73/75.

Told you the boundaries are a b1tch in Jan!!

It's because mainly F Maths students do them in Jan and then weaker students do them as their final applied exam in June.

...

Thanks for this A question about 5 - is it a valid method to find the coordinates of R instead? I forgot this method and did a similar thing, equating imaginary parts to 0, and instead found the points on the real axis to be +/- rt 3 rather than the original circle equation. Then just said these are on C as it's the coordinates of intersection with the real axis. Not sure if that would get any marks