OCR MEI FP1 Thread - AM 20th May 2016

the asymptote was 1

You had something like

$\dfrac{1}{6}n(n+1)(3n^{2}+(3-2p)n-3) = \dfrac{1}{6}(n^{2}+n)(3n^{2}+(3-2p)n-3)$

as the formula I think. The coefficient on $n^{4}$ is $\dfrac{1}{2}$ if you multiply it out, and the coefficient on $n^{3}$ is given by multiplying $\dfrac{1}{6} \times (1 \times 3 + 1 \times (3 - 2p))$ by matching up terms in either bracket to get $n^{3}$ terms.

Set $\dfrac{1}{2} = \dfrac{1}{6} \times (1 \times 3 + 1 \times (3 - 2p))$ and solve for $p$.

Anyone remember the mark-split for the individual parts of q7? Roots of the polynomial eq'n.

I did alpha*beta*given root*complex conjugate = -b/a
I got that p = 3/2 and I think another answer might have been -1 ??

If someone could post the question+answers would be v grateful

Ahhhhhhh thanks, how many marks was just that part?

For Qs 7 i think, how did you calculate alpha x beta. I didn't get that and so couldn't really do the rest of the qs and find A and B and then the w/j bit. I've done all the fp1 past papers and there has been nothing like this qs before. Also the series method of differences qs (i think it was near the end, honestly can't remember anymore) was a toughie, i think I may have gotten partly there but still hard

someone please explain the method for 9iv)????

There was one at $x = 1$. The horizontal asymptote was at $y=3$.

will I get all the marks except for that of naming the asymptote?

If that was the last part, then...

They gave you the sum $\dfrac{45}{stuff} + ... + \dfrac{105}{stuff}$ so you had to work out the limits of the sum ($2n_{1} + 5 = 45 \Rightarrow n_{1} = 20, \, 2n_{2} + 5 = 105 \Rightarrow n_{2} = 50$), then use these and the formula to get the answer...

$\displaystyle \sum_{r=20}^{50} u_{r} = \sum_{r=1}^{50} u_{r} - \sum_{r=1}^{19} u_{r}[br][br]\mathrm{plug numbers into your calculator...}[br][br]= 0.00813$

I think it was a 3 mark question, which will be 1 mark per correct asymptote, so you'll just get a mark for each correct one you wrote down.

did exactly the same method but ended up with -6.25 x 10^-4 as my answer. strange. should i only lose one mark since i did the right method?

thanks, what do you think the boundary will be for a b?

Yep if you wrote down the method and just put it into your calculator wrong you should only drop the last mark

I have absolutely no idea sorry :P

To get $\alpha \beta$ I think the easiest would have been to use the formulae for roots of equations (eg the sum of the roots is $-\dfrac{b}{a}$

The one to use in the question was the product of the roots is $\dfrac{e}{a}$ (for a quartic), so you got $\alpha \beta \gamma \delta = \dfrac{-26}{2} = -13$ and you had $\gamma$ and $\delta$ so you could multiply them together (gave you 13) and you get $\alpha \beta = -1$.

And I agree the method of differences question was tough, but I'm guessing that's why it was the last question on the paper :P

did exactly the same method but ended up with -6.25 x 10^-4 as my answer. strange. should i only lose one mark since i did the right method?

in the method of differences question I replaced 1, 2 , 3 and 4 in the formula, I wrote the equivalence of the sum of the three terms, but there wasn't a pattern of what would cancel out to be seen, so I stopped there, will I get any marks?

in the method of differences question I replaced 1, 2 , 3 and 4 in the formula, I wrote the equivalence of the sum of the three terms, but there wasn't a pattern of what would cancel out to be seen, so I stopped there, will I get any marks?

Maybe 1, 2 at most sorry.