Edexcel FP1 Thread - 20th May, 2016

Could anyone explain question 5b from the exam style paper in the textbook please https://8fd9eafbb84fdb32c73d8e44d980d7008581d86e.googledrive.com/host/0B1ZiqBksUHNYTnpyeF8xQlZweHc/REV1.pdf I can't see the link between finding the grad of PQ in part a and then PR in part b? Thanks

you're given that PR is perpendicular to QR, so find the gradients using (y2-y1/x2-x1)
The products of these gradients should be -1. then rearrange to get in the form 1/pq=9 as 1/pq is the gradient at any point on the curve.

Thanks, on solutions bank it goes straight to the gradients for PR and QR and says that it's deduced from part a but I can't see the link between them? https://644625398389466aee00633223056f55519d5fbc.googledrive.com/host/0B1ZiqBksUHNYQWY1UWtNa0NRNU0/practice%20paper.pdf I tried to work it out by finding the gradients of PR and QR but I couldn't find a way to simplify my algebra down? :/
Thanks

If the gradient of a general chord PQ is $-\frac{1}{pq}$ (the p being from the P and the q from the Q) then if you replace Q with R where R is the point that is exactly the same as Q except for the fact that we have $q=3$ (i.e: substitute $q=3$ into Q to get the point R) then the gradient of the chord PR is the gradient of PQ with $q=3$ so $-\frac{1}{pq} = -\frac{1}{3p}$.

Same for QR, except this time it's Q staying constant and R is just the point P with $p=3$ instead.

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How is induction hard?

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How about you? What did you find most difficult??

I'd have to go with co-ordinate systems as well, I guess.

That question in the 2016 IAL F1 paper on coordinate systems, WOW!

Did anyone who sat the 2016 Jan series get that right? (of the people on TSR who sat it) @Zacken @aymanzayedmannan

Hi, is anyone able to explain please the divisibility induction method used in questions 54 and 55 of rev ex 2 of the textbook https://644625398389466aee00633223056f55519d5fbc.googledrive.com/host/0B1ZiqBksUHNYQWY1UWtNa0NRNU0/REV2.pdf How do you know when to bring in this 'm' and is there an easy way to go about using it?
Thanks

Could anyone help please? :-)

It's not a required method and you could use the $f(k+1) - f(k)$ but here, all you need to know is that saying that $f(n)$ is divisible by $9$ means you can write $f(n) = 9m$ for some integer $m$. You tend to use it in induction proofs when $f(n)$ is a sum of exponential and linear terms instead of just an exponential or just a linear term.

On question 26 https://644625398389466aee00633223056f55519d5fbc.googledrive.com/host/0B1ZiqBksUHNYQWY1UWtNa0NRNU0/REV2.pdf , am I right in thinking n is 13 because r=0 and if like normal the series started from r=1 then there would be 12 terms instead? Thanks