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Edexcel FP1 Thread - 20th May, 2016

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Original post by connorbarr
Haven't done the question but it is possible for the tangent to meet and xy graph again i think


but ye the mark scheme uses normal so must be wrong
In the FP1 textbook, pg 82, there's something about linear transformations. What do we need to know about this? I dont really understand it too well
Original post by dididid
Hey could someone please link/ send me some hard proofs by induction for each of the proofs mentioned on the spec

This was a pretty cool question on an old fp2 paper - although you can only do it if you have done c3/c4 first.
Induction.png6.9KB
Original post by AmarPatel98
In the FP1 textbook, pg 82, there's something about linear transformations. What do we need to know about this? I dont really understand it too well


ye its well **** i had to look online, havent seen it in a past paper before but i think for it to be a linear transformation u can only add x and y values and times by a constant, might be wrong
Original post by iMacJack
But I was unable to get the one in the IAL 2015 paper using it.


POMI.jpg
Thats how I done the question, not sure if you'll know this method. Don't use it if you don't.
Good luck everyone!!!
Good luck for tomorrow everyone! I'm sure it won't be too bad. Well, I hope so anyway...
I was gone pull a all nighter but then I was like I'll just wake up early tomorrow to cram some last minute revision in. I hope its not one of them wake ups that you say now just to make yourself feel better.
good luck

r= 0 summations, is it just the constant that gets affected by the n+1 at the end? and not the sum of r^2 + sum of r bit?
Some one plz help
cant understand 8b even after looking at Mark Scheme

https://edba9d72a6846be4dc688090a802a423effa6b3a.googledrive.com/host/0B1ZiqBksUHNYWWItOHhJTkRBc3M/06%20Gold%202%20-%20FP1%20Edexcel.pdf
Original post by dragozox


Multiply the exression by A^-1 to the right
AAA^-1 = 7AA^-1 + 2IA^-1
Becomes
A= 7I +2A^-1
Then rearrange
Original post by Windowswind123
Multiply the exression by A^-1 to the right
AAA^-1 = 7AA^-1 + 2IA^-1
Becomes
A= 7I +2A^-1
Then rearrange


I Did do that but shouldnt it be:

A^-1AA = 7A^-1A+2A^-1I

which gives A=7 + 2A^-1I

I dont get how 7 and I get together?? Thank You!!
(edited 7 years ago)
Original post by dragozox
I Did do that but shouldnt it be:

A^-1AA = 7A^-1+2A^-1I

which gives A=7 + 2A^-1I

I dont get how 7 and I get together. Thank You!!


AA^-1 =I
Original post by anndz3007
AA^-1 =I


ooooh I C!!!

Thanks For clarifying!! :smile:
Original post by dragozox
I Did do that but shouldnt it be:

A^-1AA = 7A^-1A+2A^-1I

which gives A=7 + 2A^-1I

I dont get how 7 and I get together?? Thank You!!

A*A-1 =I
And AI = A and IA^-1 = A^-1
Original post by Windowswind123
A*A-1 =I
And AI = A and IA^-1 = A^-1


Thank you Sooo much!!

This makes so much sense now!! :smile:
Good luck everyone.
Original post by tazza ma razza
good luck

r= 0 summations, is it just the constant that gets affected by the n+1 at the end? and not the sum of r^2 + sum of r bit?


the general summations dont change because you're just adding 0, 0^2 or 0^3 to the start, but for the constants, you're multiplying it by the number of time the sum is done, so when r=0, there's one more term than you'd usually expect, so if the constant was '5', instead of making it 5n, it become 5(n+1) to account for when r=0 :smile:
Any last minute advice?
Original post by Foutre en L'air
Any last minute advice?


Get some sleep.

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