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Edexcel FP1 Thread - 20th May, 2016

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Original post by tazza ma razza
I think simultaneous equations w/ matricies will come up - never seen it before and i have no idea how to do them lol


I was thinking that! and me neither lmaoo
Original post by tazza ma razza
I think simultaneous equations w/ matricies will come up - never seen it before and i have no idea how to do them lol


Do you mean using simultaneous equations to solve matrices? I don't think that's on spec (although it was in my GCSE).
Original post by Zacken
Do you mean using simultaneous equations to solve matrices? I don't think that's on spec (although it was in my GCSE).


chapter 4 ex 4j and point 4.11 example 24

seems easy enough just want to confirm the order of each element within the matrix
Original post by tazza ma razza
chapter 4 ex 4j and point 4.11 example 24

seems easy enough just want to confirm the order of each element within the matrix


Don't have a textbook, but:

If you have the simultaneous equations:

ax+by=cax+by = c

and

dx+ey=fdx + ey = f

Then you write down:

(abde)(xy)=(cf)\displaystyle \begin{pmatrix}a &b \\ d & e\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}c \\ f \end{pmatrix}

Which makes sense, because if you multiple out the LHS you end up with:

(ax+bydx+ey)=(cf)\displaystyle \begin{pmatrix} ax + by \\ dx + ey \end{pmatrix} = \begin{pmatrix} c \\ f \end{pmatrix}

and comparing components gets you the original equation.
(edited 7 years ago)
Original post by Zacken
Don't have a textbook, but:

If you have the simultaneous equations:

ax+by=cax+by = c

and

dx+ey=fdx + ey = f

Then you write down:

(abde)(xy)=(cf)\displaystyle \begin{pmatrix}a &b \\ d & e\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix}c \\ f \end{pmatrix}


yh i'm just getting that - tbh i would just use my graphical to solve it for me and then i know which values to get lol
Original post by tazza ma razza
yh i'm just getting that - tbh i would just use my graphical to solve it for me and then i know which values to get lol


Look at my edited post, it's not hard to understand why that works.
Original post by Zacken
Look at my edited post, it's not hard to understand why that works.


ah i just expanded it on paper and i get it - working through the example in the book.

tbh i can't think of anything hard that will trip me up. it will be a silly mistake somewhere. conics most likely will be that area.
Original post by economicss
Is it part a or b you're struggling with? :smile:


I'm also stuck on the question (part b), i cant seem to do the last part f(k+1), would someone be able to write down the answer?
Thanks
Original post by Strom
I was thinking that! and me neither lmaoo


ex 1g also
example 10 ch4
Can anyone refer me to a tough proof by divisibility question if any?

Edit: I found something - no worries.
(edited 7 years ago)
For the proof of divisibility qs- I always use the following method- can someone pls check for me if it is the right way to do it/would get me full marks:

E.g. prove that f(n)=7^(2n)-48n-1 is divisible by 2304
sub in 1, make assumption n=k etc... then

f(k+1)-f(k)= 7^(2k+2) - 48(k+1) -1 - 7^(2k) +48k + 1
= 48(7^(2k)) - 48

then I rearrange f(k) to make 7^(2k) subject of the equation
i.e. 7^(2k)= f(k) + 48k + 1
Then sub it in to f(k+1)-f(k)

Is this a correct way of doing it??
Original post by thesmallman
For the proof of divisibility qs- I always use the following method- can someone pls check for me if it is the right way to do it/would get me full marks:

E.g. prove that f(n)=7^(2n)-48n-1 is divisible by 2304
sub in 1, make assumption n=k etc... then

f(k+1)-f(k)= 7^(2k+2) - 48(k+1) -1 - 7^(2k) +48k + 1
= 48(7^(2k)) - 48

then I rearrange f(k) to make 7^(2k) subject of the equation
i.e. 7^(2k)= f(k) + 48k + 1
Then sub it in to f(k+1)-f(k)

Is this a correct way of doing it??


Yes.
Original post by thesmallman
For the proof of divisibility qs- I always use the following method- can someone pls check for me if it is the right way to do it/would get me full marks:

E.g. prove that f(n)=7^(2n)-48n-1 is divisible by 2304
sub in 1, make assumption n=k etc... then

f(k+1)-f(k)= 7^(2k+2) - 48(k+1) -1 - 7^(2k) +48k + 1
= 48(7^(2k)) - 48

then I rearrange f(k) to make 7^(2k) subject of the equation
i.e. 7^(2k)= f(k) + 48k + 1
Then sub it in to f(k+1)-f(k)

Is this a correct way of doing it??


If this is when n is more than or equal to 2 - you need to sub in 2 as someone pointed out on the forum. Then do the simple induction process - finally on to your conclusion. Done.
Original post by NotNotBatman
For divisibility tests, is the conclusion different from true for n=k, k+1,n =1, so all n. Do you have to state since f(k) divisible by n, so f(k+1) is divisible by n etc.


Oh, wait - do you mean when you have find f(k) in f(k+1)? - I say since f(k) is div. by n then f(k+1) must be div. by n - then proceed with the conclusion.
Original post by Marxist
If this is when n is more than or equal to 2 - you need to sub in 2 as someone pointed out on the forum. Then do the simple induction process - finally on to your conclusion. Done.


Original post by Zacken
Yes.



Thanks guys!!!
How many got full marks on June 2012 FP1?
Is ZZ=Z2 ZZ^* =\left |Z \right |^2 For all complex numbers?
For the June 2015 IAL paper Q5 (b), will my answer get full marks? As the mark scheme says the 'use of negative lengths scores M0'. What does that mean?

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Any hard non-edexcel papers?

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