The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Edexcel FP1 Thread - 20th May, 2016

Scroll to see replies

Original post by LLk1
How is 4b 8? in jan 2016 ial


can't even find this paper, do you have a link?
Reply 541
Avatar for LLk1
LLk1
1
Original post by LewisClothier
can't even find this paper, do you have a link?


https://drive.google.com/folderview?id=0B23NLD5L6099eUg0YjVhNXZRblk&usp=sharing&tid=0B23NLD5L6099YjV3SzBabTE4NDQ
Original post by LLk1


So, you have the geometrical transformation of the matrix A, should be 135 degree clockwise rotation, then you need to find a multiple of 135 that equals a multiple of 360.
In other words, how many times would you need to apply this transformation to I to get back to I. the answer cannot be 0 because it needs to be a positive integer, although that would work... so you would need to apply the transformation 8 times to get back to I.
Original post by LLk1
How is 4b 8? in jan 2016 ial


I think you have to look at the number of times the matrix would need to be transformed in order to be in the form of the identity matrix.
Not toooo sure though, made a lucky guess I suppose
Original post by LewisClothier
So, you have the geometrical transformation of the matrix A, should be 135 degree clockwise rotation, then you need to find a multiple of 135 that equals a multiple of 360.
In other words, how many times would you need to apply this transformation to I to get back to I. the answer cannot be 0 because it needs to be a positive integer, although that would work... so you would need to apply the transformation 8 times to get back to I.


Couldn't have ever possibly put it better
Original post by LLk1


Thanks for this. Feels like there isn't much practise available.
Has anyone done the question on the June 2013 R paper where for the summation r=0 and not r=1, its also on the edexcel Gold level 2 paper as well its the last question on that one?
Im really stuck on what you are supposed to do
jan 2016 (IAL) is difficult af- is everything there in our spec??
Original post by LLk1


Can someone check Q 1(c) for me???- the mark scheme implies that (2i)^(2)=+4
Surely this is wrong as it should be -4?

FS i'm confused :frown:
Original post by thesmallman
jan 2016 (IAL) is difficult af- is everything there in our spec??


It's not difficult - there are some stuff in our spec and some that aren't.
Original post by katie*may
Has anyone done the question on the June 2013 R paper where for the summation r=0 and not r=1, its also on the edexcel Gold level 2 paper as well its the last question on that one?
Im really stuck on what you are supposed to do


see

Original post by Zacken
I'll give you two ways to look at it:

1. ∑r=0n(2n+1)=2(n)+1⏟0th   term+∑r=1n(2n+1)=(2n+1)+(2n+1)n=(2n+1)(n+1)\displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)

2. ∑r=0nf(r)=f(0)+f(1)+f(2)+⋯+f(n)⏟n terms⏞(n+1) terms\displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}.

Since in this case you have ∑r=0n(2n+1)=(2n+1)∑r=0n1\sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1

Then you have f(r)=1f(r) = 1 being added together n+1n+1 times to get a total of (2n+1)(1+1+⋯+1)=(2n+1)(n+1)(2n+1)(1 + 1 + \cdots +1) = (2n+1)(n+1).
Original post by thesmallman
jan 2016 (IAL) is difficult af- is everything there in our spec??


Nope, the roots of polynomials things aren't on your spec.
Can anyone explain what a Locus/Loci actually is? would be much appreciated.
Original post by thesmallman
Can someone check Q 1(c) for me???- the mark scheme implies that (2i)^(2)=+4
Surely this is wrong as it should be -4?

FS i'm confused :frown:


I can't see the markscheme as you haven't link it, but:

∣a+ib∣=a2+b2≠a2+(ib)2\displaystyle |a + ib| = \sqrt{a^2 + b^2} \neq \sqrt{a^2 + (ib)^2}.

You might want to lookup the definition of modulo again.

So, here: ∣(3+k)+2i∣=(3+k)2+22|(3+k) + 2i| = \sqrt{(3+k)^2 + 2^2}.

Remember ∣z∣=ℜ(z)2+Im(z)2|z| = \sqrt{\Re(z)^2 + Im(z)^2} and the imaginary part of the complex number is the number in front of the ii and not including the ii.
Original post by katie*may
Has anyone done the question on the June 2013 R paper where for the summation r=0 and not r=1, its also on the edexcel Gold level 2 paper as well its the last question on that one?
Im really stuck on what you are supposed to do


I've been stuck on this one as well...I'm sure the equivalent summation is r=1 to n+1 so you can multiply 2n by (n+1) and multiply 1 by n+1)
Original post by Zacken
Nope, the roots of polynomials things aren't on your spec.


Oh right fair- btw can you have a look at Q1(c)? because the mark scheme implies (2i)^(2)=+4 which I think is wrong?

thanks
Edit: Nvm just read your post, thanks!!
Original post by yesyesyesno
Can anyone explain what a Locus/Loci actually is? would be much appreciated.


The set of points that satisfy a certain property. So for example, the set of points that are equidistant from another point is termed a circle because every point on the circle satisfies the property that they are an equal distance (the radius) from the centre (another point).
Original post by Patrick2810
see


Thank you! :lol:
Original post by OhsoIntrovert
I've been stuck on this one as well...I'm sure the equivalent summation is r=1 to n+1 so you can multiply 2n by (n+1) and multiply 1 by n+1)


See a few posts above.

Original post by thesmallman

Edit: Nvm just read your post, thanks!!


No problem.
Original post by Zacken
I can't see the markscheme as you haven't link it, but:

∣a+ib∣=a2+b2≠a2+(ib)2\displaystyle |a + ib| = \sqrt{a^2 + b^2} \neq \sqrt{a^2 + (ib)^2}.

You might want to lookup the definition of modulo again.

So, here: ∣(3+k)+2i∣=(3+k)2+22|(3+k) + 2i| = \sqrt{(3+k)^2 + 2^2}.

Remember ∣z∣=ℜ(z)2+Im(z)2|z| = \sqrt{\Re(z)^2 + Im(z)^2} and the imaginary part of the complex number is the number in front of the ii and not including the ii.


Ahh yeah I always knew that omg, can't believe I made such a rookie error :frown:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you