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# Edexcel FP1 Thread - 20th May, 2016 watch

1. (Original post by LLk1)
How is 4b 8? in jan 2016 ial
can't even find this paper, do you have a link?
2. (Original post by LewisClothier)
can't even find this paper, do you have a link?
3. 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.
4. (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
5. (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
6. Thanks for this. Feels like there isn't much practise available.
7. 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
8. jan 2016 (IAL) is difficult af- is everything there in our spec??
9. 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
10. (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.
11. (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.

2. .

Since in this case you have

Then you have being added together times to get a total of .
12. (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.
13. Can anyone explain what a Locus/Loci actually is? would be much appreciated.
14. (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
I can't see the markscheme as you haven't link it, but:

.

You might want to lookup the definition of modulo again.

So, here: .

Remember and the imaginary part of the complex number is the number in front of the and not including the .
15. (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)
16. (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
17. (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).
18. (Original post by Patrick2810)
see
Thank you!
19. (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)
No problem.
20. (Original post by Zacken)
I can't see the markscheme as you haven't link it, but:

.

You might want to lookup the definition of modulo again.

So, here: .

Remember and the imaginary part of the complex number is the number in front of the and not including the .
Ahh yeah I always knew that omg, can't believe I made such a rookie error

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