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# MAT Prep Thread - 2nd November 2016 watch

1. There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.
2. (Original post by 11234)
did u get n has to be even
I messed that one up too. Everyone else is saying n has to be even
3. (Original post by furmat)
There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.
How did you know that about function composition?
4. How did everyone find q3 and q4
5. (Original post by RuairiMorrissey)
How did you know that about function composition?
Yeah it's not the most obscure knowledge but it's certainly not in C1 or C2. I think there must be a C1/C2 way of figuring out there's only 1 value of c
6. (Original post by furmat)
There was no need to try 10 different permutations of A and B to find c. In the first part you showed that the composition of functions A and B is commutative. Function composition is also associative, so any composition with 2 As and 3 Bs would be equivalent to AABBB. You could then check it for this value and be satisfied it was the only solution. Sadly I must have made an arithmetic blunder because I don't think I got 107 when I did that - I can't believe I made such a mistake, but I gave all my reasoning so if they're reasonable marking it maybe they'll only take off a mark for the arithmetic slip, given I explained my answer in full.
Could you explain this because the way i see it, I dont think proving that A(B(x)) = B(A(x)) implies commutativity.

I may be wrong but how do you know that A(B(B(x))) = B(A(A(x))) etc..
7. when do they start givin out interviews
8. (Original post by 11234)
How did everyone find q3 and q4
I found q3 strange but easy. I didnt do q4 cuz i did q6 instead
9. (Original post by m0.4444)
I found q3 strange but easy. I didnt do q4 cuz i did q6 instead
would u say this paper was harder than last years?
10. (Original post by sillygoose123)
could you explain this because the way i see it, i dont think proving that a(b(x)) = b(a(x)) implies commutativity.

I may be wrong but how do you know that a(b(b(x))) = b(a(a(x))) etc..
ab = ba

abb = bab

bab = baa
11. (Original post by m0.4444)
I messed that one up too. Everyone else is saying n has to be even
I got no solutions : (
12. (Original post by collmr)
ab = ba

abb = bab

bab = baa
Ahhh i see now i feel like a sillygoose
13. (Original post by RuairiMorrissey)
I got no solutions : (
I'm pretty sure it's no solutions...
14. (Original post by sillygoose123)
Could you explain this because the way i see it, I dont think proving that A(B(x)) = B(A(x)) implies commutativity.

I may be wrong but how do you know that A(B(B(x))) = B(A(A(x))) etc..
A(B(x)) = B(A(x)) is kind of the definition of commutativity for these two functions, if you treat the functions as elements and function composition as a binary operation that is associative (which is the case). The way I'm saying it makes it sound more obscure than it really is. I still made an error in the actual calculation for c
15. (Original post by RuairiMorrissey)
I got no solutions : (
Thats what i put
16. (Original post by RuairiMorrissey)
I got no solutions : (
Basically you show that if n is even, you can do difference of squares and play around with the algebra until you get a factor of x^2 + 1 which therefore proves works for all even n. To prove that it doesnt work for odd n, I just tested it with random values of odd n and found counter examples.
17. (Original post by sillygoose123)
Basically you show that if n is even, you can do difference of squares and play around with the algebra until you get a factor of x^2 + 1 which therefore proves works for all even n. To prove that it doesnt work for odd n, I just tested it with random values of odd n and found counter examples.
I just subbed in x = i

: I
18. (Original post by RuairiMorrissey)
How did you know that about function composition?
if AB = BA...
AAB = A(BA) = (BA)A
AABB = A(BA)B = (BA)AB = AB(BA) = (BA)(BA) = B(BA)A
etc.
basically for any two combinations of A and B, as long as the number of A and number of B are constant, they come out as the same answer
19. (Original post by RuairiMorrissey)
I just subbed in x = i

: I
Yeah I think that's the simplest way to do it. You can even avoid using i by just subbing in x^2 = -1 but this requires extreme reliance on the factor theorem if you haven't been taught about imaginary numbers lol
20. (Original post by boombox111)
if AB = BA...
AAB = A(BA) = (BA)A
AABB = A(BA)B = (BA)AB = AB(BA) = (BA)(BA) = B(BA)A
etc.
basically for any two combinations of A and B, as long as the number of A and number of B are constant, they come out as the same answer
Ah I see, well done.

What do you reckon the shortlisted/succesful average will be?

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