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# MAT Prep Thread - 2nd November 2016 watch

1. (Original post by joodaa)
Could someone explain q2 part IV of the 2015 paper? I dont understand how you can factorise it the way the mark scheme has
Zacken

I don't quite understand why using a = 3, b=2 and n=2014 doesn't work... it would be logical to me to use these values... Could you possible explain this?
2. (Original post by joodaa)
Could someone explain q2 part IV of the 2015 paper? I dont understand how you can factorise it the way the mark scheme has
It's a direct application of the (i); the only possible issue I can imagine is you're confused about what a and b are (but even this should be fairly obvious looking at the solution).
Spoiler:
Show
a = 3^3, b = 2^3
3. (Original post by DFranklin)
It's a direct application of the (i); the only possible issue I can imagine is you're confused about what a and b are (but even this should be fairly obvious looking at the solution).
Spoiler:
Show
a = 3^3, b = 2^3
Oh um awkward I didn't make the link that that was what they used ugh :/
Is there a reason why they didnt use a = 3 and b is 2?
4. (Original post by joodaa)
Oh um awkward I didn't make the link that that was what they used ugh :/
Is there a reason why they didnt use a = 3 and b is 2?
Part (i) tells you that (a-b) is a factor of a^n - b^n. But if (a - b) = 1, this isn't terribly helpful (or surprising!).
5. (Original post by DFranklin)
Part (i) tells you that (a-b) is a factor of a^n - b^n. But if (a - b) = 1, this isn't terribly helpful (or surprising!).
Ahh okay thanks i think i get it now cheers dfranklin
6. (Original post by Mystery.)
Thanks so much everyone.
Did you get the same multiple choice answers as me then?

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7. Please could someone check if

g(a,b)= s(b, m( g( a, P(b))), a) is a solution to question 5 part IV 2015

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8. Do we get partial credit on the longer questions? So if you started part iv and it was worth 6 marks and didn't finish it, could they still give you some of the marks for that part?
9. (Original post by Oceankarma)
Do we get partial credit on the longer questions? So if you started part iv and it was worth 6 marks and didn't finish it, could they still give you some of the marks for that part?
Yeap, it says on their website 'Part marks are available for the longer questions.'
10. (Original post by theaverage)

g(a,b)= s(b, m( g( a, P(b))), a) is a solution to question 5 part IV 2015

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I did g(a,b)= s(p(b), m(g(a, p(b))), a). The reason for this is we want to perform m |b| times on a. To do this, we can increment b up to 0, each time of which will perform m(a). Therefore the variable we compare (our 'x' in the s function) must be p(b), or b + 1, with comparison to function f(a, b).

Try out your formula with g(5, -1). You would expect 4.
g(5, -1) = s(-1, m(g(5, 0)), 5) = m(g(5, 0)) - as -1 <= 0
m(g(5, 0)) = m(s(0, m(g(5, 1)), 5) = m(m(g(5, 1))) as 0 <= 0
m(m(g(5, 1))) = m(m(s(1, m(g(5, 2)), 5))) = m(m(5)) as 1 > 0

g(5, -1) = m(m(5)) = 3.
11. (Original post by some-student)
I did g(a,b)= s(p(b), m(g(a, p(b))), a). The reason for this is we want to perform m |b| times on a. To do this, we can increment b up to 0, each time of which will perform m(a). Therefore the variable we compare (our 'x' in the s function) must be p(b), or b + 1, with comparison to function f(a, b).

Try out your formula with g(5, -1). You would expect 4.
g(5, -1) = s(-1, m(g(5, 0)), 5) = m(g(5, 0)) - as -1 <= 0
m(g(5, 0)) = m(s(0, m(g(5, 1)), 5) = m(m(g(5, 1))) as 0 <= 0
m(m(g(5, 1))) = m(m(s(1, m(g(5, 2)), 5))) = m(m(5)) as 1 > 0

g(5, -1) = m(m(5)) = 3.
Yeah makes sense, thanks )

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12. Anyone done the 2004 multiple choice?

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13. How would you go about solving sin (2x) + (sin(x))^2 =1

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14. (Original post by joodaa)
Ahh okay thanks i think i get it now cheers dfranklin
I still don't get it, do you mind explaining?
15. (Original post by theaverage)
How would you go about solving sin (2x) + (sin(x))^2 =1

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Use some trigonometric identities e.g. sin^2x + cos^x = 1
I still don't get it, do you mind explaining?
You know it isn't a prime if neither of the 2 brackets equal 1. If you simple did 3 and 3 the first bracket would equal 1 which means it could be a prime number. However if you use 3^5 and 2^5 you can work out the first bracket isn't 1, and the 2nd clearly isn't 1 hence you can conclude that the number isn't a prime.

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17. (Original post by theaverage)
Anyone done the 2004 multiple choice?

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I have done them just now, let me know if you want any help or if you want to share answers.
18. could anyone explain 2013 2 II) c) I can't follow the marking instructions at all
19. (Original post by theaverage)
You know it isn't a prime if neither of the 2 brackets equal 1. If you simple did 3 and 3 the first bracket would equal 1 which means it could be a prime number. However if you use 3^5 and 2^5 you can work out the first bracket isn't 1, and the 2nd clearly isn't 1 hence you can conclude that the number isn't a prime.

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Ahhh! I get it thanks dude!
20. (Original post by DFranklin)
It's a direct application of the (i); the only possible issue I can imagine is you're confused about what a and b are (but even this should be fairly obvious looking at the solution).
Spoiler:
Show
a = 3^3, b = 2^3
So theoretically you can use 3^n and a^n where n is any integer bigger than 1?
Because they'd all be a factor right? or am I talking nonsense...

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