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MAT Prep Thread - 2nd November 2016

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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
cheersss


@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?
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
cheersss
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

Reply 822
Avatar for joodaa
joodaa
5
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


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?
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!).
Reply 824
Avatar for joodaa
joodaa
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
Original post by Mystery.
Thanks so much everyone.


Did you get the same multiple choice answers as me then?

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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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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?
Reply 828
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Quido
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.'
Original post by theaverage
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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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.
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 :smile:)

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Anyone done the 2004 multiple choice?



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

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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?
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
Original post by Someboady
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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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.
could anyone explain 2013 2 II) c) I can't follow the marking instructions at all
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!
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



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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