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HSC 2014 MX2 Marathon (archive) (4 Viewers)

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Sy123

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Re: HSC 2014 4U Marathon

 
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Tugga

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Re: HSC 2014 4U Marathon

Squaring both and adding gives you 2*(expanded form of cos(a-b)) + 2 =8/3 (since sin^2a+cos^2a=1)

So cos(a-b)=1/3

P.s. Sorry for ugly formatting, don't have the time right now to learn latex, maybe later. D:
 

Tugga

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Re: HSC 2014 4U Marathon

A q from this year's Baulko 4u trial


screenshot tool
let that cube root thingo be a.

noting a^3 = 30+18a, it follows that a^3-18a-30=0.

therefore, P(x)=x^3-18x-30

A more rigorous proof would prove that its the lowest degree. I would probably do something like square it and show you get something you can't cancel with a degree 1 polynomial?
 

madharris

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Re: HSC 2014 4U Marathon

Got a difficult one for you guys. I saw that it was going to be in carrot's trial.

 

aDimitri

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Re: HSC 2014 4U Marathon

equivalent of letting x = 2n-1, where n is any integer.
let x = 2n-1 where n is an integer.



same shit to prove it for all negative odd but do n = k-1
but x^2 - 1 is an even function so it's sorta redundant to do it for negative numbers too
 

Stygian

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Re: HSC 2014 4U Marathon

equivalent of letting x = 2n-1, where n is any integer.
let x = 2n-1 where n is an integer.



same shit to prove it for all negative odd but do n = k-1
but x^2 - 1 is an even function so it's sorta redundant to do it for negative numbers too
alternatively u can take the approach that every odd number is congruent to

1 mod 8
3 mod 8
5 mod 8
7 mod 8

Squaring it becomes

1 mod 8
9 mod 8 or 1 mod 8
25 mod 8 or 1 mod 8
49 mod 8 or 1 mod 8

Subtracting 1 from each gives 0 mod 8 i.e. all are divisible by 8
 

aDimitri

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Re: HSC 2014 4U Marathon

alternatively u can take the approach that every odd number is congruent to

1 mod 8
3 mod 8
5 mod 8
7 mod 8

Squaring it becomes

1 mod 8
9 mod 8 or 1 mod 8
25 mod 8 or 1 mod 8
49 mod 8 or 1 mod 8

Subtracting 1 from each gives 0 mod 8 i.e. all are divisible by 8
but it's still not proven for all odd, just shown for the first 4
 

Stygian

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Re: HSC 2014 4U Marathon

oh true yeah.
derp moment T_T
haha it's ok :)

i do think the mod arithmetic approach is better in terms of efficiency but obviously your solution is more within the scope of the course and so better for this thread
 

Stygian

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Re: HSC 2014 4U Marathon

(also im having a brain fart on that problem you posted in the Advanced thread, i can't seem to think of how to approach it)
 
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