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HSC 2013 MX2 Marathon (archive) (2 Viewers)

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SpiralFlex

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Re: HSC 2013 4U Marathon

Sorry didn't see that substitution.
 

bleakarcher

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Re: HSC 2013 4U Marathon

Evaluating the definite integral is well within the scope of MX2, the substitution



kills it.
How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.
 

seanieg89

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Re: HSC 2013 4U Marathon

How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.
Wanted a substitution that mapped the interval [0,1] to [0,1] so I could express I in terms of itself and some other terms. (ie I wanted to make a substitution but I didn't want the limits of integration to change.) This was the simplest rational function I could think of satisfying this property and it worked out nicely.

The first way I solved this question though was via a longer, different, less difficult to spot method...differentiation under the integral sign. But this is not MX2 material and it was longer than the substitution anyway.
 

bleakarcher

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Re: HSC 2013 4U Marathon

Wanted a substitution that mapped the interval [0,1] to [0,1] so I could express I in terms of itself and some other terms. (ie I wanted to make a substitution but I didn't want the limits of integration to change.) This was the simplest rational function I could think of satisfying this property and it worked out nicely.

The first way I solved this question though was via a longer, different, less difficult to spot method...differentiation under the integral sign. But this is not MX2 material and it was longer than the substitution anyway.
ah okay, I see. nice.
 

bleakarcher

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Re: HSC 2013 4U Marathon

Let L=sqrt(6+sqrt(6+sqrt(6+...
<=> L^2=6+L
<=> L^2-L-6=0
<=> (L-3)(L+2)=0
=> L=3 since L>0

The continued fractions one should be the golden ratio, [1+sqrt(5)]/2. Sorry I still haven't learnt latex.
 

seanieg89

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Re: HSC 2013 4U Marathon

Much easier than most of the questions I post here, but still quite a cool result.

 

jyu

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Re: HSC 2013 4U Marathon

Much easier than most of the questions I post here, but still quite a cool result.

i) y=f(x) intersect y=x at least once

ii) no, y=f(x) can be wholely above or below y=x
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Denote as the k'th triangular number (surprise, surprise).



So we obtain:

as the sum of the multiples of 3.



So we obtain:

as the sum of multiples of 5.

Now we will double count all the multiples of 15, so using the same logic:



as the sum of multiples of 15.

Now we add and subtract some shit:



Noting that



 

Sy123

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Re: HSC 2013 4U Marathon

Denote as the k'th triangular number (surprise, surprise).



So we obtain:

as the sum of the multiples of 3.



So we obtain:

as the sum of multiples of 5.

Now we will double count all the multiples of 15, so using the same logic:



as the sum of multiples of 15.

Now we add and subtract some shit:



Noting that



Nice work. I'll try make a question with triangular number properties or something considering how much you like them heh.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Awesome question. I got the answer but I'll let others have a go first, lets try and have different people answering besides the usual few. If no one answers this by tonight I'll put up my solution though.

I'll try come up with a cool question, but I used most of my ideas in my test lol.
 
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