HSC 2013 MX2 Marathon (archive) (4 Viewers)

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seanieg89

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Re: HSC 2013 4U Marathon

First find the function f(t) such that the equality holds:





For F the primitive of f:




















I have a feeling this argument is not valid but I cant see it right now, its a bit late
Your f is indeed the unique function which gives us equality for all t >= 0 (*), but I don't know how you are deducing that any g satisfying the integral inequality must be smaller than f everywhere (which is what we are aiming for).

(*) Your method of showing this is mostly okay, but the division by something that is potentially (a priori) negative or zero does raise some niggling potential problems...thankfully none of them crop up here so don't worry about this.
 

Sy123

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Re: HSC 2013 4U Marathon





Integrating everything gives:





Take



Yep this is pretty much it, although I would suggest that for your first step you use a dummy variable such as z, then integrating from 0 to x.

EDIT: I probably should of put the conditions on x
 
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seanieg89

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Re: HSC 2013 4U Marathon



by the product rule, fundamental theorem of calculus, and our assumed integral inequality.

Replacing t with r and integrating this expression from r=0 to r=t gives:



Hence we can conclude that:



Our working is valid for all non-negative t.
 
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