HSC 2013 MX2 Marathon (archive) (3 Viewers)

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Re: HSC 2013 4U Marathon

1) [(1/a)-(1/b)]^2 + [(1/a)-(1/c)]^2 + [(1/b)-(1/c)]^2 >=0 (equality occurs iff a=b=c)
2/a^2 + 2/b^2 + 2/c^2 -2[(1/ab)+(1/bc)+(1/ac)] >=0
=> (1/a)^2 + (1/b)^2 + (1/c)^2 >= 1/bc + 1/ac + 1/ab for all real a,b,c=/=0 (*)
Hence, bc/a + ac/b + ab/c = abc[(1/a^2)+(1/b^2) +(1/c^2)]>=abc[(1/ab)+(1/bc)+(1/ac)]=a+b+c as req.
2) Refer back to (*), replace a with sqrt(a), b with sqrt(b) and c with sqrt(c) and the second inequality immediately falls out for a,b,c>0

Sorry if it's hard to read, still haven't learnt latex.










http://www.codecogs.com/latex/eqneditor.php

Make the latex there then put
 
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bleakarcher

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Re: HSC 2013 4U Marathon

Hey guys, here's a question from last year's UNSW maths competition. Let f(x) denote a strictly positive continuous function defined on all real numbers with the property that f(2012)=2012 and f(x)=f(x+f(x)) for all x. Prove that f(x)=2012 for all x.
 

Sy123

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Re: HSC 2013 4U Marathon

Hey guys, here's a question from last year's UNSW maths competition. Let f(x) denote a strictly positive continuous function defined on all real numbers with the property that f(2012)=2012 and f(x)=f(x+f(x)) for all x. Prove that f(x)=2012 for all x.


for some number k, f is continuous and defined on all real therefore a and k can be anything.
If that is the case then a and k can be such that they represent any pair of real numbers. The function of these numbers are all equal, therefore

for some constant C since f(a) = f(k)





Is that good enough?

============







 

Makematics

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Re: HSC 2013 4U Marathon

Hey guys, here's a question from last year's UNSW maths competition. Let f(x) denote a strictly positive continuous function defined on all real numbers with the property that f(2012)=2012 and f(x)=f(x+f(x)) for all x. Prove that f(x)=2012 for all x.
omg thank you so much for asking that. really wanted to know the solution :p
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Yep those are the only solutions, what was your method?
My aim was to construct a cubic polynomial which has roots x, y, z, using sum of roots/product of roots/two at a time. Then by finding the roots of this polynomial, we find the values of x, y, z, that satisfy the system of equations. The cubic is in the form:



The sum of roots is easy as it's already given:



So we have:



Now we get the value of 'b' by finding the sum of the pair roots (xy+xz+yz). We know that:







So we now have:



Now we know that x, y, z and the roots of this equation, so by substitution we arrive at:







If we add all these together we get:







So our polynomial is:



Which simplifies to:



Which has all roots equal to 1. So x, y, z are all equal to 1. And the polynomial can not have other roots and so this is the only possible solution set to the given system of equations.
 

bleakarcher

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Re: HSC 2013 4U Marathon



for some number k, f is continuous and defined on all real therefore a and k can be anything.
If that is the case then a and k can be such that they represent any pair of real numbers. The function of these numbers are all equal, therefore

for some constant C since f(a) = f(k)





Is that good enough?

============







Think it works dude, nice. I'm not so sure about my method though:

f(x)=f(x+f(x))
f '(x)=[1+f '(x)]*f '(x+f(x))=[1+f '(x)]*f '(x) => f '(x)=0 for all x. Since f(x) is continuous and f(2012)=2012, f(x)=2012 for all x.
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

Think it works dude, nice. I'm not so sure about my method though:

f(x)=f(x+f(x))
f '(x)=[1+f '(x)]*f '(x+f(x))=[1+f '(x)]*f '(x) => f '(x)=0 for all x. Since f(x) is continuous, f(x)=2012 for all x.
How did you get that ?
 
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