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HSC 2013 MX2 Marathon (archive) (8 Viewers)

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seanieg89

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Re: HSC 2013 4U Marathon

Yep HeroicPandas, that's pretty much it, that is 7 marks from a school trial (ascham 2005)

===





By the product condition, the inequality can be rewritten as:

(abc-1) + (a+b+c) - (ab+bc+ac) > 0

<=> (a-1)(b-1)(c-1) > 0.

Hence either one or all three of the LHS factors must be positive. They cannot all be positive because of the product condition, and so we can conclude that exactly one of a,b,c is greater than one.
 

Sy123

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Re: HSC 2013 4U Marathon

By the product condition, the inequality can be rewritten as:

(abc-1) + (a+b+c) - (ab+bc+ac) > 0

<=> (a-1)(b-1)(c-1) > 0.

Hence either one or all three of the LHS factors must be positive. They cannot all be positive because of the product condition, and so we can conclude that exactly one of a,b,c is greater than one.
Nice solution, I think my one might be yours 'in disguise' since it uses a similar concept

My one was: First conclude that at least 1 of a,b,c needs to be > 1 if it satisfies abc=1. And at least 1 of a,b,c < 1.
This leaves 2 cases, 2 of them > 1, and 1 of them < 1 or
2 of them < 1 and 1 of them > 1, and the second is what we need to prove.

Without loss of generality, let a > 1, b < 1.





Since a > 1, then bc < 1, so inequality sign flips when cancelling out





Since b < 1, we can divide



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









 

rural juror

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Re: HSC 2013 4U Marathon





(above results and the one in the 3U thread are verified since they come from the book, Summation of Series - Jolley)

thought of something abit dodgy for this one, looking at (1-x)^n/(1-x) = (1-x)^n(which is true) for |x|<1, if we equate x^n coefficients the result comes out, by eqauting 1/1-x as 1+ x + x^2 + x^3...

how did you guys do it????
 

Sy123

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Re: HSC 2013 4U Marathon

thought of something abit dodgy for this one, looking at (1-x)^n/(1-x) = (1-x)^n(which is true) for |x|<1, if we equate x^n coefficients the result comes out, by eqauting 1/1-x as 1+ x + x^2 + x^3...

how did you guys do it????
Yes that actually works, quite clever. (assuming you mean m when you said (1-x)^m / (1-x) = (1-x)^(m-1))

My way was:

Equating co-efficient of x^n in:

 

rural juror

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Re: HSC 2013 4U Marathon

oh right yeh, n-1 for the second part. your method is pretty neat too.

The one jsut above it, though, i have a solution for, but it was pretty long, and involved the pascal addition theorem and induction and looking at n = even and n= odd case, though it eventually came out. Did you have a quicker method????
 

obliviousninja

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Re: HSC 2013 4U Marathon

So much inequalities lately, even though it has been examined once in the last 5 years.
 

Sy123

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Re: HSC 2013 4U Marathon

oh right yeh, n-1 for the second part. your method is pretty neat too.

The one jsut above it, though, i have a solution for, but it was pretty long, and involved the pascal addition theorem and induction and looking at n = even and n= odd case, though it eventually came out. Did you have a quicker method????
Yeah I did have a quicker method than that. Post an outline of your solution if you want anyway, it doesn't need to be detailed.
 

obliviousninja

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Re: HSC 2013 4U Marathon

Pretty sure they're examined every year
lol mb, I was refering to the wrong thing.
On another note, I did the binomial last question 3u 2011, for some reason it was way harder than most of the ones in 4u.
 

Sy123

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Re: HSC 2013 4U Marathon





 
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HeroicPandas

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Re: HSC 2013 4U Marathon

Two problems I see:

1) Think about if you can actually make the transformation of

2) In one of the last lines, your inequality sign changes direction.
1) i seriously dont know
2) so u are saying this:



..is wrong?
idk man..
 
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