MedVision ad

HSC 2013 MX2 Marathon (archive) (6 Viewers)

Status
Not open for further replies.

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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

New Member
Joined
Feb 24, 2013
Messages
16
Gender
Male
HSC
2013
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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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

New Member
Joined
Feb 24, 2013
Messages
16
Gender
Male
HSC
2013
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

(╯°□°)╯━︵ ┻━┻ - - - -
Joined
Apr 7, 2012
Messages
6,624
Location
Sydney Girls
Gender
Female
HSC
2013
Uni Grad
2017
Re: HSC 2013 4U Marathon

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

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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

(╯°□°)╯━︵ ┻━┻ - - - -
Joined
Apr 7, 2012
Messages
6,624
Location
Sydney Girls
Gender
Female
HSC
2013
Uni Grad
2017
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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon





 
Last edited:

HeroicPandas

Heroic!
Joined
Mar 8, 2012
Messages
1,547
Gender
Male
HSC
2013
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..
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 6)

Top