HSC 2013 MX2 Marathon (archive) (7 Viewers)

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hamstar

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Re: HSC 2013 4U Marathon

I think the answer is

(n+1)! - 1

you can even prove by mathematical induction.
 
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Re: HSC 2013 4U Marathon

I think the answer is

(n+1)! - 1

you can even prove by mathematical induction.
This is right.
The whole point of the question is to get a telescoping sum... those come up in the HSC a lot (not really).
 

Sy123

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seanieg89

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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.
Tricky question! Am clearly rusty at contest mathematics.

 
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Sy123

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Sy123

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Re: HSC 2013 4U Marathon

I'm not sure if this counts as using the Binomial Theorem or not, but consider the following:



Then the result falls out immediately.
You could do that, but of course there are ways of proving that result without using Binomial theorem (i.e. combinatorics), iirc one of your questions on here was about that.
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

realise you are a fucking idiot
 
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Trebla

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Re: HSC 2013 4U Marathon

I got a feeling I've asked this before, but will put it in anyway.

Prove that for any set of real numbers {xi}



Also, here is a Conics question for the purposes of variety on this thread:

The points P(a cos θ, b sin θ) and Q(a cos φ, b sin φ) lie on an ellipse. If PQ is a focal chord which passes through the positive focus show that the length of PQ is



where e is the eccentricity of the ellipse.
 
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Sy123

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Re: HSC 2013 4U Marathon

I got a feeling I've asked this before, but will put it in anyway.

Prove that for any set of real numbers {xi}



Also, here is a Conics question for the purposes of variety on this thread:

The points P(a cos θ, b sin θ) and Q(a cos φ, b sin φ) lie on an ellipse. If PQ is a focal chord which passes through the positive focus show that the length of PQ is



where e is the eccentricity of the ellipse.
Yep you have lol.

======









 
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seanieg89

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Re: HSC 2013 4U Marathon

I will provide a sketch of a proof. I encourage anyone who hasn't done this question to still try it though, it is good even when viewed as a purely algebraic exercise.

1. This is a standard integration by parts induction. We use I_n to denote this integral from now on.

2. Now the infinite series in ii) can be rewritten as the infinite series:



3. For finite partial sums of this series, note that we can take the integral outside the sum. Use what we know about geometric series to find an expression for the thing inside this integral.

4. Split this integral into the sum of two integrals, one of which tends to zero as N -> infinity, and the other of which is independent of N. (Here N is the number of terms we are taking in our partial sum).

5. Flex our integration muscles. Its actually quite an easy integral, we have already done the hard work really.

6. QED bitches.

Note that this is an MX2 level proof, which could be shortened with more powerful tools. For example, the manual handling of the "error term" could be bypassed by using the monotone convergence theorem or some weaker result about the interchange of infinite summation and integration.
 

seanieg89

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Re: HSC 2013 4U Marathon

(Sorry for being so lazy, but I guess this thread is meant for current students. Hopefully my sketch will help someone actually doing their HSC to do the question!) If no-one uploads an actual answer I'll upload my handwritten solution later.
 

Sy123

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Re: HSC 2013 4U Marathon





(please inform me if there is a mistake)
 
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