HSC 2013 MX2 Marathon (archive) (4 Viewers)

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Sy123

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Re: HSC 2013 4U Marathon

Fixing faulty question
 
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seanieg89

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Re: HSC 2013 4U Marathon

That is from the IMO.
From the early 70s. Those problems are often doable by good mx2 students with no prior olympiad training and this one is no exception.
 

Sy123

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Re: HSC 2013 4U Marathon

From the early 70s. Those problems are often doable by good mx2 students with no prior olympiad training and this one is no exception.
So would you say that most of the 1970/prior problems are good practise for problem solving skills for mx2 students?

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seanieg89

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Re: HSC 2013 4U Marathon

So would you say that most of the 1970/prior problems are good practise for problem solving skills for mx2 students?

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Yep although not directly relevant to the HSC syllabus they are good practice for problem solving and rigorous argument, especially for those who want to pursue mathematics further. You certainly don't have to be of modern olympiad standard to solve most of them.
 

Exodia

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Re: HSC 2013 4U Marathon

If it has at least 1 real root doesn't that mean it should have 2 real roots due to the conjugate complex root theorem and the fact that it is a quartic? Or are we not counting multiplicities here? I am sure my solution is wrong because when I sub a=1 and b=0 I get x=-1 as a real solution and 1<4.
What is the proper way to doing this question?
 
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seanieg89

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Re: HSC 2013 4U Marathon

I won't spoil the question yet for those who are working on it, but there are likely several approaches.

A hint to doing it the way that I did is dividing the poly by x^2, (observing first that 0 cannot be a root of the poly).

Having at least one real root does not mean it must have two real roots, it could just as well have all four roots real.

And your first solution is wishful thinking really, we cannot arbitrarily set a=0, it just changes the problem you are solving.
 

Lieutenant_21

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Re: HSC 2013 4U Marathon

@Sy123: I can do but couldn't do your one :( give me a hint please!
By the way, does it have to do with expressing the sum in terms of euler's identity? i.e. the real part of: e^2ix+e^4ix + ... + e^4nix (which I don't think is within the MX2 syllabus).
 
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Lieutenant_21

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Re: HSC 2013 4U Marathon

The only geometric series I can think of are:

and (cis2x)+(cis2x)^2+...+(cis2x)^n

By the way, do we have to memorise the sums to products formulas? I have seen many questions that require them but I think they give the formula or ask to prove that i.e. 2sinacosb=(sin(a+b)+sin(a-b))

I am very interested in knowing the solution to your palindromic polynomial question. So far I am stuck but I found that the roots have to be 1 and/or -1.
 
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seanieg89

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Re: HSC 2013 4U Marathon

1. You are on the right track. Forget about Euler's formula though. From the perspective of pretty much any MX2 level question it is just an alternative way of writing cis, and one that will cost you marks / potentially confuse you.

2. I didn't but you may find it helpful. What I did memorise was all the ideas though, if I needed the sums to product formula I would be able to figure it out very quickly. I think this is a more efficient use of memory.

3. I will not ruin it for others yet, but the roots certainly don't have to be 1 and/or -1.
 

HeroicPandas

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Re: HSC 2013 4U Marathon

So would you say that most of the 1970/prior problems are good practise for problem solving skills for mx2 students?

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I got ur integral if and only if n is even :O

Also is ?
 

Sy123

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Re: HSC 2013 4U Marathon

The only geometric series I can think of are:

and (cis2x)+(cis2x)^2+...+(cis2x)^n

By the way, do we have to memorise the sums to products formulas? I have seen many questions that require them but I think they give the formula or ask to prove that i.e. 2sinacosb=(sin(a+b)+sin(a-b))

I am very interested in knowing the solution to your palindromic polynomial question. So far I am stuck but I found that the roots have to be 1 and/or -1.
You are on the right track with geometric series. Make some strong manipulations so you can simplify your sum much more easily (or just grind out the result)

I got ur integral if and only if n is even :O

Also is ?
That result is correct, it should be straightforward from there, I'm not sure how it would prove it for only n even though
 

Lieutenant_21

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Re: HSC 2013 4U Marathon

What is the fastest way to prove the AM:GM inequality? My teacher uses induction + calculus which is a tedious method.
 

Sy123

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Re: HSC 2013 4U Marathon

What is the fastest way to prove the AM:GM inequality? My teacher uses induction + calculus which is a tedious method.
In my opinion this is one of the few problems where induction isn't "cheap". Its not like we are getting the inequality out of nowhere since we can observe that its true for n=2 n=3 and n=4 rather easily. So its feasible to 'guess' the general case and prove by induction.

i.e. I dislike proofs of induction for identities where you can't possibly come guess the result.
 
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