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1. Are you marking these? 2. Is it open for non-students (such as myself)?

Did anyone ever play the old school game myth2? It is pretty deep strategically.

$A group of $23$ people are in a room.\\ \\ Prove that it is more likely than not that there is a common birthday within the group. (Ie that two or more of them have the same birthday). You may assume that there are 365 equally likely possibilities for each person's birthday.$

Re: MX2 Integration Marathon It's wrong.

Definitely falls into the category of cool! Especially in light of its generalisations in algebraic topology. How are you finding university larg?

Haha yep. This sort of thing is why we need rigour in mathematics and can't just believe things because they look true.

The problem is your assumption that the normal will "cross" to the right side of the focus. This does not happen, nor is the diagram any guarantee that it should. The same sort of "argument by pictures" is what can lead to fallacies like the equilateral triangle thing.

Why would the intuitive answer be three? Cubics don't necessarily have three real roots. As soon as you start to get close to the vertex of the parabola, you reach a bounding curve. When you pass this curve you can only draw one normal to the parabola from your point. You can actually...

Surely a typo, the numerator dominates the hell out of the denom...

Re: The Nonsense Thread MC's 1 mark each?

Re: The Nonsense Thread Haha fucking hipsters. Cool yeah, I might do it then. What's the new mx2 format again? 10 MC 6 normal? How many marks each question?

Re: The Nonsense Thread It seems rather trendy to write practice papers now. If I wrote an mx2 paper that was relatively easy apart from the last question would people attempt it?

Re: 2012 HSC MX2 Marathon From your second line to your third line of displayed equations.

Related: Think about what (-1)^pi is and you will notice an often glossed over gap in the high school definitions of powers.

I read what you wrote, the second RHS isn't the first RHS multiplied by 2n, there is an inequality involved.

Re: 2012 HSC MX2 Marathon I don't think q can possibly be that, that would imply that the product of roots of p has modulus 3.

It isn't quite the same. But 2n*(RHS of i) is strictly smaller than RHS of ii.

From previous part, LHS<4n^2/pi=n^2*(4/pi) < n^2*(pi/2).

Re: 2012 HSC MX2 Marathon .

Re: 2012 HSC MX2 Marathon "complex polynomial" = polynomial with complex coefficients.