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Re: HSC 2014 4U Marathon - Advanced Level Nice solution, I don't know why my final expression turned out so ugly.

Think about this graphically. When we find the points of intersection, we equate the y's. The new cubic polynomial formed is x^3-6x^2-(m+2)x+1-b=0. We are given that there exists a triple root. Hence, this implies that the line y=mx+b must of intersected the original polynomial at its point of...

Re: HSC 2014 4U Marathon - Advanced Level Here's what I got

Re: HSC 2014 4U Marathon - Advanced Level Just asking, are the numbers nice, because I got something really ugly.

Please for the love of god, not complex/poly for the 3rd year in a row. I'll probably stand a better chance if it is something else. If it's inequalities and probability, there will be many tears shed

Re: HSC 2014 2U Marathon

Anyone have a digital copy that they would like to share? :D

inb4 last question probability

Re: HSC 2014 4U Marathon

Re: HSC 2014 4U Marathon exactly the same as mine

That Q16 (b) integral was one of the nicest crafted integrals I've ever seen, so beautiful...

yay, i've been looking everywhere for this paper

You can always verify your answer by doing sum and product of roots. Here is how you do 4(iii) . Parts (i) and (ii) are basic operations. To avoid confusion, let z(1)=x+iy and z(2)=a+ib and try working from there.

Re: HSC 2014 4U Marathon Since nobody has answered it, I might as well do so . Strong induction in disguise

Yeah, dem feels

Re: MX2 Integration Marathon it's not your simple limiting sum nor is it and indefinite integral, it's a special type of series

Re: MX2 Integration Marathon in your case, it'll be in terms of 'k' [i.e. (2k+1)/2^(2k+1)], since that's what you chose to represent as the index of summation

Yep, that's what I did using the insertion method

Re: MX2 Integration Marathon yeah there should also be a (2n+1)/2^(2n+1) outside and expand the sigma

my bad, typo