Search results

yep, carrotsticks did: https://www.dropbox.com/s/h93cco9ax9aywyk/2%20Unit.pdf?m

Yep, that makes perfect sense then, thanks.

A bit of nitpicking, for q14biv, how can you switch m and x when 0 < m < 1 but x >= 1?

Re: HSC 2013 4U Marathon Here's a doable one, four lines of working and one of my favourites. $if$ \, \frac{sin(\pi x)}{\pi x} = \prod_{n = 1}^{\infty}(1 - \frac{x^2}{n^2}) $show that$ \; \pi \cot(\pi x) = \frac{1}{x} + \sum_{n = 1}^{\infty}\left(\frac{2x}{x^2 - n^2}\right) where...

Re: MX2 Integration Marathon Feel like some math today, brute forced also. let\,u=\sqrt {a \sqrt {b \sqrt x}} squaring, x=\frac{u^8}{a^4 b^2} \frac{\mathrm{d} x}{\mathrm{d} u}=\frac{8u^7}{a^4 b^2} then, \int \sqrt{a \sqrt{b \sqrt{x}}}dx = \int \frac{8u^8}{a^4 b^2}du = \frac{8u^9}{9a^4 b^2}...

It's funny cause it's true.

So while I was procrastinating English, I wrote up these solutions complete with shoddy working :) Hope it's legible and please tell me if I made mistakes haha. https://drive.google.com/folderview?id=0B4D3wNobxSuvMG93NHRCNS14UTA&usp=sharing

I got B. If one but not both f(x) or g(x) is negative, then it flips the inequality around doesn't it?

bottleofyarn will be attending.

Count me in. Hard tests are appreciated :)