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he only made 2 so far so you can imagine the 2012 guys actually shitting themselves cause they wouldn't anticipate these kind of questions ever.

\frac{\pi^2}{6} :smile:

Can you PM me the Paper lol

Re: MX2 Integration Marathon I wanna do it \int_{-1}^{1}\frac{2x}{(((x+1)^2)+4)^2} \\ \\ \\ $Let x+1 = tan\theta$ \\ \\ 2\int_{0}^\frac{\pi}{4}\frac{(2tan\theta-1)sec^2\theta}{16sec^4\theta} \\ \\ = \int_{0}^\frac{\pi}{4}\frac {2tan\theta-1}{8sec^2\theta} =...

Re: MX2 Integration Marathon Well the hard part is the integral really so i'll just do that :P \frac{2}{2cos^2x-sin2x} = \frac{1}{cos^2x - sinxcosx} \\ \\ \\ = \frac{1}{cosx(cosx-sinx)} = \frac{sin^2x+cos^2x}{{cosx(cosx-sinx)}} = \frac{cosx +sinx}{cosx - sin x} - \frac{sinx}{cosx} \\ \\...

probability and inequalities IMO

Yup I remember doing the 4U exam last year, had a massive struggle in Q16 (mechanics and conics combined), but after carefully deconstructing it and getting the right answers (they're proof questions) i was fist pumping to myself saying "YEAH" , I think one of the supervisors looked at me as if...

Me too :D

HSC maths doesn't even teach the concept of 2 sided limits.........

Re: HSC 2014 2U Marathon Heh, it can be solved by inspection :P

got it memorised?

Re: HSC 2013 3U Marathon Thread a = x \\ \ \frac{d}{dx} (\frac{1}{2} v^2) = x \\ \\ \ \frac{1}{2} v^2 = \int x dx = x^2/2 + C \\ \\ v^2 = x^2 +4 \\ v = \sqrt{x^2+4} \\ \\ \ \frac{dx}{dt} = \sqrt{x^2+4} \\ \\ \ \frac{dt}{dx} = \frac{1}{\sqrt{x^2+4}} and yeah ceebs

Re: HSC 2014 2U Marathon Lol i remember seeing that question in the Kings 4U paper, funny how it could actually be done with 2U technique.

ahhhhh going to uni...... what is detention?

Re: HSC 2014 2U Marathon i)\ \ $By drawing the graph$ \ \ \ f(x) = \sqrt{1-x^2} $ and considering the area, find$ \\ \int_{0}^{1}\sqrt{1-x^2}\ dx \\ \\ ii) $Find the approximation of$ \int_{0}^{1}\sqrt{1-x^2}\ dx \ $using the Simpson's Method$ \\ iii) $Hence, find the approximate value of \pi$

Re: HSC 2014 2U Marathon Bump... lol how did that guy get banned.

Re: HSC 2014 2U Marathon y = b^x \\ ln(y) = xln(b)\\ \frac{ln(y)}{ln(b)}= x \\ \frac{dx}{dy} = \frac{1}{yln(b)} = \frac{1}{b^xln(b)} \\\therefore \frac{dy}{dx}= b^xln(b)

What about dem CAGs?

Re: HSC 2014 2U Marathon \int \frac{1}{xlnx} \ \ dx \\ \\ \\ \int tan^2x \ \ dx

Re: HSC 2014 2U Marathon The result in part iii is only true for \ \ \ -1<x<1