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Re: MX2 2016 Integration Marathon I didn't use wolfram alpha to guess it, its something that comes with practice. As I said, it's similar to the last question of the 2014 HSC where you can guess it and solve the integral in a couple of steps. Guessing a term and differentiating it to make...

Re: MX2 2016 Integration Marathon \frac{d}{dx} \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) = \frac{e^x(\sin{x} - 1)}{1-\cos{x}} \therefore \int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx = - \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) + c (This question...

Re: HSC 2016 4U Marathon Key word there is conventionally, however this isn't a conventional question, since when did integrand go to prison, come out, go to uni to only quit later and get hired by UPS, and still have time to answer questions on bos

Re: 2016 HSC Economics marathon In economics you should always assume there is always interest whenever loans are taken out. So when Australia takes out a loan from another country, its known as an inflow of cash, recorded as a credit on the KAFA. The interest that is repaid when Australia...

It does work, you should be able to get to: |z^2| - 4 \leq 8|z| Which will lead you to |z| = 4+2\sqrt{5} as the max value

I just copied and pasted it onto google translate, and I let it handle the rest

"Anyway, you can still finish on the line HSC" Done :)

One medium-sized cardboard box

Re: HSC 2016 4U Marathon Just to give a visual representation of this:

What? :confused:

Solid name reveal.

Are they replacing the standard integral sheet? Because some of the integrals such as: \int \frac{dx}{\sqrt{x^2+a^2}} = \ln{(x + \sqrt{x^2+a^2})} + c are not on the reference sheet.

Yes we have been archived, it was a fun ride guys! Probably the next time we see a 2015ers thread, it will be the HSC results thread. Enjoy the break everyone, and good luck with your results! :)

I was just correcting your statement about |z| = \sqrt{x^2+y^2} just so there was no confusion. In terms of this question, it doesn't matter, since the square root of 1 is still 1, however for cases where the radius is not 1, then you would have got it wrong.

Actually no. |z| = \sqrt{x^2 + y^2} = \sqrt{ z \cdot \bar{z} } = 1

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Can confirm

The numbers will be between 0 to 9

Re: MX2 2016 Integration Marathon Then lets post them here, as the other one will most likely be archived

Guys I just realised, this thread is going to be archived soon. Are we all going to move to the confessions thread when that happens?