Search results

Re: 2012 HSC MX2 Marathon Yeah that's probably a good idea. This one is for people who are doing their HSC, and not done it already, $Show $ \int^a_0 f(x)dx = \int^a_0 f(a-x)dx

We had 1 girl and 8 guys in our class.

The discriminant is >0

Most likely. Multiple choice will be just a waste of time. No need to put it in exams.

Multiple choice...? They planning to bring that in?

Paid mine

Ok that's good, I think my lectures and such end at 3 on fridays.

Do you know what time on friday?

Re: 2012 HSC MX2 Marathon I thought I posted by solution for that, guess I'll do it now :p. \int\frac{1+sinx}{cos}dx \\\\ let u = sinx \\\\ \int\frac{1+sinx}{cos}dx = \int \frac{1+u}{1-u^2}du \\\\ = \int\frac{1}{1-u}du \\\\ = -ln|1-u| + c \\\\ =-ln|1-sinx|+c $Using $ lnx =...

You mean evolve? Lol DO EIT!

Re: 2012 HSC MX2 Marathon First graph f(x) then using that graph y=1/f(x)

Re: 2012 HSC MX2 Marathon m<0 only applies for 0 < x < a which isn't a condition, so why is m<0?

Re: 2012 HSC MX2 Marathon Lol I'm stuck at this point too, can't work that out.

My school rank was >9000

Re: 2012 HSC MX2 Marathon That does make sense, though you wouldn't be able to figure it out using conventional methods, (as in you can't just sub the number in), which what I was trying to point out lol.

Re: 2012 HSC MX2 Marathon Trebla's question didn't work as at pi/2 it is undefined.

What degree are you doing?

I got this preyty much, though slightly closer to the center.

Re: 2012 HSC MX2 Marathon You should recognise it's almost f'(x)/sqrt[f(x)], and with a little manipulation you can get it there, once it's like that you can just quickly integrate. But as Carrotsticks said, you need to do a lot of questions to get yourself familiar to this.

I'll be at the one on Wednesday. And mathsoc needs to have dodgeball events!