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I don't think I ever had one

Yes, I'd always leave it in the exact form unless it is specified not to.

Re: HSC 4U Integration Marathon 2017 I know, but it's the right answer and I can't be bothered to put it in it's simplest form. Edit: stuff it ln(3^(5/4)/2^(23/12))

Re: HSC 4U Integration Marathon 2017 Also sorry if my handwriting poses any issues

Re: HSC 4U Integration Marathon 2017 http://imgur.com/d7oqSIR

Re: HSC 4U Integration Marathon 2017 Also this is another method: http://imgur.com/h2IBmvL

Re: HSC 4U Integration Marathon 2017 http://imgur.com/a/LQn6t Checked it on wolfram so it's all good.

Oh yeah I got that about f''(x), so I did narrow it down to A and B, but then I just measured the gradient of f'(1) to make sure it was greater than 1. Yeah that makes sense, with the secant line, didn't think of that.

Also for 9, am I the only person who literally measured the gradient of the tangent with a ruler?

Yeah I'm agreeing with this.

That seems about right.

http://rawmarks.info/wiki/Mathematics_Extension_1 This might help.

Well, your answer for iii) might just count as a carry on error, so I'm guessing you'd get the mark.

Me and my friend got a/4.

Same here. I was scared as my friend said plus or minus root 2 on 2 but yeah if you confirmed it there that's good. But he beat me internally :(

Yeah one of my friends did 4 unit, and his 3 unit marks became a lot higher afterwards. I'm guessing it's got to do with problem solving ability, as we do much harder questions that require a higher problem solving ability. I'm mainly talking about harder 3 unit anyway.

we've been found out guys!!!

What I did was let angle CBA be theta, and that was equal to angle CDA as they are angles in the same segment. Angle CDA will be equal to angle CMF for the same reason as CMDE is a cyclic quadrilateral. Also, as angle ACB is 90 degrees, angle CAB will be 90 - theta. So now we can find angle MFA...