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The function looks something like And since the definite integral is the area under the curve, you can find it by finding A1, A2, A3, A4 and adding them together: A_{total}=A_1+A_2+A_3+A_4 Now, they're all trapezium's and the area of a trapezium is \frac{h}{2}(a+b) (Where h is the height, a is...

Thanks for that! Also, I love your new signature :p

Re: 2012 HSC MX1 Marathon You've made an error in the expression for Tr+1 of the second binomial, although coincidentally one of your answers is the correct one.

I just got myself a job tutoring (by which I mean providing help to a person that's falling behind in class - not tutoring as in "I NEED TO BE PERFECT AT EVERYTHING TO GET INTO MEDICINE AT UNSW SO I NEED TUTORING") a year 10 for maths. Two questions: 1- What can I expect? Both in regards to...

Re: 2012 HSC MX1 Marathon <a href="http://www.codecogs.com/eqnedit.php?latex=\\ f(x)=\sin x -x@plus; \frac{x^3}{6}\\ f'(x)=\cos x -1@plus;\frac{x^2}{2}\\ f''(x)=-\sin x @plus;x\\ f'''(x)=-\cos x @plus;1\\ ~\\ f(0)=\sin 0 -0@plus; \frac{0^3}{6}=0\\ f'(0)=\cos 0 -1@plus;\frac{0^2}{2}=0\\...

Re: 2012 HSC MX1 Marathon I think you've written that wrong.

Thanks! If you don't mind me asking, what was your final mark?

THIS. My teacher refused to explain where the formulae come from (for some reason he hates calculus even though he's got a physics Phd) and so I struggled to remember them, although applying them was relatively easy. The other evening I was lying in bed doing maths in my head while trying to get...

You wouldn't get a question like that since the syllabus only uses maths up to a year 10 level (which I think is ridiculous). As the above post said, you'd have to take the second derivative of the graph which you don't learn to do until year 12 2 unit maths. However, you should be able to...

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \textrm{Let}~z=x@plus;iy\\ \frac{z-i}{z@plus;1}\\ =\frac{x@plus;i(y-1)}{(x@plus;1)@plus;iy}\\ =\frac{[\left x@plus;i(y-1)\right ][\left(x@plus;1)-iy ]\right}{[\left(x@plus;1)@plus;iy]\right [\left(x@plus;1)-iy ]\right}\\ ~\\ \textrm{If a...

<a href="http://www.codecogs.com/eqnedit.php?latex=\int^{\frac{\pi}{4}}_0 \tan xdx=\int^{\frac{\pi}{4}}_0 \frac{\sin x}{\cos x}dx=-\int^{\frac{\pi}{4}}_0 \frac{-\sin x}{\cos x}dx \\ \textrm{Let}~u=\cos x\\ \textrm{When}~x=\frac{\pi}{4}~u=\frac{1}{\sqrt{2}}\\ \textrm{When}~x=0~u=1\\ ~\\...

Oh lord. This made my evening.

What do you define as "hard"? Even ignoring that, you can hardly compare the difficulty of subjects that are entirely different. What makes MX2 any harder than EX2 - they have absolutely nothing in common that you can compare between them.

Re: 2012 HSC MX1 Marathon Can you describe your approach? I'm curious :p

Re: 2012 HSC MX1 Marathon I've got it. Give me about an hour and a half to write it up all nice and neat (I've got an ortho appointment in about 10 minutes and my current working out is hardly legible). Thoroughly enjoyed that problem :) EDIT: My scanner has decided it doesn't want to work...

FFUUU. It's these kind of stupid mistakes that are going to make me end up with an E2 in maths.

Re: 2012 HSC MX1 Marathon But that theorem is easily proved without assuming the result of the question is true. I would've proved it in my answer but (as far as I'm aware) that result is assumed knowledge for the HSC Course. Not entirely sure what you mean by that. As far as I can see, the...

I'm pretty sure those are correct (Perms and combs is my weakness)

Re: 2012 HSC MX1 Marathon I can't see how... can you please elaborate? EDIT: I'm going to bed, I'll check back here when I get home tomorrow.

Re: 2012 HSC MX1 Marathon...