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Re: 2012 HSC MX2 Marathon Good. That's what I got. Thanks Mr. Carrotsticks :)

Re: 2012 HSC MX2 Marathon Hey guys can you implicitly differentiate this: <a href="http://www.codecogs.com/eqnedit.php?latex=(\frac{y}{x})^{\frac{1}{2}}@plus;(\frac{x}{y})^{\frac{1}{2}}=9" target="_blank"><img...

Re: 2012 HSC MX2 Marathon I haven't seen many integration questions... This is from someone at school (you know who you are :)) <a href="http://www.codecogs.com/eqnedit.php?latex=\int sin(lnx)dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int sin(lnx)dx" title="\int...

What other method?

Re: 2012 HSC MX1 Marathon You have real roots when the disriminant is greater than or equal to zero... <a href="http://www.codecogs.com/eqnedit.php?latex=\Delta =m^2-4(4)(9) \\ \Delta \geq 0\\ \therefore m^2-144\geq 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\Delta...

Use double angles and factorise... sin2x-cosx=0%5C%5C%202sinxcosx-cosx=0%5C%5C%20cosx%282sinx-1%29=0%5C%5C%20cosx=0,%20sinx=%5Cfrac%7B1%7D%7B2%7D And then solve for the relevant domain. Good luck:) Edit: Beaten :(

Re: 2012 HSC MX1 Marathon <a href="http://www.codecogs.com/eqnedit.php?latex=ln(3x@plus;1)^2-ln(x@plus;1)=ln(7x@plus;4) \\ ln(\frac{(3x@plus;1)^2}{x@plus;1})=ln(7x@plus;4) \\ (3x@plus;1)^2=(7x@plus;4)(x@plus;1) \\9x^2@plus;6x@plus;1=7x^2@plus;11x@plus;4 \\2x^2-5x-3=0 \\(2x@plus;1)(x-3)=0 \\x=3...

Re: 2012 HSC MX1 Marathon This is a question that I just saw at school that you guys might appreciate (excuse the dodgy paint skills...)

Mate you are my hero :)

Sum of roots is -b/a. As the coefficient of x^2 is 0, the sum of roots is zero.

<a href="http://www.codecogs.com/eqnedit.php?latex=\alpha @plus;\beta @plus;\gamma =0 \\ \therefore \alpha @plus; \beta=-\gamma \\ \alpha @plus; \gamma=-\beta \\ \gamma@plus; \beta=-\alpha \\ (\alpha@plus;\beta)^2(\gamma@plus;\beta)^2(\alpha@plus;\gamma)^2=(-\gamma)^2(-\beta)^2(-\alpha)^2 \\...

Re: 2012 HSC MX1 Marathon Same. I just checked it on Wolfram Alpha and it said it was undefined.

Re: 2012 HSC MX1 Marathon Here are some new questions on maximisation and minimisation. Enjoy :)

Re: 2012 HSC MX1 Marathon The easiest way to do that is to recognise that the function is odd, therefore the definte integral has to equal zero (between symmetrical x values). The question that you posted Timske has a very messy answer (assuming I read it correctly as the format is difficult to...

Try these two websites: Differentiation: http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/trigderivdirectory/TrigDerivatives.html Integration: http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/trigintdirectory/TrigInt.html The layout on the first one isn't that good but maybe it'll help as...

Like AAEldar said, we need the rest of the question. Probably the easiest way to do it however (I'm guessing as I don't know the specific question), would be to manipulate the sums of roots inside each of the brackets. For example, if alpha+beta+gamma=9, then alpha+beta=9-gamma and then you can...

The mistake is in the first line. From 10:15->1:45 is only 3.5 hours not 4 hours. Edit: Beaten

Re: 2012 HSC MX2 Marathon From memory, the only way to do this involves a lot of trig manipulation so I don't think there's a faster way. I'm happy to be corrected though

The way to do this question is: After 1 hour, A will have travelled 30km and B will have travelled 40km. Extending this, after time T, A will have travelled 30T km and B will have travelled 40T km. Therefore the distance from the intersection will be 100-30T for A and 100-40T for B. To find...

Re: 2012 HSC MX1 Marathon Its the last extension question in the Cambridge 3U book so you can do it, its just way harder than anything you'll ever see in the HSC. I won't do the whole thing for you, but I'll try and give you a hint and see if you can do the rest... Because I can't draw a...