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2. $Let $ P(x) = x^3 + px + r P'(x) = 3x^2 + p $By multiple root theorem, if it has a root of multiplicity two, $ P(\alpha) = P'(\alpha) = 0 $ where $\alpha $is the double root$ P'(\alpha) = 3\alpha^2 + p = 0 \alpha = \pm\sqrt{\frac{-p}{3}} P(\alpha) = \alpha^3 + p\alpha + r = 0...

Oh wow. Similar to the one that Carrotsticks posted earlier today haha. |z - i| starts from 1 on the imaginary axis. |z - 1| starts from 1 on the real axis. Perpendicular bisector of the chord joining those two points is y = x For |z - i| <= |z - 1|, has to be the region below and including...

I know, but multiple choice questions can sometimes give two or more correct answers. It's about selecting the more correct answer.

Re: 2012 HSC MX2 Marathon I thought that they rolled the dice one at a time. For X to win on the second go, he has to miss the first time, Y then has to miss, and then he gets 6 on his next roll. Apply the same thing for the next go.

Maybe stating that the nuclei mass is too large is a bit too general where as saying it's radioactive due to the proton-neutron ratio is more specific.

mm yeah sorry about that, forgot there was another negative.

Re: 2012 HSC MX2 Marathon lol.. I hope whatever locus you were looking at in your tutorial was more complex than the one you gave.

Second line. Supposed to be -8/3.

Re: 2012 HSC MX2 Marathon P(winning on first go) = 1/6 P(winning on second go) = 5/6 x 5/6 x 1/6 P(winning on third go) = 5/6 x 5/6 x 5/6 x 5/6 x 1/6 By continuing the trend and adding all the possible cases.. P(X winning) = 1/6 + 1/6 x (5/6)^2 + 1/6 x (5/6)^4 + ... Common ratio of...

Re: 2012 HSC MX2 Marathon |z - i| starts from 1 on the imaginary axis. |z - 1| starts from 1 on the real axis. Perpendicular bisector of the chord joining those two points is y = x For |z - i| >= |z - 1|, has to be the region below and including y = x, therefore y \leq x

Re: 2012 HSC MX2 Marathon The region of y \leq x

You dont know what the constants are on each side, you can't assume theyre both c. To prevent this problem, just add c to one side which can account for the constants on both side.

After you integrate, leave the c there. Sub x = 0 to obtain c and you get your answer

Re: HSC 2012 Chemistry Marathon In (a), explain what you mean by 'efficient'. The rest is fine though. zeebobDD, i doubt many people have started chemical monitoring and management lol.

Re: 2012 HSC MX2 Marathon $Let z = rcis\theta, w = Rcis\phi$ |Re(Rrcis(\theta + \phi)| = Rrcos(\theta + \phi) Rrcos(\theta + \phi) = Rr $ when \theta + \phi = $ \frac{\pi}{2} cos(\theta + \phi) = 1 $ which is its maximum only when \theta + \phi = $ \pm \frac{n\pi}{2}, n \in \mathbb{Z}...

Pick it up now before you don't have the chance to. Even if you struggle, you could always drop it.

I don't think you can.. y = ln(sinx) gives a graph thats <=0 so you cannot graph y = ln(ln(sinx)) as there won't be any defined points.

Re: 2012 HSC MX1 Marathon I don't need this, i'm posting it up for people to do for practise.. lol. Scroll down a bit further after that question to find Carrotsticks addition to the question.

Re: HSC 2012 Chemistry Marathon Sweet, thanks for the feedback. New question:

Re: HSC 2012 Chemistry Marathon Content sounds pretty good, expression could be better (but this isn't english so i'm not going to criticise that lol). Lavoisier defined acids as containing oxygen as he found non-metal oxides which reacted to form acids in water, e.g. sulfuric acid. Then...