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re: HSC Chemistry Marathon Archive If the test is ever easy, it's compensated for by crazy harsh marking haha.

Re: HSC 2015 4U Marathon Saves time in the HSC exam having to prove the AM-GM inequality by using the parabola way haha. (Both ways are valid though of course.)

Re: HSC 2015 4U Marathon $Also, provided all steps were biconditional in order to get from the required inequality to $ab\leq 4$, logically speaking, it is valid to simply prove $ab\leq 4$ and this proves the requirement, no need to retrace the steps. However, I'm not sure the HSC markers...

Re: HSC 2015 4U Marathon $One can prove that $ab\leq 4$ by noting that $b=1-a$, so $ab=a(1-a)$, and this is a concave down quadratic which is maximised at its vertex at $a=\frac{1}{2}$, where the value is $\frac{1}{4}$.$

$We know $g=\frac{GM}{R^2}$ by equating weight force $mg$ on the planet's surface with gravitational force on the planet's surface $F_G = \frac{GmM}{R^2}$ (Newton's Law of Gravitation). So $R = \sqrt{\frac{GM}{g}}$ for a planet of mass $M$. Since we know the Earth's mass from the formula sheet...

Re: HSC 2015 4U Marathon $It is assumed from the summation index that $k\geq 1$, so you won't lose marks if you don't state it.$

Re: HSC 2015 4U Marathon With these inequalities, if you can't see how to do it, it may help to start from the statement you want to prove and use biconditional steps (basically if and only if steps) to simplify it: $$\sum_{k=1}^n \frac{1}{k} > \ln (1+n) \Longleftrightarrow \exp \left(...

Re: HSC 2015 4U Marathon $Note that$ $$\ln (n+1) = \ln\left(\frac{2}{1}\times\frac{3}{2}\times\frac{4}{3}\times \cdots \times \frac{n}{n-1}\times \frac{n+1}{n}\right)$$ $$=\ln \left( \prod _{k=1}^n \frac{k+1}{k}\right)$$ $$= \ln \left(\prod _{k=1}^n...

Or it could be delayed.

Re: Announcement from BOSTES - significant change to calculus courses So has simple harmonic motion been transferred from MX1 to MX2? I can see it in the latter but not in the former on the post on the previous page (though it could be 'hidden' amongst the MX1 topics).

Re: HSC 2015 4U Marathon Don't worry, it's not.

Re: HSC 2015 4U Marathon $Well this was my attempt for the general case.$ $(ii) Note that the clock has $np$ minute marks, which we can label $0,1,2,...,np-1$ going clockwise from the top. Also, we should have $p\geq 2$; if $p=1$, the clock doesn't really make sense lol. Also note that it is...

Yeah that's correct.

$Some common ones are $\frac{\mathrm{d}x}{x}=\mathrm{d}(\ln x)$, $\cos x \text{ d}x = \text{d}(\sin x)$, etc..$

The good old reverse chain rule (aka inspection).

I'm guessing they just did it by inspection (without a substitution)?

Ozone is actually a resonance structure, which basically means it's changing between a few states (the ones on the ChemWiki page). So you'd actually be entirely wrong if you said it was a simple coordinate covalent bond. But then again, there are quite a lot of things in HSC Science syllabi that...

$You can do it using the substitution $u=\sin \theta$.$

Re: HSC 2015 2U Marathon These two methods are actually exactly the same. Integrating that line over the right bounds just gives us that triangle's area. If you're getting a wrong answer doing the latter, you might have made a silly mistake somewhere.

Re: HSC 2015 2U Marathon $Just note that the intersection occurs at $x=1$. Draw a vertical line at $x=1$, so this splits the area into two by symmetry, since the parabolas are just horizontal shifts of each other (because $4x-x^2 \equiv 4-(x-2)^2 $, so by our rules for shifting graphs, we know...