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Re: HSC 2013 4U Marathon 1)\ (\frac{1}{a}-\frac{1}{b})^{2} + (\frac{1}{a}-\frac{1}{c})^{2} + (\frac{1}{b}-\frac{1}{c})^{2} > 0 $(equality occurs iff a=b=c)$ \frac{2}{a^{2}} + \frac{2}{b^{2}} + \frac{2}{c^{2}} -2\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right ) \geq 0 \Rightarrow \left...

Re: HSC 2013 4U Marathon $Prove that$ \left ( \frac{1}{2}+cos\frac{\pi }{20} \right )\left ( \frac{1}{2}+cos\frac{3\pi }{20} \right )\left ( \frac{1}{2}+cos\frac{9\pi }{20} \right )\left ( \frac{1}{2}+cos\frac{27\pi }{20} \right )=\frac{1}{16}

Re: HSC 2013 4U Marathon $Prove that$ \prod_{k=1}^{n}\left ( 1+2cos\frac{2\pi .3^{k}}{3^{n}+1} \right )=1

I edited my question. I think it is a very good question, can be difficult though. Could you please help me with: http://community.boredofstudies.org/showthread.php?t=304024

:)

Re: MX2 Integration Marathon I did it using the trig substitution, but I'm just wondering if it can be done using the Abel transform (u=\sqrt{X}')

Re: HSC 2013 4U Marathon $Prove the inequality:$ {a^{2}}+b^2+c^2+\frac{2}{5}abc<50 $where:$ $a,b,c are the lengths of triangle's sides$ $and the perimeter of the triangle is 10.$

Re: MX2 Integration Marathon Is this correct?

Please help me do this question: I can't really see any pattern. I hate this type of questions, it's like 'guess what I thought when I made up he question'...

Re: HSC 2013 4U Marathon $Prove that$ 1) \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geqslant (a+b+c) 2) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}} +\frac{1}{\sqrt{ab}}

Re: MX2 Integration Marathon Do you think the shorter method is doable?

Re: MX2 Integration Marathon Do you guys think it's a good idea to make a "Harder 3U marathon"? I think I have done enough integration, need to do some harder 3U.

Re: MX2 Integration Marathon I changed the integral to: \int \frac{e^{x}}{(1+e^{2x})\sqrt{(1-e^{2x})}}dx The let u=e^x The integral becomes: \int \frac{du}{(1+u^{2})\sqrt{(1-u^{2})}} I can do this the long way, such as on wolfram alpha (let u=\sin\theta) but I think we can use a short cut to...

Re: MX2 Integration Marathon True.

Re: HSC 2013 4U Marathon $For$ n\geq m \sum_{k=0}^{m}\frac{m!(n-k)!}{n!(m-k)!}=\frac{m!(n-m)!}{n!}\sum_{k=0}^{m}\frac{(n-k!)}{(n-m)!(m-k)!} =\frac{1}{\binom{n}{m}}\sum_{k=0}^{m}\binom{n-k}{n-m} \sum_{k=0}^m\frac{\binom {{1+}m}{k}}{\binom {{1+}n}{k}} &=\frac1{\binom...

Re: MX2 Integration Marathon It can be done using elementary functions but it is a hard one... I will try the one with the new limits. ========================= \int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \ dx Evaluate the above integral WITHOUT using the substitutions x=tan\theta or x=\frac{1-u}{1+u}

Re: MX2 Integration Marathon Awesome question! $Now I need to evaluate$ \frac{1}{2}\int_{0}^{\frac{\pi }{2}}ln(\sin u )du

Re: MX2 Integration Marathon Lol that was an easy one but I though it can be done by IBP be making u=x/2 etc. which is a bit harder

Re: MX2 Integration Marathon \int cot^{-1}(x^{2}+x+1)dx

Re: MX2 Integration Marathon From now on I will try to post only questions that are done by very tricky substitutions, IBP or very tedious. \int \frac{x^{2}}{(1+x^{2})^{2}}dx