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Re: HSC 2014 4U Marathon - Advanced Level yep that's right

Re: HSC 2014 4U Marathon - Advanced Level part ii)?

Re: HSC 2014 4U Marathon - Advanced Level not sure which forum to post this. Hope this is ok

Re: HSC 2014 4U Marathon - Advanced Level Its not a typo. And yeah, the answer is 1. I had to use exponents to do it, though happy to be shown an easier way

Re: HSC 2014 4U Marathon - Advanced Level \lim_{x\rightarrow \infty} \left( \tan \frac{\pi x}{2x+1}\right)^{\frac{1}{x}}\textrm{ Hint: uses the definition } e^x = \lim_{n\rightarrow \infty} \left(1+\frac{x}{n}\right)^n

Re: HSC 2014 4U Marathon - Advanced Level OK, I'm going to need a hint

Re: HSC 2014 4U Marathon - Advanced Level The USSR Olympiad Problem book by Shklarsky has a small but deeply insightful chapter on the algebra of polynomials. Chapter 8, i think. All questions have worked solutions too.

Re: HSC 2014 4U Marathon - Advanced Level \begin{align*}\frac{a^2}{a+b}-\frac{a}{2}&= \frac{2a^2-a(a+b)}{2(a+b)}\\&= \frac{a^2-ab}{2(a+b)}\\&\geq \frac{a^2-(\frac{a^2}{2}+\frac{b^2}{2})}{2(a+b)}\\&= \frac{a^2-b^2}{4(a+b)}\\&=\frac{a-b}{4}, a+b>0\\\therefore...

Re: MX2 Integration Marathon \int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx} (without using complex numbers)

I am not sure how helpful this is for you, but I think that being prepared to take a break from a question with the intention of coming back to it later is important. Especially with the latter questions e.g. circle geometry involving multiple circles or questions that are flooded with \sum 's...

I am sure that says something about the lack of elegance in my solution. Here it is without the tex \begin{align*} abc &= 1 \Rightarrow ((a+1)-1)((b+1)-1)((c+1)-1) &= 1\\ (a+1)(b+1)(c+1) - [(a+1)(b+1)+(a+1)(c+1)+(b+1)(c+1)] + [(a+1)+(b+1)+(c+1)] -1 &= 1\\ \textrm{Dividing through by...

\begin{align*} abc &= 1 \Rightarrow ((a+1)-1)((b+1)-1)((c+1)-1) &= 1\\ (a+1)(b+1)(c+1) - [(a+1)(b+1)+(a+1)(c+1)+(b+1)(c+1)] + [(a+1)+(b+1)+(c+1)] -1 &= 1\\ \textrm{Dividing through by }(a+1)(b+1)(c+1) \textrm{ and rearranging gives}\\ \left(\frac{1}{(b+1)(c+1)} -...

hey rolpsy, what program do u use for the graph? Is it free to download? Thanks

Pappus' Centroid Theorem as a check on volumes. Which is here: http://mathworld.wolfram.com/PappussCentroidTheorem.html. But I have a feeling that if you can apply correctly in exam conditions, you are probably good enough to get volumes right anyway.

Re: MX2 Integration Marathon \begin{align*}\textrm{Let} \;x &= a\tan{\theta}\\I:= \int_{0}^{\infty}\frac{\ln{x}}{x^2 + a^2} \textrm{d}x &=\int_{0}^{\frac{\pi}{2}}\frac{\ln(a\tan{\theta})}{a^2\tan^2{\theta}+a^2} a\sec^2{\theta} d\theta\\& = \int_{0}^{\frac{\pi}{2}}\frac{\ln(a\tan{\theta})}{a}...

Re: MX2 Integration Marathon \int_2^4\ \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}

Well, the trivial case of a = 2 and b = 3, definitely sits the function between -5 and 4. Should that 4 be -4?

Therse are really helpful comments that you add, Carrotsticks. Thanks

Re: MX2 Integration Marathon yep