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correct solution $ area $=\int_1^3(x^2-x^{-2})dx=\left[\frac{x^3}{3}+\frac{1}{x}\right]_1^3=[(9+\frac{1}{3})-(\frac{1}{3}+1)]=8

Re: HSC 2015 4U Marathon Here is my solution: z^n-1=(z-1)(z-z_1)(z-z_2)\codts(z-z_{n-1}), $ differentiating both sides $ nz^{n-1}=(z-1)^{\prime}(z-z_1)(z-z_2)\cdots(z-z_{n-1})+(z-1)[(z-z_1)(z-z_2)\cdots(z-z_{n-1})]^\prime, $ substituting $ z=1, $ the desired result follows. $

Re: HSC 2015 4U Marathon \cos3\theta=4\cos^3-3\cos\theta, $ let $ x=\cos\theta, $ then $ 144x^3-108x=36\cos3\theta. $ so $ 36\cos3\theta=17, \cos3\theta=\frac{17}{36}, 3\theta=\cos^{-1}\frac{17}{36}, \theta=\frac{1}{3}\cos^{-1}\frac{17}{36}. $ therefore $...

Re: HSC 2015 4U Marathon yes only good for even number n and k is half of n. the idea is absolutely neat. well done.

Re: HSC 2015 4U Marathon this makes more sense, your question is actually asking the product of real parts of all nth roots of 1

Re: HSC 2015 4U Marathon is k any integer>=1, like can it be greater than n ?

Re: HSC 2015 4U Marathon There's still a serious problem in the question: say, if n=6, \alpha=cis\frac{2\pi}{3}, then the LHS will be zero. So I modify this question into $ If $ z_1, z_2, \cdots, z_{n-1} $ are distinct $ n $-th root of $ 1 $ other than $ 1, $ show that $...

Re: HSC 2015 4U Marathon the third last line is wrong. refer to my previous post, the RHS of the question should be n.

Re: HSC 2015 4U Marathon you do not really expand term by term, you can expand by inspection, say, you know that the term x^4 comes from products of x^3 and x, x^2 and x^2, and x and x^3. meanwhile, for p(x)p(-x), if you group even power terms and odd power terms, you will have sum times...

Re: HSC 2015 4U Marathon wrong question, RHS should be n.

Re: HSC 2015 4U Marathon good one carrotsticks. sy, i got same as yours.

Re: HSC 2015 4U Marathon there is a sec^2 coming out from dx, so root of 1+tan^4 will become sin^4+cos^4, and complete square and double angle I have already tried x=1/t, and the idea same as yours, but it seems not work, maybe i ignored something

Re: HSC 2015 4U Marathon good question. bump

Re: HSC 2015 4U Marathon $ substitution $ x=\tan\frac{\theta}{2}, $ then $ dx=\frac{1}{2}\sec^2\frac{\theta}{2}d\theta, \frac{x^2-1}{x^2+1}=-\cos\theta $ (which follows from the t-formula) $ $ So the integral...

you never drew a graph? remember, the most important thing of all when comuputing an area is to sketch the graph. From the graph, you would know which part is above or below the x-axis.

You can always split up. But for this one you do not need to split up because the whole region is above x-axis.

Re: HSC 2015 4U Marathon what is MATH1131, in which is Roll's theorem going to be covered?

Re: MX2 2015 Integration Marathon well, i would say x is automatically treated as nonnegative. for negative and even imaginary x, that involves multi-valued function, which I don't think talking about convergence or divergence is appropriate. like what is \sqrt{i+\sqrt{i}}? there are four...

Re: HSC 2015 4U Marathon lol, isn't z=0 a root, which is real?

Re: MX2 2015 Integration Marathon yes, I already figured it out, when x=0, f(x)=0, when x>0, f(x) is what you derived, and for x<0, f(x) is not defined. besides, for any positive x, we can use monotonic convergence to prove f(x) is well defined. Interesting question and good dicussion :D