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Re: HSC 2014 4U Marathon Yep Alternatively just use \frac{a+b+c+d}{4} \geq \sqrt[4]{abcd} and just a bit more. Here is a good sum: \\ $Find$ \ \ \ \ \sum_{k=m}^n \binom{n}{k} \binom{n}{k-m}

Re: HSC 2014 4U Marathon \\ $The area of a triangle with sides$ \ a,b,c \ $is given by the formula$ \ A = \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4} Why is latex not working?

Re: HSC 2014 4U Marathon Yea something like that \\ $The area of a triangle with sides$ \ a,b,c \ $is given by the formula$ \ A = \sqrt{s(s-a)(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4} Something wrong with latex?

It really isn't High school maths is quite easy once you learn to like Maths in general

Ah alright this makes sense, thank you.

These are the marks for 2U and MX1 in 2012 for someone I know Why does one round up to 99 and the other to 98 if both (according to most people) apparently do the operation (Asses + Exam)/2 = HSC

Re: HSC 2014 4U Marathon Sketch the graph of: y = \frac{1}{\sqrt{2nx - x^2}} \ \ $for integer$ \ n Its easy enough to graph, then consider the upper and lower rectangles at, x=1,2,3, \dots , n Considering the lower rectangles first \sum_{k=2}^n \frac{1}{\sqrt{2nk - k^2}} < \int_1^n...

Re: HSC 2014 4U Marathon That is the correct answer, but I was able to do it without resorting to Riemann Sums (indeed a rather crude version of it, try lower and upper rectangles of fixed length)

Re: HSC 2014 4U Marathon I think I've found the problem, when we consider the final result, z = \frac{1}{3} (u+v) Substituting this into the first 2 equations, we get: k= l = \frac{3}{2} Meaning the division or multiplication by (k-l) cannot be done, since its zero.

Re: HSC 2014 4U Marathon Take the 2 equations from part (i), and re-arrange both so that \frac{1}{2} u is the subject, equate the equations, to yield an equation only in z,v which re-arranged gives you the required result, I did it and the algebra matches with the Show that

Re: HSC 2014 4U Marathon My solution: $Let$ \ x = \frac{1}{a^2-1} \ \Rightarrow a = \sqrt{\frac{x+1}{x}} , \ b = \sqrt{\frac{y+1}{y}} , \ c = \sqrt{\frac{z+1}{z}} \therefore \ x+y+z = 1 \\ \\ $We must prove that$ \ \ \ \frac{\sqrt{x}}{\sqrt{x} + \sqrt{x+1}} + \frac{\sqrt{y}}{\sqrt{y} +...

Re: HSC 2014 4U Marathon A similar method can be used with (1+Ai)(1+Bi) to generate the more general version; \tan^{-1} A + \tan^{-1}B = \tan^{-1} \frac{A+B}{1-AB}

Re: HSC 2014 4U Marathon $Recall$ \ \ \cos(2\theta) = 2\cos^2 \theta - 1 \\ \\ \Rightarrow \ \cos \frac{\pi}{12} = 2\cos^2\frac{\pi}{24} - 1 \\ \\ \Rightarrow \ \cos \frac{\pi}{24} = \frac{1}{2} \sqrt{1+\frac{1+\sqrt{3}}{2\sqrt{2}}} Which after a bit of squaring and manipulation, should...

Re: HSC 2014 4U Marathon First step \sin \frac{7\pi}{12} = \sin \left(\frac{\pi}{2} + \frac{\pi}{12} \right) = \cos \left(-\frac{\pi}{12} \right) = \cos \frac{\pi}{12} Next step is to get from cos(pi/12) to cos(pi/24)

Re: HSC 2014 4U Marathon Make sure to say so if this one is too hard/open \\ $Using complex numbers find the sum$ \\ \\ \cos(\theta) + \cos(\theta + \alpha) + \cos(\theta + 2\alpha) + \dots + \cos(\theta + (n-1)\alpha)

Re: HSC 2014 4U Marathon Therein lies the problem with relying on arctan, it forces the solution to be within the domain, -\frac{\pi}{2} < x < \frac{\pi}{2} (this will become clearer when you learn inverse functions properly). This is what is supposed to happen: \\ (1+i\sqrt{3})(1+i) =...

Re: HSC 2014 4U Marathon What did you put in mod-arg form? How did you know the angle was -5pi/12? It is not an exact ratio like pi/3 or pi/6 that people remember

Re: HSC 2014 4U Marathon Ah yes my bad, this is what I did as well, because starting with cos(7pi/12) leads to sin(pi/12)

Re: HSC 2014 4U Marathon Yes the version in the form (a+ib) is correct, how did you get the De Moivere's theorem version from that though? As far as how to get cos(pi/24) is concerned, you need to find how to get from cos(7pi/12) to cos(pi/24)

Re: HSC 2014 4U Marathon Nice solution, here is mine \\ $By a little manipulation we simply need to prove:$ \frac{1}{n+1} \geq \sqrt[n]{\prod_1^n \frac{a_k}{a_k+1}} \\ $By AM-GM$ \\ \\ \sqrt[n]{\prod_1^n \frac{a_k}{1+a_k}} \leq \frac{1}{n} \sum_1^n \frac{a_k}{1+a_k} \\ \\ \Rightarrow \...