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Re: HSC 2013 4U Marathon I'm not sure why its untrue either. |z|=1 z^n+z^{-n}=\cos n\theta + \sin n\theta + \cos (-n\theta) + \sin (-n\theta) = 2\cos n\theta

Re: HSC 2013 4U Marathon Bolded what was wrong, when you add the two equations, you get: \alpha^2+\alpha^{-2}+A(\alpha+1/\alpha)+2 = 0 when you use that formula, you get: 2\cos 2\theta+ A2\cos \theta +2 = 0 \\ \\ \cos 2 \theta + A \cos \theta + 1 =0 Intuitive method though

Re: HSC 2013 4U Marathon Great question there, here is my attempt: iii) Two letters k, and j, can only be arranged in a set of n letters, either that k is BEFORE j, or k is AFTER j. The number of arrangements of k after j and k before j are equal. Hence the number of arrangements are...

Re: HSC 2013 4U Marathon Sure go ahead.

Re: HSC 2013 4U Marathon $Consider the Triangular Numbers, defined by the recurrence sequence $ \\ \\ T_n=n+T_{n-1} \ \ \ \ T_1=1 $i) By considering the identity $ \\ \\ j^2(j+1)^2-j^2(j-1)^2 = 4j^3 \\ \\ $Show that$ \ \ \ \ \sum_{k=1}^n k^3=\frac{n^2}{4}(n+1)^2 $ii) Hence show that...

Re: HSC 2013 4U Marathon Nice work, its easier than my method as well, I let the roots be -1, alpha, 1/alpha then used sum of roots, found the quadratic in alpha and discriminant.

Re: HSC 2013 4U Marathon Setting A=2 B=7 Gives an equation with 3 real roots according to Geogebra. What are the steps of your working?

Re: University. Thanks for the advice :) I really hope I achieve my dream now

Re: What absurd things have people done in your school due to competition and rivalry hahahahahahhahahahahahhahahahaha

Re: HSC 2013 4U Marathon $Consider the polynomial$ \\ \\ S(x)=Ax^3+Bx^2+Bx+A $Show that if all roots are real, then $ \\ \\ \frac{B}{A} \geq 3 \ \ \ \ \frac{B}{A} \leq -1 (keep in mind, that the astroid question is still left unanswered, it can be answered with only knowledge of implicit...

Re: University. Ahhh man :/ I'll still work at it though

Re: University. Policing at UWS Atar aim is a bit high but I believe in myself.

Ah yes, well then it would be the same effect as Case 1 part 1 but in reverse. If Jeremy cuts it into an infinitely small piece, Marie is forced to choose first. I will see if I can figure out an exact optimum ratio.

Are we assuming Marie is a great logician? My attempt is: Case 1: Jeremy should cut the cake into two equal pieces at first, then the next bit depends on whether Marie chooses first: - If she chooses first, cut the second cake into 1 infinitely small piece so Jeremy keeps the large full...

x^2+4x+4=0 (x+2)^2=0 x=-2, -2 There is something called multiplicity, that is if a root is a solution TWO times it has a multiplicity of TWO, if a solution 'appears' 3 times, i.e. (x-\alpha)^3=0 There is a 'triple root' at x= alpha, it has a multiplicity of THREE. In 2U you don't need to...

You will never need to use it, TECHNICALLY you can use 4U knowledge but it is heavily disliked (you may lose a mark) and there is most likely a much easier way to do it using normal ways.

+1 Except I would say Physics top 6 or so

I'm not saying that I only learn stuff for the pleasure of learning (I do sometimes). But sometimes I learn, yes for the examination, but I am interested in the work I study, I do want to know much more about it and I do extra research a lot of the time but in the end, even if I fully understand...

So we need people to fair to people who skip the most important part of education, and on occasion give them higher marks than those who are thinking? (Yes, rote learners get higher sometimes due to the extreme strictness of the sciences) If we don't initially have natural talent, we can...

That is just defeating the purpose of education, we go to school to learn, to evolve our minds. There are some advantages in terms of natural talent, but really critical thinking is part of being human, and most people are quite good at it, we just need to learn to develop it and not throw this...