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I haven't come across a question in 4u which required differentiating / integrating complex numbers, but you most certainly can. The 'i' is actually treated as a constant value.

Re: 2014 BOS Trial Exam Results Thread It'd be interesting to see up to which rank on this exam would result in a state rank for MX2. I'm guessing up to rank 3 will state rank though.

I may have worded the bolded weirdly but what I meant was that when you explicitly solve for the quadratic you get two complex roots. Yes I do understand that this only applies for quadratics but nonetheless still a valid proof? Also when I did the question persnally I proved it using conjugates...

Started complex today, I think this works an alternative though ax^2 + bx+ c. Since one of the roots are complex, the other must complex also. Then the complex root alpha = (-b +- i(b^2 - 4ac)^(1/2))/2 Obviously then the other root has the sign switched in front of the 'i', which is its...

Re: HSC 2014 4U Marathon - Advanced Level whoops i just realised the previous method didn't quite work. Thanks Realise. But I just thought of another method. Change the LHS into (3-c)^2 + (3-a) ^2 + (3-b)^2 The goal is to prove that this is greater than 12 After expanding and cancelling...

Re: HSC 2014 4U Marathon - Advanced Level A fair criticism. However if you look at Sys previous post he didnt include a proof for am gm. So I assumed on this forum I was could assume the really common theorems. Of course in an exam i wouldve most definity included proofs. But i didnt think...

Re: HSC 2014 4U Marathon - Advanced Level Just an alternative, from (a+b+c)(1/a + 1/b+ 1/c) >=9, 1/a + 1/b + 1/c >=3 From AM GM, a+b+c/3 = 1> (abc)^1/3. Thus abc < 1 transform LHS into (a+b+c)^2 - ab -bc -ca. which equals 9 - abc(1/a + 1/b + 1/c) >= 9-3 =6

Um who was the imo guy?

Re: HSC 2014 4U Marathon - Advanced Level Given that a + b + c= 3 Prove that a^2 + b^2 + c^2 + ab+ bc+ca>=6

you can because CB = 2 x GC (since OG is perpendicular to CB) and similarly AD = 2 x AF same reason Therefore CB / AD = (2x GC) / (2x AF) = GC/AF this is becasue the 2s cancel out. Do you understand? Or do you mean that I should've posted this working?