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Re: HSC 2014 4U Marathon Lol for part iv) i think you meant the points (1,1) and (e,0)

Re: HSC 2014 2U Marathon my bad :P

Re: HSC 2014 2U Marathon There's a similar question like that in the one of the past 2U HSC paper (don't remember which one haha)

Re: HSC 2014 2U Marathon It's cos i ceebs :P but ok it's \pi(e^4 - \frac{1}{4}e^4+1/4) = \pi(\frac{3}{4} +1/4)

Re: HSC 2014 2U Marathon \pi (e^4 - \int_{0}^{1}e^{4x} \ \ dx)

yeah pretty much before the exam you should practice the easy questions slowly, so that you're more aware about the kind of mistakes you usually make. During the test/exam, only think about the question you're doing and not the whole paper as a whole i.e don't be like "I fricken need to score...

wow time flies

Re: HSC 2014 2U Marathon Hint P = 5e^{\frac{t}{10}}

I don't know if there is a flaw in this (a^2-b^2)\geq 0 \Rightarrow a^4+b^4 \geq 2a^2b^2 \\ \\ $Similarly$ \ \ c^4+d^4 \geq2c^2d^2 \\ \\ \therefore a^4+b^4+c^4+d^4\geq 2a^2b^2+2c^2d^2 \\ \\ $now consider$ \ \ (x+y)^2\geq 0 \Rightarrow x^2 +y^2 \geq 2xy \\ $Let x = ab and y = cd \\ \\...

$From the first equation, take nth root of both sides $\\ \\ n!^{\frac{1}{n}} < n+1 \\ \\ n!*n!^{\frac{1}{n}} < n!*(n+1)= (n+1)! \\ \\ (n!)^{\frac{n+1}{n}}< (n+1)! \\ \\ $then take the (n+1)th root from both sides $

PM me pls lol

RAWR

Post edited!..... that's just how the question is given.

why?

Ok I maybe I'm derping and stuff but i can't seem to get this one $Suppose that function f is a differentiable function of a single variable and F(x,y) is defined by$ \ \ F(x,y) = \ f(x^2-y) \\ $a) Show that$ \frac{\partial F}{\partial x} + 2x\frac{\partial F}{\partial y} = 0 \\ \\ b) \ \...

cub3root derailing again

lol the concept of telescoping sum isn't explicitly taught in high school, but in uni.

81-->95 aligned

heh the 4U 2013 paper was probably one of the easiest CSSA papers that it received some criticism (not really).

Great memory indeed, lol aren't you in uni?, how did you get to see the paper?