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i liked module c. module a was meh, module b was the killer. :\... ugh.. had to be deane...

yes please!!! thank you for your time!

Re: HSC 2013 3U Marathon Thread god this Q <.<. im going to call the magnetic flux symbol "A" and the Telsa "B" [(x-A)(x-B)]^2n = [(x(x-1)-1)]^2n RHS = (x-A)^2n . (x-B)^2n expanding both brackets of (x-A)^2n . (-B+x)^2n (2nC2n-2.(-A)^2n-2.(x^2) + 2nC2n-1.(-A)^2n-1.(x) + 2nC2n(-A)^2n) ...

Re: HSC 2014 2U Marathon Hmm, how do you find the volume in the y=10 thing? i had 40pi i don't really know how to get 400pi edit: nvm i think i confused my r and h :X

Re: HSC 2013 3U Marathon Thread i don't want to use latex to type up everything again ): took way too long last time. i) (1+x) +x(1+x)+x(1+x)^2...+x(1+x)^n-1 <-- if we ignore the first term, second to last is a GP series, so a = x(1+x) r = (1+x). sub in formula, cancel you would end up with...

Re: HSC 2014 2U Marathon Ah sorry! is it 88pi/15?

Re: HSC 2014 2U Marathon Ah that explains everything then :P also do you mean y=-10?

Re: HSC 2014 2U Marathon have already tried assuming k=1, ended up getting a result that was opposite in sign compared to II so ))))): sadday

Re: HSC 2014 2U Marathon I) it's a gp with r = -x ii) integrate x^2n+1/1+x = 1-1/2+1/3-1/4.....+1/2n+1 - ln2 iii) since n approaches infinity, 1 /2(n+1) approaches 0. And since I is less than 1/2(n+1), hence I < or equal to 0 Iv) I don't really know how to do this one, but I feel like it's...

Re: HSC 2014 2U Marathon Assuming it's 1 dx instead of two, it would eventually turn out like 1/x/lnx/ln (lnx) which integrated would be ln (ln (lnx))) + C. If it's two 2dx I'd have no clue

Re: HSC 2014 2U Marathon i think its just a 1 mark question, because 1/(xlnx) can be seen as (1/x)/(lnx), hence, ln(lnx)+c. Also the answer to this question http://postimg.org/image/id86iq0qb/ = ((pi^pi)-1)/ln(pi). (working, you just move the lnpi from the rhs to lhs and bingo) also guys...

Re: HSC 2014 2U Marathon answer is 0 (i think). its tedious i may have made a mistake T_T!!!!!

Re: HSC 2013 3U Marathon Thread 2

Re: HSC 2014 2U Marathon cbb typing all the working, but with expanding, seperating the numbers you get something like: 1+(3-2x)/(x-1)^2 --> 1-(2x-2)/(x-2)^2+1/(x-1)^2 --> 1-2(1/x-1)+(x-1)^-2 through integrating normally x-2ln(x-1)-(x-1)^-1 and then you sub in x = 10 and 2 for the integral...

Re: HSC 2013 3U Marathon Thread Consider \ \sum_{r=0}^n \binom{n}{r}(-1)^r(x)^r = \(1-x)^n Integrate both sides \(-1)\frac{(1-x)^{n+1}}{n+1}+c=\sum_{r=0}^n \binom{n}{r}(-1)^{r}(x)^{r+1}\frac{1}{r+1} \At x = 0, c =\frac{1}{n+1} Therefore: \...

Re: HSC 2014 2U Marathon i) For Linear functions, trapezoid rule ii) simpsons rule reasoning: trapezoid rule works in a way that it seperates the function into different shapes with a line at the end (not at the x axis). simpsons rule is similar, except at the end of the shape its curved...

Re: HSC 2014 2U Marathon for i) t=0 p=5 lions = 150-e^t dt/dp = 2/p (rearranged) t = 2ln(p)+c c = -2ln5 t=2ln(p/5) therefore p=5e^(t/2) 5e^(t/2) = 150-e^t set e^(t/2) as "a" 5a = 150-a^2 a^2 +5a -150 therefore a = 10, -15 e(t/2) = 10, -15 but we cannot use the negative 15 because we cannot...

Sorry i cannot add to the OP's discussion but im curious to the bolded and underlined statement. Does that mean that purely your RAW MARK determines your band within that subject, not HSC mark which takes into account your internal marks?

Bump. Subjects: Advanced and Extension 1 English, 2u and 3u maths, business, economics and physics.

Hi guys Can anyone here give me some 2u/3u past papers from various schools to practice on? it would be greatly appreciated