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wow I made such a noob error what have I become

Just noticed options 5 and 6 are the same on account of a typo Ah well people know what it means Also who was the one who was neither 2014/2013 yet thought 2014 was harder? (Carrot?)

Bad luck man :( Just don't let it get you down for other subjects! I think 88-90 would scale to 97, but I am no expert

For some reason in my 3000 posts here I've never made a poll thread Thanks!

Alternate solution for Q16 (c) \int \frac{\ln x}{(1+\ln x)^2} \ dx = \int \frac{x\ln x}{x(1+\ln x)^2} dx \\ u = x\ln x \ , \ du = (\ln x + 1) dx \\ dv = \frac{1}{x(1+\ln x)^2} dx \ , \ v = -\frac{1}{1+\ln x} I might try to look for alternate solutions than Carrot's and post them here

Pick whether you found the 2014 paper harder or easier (separated into categories so we can observe bias) EDIT: Why won't it let me make a poll? I selected the option when making the thread

Need to see the paper Please someone send me a link as soon as possible

Good luck everyone!

Just relax, think clearly, have common sense, do everything properly, if you are looking for band 6 only, prioritize questions that will give you a higher mark/time ratio, otherwise, just gun everything Good luck!

Prediction: Q16: Mixture of (Integration + Inequalities + Series), for a 2010 esque thing Q15: Polynomials + Inequalities, Probability . . .

I honestly thought that this year's exam was easier than last year's, did anyone else feel the same? Perhaps I'm looking at it through a lens 1 year later, but apart from the Circle Geometry (that I failed miserably), I thought the test was quite doable, no?

Hey, I have made some room for PMs, so if you want to send a private message, go ahead As for your question, I didn't really use a textbook for 4U, the only textbook I had was the Fitzpatrick one but that's because the school told me to buy it. I never used it after the middle of Complex...

Re: HSC 2014 4U Marathon \\ $Given$ \ a,b,c \ $are real$ \\ $Show that at least one of the following equations has at least one real root$ \\ \\ x^2 + (a-b)x + (b-c) = 0 \\ x^2 + (b-c)x + (c-a) = 0 \\ x^2 + (c-a)x + (a-b) = 0

Re: HSC 2014 4U Marathon - Advanced Level Nice My alternate solution: \\ LHS = \sqrt[3]{a^3(a+2b)} + \sqrt[3]{b^3(b+2c)} + \sqrt[3]{c^3(c+2a)} \leq \frac{1}{3}(a(a+2b) + 2a) + \frac{1}{3}(b(b+2c) + 2b) + \frac{1}{3}(c(c+2a) + 2c) \\ \\ = \frac{1}{3}(2(a+b+c) + (a+b+c)^2) = 1 \ \square

Re: HSC 2014 4U Marathon - Advanced Level \\ $Given$ \ a+b+c = 1 \ \\ \\ $Prove that$ \ a\sqrt[3]{1+b-c} + b\sqrt[3]{1+c-a} + c\sqrt[3]{1+a-b} \leq 1

With 7th I got 99 in HSC but I don't know my actual raw mark, if its worth anything, after the exam I felt like I lost no marks, so I may have made a couple mistakes (i.e. 2-3 marks lost perhaps)

Anything goes, provided it is mathematically correct and proven (and within reasonable confines of the syllabus) So yes they will accept that method

Re: HSC 2014 4U Marathon - Advanced Level (Assuming a,b,c are non-negative) \\ $Make the substitution$ \ x = a/3, y = b/3, z = c/3 \\ $So$ \ x+y+z = 1 \\ $And the problem is transformed into proving that$ \\ \\ xy+ xz + yz \leq \frac{1}{3} \ \ $(after using expansion of$ \ (x+y+z)^2 \ $)$ \\...

Good luck guys

¯\_(ツ)_/¯