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Re: HSC 2014 4U Marathon - Advanced Level I tried it with induction, heres what i did: From part i, we know it is true for n=2. Let us assume true for n=k \sum_{j=1}^k \omega_j f(x_j) \geq f(\sum_{j=1}^k \omega_j x_j) Prove true for n=k+1 RHS= f(w_{k+1} x_{k+1}...

Re: HSC 2014 4U Marathon - Advanced Level A property of convex functions is that the line passing through the points (x_1,f(x_1)) and (x_2,f(x_2)) is greater than or equal to the function at every point within the closed interval [x_1,x_2] . \therefore y=f(x_1)+\dfrac{f(x_2)-f(x_1)}{x_2-x_1}...

Re: HSC 2014 4U Marathon Nice work :D

Re: HSC 2014 4U Marathon Prove that A_n=8+88+888+...+888...8=\dfrac{8}{81} (10^{n+1}-9n-10) Where A_n contains n terms, with the term T_k having k digits, each of them being 8.

Re: HSC 2014 4U Marathon - Advanced Level Okay so this is what i came up with, not fully confident but here it is: \dfrac{n(n-1)}{2}=0+1+2+...+(n-1) Using the fact that an odd+even=odd and odd+odd=even The sum gives o, then an odd number, then another odd, then an even twice and an odd...

Re: HSC 2014 4U Marathon - Advanced Level Prove that the average value of a function between the closed interval [a,b] is given by \dfrac{\int_a^b f(x)\cdot dx}{b-a}

Re: HSC 2014 4U Marathon - Advanced Level Nice question heres my attempt \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n + ik} = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{n-ik}{n^2 + k^2}= \lim_{n \to \infty} \sum_{k=1}^n \dfrac{n}{n^2 + k^2}- i\lim_{n \to \infty} \sum_{k=1}^n \dfrac{k}{n^2 +...

Re: MX2 Integration Marathon Nice question, works out very neatly :) \int\frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}dx Let t=\tan{x} \therefore \dfrac{dt}{1+t^2}=dx =\int\frac{(\dfrac{2t}{\sqrt{1+t^2}})^2 (\dfrac{1}{\sqrt{1+t^2}})2 }{((\dfrac{2t}{\sqrt{1+t^2}})^3...

Re: HSC 2014 4U Marathon - Advanced Level Thats pretty much what i did. For the converse, I guess an alternate way of doing it is assuming the result holds, then proving A, C and F are collinear and if it isn't, then the intersection does not occur. Other than that, in my reasoning I try to...

Re: HSC 2014 4U Marathon - Advanced Level Refer to attached diagram: Some results we can prove using similarity: \dfrac{AD}{DC}=\dfrac{FA}{FD} \dfrac{BC}{AB}=\dfrac{FC}{FB} FC \cdot FA = FD^2=FB^2 = FD \cdot FB \therefore \dfrac{AD \cdot BC}{DC \cdot AB}=\dfrac{FC \cdot FA}{FD...

Re: MX2 Integration Marathon Yeh sweet :)

Re: MX2 Integration Marathon Using the substitution u=\frac{ln(x)}{x} yields the integral: \int \dfrac{du}{u^4+1} =\dfrac{\sqrt{2}}{8}\int \dfrac{2u+2\sqrt{2}}{u^2+u\sqrt{2}+1}-\dfrac{2u-2\sqrt{2}}{u^2-u\sqrt{2}+1} \cdot du (Partial Fractions)...

Re: HSC 2014 4U Marathon - Advanced Level Nice, very clever :)

Re: MX2 Integration Marathon \dfrac{(x+1)^2 (x^2-1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}} Dividing by x^3 (Note this is assuming x>0) \dfrac{(\dfrac{x^2+2x+1}{x}) (1-\dfrac{1}{x^2})}{(x+\dfrac{1}{x})^2 \sqrt{x^2+1+\dfrac{1}{x^2}}} \therefore \int \dfrac{(x+1)^2 (x^2-1)}{(x^2+1)^2...

Re: MX2 Integration Marathon \tan (x) \cdot \tan (2x) \cdot \tan (3x) =\tan (x) \cdot \tan (2x) \cdot (\dfrac{\tan(x)+\tan(2x)}{1-\tan (x) \cdot \tan (2x) }) =-(1-\tan (x) \cdot \tan (2x) \cdot) (\dfrac{\tan(x)+\tan(2x)}{1-\tan (x) \cdot \tan (2x) }+\tan(3x)...

Re: HSC 2014 4U Marathon - Advanced Level ab(a+b)\geq 2ab\sqrt{ab} \therefore ab(a+b)+ac(a+c)+bc(b+c) \geq 2(ab\sqrt{ab}+ac\sqrt{ac}+bc\sqrt{bc}) \geq 6abc 4(ab(a+b)+ac(a+c)+bc(b+c)) \geq 3(ab(a+b)+ac(a+c)+bc(b+c)+2abc) \dfrac{a^2 (b+c)+b^2 (a+c)+c^2 (a+b)}{ab(a+b)+c^2...

Re: HSC 2014 4U Marathon Very nice, alternatively the first one could be done using sum of roots and the third one could be done by letting x=1: 1+1+1=8(1-\cos{\frac{2\pi}{9}})(1-\cos{\frac{4\pi}{9}})(1-\cos{\frac{8\pi}{9}})...

Re: HSC 2014 4U Marathon Yep very good:

Re: HSC 2014 4U Marathon Oh sorry about that

Re: HSC 2014 4U Marathon (i) Show that: x^6+x^3+1=(x^2-2x\cos{\frac{2\pi}{9}}+1)(x^2-2x\cos{\frac{4\pi}{9}}+1)(x^2-2x\cos{\frac{8\pi}{9}}+1) (ii) Hence evaluate: \cos{\frac{2\pi}{9}}+\cos{\frac{4\pi}{9}}+ \cos{\frac{8\pi}{9}} \cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}+...