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Re: HSC 2014 4U Marathon - Advanced Level \\ $Find all polynomials$ \ f \ $such that$ \ \ f(x) + f(1-x) = 1 \ $for all$ \ x

Re: HSC 2014 4U Marathon \\ $i) Explain why for any complex$ \ z \ $then$ \ |Re(z)| \leq |z| \\ $ii) Hence show that$ \ \ |ax-by| \leq \sqrt{a^2+b^2}\sqrt{x^2+y^2}

Re: HSC 2014 4U Marathon - Advanced Level Correct, the original question had 0 \leq x \leq \frac{\pi}{2} , and I forgot to write that.

Re: HSC 2014 4U Marathon - Advanced Level \\ $Find$ \ f(x) = \lim_{n \to \infty} \frac{\sin^{n+2}x + \cos^{n+2}x}{\sin^nx + \cos^nx} Someone else post a question please

Re: MX2 Integration Marathon Yep that is correct. The method I had in mind was recognising that the argument inside the logarithm looks like a solution to a quadratic equation, namely z^2+z - x = 0 So it makes sense then to just make the substitution x = u^2+u dx = (2u+1) \ du...

Re: HSC 2014 4U Marathon Yep

Re: MX2 Integration Marathon I think you have made a mistake

Re: MX2 Integration Marathon \int_2^6 \ln \left(\frac{-1+\sqrt{1+4x}}{2} \right) \ dx

Re: HSC 2014 4U Marathon If you mean I_(n-1) + I_n = 1/(2n-1) then this is fine since if we just make n -> n+1 we get I_((n+1) - 1) + I_(n+1) = 1/(2(n+1)-1) I_n + I_(n+1) = 1/(2n+1)

Re: MX2 Integration Marathon \\ $For the positive real$ \ x \ $find the minimum value of$ \ \ f(x) = \int_x^{2x} t\ln t - t \ dt

Re: HSC 2014 4U Marathon Its a little brute force but this is what I came up with: http://imgur.com/5t9AOOB

oops Did you get the answer using my suggested method?

Re: HSC 2014 4U Marathon z^2 - z(x+y) + (x^2+y^2-xy - 8) = 0 \ \ \triangle \geq 0 \ \ $since$ \ x,y,z \ $real$ \\ \\ (x+y)^2 - 4(x^2+y^2-xy-8) \geq 0 \\ 32 - 3(x-y)^2 \geq 0 \Rightarrow \ \sqrt{\frac{32}{3}} \geq |x -y| Also if you're question if Q16 difficulty or beyond, please post it in...

Not up to matrices yet :/ > USYD has best maths department > Slower than UNSW > lecturer spends 30 min on numerical examples gg

\\ $Ok, so the parametric representation of a plane$ \ \mathbf{r} \ $is$ \ \ \mathbf{r} = \mathbf{r_0} + t \mathbf{v} + s \mathbf{u} \\ $The variables that need to be found are the 3 vectors on the right hand side, all one needs to do is to find 3 points on the plane, transform them to vector...

Just imagine those equations as: 0x + y + 6z = -1 and 0x + 0y + z = 2 and apply the same method

That is essentially IVT, since P(-n) and P(3) yield opposite signs for sufficiently large n (i.e. when n goes to inifinty), then there is some root in (-n, 3), similarly, P(3) and P(5) yield opposite signs, and P(5) and P(n), for sufficiently large n is yield opposite signs.

Try differentiating and looking at the stationary points, they should tell you something about the way the curve moves.

Similarly: g(x) = f(x + 1/n) - f(x) g(0) = f(1/n) - f(0) g(1/n) = f(2/n) - f(1/n) g(2/n) = f(3/n) - f(2/n) . . . g((n-1)/n) = f(1) - f(0) Add them side by side: (1): \ \ \ \sum_{k=0}^{n-1} g\left(\frac{k}{n} \right) = 0 \ \ \ $(telescoping and cancellation of$ \ f(1) \ $and$ \ f(0)...

But using the hint: g(x) = f\left( x + \frac{1}{2} \right) - f(x) \\g\left(\frac{1}{2} \right) + g(0) = f\left(1\right) - f \left(\frac{1}{2}\right) + f\left(\frac{1}{2} \right) - f(0) = 0 \\ $Hence$ \ g\left(\frac{1}{2} \right) = - g(0) \\ g \ $is continuous over$ \ [0, 1/2] \ $and since$...