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Re: HSC 2013 4U Marathon Nice work. ================================ $Evaluate$ \int_{1}^{\sqrt{3}} \sqrt{\frac{1-x}{3+x}} \ dx

Yeah well some people can think logically without having to devote so much time to rote learning UMAT questions.

Re: Rape! It's ok if you're a Muslim! He should be given a lifetime in jail/castration in my opinion. Less 'Muslims' like him, the better. The mere fact that he has been given an appeal is revolting and insulting.

Re: HSC 2013 4U Marathon $i) Prove that$ \tan^{-1} \left (\frac{x}{x+1} \right ) - \tan^{-1} \left (\frac{x-1}{x} \right ) = \tan^{-1} \left (\frac{1}{2x^2}\right) $ii) Hence prove that$ \frac{\pi}{4} = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{18} + \tan^{-1}...

re: HSC Physics Marathon Archive That looks correct to me, nice work.

First I will form the subjects then the structure of schooling. Mathematics: I disagree that it solely should be divided into Pure and Applied Mathematics. I think everyone should have proper access to a good Mathematics education which is why at least 2 units of Mathematics in my view...

Re: HSC 2013 4U Marathon Unfortunately that integral evaluates to \pi , not 2. What needs to be done is: \int_{3}^{\infty} \frac{dx}{\sqrt{x} (x+1)} > \sum_{k=4}^{\infty} \frac{1}{(k+1)\sqrt{k}} \frac{\pi}{3} > \sum_{k=4}^{\infty} \frac{1}{(k+1)\sqrt{k}} Now lets add to both sides the...

|a-b|^2 = (a-b) \overline{(a-b)} = (a-b)(\bar{a} - \bar{b}) = a\bar{a} - a\bar{b} - b\bar{a} + b\bar{b} \Rightarrow \ |a-b|^2 = |a^2| - (a\bar{b} + b\bar{a}) + |b|^2 |a|^2 + |b|^2 - |a-b|^2 = a\bar{b} + b\bar{a} if we let a=x+iy b=u+iv It becomes clear that, a\bar{b} + b\bar{a} =...

Re: HSC 2013 4U Marathon Nice job. $Find$ \int \sin(\ln(x)) + \cos (\ln(x)) \ dx

Re: HSC 2013 4U Marathon $Find$ \int \frac{dx}{x\sqrt{2x-1}}

Re: HSC 2013 4U Marathon Well I guess, but I'm not spoon feeding :P, and its very much reasonable to do this without the spoon feeding I think.

re: HSC Physics Marathon Archive $A charge has charge-mass ratio$ \ \ 1 \times 10^{5} $The charge travels in a magnetic field of strength$ \ \ 10 T \ \ $ perpendicular to its trajectory, the charge moves in a circle$ $Calculate the time the charge takes to complete a full revolution of the...

Re: HSC 2013 4U Marathon $Consider the integral$ I_n(t) = \int_{0}^{t} x^n e^{-x} \ dx $i) Prove that$ I_n(t) = n! \left (1 - e^{-t} \sum_{k=0}^{n} \frac{t^{k}}{k!} \right ) $ii) Prove that$ e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} +...

If the preferred solution is by induction, then they will normally ask for induction. For instance, they may ask to prove: 1^2+2^2+ \dots + n^2 = \frac{n}{6} (n+1)(2n+1) It is possible to prove so without induction by considering a telescoping sum, but they will say 'by using induction' if...

Re: HSC 2013 4U Marathon f''(x) = k f(x) f'(x) \ \ \ \ (1) f'(x) = k \int f(x) f'(x) \ dx \ \ \ \ (2) f'(x) = \frac{k}{2} [f(x)]^2 \ \ \ \ (3) I= \int f'(x)^2 f(x)^n \ dx Using IBP u=f'(x) f(x)^{n-1} du = k f'(x) f(x)^n + (n-1) \frac{k}{2} f(x)^n due to identity...

Re: HSC 2013 4U Marathon Oops :/

Re: HSC 2013 4U Marathon My attempt: Part 1 First provide the counter example Consider the series \sum_{n=1}^{\infty} \frac{1}{n} = S By drawing the graph y=1/x, and constructing the upper rectangles of width 1 from 1 to some integer m, it follows that: \sum_{n=1}^{m} \frac{1}{n} >...

Re: HSC 2013 4U Marathon Ok so the sequence that we have to work with is only for real numbers? Because the assumption you gave us before said a is a complex number, so I just got a little confused. Thank you for clearing it up

Re: HSC 2013 4U Marathon A couple of questions: When you mean decreasing, you mean |a| is decreasing right? And in difficulty, what difficulty would you rate Q8 2010 HSC? (Just so I can compare how YOU view difficulty)

Re: HSC 2013 4U Marathon I just made this: $Let$ I_n = \int_{0}^{\pi/4}\tan^{2n}x \ dx $i) Prove that$ \ \ \ I_{n} + I_{n-1} = \frac{1}{2n-1} \ \ \ n\geq 1 $ii) Hence prove that$ \frac{\pi}{4} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \dots