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I don't know for other subjects but I know maths is trail 1 15% trial 2 25%, so in total 40% Other subjects are similar, around the 35-40 percentage mark

Re: HSC 2013 4U Marathon $Sum the series by expanding out the sum$ \ \ \sum_{k=1}^{n} (k - (k-1)) $Do something similar to prove without induction$ \sum_{j=1}^{n} \tan^{-1} \left(\frac{1}{2j^2}\right) = \tan^{-1} \left(\frac{n}{n+1} \right) $Hence find the limiting sum as$ \ n \...

I am wondering whether there is a differential equation type solution. If: f(h) = \frac{V}{g} \sqrt{V^2 + 2gh} Then we can see that: f(h) \cdot f'(h) = f(0) So that means, if we can prove the above differential equation perhaps by thinking physically, using forces and whatnot then...

Re: HSC 2013 4U Marathon Yep this is pretty much it, although I would suggest that for your first step you use a dummy variable such as z, then integrating from 0 to x. EDIT: I probably should of put the conditions on x

After quadratic formula and that business: R(a) = \frac{v^2 \cos a}{g} (\sin a + \sqrt{\sin^2 a + \frac{2gh}{v^2}}) Taking the natural logarithm of both sides then differntiating gives us an easier time, also notice that when we do so the second bracket becomes a standard integral...

We do them at the times when everybody does them, the 2 sets of trials are separated by 2 weeks anyway. Its the fault of the school to do the CSSA papers after the security period is up/a significant time after the papers have been conditioned for release. ===== Goals Perfect the CSSA MX2...

2 full trials both full content, first set of trials is CSSA and second is Independent for Maths and I think other subjects combine trials or use independent or something

Re: HSC 2013 4U Marathon $Prove$ x - \frac{1}{3}x^3+ \frac{1}{5}x^5 - \dots - \frac{1}{4k-1}x^{4k-1} < \tan^{-1} x < x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \dots + \frac{1}{4k+1}x^{4k+1} $And hence prove that$ \tan^{-1} x = \sum_{n=1}^{\infty} \frac{(-1)^{k-1}x^{2k-1}}{2k-1}

Yep, I honestly prefer it though, not sure why. I guess I'd rather have it spread over 2 exams than a whole chunk of the year concentrated on one exam.

Re: HSC 2013 3U Marathon Thread Yep well done, alternatively you can consider the co-efficient of 1/x^2 in the expansion: (1+x)^n(1+ \frac{1}{x})^n \equiv \frac{(1+x)^{2n}}{x^n}

Re: MX2 Integration Marathon $Prove$ \ \ \int_0^1 (1-\sqrt{x})^n \ dx = \frac{2}{(n+1)(n+2)}

Does anyone here have 2 set of trials instead of just 1?

A bracelet can be flipped to the other side, making every permutation identical to the one you get when you flip it, and thus you divide by 2 (i think)

Re: HSC 2013 4U Marathon First find the function f(t) such that the equality holds: f(t) = A + B\int_0^t f(s) \ ds f'(t) = B(F'(t) - F'(0)) For F the primitive of f: \frac{f'(t)}{f(t)} = B \ln f(t) = Bt + c f(0) = A + B \int_0^0 f(s) \ ds = A c = \ln A...

most to least comfortable: MX2 Economics Chemistry English X1 English Adv

Re: MX2 Integration Marathon Using the identity: \int_0^a f(t) \ dt = \int_0^a f(a-t) \ dt I = \int_0^{\pi } xf(\sin x) \ dx = \int_0^{\pi }(\pi -x) f(\sin x) \ dx Therefore: I = \frac{\pi}{2} \int_0^{\pi} f(\sin x) \ dx Now, upon the generic substitution: t= \tan(x/2) , we arrive at...

Its not the fact that it hurts anybody Its just that they have such low self-esteem that a whole year of education is based off of a small cosmetic gain, and what that says about the modern western culture.

Re: HSC 2013 3U Marathon Thread 4U people please refrain from answering this one unless considerable time has passed. $Using the substitution$ \ \ u = \frac{\pi}{2} -x $Evaluate$ \int_0^{\pi /2}x \sin^2 x \ dx

Re: HSC 2013 3U Marathon Thread It may open the door to rigour issues, I'm not sure whether it is allowed or not to do so, and then make the denominator of the RHS zero (i.e. ab=1) An alternative is to add \tan^{-1}(1) to both sides, so on the right hand side you have pi/2 + pi/4 = 3pi/4 On...

Re: HSC 2013 3U Marathon Thread Contructing a right angled triangle with short sides: x and 1 can always be done. (x =/= 0 ) One angle will be atan(x), and the other one will be atan(1/x), by definition of what inverse tan means