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Yeah, it was a typo on my part lol. I've edited the working accordingly. In I[0,pi/2]log(sin2x)dx, u let x=t/2 dx=0.5dt I[0,pi/2]log(sin2x)dx=0.5I[0,pi]log(sint)dt=0.5I[0,pi]log(sinx)dx

This integral is just an exercise in substitution lol I'll use the notation I[a,b]...dx to represent the definite integral from a to b wrtx. Let J=I[0,pi/2]log(sinx)dx Let x=pi/2-t in the integral dx=-dt I[0,pi/2]log(sinx)dx=I[pi/2,0]-log(cost)dt =I[0,pi/2]log(cost)dt=I[0,pi/2]log(cosx)dx...

Actually there are many configurations that are unsolvable (only 1 in 12 of the permutations of the cube will actually be solvable). The problem lies in the fact that there are no algorithms that can flip one edge or one corner only.

For interest, here are 14 different proofs for zeta(2): http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf Proof 11 is quite a nice proof imo.

Current course: Commerce (Actuarial studies)/Science (maths) UAI: 99.65 GPA: 6.375 Usyd combined law transfer offer

2006 student: Biology: 91 Chemistry: 94 English Advanced: 84 Japanese 2u: 95 Japanese 3u: 44 Math 3u: 100 (6th in state) Math 4u: 97 UAI: 99.65 English pulled me down quite a bit... -_-''

My mechanics are a bit rusty, but this is my attempt. Denote the tension in the string connected to O as T1, and the tension connected to the sliding ring as T2. Since the particle is moving in circular motion with constant angular velocity w, horizontal resolution of forces yields...

This is a rather nice question that involves proving the irrationality of e and pi without the usual proof by contradiction. (a) Show that y=1+x/[(1!*a)]+(x^2)/[a(a+1)*2!]+... satisfies the differential equation xy''+ay'=y (b) Hence or otherwise, show that y/y' is irrational (c) Show that...

lol I wonder why no one tried brute forcing it XD Here is an alternate brute force approach: arg(z)+arg(z+3)=pi/2 arctan(y/x)+arctan[y/(x+3)]=pi/2 Take the cosine of both sides (note that after cosining both sides, the trigonometric addition formula cos(x+y)=cosxcosy-sinxsiny must be used)...

buchanan, the "2u proof" of the irrationality of e on your website requires knowledge of the taylor series expansion of e, which is not even in the 4u syllabus. The reasoning used in that proof seems more in the level of harder 3u imo. I also think you go a bit far in criticizing these...

======================================================== Session Course Title Result ======================================================== S1 ACCT1501 Accounting & Financial Mgt 1A....75 DN S1 ECON1101 Microeconomics 1.................76 DN S1...

Re: General Thoughts - Maths Extension 2 Didn't Stephanie Wang get a 120 last year? I also know another guy, Ning, who got an exam mark of 100 for 4u, which implies a 120 as well? (I was under the impression that u only get 100 exam only if u actually get full marks for the paper)

Just to clear up some confusion, the indefinite integral of xtanx has no answer expressible in terms of simple functions. Golbez's approach was not correct, in the second IBP, he was simply reversing what he did in the first IBP, which results in the rather useless equality I=I. The integrals I...

A more plausible version of kony's question is integrate e^(-x²) from 0 to infinity, which does have a very nice closed form answer. A slightly more nasty version would be to integrate (x^n)[e^(-x²)] from 0 to infinity, where n is a positive integer. Another nice q is to show that the integral...

here is a cheap way of drawing inverses: by definition, if u have a function like y=x^2, in the inverse, u just swap the ys and xs to get x=y^2 Therefore, just swap the x and y labels on ur axis so ur x becomes the vertical axis and y is the horizontal. Then, to appease the marker, just rotate...

X represents the angle between the x-axis and the asymptote, not the angle between the tangent and the x-axis. Its quite clear if u draw the diagram. For your interest, there is actually a third method to do it, but it involves university mathematics, more specific, it involves the use of...

90s/120 would get u around 91 or so, low 100s prolly 94ish. Just as a guide for last year's paper, I reckon I got around 109-110/120, which ended up as a 97. Trebla, shouldnt the proportion of E4s be very similar for each year since scaled marks are determined on ur ranking among the cohort and...

Even more neat is the following problem, which uses the above identity Evaluate the series: [[1+(1/1^2)+(1/2^2)]^(1/2)]+[[1+(1/2^2)+(1/3^2)]^(1/2)]+...+[[1+(1/2006^2)+(1/2007^2)]^(1/2)]

For the complex number question, just compare the mod arg form (which has an argument of 5pi/12), with the cartesian form. Equating imaginary parts will give u the answer. For the integrals, an odd/even argument for the first 2, and stating that since the function is always smaller than the...

It was quite an interesting paper, less routine questions, which was a refreshing change. It was harder than the past papers, though not by a lot. Time was sorta tight though as usual for 3u, I had around 10-15 mins or so left i think