Recent content by adzy

Anybody have past papers with solutions to them (Not including past HSC papers)? Or any interesting/challenging Question 7's?

y=x^x Bunch of dots? More or less. It'll still be hell to draw. Plus, the hole at y-1 at x=0.

It'll be a graph question of y=x^x for x<=0

This is SGS 1994 4U Trial Paper There is no follow up question but the preevious questions before that were: a) If I_n = integral x^n * e^(x^2) dx show that I_n + (n-1) I_(n-2) = 2e (n>=2) Edit: Shoulw the answer read I_n + (n-1)/2 (or times two?) * I_(n-2) = 2e? What they have is what...

A question 8 from a trial paper Evaluate summation from n=1 to infinity of 1/(1+u_n) given that (u_n+1) = u_n + (u_n)^2 and u_1 = 1/3 I go to as far as summation of n=1 to infinity of: U_n / U_n+1 but uh...I don't think that really does anything.

The lengths of the sides of a triangle from an arithmetic progression and the largest angle of the triangle exceeds the smallest by 90 degrees. Find the ration of the lengths of the sides. I dont want to trig bash anymore. Can anybody find an elegant solution?

Thanks dude. ii) Deduce that U_k+1 = k(U_k + U_(k-1)), k=4,5,6... Still any headway on this?

Thanks dude. Haha, if I see that q8 I think i just panic and not even try it. ii) Deduce that U_k+1 = k(U_k + U_(k-1)), k=4,5,6... Any headway on this?

Found this in my brother's stack of old trial papers. I found this test a higher standard than anything else I've ever done (consistent high standard throughout) Just a note, the examiners are Graham Arnold and Denise Arnold. Heh. Q8 n letters L1, L2, L3, ..., Ln are to be placed at...

Yes it was quite easy, and one of the top students ranted about it. If you guys want something interesting, apparently 2004 Ruse Trial is much harder. Then again most papers are much harder than 05 Ruse Trial

Here's what wikipedia has on it: The Riemann hypothesis is a conjecture about the distribution of the zeros of the Riemann zeta function ζ(s). The Riemann zeta function is defined for all complex numbers s * 1. It has certain so-called "trivial" zeros for s = −2, s = −4, s = −6, ... The...

Thanks, quite elegant.

Can anybody do this question? Show (Nc1)^2 + 2(Nc2)^2 + 3(Nc3)^2 + .... n(NcN)^2 = (2n-1)! / [(n-1!)]^2 where NcR = N choose R

What would that look like on cartesian graph?

So answer is just -1/(x^2+3x+2) + c using acmilan's subsitution