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No 'real' solutions ^_^ But yeah that's called reducible to quadratics.

LOL question done to death haha

I second that

Re: MX2 Integration Marathon $Well isn't that nice... $ \\ \\ \int_{0}^{2014\pi} \sqrt{1-(1-2sin^2x)} \\ =\int_{0}^{2014\pi} \sqrt{2}sinx $ dx $ \\ =\sqrt{2} \left [ -cosx \right ]_{0}^{2014\pi}$ dx $ \\ =\sqrt{2}(-1+1) \\ =0 But i feel i remember doing something like this before, and its...

Re: HSC 2014 4U Marathon - Advanced Level Ah k cool thanks :)

Re: MX2 Integration Marathon On the third last line it's meant to be a u not x but other than that thanks :) @SYAMAN: Brah that frog is so distracting LOL.

Re: HSC 2014 4U Marathon - Advanced Level Ooh, = \lim_{x \rightarrow \infty}\frac{x+m-x-n}{(\sqrt{x+m}+\sqrt{x+n})} \\ =\lim_{x \rightarrow \infty}\frac{m-n}{(\sqrt{x+m}+\sqrt{x+n})} \\ $So does it$ \rightarrow 0? How do you guys know if its 0+ or 0-?

Easiest way i can think of is: RHS= \\ \\ \frac{2tanx}{1-tan^2x}-\frac{1}{cos^2x-sin^2x} \\ \\ $Then for the second fraction multiply top and bottom by sec^2x.$ \\ =\frac{2tanx}{1-tan^2x}-\frac{sec^2x}{1-tan^2x} \\ \\ $Common denominator$ \\ =-\frac{tan^2x-2tanx+1}{(1-tanx)(1+tanx)} \\ \\ $Now...

Re: HSC 2014 4U Marathon i) Hmm with this one i simply drew up the unit circle and placed z on the argand diagram between 0 and pi/2. Similarly, for z^2 the domain extends to pi so I placed z^2 in the second quadrant which also lies in the unit circle. By the modulus of z is clearly 1 for both...

Re: HSC 2014 4U Marathon Ah like parametrics aye, yeh the t thing helped alot TREBLAAA hi :)

Re: HSC 2014 4U Marathon If x=1: z^2=2i, If x=2:z^2=3+4i, If x=3: z^2=8+6i If i let z=x+iy but also x+i then x^2-y^2+2xyi=x^2-1+2xi Equating imaginaries, y=1 right? But clearly not the case by substituting... EDIT: We could just take them as points (0,2) (3,4) and (8,6). SYYYYYYYYYYYYYY Mind...

sounds about right :) Thanks guys.

Re: HSC 2014 4U Marathon Facedesk* I just recapped, I kept trying to think visually but i guess algebraic would be best... Its just letting z=x+i that threw me off. So i let z=x+iy. for i) it is clear that locus is y=1 by equating coeff ii) Similarly from your expansion y=1 should also be locus...

Re: HSC 2014 4U Marathon Ok, z^2=(x+i)^2 Using the same logic, it is the line squared so a parabola with vertex at y=1?

I don't know maybe someone is good at these here which is why i posted it.

Re: HSC 2014 4U Marathon - Advanced Level What does that pi looking thing mean? (Noob alert hahaha)

I was thinking more along the lines of: \frac{no.+of ways+with+couple+together}{no.+of+ways+total$} to make use of i) PS: Forgot how to space in latex.

K looked back and copied out the exact question. And no a couple is not a single man and a single woman, that is something that I must have included even in my paraphrasing. Edited original post. But yeah all good now, anyone?

Yep paraphrasing. 1. Yes 2. No. 3. I presume adjacent 4. One Sorry for the confusion

Re: HSC 2014 4U Marathon I'm probably going to sound stupid, but ok z=x+i. Locus of z=x+i, is that just all of the plane? 0.0 Liek all of x values right, so adding i doesn't really do much... Man i need to revise my complex. K, sub any x Then will lie in y=i. So just line y=i? Horizontal line...