Search results

Re: HSC 2014 4U Marathon Find \lim_{x \to -\infty}\sqrt{x^2+4x+7}+x

Re: HSC 2014 4U Marathon The only case where it could be defined is if a=b=0 which i eliminated with the given condition

Re: HSC 2014 4U Marathon No its the fact that the range of secx is outside the domain of inverse sin, so therefore the integral is actually undefined!

Re: HSC 2014 4U Marathon Its nothing to do with that, what is the domain of Inverse Sin and the range of secx?

Re: HSC 2014 4U Marathon Well its kinda right but there is something wrong about it, can you figure it out?

Re: HSC 2014 4U Marathon Are you sure?

Re: HSC 2014 4U Marathon \int_a^b{ \dfrac{secxtanx}{\sqrt{1-sec^2{x}}}}dx \;\;\; (0<a<b<\frac{\pi}{2} )

Re: HSC 2014 4U Marathon Very good but there is one detail in part ii you missed, The graph (on the right branch) actually intercepts the asymtote at (1,1) and approaches y=x from the top side. Other than that well done. And also you accidentally labeled asymtotes as x=\pm a rather than...

Re: HSC 2014 4U Marathon Well consider the graph y=x^{1/2} and y=x^{1/3} , can you see how for x>1 , y=x^{1/2} is greater, while for 0<x<1 , y=x^{1/3} is actually greater. What it means for the graph is where the x-intercept lies, between the asymtotes (x=\pm \sqrt{a}) or towards the...

Re: HSC 2014 4U Marathon - Advanced Level oh ok ima come back to this later, gonna move on to something else.

Re: HSC 2014 4U Marathon - Advanced Level yeh i just realised and found 3rd combination, im trying to fill in the holes but not finished, have you solved it?

Re: HSC 2014 4U Marathon - Advanced Level yeh so my answer is 2 now, can you check my reasoning i posted above

Re: HSC 2014 4U Marathon - Advanced Level I got 97 for 3u and 4u and overall 99.4, which im super happy with but i was aiming much higher leading towards hsc (long story). And thanks! :) Not just returning the compliment but you seem very capable yourself!

Re: HSC 2014 4U Marathon - Advanced Level Prove the following expression: (2+\sqrt{2})^n\cos{\frac{n\pi}{4}}=\sum\limits_{k=0}^{2n} \binom {2n} {k}\cos{\frac{k\pi}{4}} and similarly prove: (2+\sqrt{2})^n(\cos{\frac{n\pi}{4}} +\sin{\frac{n\pi}{4}})=\sum\limits_{k=0}^{2n} \binom {2n}...

Re: HSC 2014 4U Marathon - Advanced Level Woah! Thats a crazy smart solution! Had no clue how you would use cube roots of unity. one thing though, i think there should be an x in front of PQR?

Re: HSC 2014 4U Marathon - Advanced Level Something I made up

Re: HSC 2014 4U Marathon I) \vert z \vert =\vert Re(z)+iIm(z) \vert =\sqrt{Re^2(z)+Im^2(z)} \geq\sqrt{Re^2(z)}=\vert Re(z) \vert II) Let z=(x+iy)(a+ib) \longrightarrow \vert z \vert= \sqrt{a^2+b^2}\sqrt{x^2+y^2} and \vert Re(z) \vert=\vert ax-by \vert Hence using part (i) |ax-by| \leq...

Re: HSC 2014 4U Marathon You would be close if a=1. Hint: You should have two vertical asymtotes and one oblique.

Re: HSC 2014 4U Marathon - Advanced Level Prove that \pi=lim_{n\to\infty} \; 2^n\sqrt{2-\sqrt{2+\sqrt{2+...\sqrt{2}}}} (Where there are a total of n square root brackets)

Re: HSC 2014 4U Marathon - Advanced Level Hahaha I'm kidding I was meant to do it this year but was accelerated so last year