HSC 2012 MX1 Marathon #1 (archive) (3 Viewers)

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

 

[largarithmic]

Re: 2012 HSC MX1 Marathon

how would i find the inverse of y = e^x/(3 + e^x)

i get stuck on x(3+e^y) = e^y

You do polynomial division.

x = (e^y)/(3+e^y) = 1 - 3/(3+e^y)
so 3/(3 + e^y) = 1 - x
so 3/(1-x) = 3+e^y
so y = ln[3/(1-x) - 3]

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

how would i find the inverse of y = e^x/(3 + e^x)

i get stuck on x(3+e^y) = e^y

I do not understand why you would need to find the inverse of the function.

EDIT: Oh it is unrelated to the differentiation question haha.

 

[Timske]

Re: 2012 HSC MX1 Marathon

I do not understand why you would need to find the inverse of the function.

EDIT: Oh it is unrelated to the differentiation question haha.

dont know just trying to do what this sheet is asking

 

[deswa1]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textsl{Prove that the series } 1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}@plus;... \textsl{ diverges}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textsl{Prove that the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+... \textsl{ diverges}" title="\textsl{Prove that the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+... \textsl{ diverges}" /></a>

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textsl{Prove that the series } 1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}@plus;... \textsl{ diverges}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textsl{Prove that the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+... \textsl{ diverges}" title="\textsl{Prove that the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+... \textsl{ diverges}" /></a>

Umm, I don't think the average MX1 student will be able to *rigorously* prove the divergence of the Harmonic Series.

A good Extension 2 student may be able to relate it to upper rectangles for the curve y=1/x and show that since the area is unbounded as x --> infinity, so is the series.

However, I don't think this will happen for an Extension 1 student.

 

[math man]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textsl{Prove that the series } 1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}@plus;... \textsl{ diverges}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textsl{Prove that the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+... \textsl{ diverges}" title="\textsl{Prove that the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+... \textsl{ diverges}" /></a>

This question is best suited to harder 3u, which is apart of the 4u course, not 3u, and yes as carrot said

 

[deswa1]

Re: 2012 HSC MX1 Marathon

This question is best suited to harder 3u, which is apart of the 4u course, not 3u, and yes as carrot said

What about now:
<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{It is known that the series }1@plus;\frac{1}{2}@plus;\frac{1}{4}@plus;\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series } 1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}... \textup{ to a geometric sequence where the ratio is not less than one, prove} \\\textup{that }1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}... \textup{diverges}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{It is known that the series }1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{ to a geometric sequence where the ratio is not less than one, prove} \\\textup{that }1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{diverges}" title="\textup{It is known that the series }1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{ to a geometric sequence where the ratio is not less than one, prove} \\\textup{that }1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{diverges}" /></a>

With a bit of thinking, a good 3U student (I think) should be able to get it but this proof might not be fully rigourous...

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

What about now:
<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{It is known that the series }1@plus;\frac{1}{2}@plus;\frac{1}{4}@plus;\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series } 1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}... \textup{ to a geometric sequence where the ratio is not less than one, prove} \\\textup{that }1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}... \textup{diverges}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{It is known that the series }1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{ to a geometric sequence where the ratio is not less than one, prove} \\\textup{that }1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{diverges}" title="\textup{It is known that the series }1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series } 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{ to a geometric sequence where the ratio is not less than one, prove} \\\textup{that }1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... \textup{diverges}" /></a>

With a bit of thinking, a good 3U student (I think) should be able to get it but this proof might not be fully rigourous...

The proof is valid.

If there is some series A_n that diverges and there exists another series B_n such that B_n > A_n, then B_n also diverges.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Consider the equation y=x^n + nx + 1.

Find the condition such that the curve has a stationary point with integer co-ordinates.

Find the co-ordinates of the stationary point in terms of n.

 

[largarithmic]

Re: 2012 HSC MX1 Marathon

The proof is valid.

If there is some series A_n that diverges and there exists another series B_n such that B_n > A_n, then B_n also diverges.

Is that sorta stuff like, allowed/used in 3u, when people dont really even know what converges/diverges mean?

Also, sorry man for not talking to you today outside the timetabling unit

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Is that sorta stuff like, allowed/used in 3u, when people dont really even know what converges/diverges mean?

Also, sorry man for not talking to you today outside the timetabling unit

Haha nws =)

I was actually going through the Analysis course notes whilst waiting for another BOS'er you are familiar with.

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

Consider the equation y=x^n + nx + 1.

Find the condition such that the curve has a stationary point with integer co-ordinates.

Find the co-ordinates of the stationary point in terms of n.

<a href="http://www.codecogs.com/eqnedit.php?latex=y=x^n@plus;nx@plus;1\\ y'=nx^{n-1}@plus;n=n(x^{n-1}@plus;1)\\ $Stationary points occur when $y'=0\\ n(x^{n-1}@plus;1)=0\\ \therefore x^{n-1}=-1\\ $Hence, for there to be a stationary point with integer coordinates $n$ must be even since $n-1$ must be odd$\\ \therfore x=-1\\ y=(-1)^n@plus;n(-1)@plus;1\\ ~~~=1-n@plus;1$ (Since n is even)$\\ ~~~=2-n\\ \therefore $ the coordinates of the stationary point are $(-1,2-n)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=x^n+nx+1\\ y'=nx^{n-1}+n=n(x^{n-1}+1)\\ $Stationary points occur when $y'=0\\ n(x^{n-1}+1)=0\\ \therefore x^{n-1}=-1\\ $Hence, for there to be a stationary point with integer coordinates $n$ must be even since $n-1$ must be odd$\\ \therfore x=-1\\ y=(-1)^n+n(-1)+1\\ ~~~=1-n+1$ (Since n is even)$\\ ~~~=2-n\\ \therefore $ the coordinates of the stationary point are $(-1,2-n)" title="y=x^n+nx+1\\ y'=nx^{n-1}+n=n(x^{n-1}+1)\\ $Stationary points occur when $y'=0\\ n(x^{n-1}+1)=0\\ \therefore x^{n-1}=-1\\ $Hence, for there to be a stationary point with integer coordinates $n$ must be even since $n-1$ must be odd$\\ \therfore x=-1\\ y=(-1)^n+n(-1)+1\\ ~~~=1-n+1$ (Since n is even)$\\ ~~~=2-n\\ \therefore $ the coordinates of the stationary point are $(-1,2-n)" /></a>

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

Also adding on to part iii), find the percentage error of the approximate and actual area.

 

[kingkong123]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{A function is }y=f(x)\textup{ is given in parimetric form by }x=tan\theta \textup{ and }\\y=4sin2\theta\textup{ for }\frac{-\pi}{2}<\theta<\frac{\pi}{2}.\textup{ State the domain and range of the function and show that }\\y=\frac{8x}{1@plus;x^{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{A function is }y=f(x)\textup{ is given in parimetric form by }x=tan\theta \textup{ and }\\y=4sin2\theta\textup{ for }\frac{-\pi}{2}<\theta<\frac{\pi}{2}.\textup{ State the domain and range of the function and show that }\\y=\frac{8x}{1+x^{2}}" title="\textup{A function is }y=f(x)\textup{ is given in parimetric form by }x=tan\theta \textup{ and }\\y=4sin2\theta\textup{ for }\frac{-\pi}{2}<\theta<\frac{\pi}{2}.\textup{ State the domain and range of the function and show that }\\y=\frac{8x}{1+x^{2}}" /></a>

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Also adding on to part iii), find the percentage error of the approximate and actual area.

Adding to that, show that the percentage error --> 0 as n approaches infinity.

 

[zeebobDD]

Re: 2012 HSC MX1 Marathon

Also adding on to part iii), find the percentage error of the approximate and actual area.

coulda done it for my test coming up next week if it wasnt cos inverse

 

[Timske]

Re: 2012 HSC MX1 Marathon

zomg yallah habib cant do this

int0~1 2x/(2x + 1)^2 using u = 2x + 1 , x = (u-1)/2
x: 0 ~ 1
u: 3~4

du/dx = 2
dx = du/2
int3~4 2((u-1)/2))/u^2 * du/2
int3~4 ((u-1)/2)/u^2 * du
int3~4 (u-1)/2u^2 * du

i get stuck here..

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Split the fraction so you get two integrals.

 

[Timske]

Re: 2012 HSC MX1 Marathon

integral 1/(sqrt(x(1-x)) using y = sqrt(x)