HSC 2012 MX2 Marathon (archive) (1 Viewer)

[seanieg89]

Re: 2012 HSC MX2 Marathon

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

I'm not sure why you added the part for, but that's right.

Reaction haha.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Does it have inflections?

Yes. They occur when you e^f(x) the left and right branches.

 

[deswa1]

Re: 2012 HSC MX2 Marathon

Using the fact that <a href="http://www.codecogs.com/eqnedit.php?latex=arg\frac{z_{1}}{z_{2}}=argz_{1}-argz_{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" title="arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" /></a>

Prove that <a href="http://www.codecogs.com/eqnedit.php?latex=tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" title="tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" /></a>

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Using the fact that <a href="http://www.codecogs.com/eqnedit.php?latex=arg\frac{z_{1}}{z_{2}}=argz_{1}-argz_{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" title="arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" /></a>

Prove that <a href="http://www.codecogs.com/eqnedit.php?latex=tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" title="tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" /></a>

Haha I made a very similar question to this a couple of months ago, but the first question was:

Show that:



Then I got them to deduce a similar identity to yours.

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

An easy polynomials question:

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

Using the fact that

Prove that

let z1 = 1 + 2i and z2 = 3 +i
therefore arg(z1) = arctan(2) and arg(z2)=arctan(1/3)
we want to prove arctan(2) - arctan(1/) = pi/4 ie arg(z1)-arg(z2)=pi/4

arg(z1)-arg(z2)=arg(z1/z2) where z1/z2 = (1+i)/2

arg( (1+i)/2 ) = pi/4.

therefore as required. were u asking this because u didnt know how or just as a forum question? nice question

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

Using the fact that <a href="http://www.codecogs.com/eqnedit.php?latex=arg\frac{z_{1}}{z_{2}}=argz_{1}-argz_{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" title="arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" /></a>

Prove that <a href="http://www.codecogs.com/eqnedit.php?latex=tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" title="tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\\\textrm{Let}~z_1=1@plus;2i~\textrm{and}~z_2=3@plus;i\\ \arg z_1=\frac{2}{1}=2\\ \arg z_2=\frac{1}{3}\\ ~\\\textrm{Now,}\\ \frac{z_1}{z_2} =\frac{(1@plus;2i)(3-i)}{(3@plus;i)(3-i))}=\frac{5@plus;5i}{10}\\~\\ \therefore \arg\frac{z_1}{z_2}=\frac{\frac{1}{2}}{\frac{1}{2}}=1\\ \textrm{But}~\arg\frac{z_1}{z_2}=\arg z_1-\arg z_2\\ \therefore \tan^{-1}2-\tan^{-1}\frac{1}{3}=\tan^{-1}1=\frac{\pi}{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\\textrm{Let}~z_1=1+2i~\textrm{and}~z_2=3+i\\ \arg z_1=\frac{2}{1}=2\\ \arg z_2=\frac{1}{3}\\ ~\\\textrm{Now,}\\ \frac{z_1}{z_2} =\frac{(1+2i)(3-i)}{(3+i)(3-i))}=\frac{5+5i}{10}\\~\\ \therefore \arg\frac{z_1}{z_2}=\frac{\frac{1}{2}}{\frac{1}{2}}=1\\ \textrm{But}~\arg\frac{z_1}{z_2}=\arg z_1-\arg z_2\\ \therefore \tan^{-1}2-\tan^{-1}\frac{1}{3}=\tan^{-1}1=\frac{\pi}{4}" title="\\\textrm{Let}~z_1=1+2i~\textrm{and}~z_2=3+i\\ \arg z_1=\frac{2}{1}=2\\ \arg z_2=\frac{1}{3}\\ ~\\\textrm{Now,}\\ \frac{z_1}{z_2} =\frac{(1+2i)(3-i)}{(3+i)(3-i))}=\frac{5+5i}{10}\\~\\ \therefore \arg\frac{z_1}{z_2}=\frac{\frac{1}{2}}{\frac{1}{2}}=1\\ \textrm{But}~\arg\frac{z_1}{z_2}=\arg z_1-\arg z_2\\ \therefore \tan^{-1}2-\tan^{-1}\frac{1}{3}=\tan^{-1}1=\frac{\pi}{4}" /></a>

 

[deswa1]

Re: 2012 HSC MX2 Marathon

were u asking this because u didnt know how or just as a forum question? nice question

Nah, I already did it. It was in a question by topic book but I thought it was a neat question so I posted it

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

This problem can be further generalised.

Here is an extension to deswa1's problem:

Show that all the values satisfying the condition...



... lie on the hyperbola:

 

[deswa1]

Re: 2012 HSC MX2 Marathon

Before I have a go, is it meant to be inverse tan(x) MINUS inverse tan(y) or is the question you posted correct?

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

This problem can be further generalised.

Here is an extension to deswa1's problem:

Show that all the values satisfying the condition...



... lie on the hyperbola:

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Before I have a go, is it meant to be inverse tan(x) MINUS inverse tan(y) or is the question you posted correct?

The formula you have with the minus sign is a specific case for the generalised version:



Good solution nightweaver066.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Here's a bit of a tricky question, similar to that in the HSC (forgot which year, I think 2007?)

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

Dont you mean the limit as n approaches infinity?

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon









Edit: Damn BOS NEVER LINKS ME TO THE LAST PAGE. >_>

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Dont you mean the limit as n approaches infinity?

Thanks for that correction. It's become habit for me to type x -> infinity now haha.

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

Here's a bit of a tricky question, similar to that in the HSC (forgot which year, I think 2007?)

After tedious working out,



As n -> infinity,





This means that as n approaches an infinitely large number, the stationary point and inflexion point will coincide (ignoring x = 0) and there will be a horizontal inflexion point. (is this right? lol)

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

Geometrically it means that as n approaches positive infinity the limiting ratio of the x-coordinates of the inflexion point to the stationary point of the curve y=x^n*log[e][x^n] is one?

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

Disregard my answer, this.