HSC 2012 MX2 Marathon (archive) (4 Viewers)

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

oh poor carrot, so long lol.... Here is the shorter way:
View attachment 24402

Haha thank you Math Man. Indeed the method I posted above is very ugly.

There's always a faster way!

 

[deswa1]

Re: 2012 HSC MX2 Marathon

Hey guys can you implicitly differentiate this:

<a href="http://www.codecogs.com/eqnedit.php?latex=(\frac{y}{x})^{\frac{1}{2}}@plus;(\frac{x}{y})^{\frac{1}{2}}=9" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" title="(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" /></a>

I did it but I want to confirm with someone. Thanks

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Hey guys can you implicitly differentiate this:

<a href="http://www.codecogs.com/eqnedit.php?latex=(\frac{y}{x})^{\frac{1}{2}}@plus;(\frac{x}{y})^{\frac{1}{2}}=9" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" title="(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" /></a>

I did it but I want to confirm with someone. Thanks

I got y' = y/x.

 

[deswa1]

Re: 2012 HSC MX2 Marathon

I got y' = y/x.

Good. That's what I got. Thanks Mr. Carrotsticks

 

[jet]

Re: 2012 HSC MX2 Marathon

Here you go. I did it in as many steps as possible so you don't miss out on anything:

Fairly sure you don't need m < 0 considering that when n is even, n - 1 is odd and you can take odd nth roots of negative numbers.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Fairly sure you don't need m < 0 considering that when n is even, n - 1 is odd and you can take odd nth roots of negative numbers.

Your assertion would be correct if n were allowed to be negative.

However, a polynomial is defined by addition or subtraction powers of increasing non-negative integer exponents of some variable, implying that n must be positive.

Whether n is odd or even, the n-1th root will still be positive.

Hence the sign is determined by m, implying that m must be positive.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

^^ sorry, I meant m must be negative in the last line.

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

I see blank pictures.

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

http://community.boredofstudies.org/attachment.php?attachmentid=24460&d=1329975443 heres the question again

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

http://community.boredofstudies.org/attachment.php?attachmentid=24460&d=1329975443 heres the question again

Not working. Reupload it.

 

[deswa1]

Re: 2012 HSC MX2 Marathon

Still doesn't work...

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

It works, but half of the question is cut off haha.

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

... this better work

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

From where did you get these questions, out of curiosity?

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

past catholic CSSA trial

 

[math man]

Re: 2012 HSC MX2 Marathon

ive seen that question in some 2011 trial, can't remember which though, fairly easy.
I'll outline the method so you can try it yourself:
(i) way obvious, if you cant do that by now 4u isn't for you.
(ii) using the info they gave you for z1 and z2 make z1 the subject then take the mod of each side and that will be it.
(iii) now using your diagram from (i) since the mods are equal it is a rhombus so the diagonals bisect the vertice angles.
Now the diagonals bisect at 90 so using simple yr 9 trig in that little triangle and the original info you can obtain that result.
(iv) the trick here is to realise z2= cis(alpha) z1 as you can rotate z1 alpha degrees to get to get to z2.
Then you use (iii) and your triangle plus double angles to work out sin(alpha) and cos(alpha)

This question is medium level, not to hard, hopefully you can do it now

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

ive seen that question in some 2011 trial, can't remember which though, fairly easy.
I'll outline the method so you can try it yourself:
(i) way obvious, if you cant do that by now 4u isn't for you.
(ii) using the info they gave you for z1 and z2 make z1 the subject then take the mod of each side and that will be it.
(iii) now using your diagram from (i) since the mods are equal it is a rhombus so the diagonals bisect the vertice angles.
Now the diagonals bisect at 90 so using simple yr 9 trig in that little triangle and the original info you can obtain that result.
(iv) the trick here is to realise z2= cis(alpha) z1 as you can rotate z1 alpha degrees to get to get to z2.
Then you use (iii) and your triangle plus double angles to work out sin(alpha) and cos(alpha)

This question is medium level, not to hard, hopefully you can do it now

thanks champ; i was just having trouble with part (ii). it turned out to be pretty easy hahaha. +rep

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

ive seen that question in some 2011 trial, can't remember which though, fairly easy.
I'll outline the method so you can try it yourself:
(i) way obvious, if you cant do that by now 4u isn't for you.
(ii) using the info they gave you for z1 and z2 make z1 the subject then take the mod of each side and that will be it.
(iii) now using your diagram from (i) since the mods are equal it is a rhombus so the diagonals bisect the vertice angles.
Now the diagonals bisect at 90 so using simple yr 9 trig in that little triangle and the original info you can obtain that result.
(iv) the trick here is to realise z2= cis(alpha) z1 as you can rotate z1 alpha degrees to get to get to z2.
Then you use (iii) and your triangle plus double angles to work out sin(alpha) and cos(alpha)

This question is medium level, not to hard, hopefully you can do it now

for ii) I used the fact that it's a rhombus and that the sides of a rhombus are equal
Also, for iv) I used algebraic manipulation (which you've done in ii))

Obviously your way works for both, but personally I think my way would be faster.

 

[math man]

Re: 2012 HSC MX2 Marathon

for ii) I used the fact that it's a rhombus and that the sides of a rhombus are equal
Also, for iv) I used algebraic manipulation (which you've done in ii))

Obviously your way works for both, but personally I think my way would be faster.

yes for ii) you first have to prove it is a rhombus, which can be done by taking the arg of the info given which shows the
angle between the two diagonals is 90 hence it is a rhombus

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

yes for ii) you first have to prove it is a rhombus, which can be done by taking the arg of the info given which shows the
angle between the two diagonals is 90 hence it is a rhombus

Yep, of course, I'm just too lazy at the moment to type it all out