HSC 2012 MX2 Marathon (archive) (1 Viewer)

Carrotsticks · Feb 23, 2012

Re: 2012 HSC MX2 Marathon

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

kingkong123 · Feb 23, 2012

Re: 2012 HSC MX2 Marathon

past catholic CSSA trial

math man · Feb 23, 2012

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 · Feb 23, 2012

Re: 2012 HSC MX2 Marathon

math man said:
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 · Feb 23, 2012

Re: 2012 HSC MX2 Marathon

math man said:
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 · Feb 23, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
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 · Feb 23, 2012

Re: 2012 HSC MX2 Marathon

math man said:
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

OMGITzJustin · Feb 23, 2012

Re: 2012 HSC MX2 Marathon

Some of the questions at the start of this thread (e.g spiral's post on page 1) are extremely hard. Having done cambridge complex questions (just homework, in class etc etc), they were fine and honestly werent that hard - it just required a bit of thinking + looking at 1 or 2 solutions for the harder ones. They are do-able, they just seem a level up from cambridge.... am I missing out on something??

nightweaver066 · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

OMGITzJustin said:
Some of the questions at the start of this thread (e.g spiral's post on page 1) are extremely hard. Having done cambridge complex questions (just homework, in class etc etc), they were fine and honestly werent that hard - it just required a bit of thinking + looking at 1 or 2 solutions for the harder ones. They are do-able, they just seem a level up from cambridge.... am I missing out on something??

I found the complex number questions in Cambridge somewhat easy-average..

You should practise complex number questions from past papers to gain a general idea as to what questions to expect in a test.

Carrotsticks · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

As nightweaver wisely said, past paper questions are the way to go.

Furthermore, it doesn't hurt to look up a few out-of-syllabus things, as they often appear in the HSC.

nightweaver066 · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

Neat question:
$Let $ P(x) = x^4 - 4A^2x + 3$ , where A is a real number.$

$By considering P'(x), or otherwise, find the values for A for which P(x) = 0 has 4 complex roots$

IamBread · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

$$Find $ \\\\ \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx$

Carrotsticks · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Find $ \\\\ \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx$

Nice question. Had some fun doing that.

largarithmic · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Find $ \\\\ \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx$

Neat question ^^

First we make the substitution u = x^2.

$u = x^2 $ implies $ du = 2x dx$

and thus

$I = \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx = \frac{1}{2}\int\frac{u e^u}{(u+1)^2)}du$

And now we note that u = (u+1) - 1 so we can split up the integral:

$I = \frac{1}{2}\int\frac{u e^u}{(u+1)^2)}du = \frac{1}{2}\left[\int\frac{(u+1)e^u}{(u+1)^2} - \frac{e^u}{(u+1)^2} du right]$
$= \frac{1}{2} \left[ \int\frac{e^u}{u+1}du - \int\frac{e^u}{(u+1)^2}du \right]$

Now we use integration by parts on the second integral in the subtraction, noting that the integral of 1/(u+1)^2 is -1/(u+1) and the derivative of e^u is just e^u:

$I = \frac{1}{2} \left[ \int\frac{e^u}{u+1}du - \left(-\frac{e^u}{u+1} - \int -\frac{e^u}{u+1}du\right)\right]$

And now the last thing cancels with the first after you sort out the minus signs, up at a constant factor at least, leaving you just with:

$I = \frac{e^u}{2(u+1)} + C = \frac{e^{x^2}}{2(x^2+1)} + C$

IamBread · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

$$Next one, Solve:$ \\\\ \int^\frac{\pi}{2}_0\frac{x}{tanx}dx$

Carrotsticks · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Next one, Solve:$ \\\\ \int^\frac{\pi}{2}_0\frac{x}{tanx}dx$

This was a bit tricky. Is the answer:

$\frac{\pi}{2}\ln(2)$

IamBread · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
This was a bit tricky. Is the answer:

$\frac{\pi}{2}\ln(2)$

You did that fast, took me ~10mins to get it. How did you do it?

Carrotsticks · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
You did that fast, took me ~10mins to get it. How did you do it?

I'll type up a solution tomorrow if I have time, too tired now for large amounts of LaTeX. Unless somebody does it for me.

Trebla · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

Find

$\displaystyle\int\dfrac{1-\sin^{12}x}{\cos x}\,dx$

IamBread · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

I'll post what I did.

$$Let $I = \int^\frac{\pi}{2}_0\frac{x}{tanx}dx \\\\ = xln[sinx]|^{\frac{\pi}{2}}_0 - \int^\frac{\pi}{2}_0ln[sinx]dx \\\\ = -\int^\frac{\pi}{2}_0ln[sinx]dx \\\\ = -\int^\frac{\pi}{2}_0 ln[cosx]dx \\\\ 2I = -\int^\frac{\pi}{2}_0 ln[sinxcosx]dx$
$= -\int^\frac{\pi}{2}_0ln[\frac{sin2x}{2}dx \\\\ = \int^\frac{\pi}{2}_0ln2 - ln[sin2x]dx \\\\ = \frac{\pi ln2}{2} - \int^\frac{\pi}{2}_0 ln[sin2x]dx \\\\ $Let $ u = 2x \\\\ \int^\frac{\pi}{2}_0 ln[sin2x]dx = \frac{1}{2}\int^\pi_0 ln[sinu]du$
$\int^\pi_0 ln[sinu]du=\int^\frac{\pi}{2}_0ln[sinu]du + \int^\pi_\frac{\pi}{2}ln[sinu]du \\\\ = \int^\frac{\pi}{2}_0ln[sinu]du +\int^\frac{\pi}{2}_0ln[sinu]du \\\\ = 2I \\\\ \therefore 2I = \frac{\pi ln2}{2} - I \\\\ I = \frac{\pi ln2}{2}$

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