HSC 2012 MX2 Marathon (archive) (2 Viewers)

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$
$\int^\pi_0 ln[sinu]du=\int^\frac{\pi}{2}_0ln[sinu]du + \int^\pi_\frac{\pi}{2}ln[sinu]du \\\\ = \int^\frac{\pi}{2}_0ln[sinu]du +\int^\frac{\pi}{2}_0ln[sinu]du \\\\ = 2I \\\\ \therefore 2I = \frac{\pi ln2}{2} - I \\\\ I = \frac{\pi ln2}{2}$

IamBread · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
Find

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

$\int\frac{1-\sin^{12}x}{\cos x}\,dx = \int secx - \frac{\sin^{12}x}{cosx}dx \\\\ = \int secx - \frac{(1-\cos^2x)\sin^{10}x}{cosx} dx \\\\ = \int cosx\sin^{10}x dx = \frac{1}{11}\sin^{11}x + c$

Nooblet94 · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

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

I occasionally do this just for fun, but obviously it's impossible cover everything that might come up. Is there anything specific you'd recommend having a look at?

seanieg89 · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

A simpler solution to the integration question:

https://docs.google.com/open?id=0B5x34d2OjtrQWlgwblItMGFUc0MzYUw3eG4zLWNodw

someth1ng · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

All of you guys are so..."keen"! I can't do any of these marathon questions...

IamBread · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

someth1ng said:
All of you guys are so..."keen"! I can't do any of these marathon questions...

None of those integration questions I put up? D:

SpiralFlex · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

Remember Spiral's three levels? Easy, Medium, Hard, Elite. That's 4...

IamBread · Feb 24, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Remember Spiral's three levels? Easy, Medium, Hard, Elite. That's 4...

Are you saying they were elite?

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