HSC 2012 MX2 Marathon (archive) (1 Viewer)

barbernator · May 8, 2012

Re: 2012 HSC MX2 Marathon

ismeta said:
Would it be possible to see your working? I'm having trouble with the question.

integrate: (2x^2+13x+16)/((2x+1)^2(x-2))

put that statement into wolframalpha, and you will get a perfect solution

Fus Ro Dah · May 8, 2012

Re: 2012 HSC MX2 Marathon

Evaluate

$\int \frac{dx}{(r^2+x^2)^{1.5}}$

cutemouse · May 9, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
Evaluate

$\int \frac{dx}{(r^2+x^2)^{1.5}}$

use the sub. x=rtan theta...

Carrotsticks · May 9, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
use the sub. x=rtan theta...

Haha, I think he wanted the MX2 students to give it a shot...

Do you have any interesting integrals/questions to post?

IamBread · May 9, 2012

Re: 2012 HSC MX2 Marathon

$\int\frac{1+x^2}{1+x^4}dx$

Fus Ro Dah · May 9, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$\int\frac{1+x^2}{1+x^4}dx$

$\frac{1}{\sqrt{2}} \tan^{-1} \left ( \frac{x \sqrt{2}}{1-x^2} \right )$

Carrotsticks · May 9, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
$\frac{1}{\sqrt{2}} \tan^{-1} \left ( \frac{x \sqrt{2}}{1-x^2} \right )$

Impressive, but I think the idea is to show everybody what you did so then people reading the thread can learn...

barbernator · May 9, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Impressive, but I think the idea is to show everybody what you did so then people reading the thread can learn...

i agree, impressive. I rate that question hard/10

juantheron · May 10, 2012

Re: 2012 HSC MX2 Marathon

$\hspace{-16}\bf{\int \frac{1+x^2}{1+x^4}dx}$\\\\\\ Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\bf{x^2}$\\\\\\ $\bf{\int \frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}dx=\int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{2}\right)^2}dx}$\\\\\\ Now Put $\bf{x-\frac{1}{x}=t\Leftrightarrow \left(1+\frac{1}{x^2}\right)dx=dt}$\\\\\\ $\bf{=\int \frac{1}{t^2+\left(\sqrt{2}\right)^2}dt=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)+\mathbb{C}}$\\\\\\ So $\bf{\int \frac{1+x^2}{1+x^4}dx = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+\mathbb{C}}$$

juantheron · May 10, 2012

Re: 2012 HSC MX2 Marathon

another solution Using Trigonometry:

$\hspace{-16}\bf{\int \frac{1+x^2}{1+x^4}dx}$\\\\\\ Let $\bf{x^2=\tan \theta\Leftrightarrow 2xdx=\sec^2\theta.d\theta}$\\\\\\ $\bf{dx=\frac{\sec^2\theta}{2x}d\theta=\frac{\sec^2\theta}{2\sqrt{\tan \theta}}.d\theta}$\\\\\\ $\bf{=\int\frac{1+\tan \theta}{1+\tan^2 \theta}.\frac{\sec^2\theta}{2\sqrt{\tan \theta}}.d\theta}$\\\\\\ $\bf{=\frac{1}{2}\int \frac{1+\tan \theta}{\sqrt{\tan \theta}}d\theta=\frac{1}{2}\int \left(\sqrt{\tan \theta}+\sqrt{\cot \theta}\right)d\theta}$\\\\\\ $\bf{=\frac{1}{2}\int \frac{\sin \theta+\cos \theta}{\sqrt{\sin \theta.\cos \theta}}d\theta=\frac{1}{\sqrt{2}}\int \frac{\sin \theta+\cos \theta}{\sqrt{\sin 2\theta}}d\theta}$\\\\\\ Now Let $\bf{\sin \theta-\cos \theta =t \Leftrightarrow \left(\cos \theta +\sin \theta\right)d\theta=dt }$\\\\\\ Now $\bf{1-\sin 2\theta=t^2\Leftrightarrow \sin 2 \theta = (1-t^2)}$\\\\\\ So $\bf{=\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{1-t^2}}dt=\frac{1}{\sqrt{2}}\sin^{-1}\left(t\right)+\mathbb{C}}$\\\\\\$

$\hspace{-16}\bf{=\frac{1}{\sqrt{2}}\sin^{-1}\left(\sin \theta -\cos \theta\right)+\mathbb{C}}$\\\\\\ $\bf{\int \frac{1+x^2}{1+x^4}dx = \frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{x^2-1}{\sqrt{1+x^4}}\right)+\mathbb{C}}$$

juantheron · May 10, 2012

Re: 2012 HSC MX2 Marathon

$\hspace{-16}\bf{\int\frac{1}{1+x^4}dx}$

asianese · May 10, 2012

Re: 2012 HSC MX2 Marathon

I just wolfram alphaed that...WTF LOL

seanieg89 · May 10, 2012

Re: 2012 HSC MX2 Marathon

asianese said:
I just wolfram alphaed that...WTF LOL

If you have trouble spotting any clever substitution to kill such integrals, always remember the brute-force methods. Partial fractions works fine here, as it does for the previous integral.

nightweaver066 · May 19, 2012

Re: 2012 HSC MX2 Marathon

I like this question.

$$The polynomial equation $ x^4 + 4x^3 - 2x^2 - 12x - 3 = 0 $ has roots $ \alpha, \beta, \gamma~and~\delta. $ Find the polynomial equation whose roots are $ \alpha + 1, \beta + 1, \gamma + 1, \delta +1$

$$Hence or otherwise, Solve $ x^4 + 4x^3 - 2x^2 - 12x - 3 = 0$

asianese · May 19, 2012

Re: 2012 HSC MX2 Marathon

$\text{Let}\, y=x+1\\ \therefore x=y-1.\\ \begin{align*}\therefore (y-1)^4+4(y-1)^3-2(y-1)^2-12(y-1)-3&=0\\ y^4-8y^2+4&=0\\(y^2-2y-2)(y^2+2y-2)&=0\\\\ \therefore (x+1)^2-2(x+1)-2&=0\\ x^2 &=3\\\therefore x&=\pm \sqrt{3}.\\\\(x+1)^2+2(x+1)-2&=0\\(x+2)^2&=3\\ \therefore x&=-2\pm \sqrt{3}\end{align*}$

I've spared the messy algebra...

The addition/product of roots would work also I think.

Inference · May 22, 2012

Re: 2012 HSC MX2 Marathon

Why don't we add some pure maths questions in here,

here's a fave one of mine, combinatorics

A permutation $\{x_1, \ldots, x_{2n}\}$ of the set $\{1,2, \ldots, 2n\}$ where $\mbox{n}$ is a positive integer, is said to have property $T$ if $|x_i - x_{i + 1}| = n$ for at least one $i \in \{1,2, \ldots, 2n - 1\}$ . Show that, for each $\mbox{n}$ , there are more permutations with property $T$ than without.

Hint: Indicator functions will help, and think about counting sets

Carrotsticks · May 22, 2012

Re: 2012 HSC MX2 Marathon

Inference said:
Why don't we add some pure maths questions in here,

here's a fave one of mine, combinatorics

A permutation $\{x_1, \ldots, x_{2n}\}$ of the set $\{1,2, \ldots, 2n\}$ where $\mbox{n}$ is a positive integer, is said to have property $T$ if $|x_i - x_{i + 1}| = n$ for at least one $i \in \{1,2, \ldots, 2n - 1\}$ . Show that, for each $\mbox{n}$ , there are more permutations with property $T$ than without.

Hint: Indicator functions will help, and think about counting sets

If you want some Pure Maths questions unrelated to the HSC, then feel free to make your own thread in the 'Extracurricular Maths' section: http://community.boredofstudies.org/forumdisplay.php?f=238

Inference · May 22, 2012

Re: 2012 HSC MX2 Marathon

Thanks, but it is related to the HSC albeit a fair extension.

You guys do combinatorics in the HSC right? The above question can be solved purely combinatorially.

barbernator · May 22, 2012

Re: 2012 HSC MX2 Marathon

Inference said:
Thanks, but it is related to the HSC albeit a fair extension.

You guys do combinatorics in the HSC right? The above question can be solved purely combinatorially.

we learn factorial notation for combs and perms and how to apply them, that is about all lol.

Inference · May 22, 2012

Re: 2012 HSC MX2 Marathon

Ah right I see, okay I guess it is a fair stretch then, maybe a bit of wishful thinking can lead somewhere

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