HSC 2012 MX2 Marathon (archive) (1 Viewer)

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Congruent? Lol

whoops auto correct hahahaha.

SpiralFlex · May 6, 2012

Re: 2012 HSC MX2 Marathon

Graphically if you plot some points (or if you already know what the square root function looks like)

Then you can immediately tell the range/domain. Try it.

Fus Ro Dah · May 6, 2012

Re: 2012 HSC MX2 Marathon

Did you mean 'Conjugate' instead of 'Congruent'?

A congruent number is an integer that is the area of the triangle with 3 rational sides.

A cool theorem related to this is "There does NOT exist a square number that is congruent". It is called Fermat's Right Triangle Theorem.

barbernator · May 7, 2012

Re: 2012 HSC MX2 Marathon

to get the ball rolling again, <a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{2x^2@plus;13x@plus;16}{(2x@plus;1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" title="\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" /></a>

Fus Ro Dah · May 7, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
to get the ball rolling again, <a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{2x^2@plus;13x@plus;16}{(2x@plus;1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" title="\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" /></a>

$\frac{2}{2x+1} + 2 \ln(x-2) - \frac{3}{2} \ln(2x+1) + C$

Find the area between the curve $y=x e^{-x^2}$ and the x axis over the span $[- \infty,+\infty]$

barbernator · May 7, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
$\frac{2}{2x+1} + 2 \ln(x-2) - \frac{3}{2} \ln(2x+1) + C$

Find the area between the curve $y=x e^{-x^2}$ and the x axis over the span $[- \infty,+\infty]$

1 units squared. find the inverse function of <a href="http://www.codecogs.com/eqnedit.php?latex=y=2e^{2x} -14e^x @plus;6" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=2e^{2x} -14e^x +6" title="y=2e^{2x} -14e^x +6" /></a> and state its new domain and range.

ismeta · May 7, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
$\frac{2}{2x+1} + 2 \ln(x-2) - \frac{3}{2} \ln(2x+1) + C$

Would it be possible to see your working?

I'm having trouble with the question.

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.

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