HSC 2012 MX1 Marathon #1 (archive) (3 Viewers)

nightweaver066 · Jan 4, 2012

Let's get the ball rolling.

Remember to post a question after solving one.

tywebb · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

(i) -1

(ii) 0

(iii) (-1)²-2(0)=1

Now find the roots.

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
(i) -1

(ii) 0

(iii) (-1)²-2(0)=1

Now find the roots.

One is real and the other two are complex.

Isn't this a MX1 marathon?

deswa1 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

The real one isn't pretty as well...

tywebb · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Yes. I wolframalpha.com-ised it and got

deswa1 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

While we're doing sums and products of roots:

EDIT: How do I get the picture to display without it being an attachment?

SpiralFlex · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

deswa1 said:
While we're doing sums and products of roots:
View attachment 24042

EDIT: How do I get the picture to display without it being an attachment?

$2(\alpha+\beta)=-p...(1)$

$(\alpha+\beta)^2+\alpha \beta =q...(2)$

$\alpha \beta (\alpha+\beta)=-r...(3)$

(1) into (3)

$\alpha \beta-\frac{2r}{p}....(4)$

(1) and (4) into (2)

$\frac{2r}{p}+\frac{p^2}{4}=q$

$p^3=4pq-8r$

tywebb · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

to display it type without spaces

[ img ]http://community.boredofstudies.org/attachment.php?attachmentid=24042&d=1325640936[ /img ]

It should come out like this:

SpiralFlex · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Alkanes · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

$Find$

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

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

deswa1 said:
While we're doing sums and products of roots:
View attachment 24042

EDIT: How do I get the picture to display without it being an attachment?

It has to already be available on the web (ie: has its own URL link) so you will need to use something like Photobucket.

If you upload a file directly from your hard drive, it will appear as an attachment.

It is known that all triangles can be circumscribed within a circle.

Prove that the smallest ratio Circle (the one circumscribing the triangle) : Triangle occurs when the triangle is an equilateral triangle.

EDIT: Wait wtf why am I doing questions? I forgot that this is a Marathon and not a help thread lol.

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
to display it type without spaces

[ img ]http://community.boredofstudies.org/attachment.php?attachmentid=24042&d=1325640936[ /img ]

It should come out like this:

But this example has already been posted on the web. Hence you can use the img tags.

How would you do it directly from the hard drive? Notice that your text between the img tags is a URL already created by the fact that he posted it up on the web.

Drongoski · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Alkanes said:
$Find$

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

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

nightweaver066 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

SpiralFlex said:

$x = 20tcos\alpha$

$t = \frac{xsec\alpha}{20}$

Subbing t in to $y = -5t^2 - 20tsin\alpha$

$y = -5(\frac{xsec\alpha}{20})^2 - 20sin\alpha(\frac{xsec\alpha}{20})$

$= -\frac{x^2sec^2\alpha}{80} + xtan\alpha$

ii) When x = 20, $20 = 20tcos\alpha$

$t = sec\alpha$

When $t = sec\alpha$ , $y = -5sec^2\alpha + 20sin\alpha sec\alpha$

$y = 20tan\alpha - 5(1 + tan^2\alpha)$

$\therefore h = 20tan\alpha - 5(1 + tan^2\alpha)$

iii) $h = -5tan^2\alpha + 20tan\alpha - 5$

$= -5(tan^2\alpha -4tan\alpha + 1)$

$= -5(tan\alpha -2)^2 + 15$

$\therefore \text{Maximum value of h occurs when } tan\alpha = 2 \text{ and the maximum height is 15m}$

bleakarcher · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
One is real and the other two are complex.

Isn't this a MX1 marathon?

How did you know that one root is real and two are complex?

nightweaver066 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Drongoski said:
$= \int (\frac {x}{x^2+4} +\frac {1}{x^2+4})dx$

$=\frac{1}{2}ln(x^2 + 4) + \frac{1}{2}tan^{-1}\frac{x}{2} + c$

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

bleakarcher said:
How did you know that one root is real and two are complex?

Because sum of roots (squared) was a very small positive value.

It is unlikely that such an ordinary polynomial (look at the coefficients, nothing funky there) has all 3 real roots within the interval [0,1]

EDIT: Woops, I meant (0,1), it is trivial that 0 and 1 are not solutions.

nightweaver066 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

New question,

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
New question,

Realistically in a 3U test (at an ordinary school), they would tell you to use the substitution u=cos(x).

Alternatively, you could turn 2sin(x)cos^2(x) into sin(2x)cos(x) and then proceed to use IBP lol.

nightweaver066 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
Realistically in a 3U test (at an ordinary school), they would tell you to use the substitution u=cos(x).

Alternatively, you could turn 2sin(x)cos^2(x) into sin(2x)cos(x) and then proceed to use IBP lol.

Reverse chain rule lol.

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