HSC 2012 MX1 Marathon #1 (archive) (2 Viewers)

Examine · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
1. Using Latex. You use this to write up equations:
www.codecogs.com/latex/eqneditor.php

Then you input them in to this forum by inserting whatever you typed in to these brackets $*insert latex stuff here* [./tex] <- without the dot. 2. It shouldn't just be thought of just like that. Imagine the graph of y = |x|. It is less than 9 when -9 < x < 9.[/QUOTE] Oh yes I see now, thanks.$

Examine · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

All this maths seems rather interesting. Can't wait for Extension 1 Prelims and HSC.

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
All this maths seems rather interesting. Can't wait for Extension 1 Prelims and HSC.

*is

hayabusaboston · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
*is

lol right on man

deswa1 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
lol nobody has yet tried my question ='(

Don't worry Mr. Carrotsticks. I just have to finish my belonging essay (I know

...) and then I'll have a go.

SpiralFlex · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

I leave you with more things to ponder.

$L'Hopital's Rule states that for a given function f, g:$

$\lim_{x \to a }f(x)=\lim_{x \to a }g(x)=0, \pm \infty$

$\textup{and if }\lim_{x \to a }\frac{f'(x)}{g'(x)} \ \textup{exists then}$

$\fbox{\lim_{x \to a }\frac{f(x)}{g(x)}=\lim_{x \to a }\frac{f'(x)}{g'(x)}}$

$By using L'Hopital's Rule,$

$\textup{Evaluate:} \lim_{x\rightarrow 0}\frac{x\sin x}{1-\cos x}$

$\textup{Hence evaluate:} \lim_{x\rightarrow 0}(1-\cos x)^\frac{1}{\ln x}$

Have to dash and do some Physics, I will attempt the questions afterwards. (Be back at 9 PM) Keep them coming!

tywebb · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

2 and e²

Next question:

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Here you go

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
Next question:

Wait a second... this looks like the Ruse Extension 2 2011 paper lol.

tywebb · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Correct! But Ext. 1 students can do it.

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
Correct! But Ext. 1 students can do it.

Use the tan(A+B) thing where tanA=5 and tanB = 3, but twice. After some simplification, we have tan(A+B+C) = 0.

Thus A + B + C = arctan(0) = pi

I'm only saying this because my tablet just broke down lol.

jdnRof · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

I'd be very careful about tanning and then arc tanning it since you need to consider the restrictions between -pi/2 and pi/2. Probably should include an argument about it.

Just to clarify carrotsticks, is one of your steps:

arctan (-4/7) + arctan (4/7)?

RishBonjour · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Did you guys finish the whole course already? FUCK I'm SO BEHIND

ohh yeah, and lazy

tywebb · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Here's James Ruse's solution:

But it can be generalised as follows:

$\ \newline\textstyle\text{Show that for }x>\sqrt{y^2+1},\ \ \tan^{-1}(x-y)+\tan^{-1}(x+y)+\tan^{-1}{2x\over{x^2-y^2-1}}=\pi$

And of course, James Ruse's result follows by letting x=4, y=1.

zeebobDD · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Last Q of one of our mx1 papers ;

Two circles intersect at A and B. P is a point on the first circle and Q is a point on the second circle such that PAQ is a straight line. C is a point on the second circle. The tangent to the first circle at P meets QC produced to R.

Prove that PBCR is a cyclic quadrilateral

Sanical · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

zeebobDD said:
Last Q of one of our mx1 papers ;

Two circles intersect at A and B. P is a point on the first circle and Q is a point on the second circle such that PAQ is a straight line. C is a point on the second circle. The tangent to the first circle at P meets QC produced to R.

Prove that PBCR is a cyclic quadrilateral

Construct BA and point D (I forgot to write point Q which is produced from PA and lies on second circle)
let /_DPB = x
/_PAB = x (angle between tangent and chord equal to angle in alternate segment)
/_BAQ = 180 - x (adjacent supplementary angles)
/_BCQ = 180 - x (angles equal on the same arc)
/_RCB = x (adjacent supplementary angles)

Therefore PBCR is cyclic since angle is equal to opposite external angle /_RCB=/_DPB

Sanical · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

New question:

cutemouse · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

tywebb said:
2 and e²

Should really justify the 3rd last line being because the exponential function is continuous.

DL9559 · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

Yahoo hoo! This one looks fun.

Hmmm...

1/2(ln(x^2+4)) + tan*-1 (x/2) + c (BTW that tan thing is meant to be inverse tan.)

Carrotsticks · Jan 4, 2012

Re: 2012 HSC MX1 Marathon

DL9559 said:
Yahoo hoo! This one looks fun.

Hmmm...

1/2(ln(x^2+4)) + tan*-1 (x/2) + c (BTW that tan thing is meant to be inverse tan.)

For the record, to indicate inverse tan, we use arctan (x)

ie: arctan (1) = pi/4

EDIT: I'm a sped and typed it wrong way around.

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