HSC 2012 MX2 Marathon (archive) (2 Viewers)

Nooblet94 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Using the fact that <a href="http://www.codecogs.com/eqnedit.php?latex=arg\frac{z_{1}}{z_{2}}=argz_{1}-argz_{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" title="arg\frac{z_{1}}{z_{2}}=argz^{_{1}}-argz_{2}" /></a>

Prove that <a href="http://www.codecogs.com/eqnedit.php?latex=tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" title="tan^{-1}2-tan^{-1}\frac{1}{3}=\frac{\pi }{4}" /></a>

deswa1 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
were u asking this because u didnt know how or just as a forum question? nice question

Nah, I already did it. It was in a question by topic book but I thought it was a neat question so I posted it

Carrotsticks · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

This problem can be further generalised.

Here is an extension to deswa1's problem:

Show that all the values satisfying the condition...

$\tan^{-1} (x) + \tan^{-1} (y) = \frac{\pi}{4}$

... lie on the hyperbola:

$y= \frac{1-x}{1+x}$

deswa1 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Before I have a go, is it meant to be inverse tan(x) MINUS inverse tan(y) or is the question you posted correct?

nightweaver066 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
This problem can be further generalised.

Here is an extension to deswa1's problem:

Show that all the values satisfying the condition...

$\tan^{-1} (x) + \tan^{-1} (y) = \frac{\pi}{4}$

... lie on the hyperbola:

$y= \frac{1-x}{1+x}$

$\tan^{-1} (x) + \tan^{-1} (y) = \frac{\pi}{4}$

$\tan^{-1}(\frac{x + y}{1 - xy}) = \frac{\pi}{4}$

$\frac{x + y}{1 - xy} = 1$

$x + y = 1 - xy$

$y(1 + x) = 1 - x$

$y = \frac{1 - x}{1 + x}$

Carrotsticks · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Before I have a go, is it meant to be inverse tan(x) MINUS inverse tan(y) or is the question you posted correct?

The formula you have with the minus sign is a specific case for the generalised version:

$\tan^{-1} (x) + \tan^{-1} (y) = \frac{\pi}{4} \\\\\\ $Let $ x=2 $ and $ y= -\frac{1}{3} $ (which lies on the curve in the question):$ \\\\\\ \tan^{-1} (2) + \tan^{-1} \left (-\frac{1}{3} \right) = \frac{\pi}{4} \\\\\\ \tan^{-1} (2) - \tan^{-1} \left (\frac{1}{3} \right ) = \frac{\pi}{4} $ since $ \tan(-x) = -\tan(x) $ because the tan function is odd.$$

Good solution nightweaver066.

Carrotsticks · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Here's a bit of a tricky question, similar to that in the HSC (forgot which year, I think 2007?)

$Consider the curve $ y=x^{n}ln(x^{n}) $ for $ x>0 \\\\ $Let its stationary point be $ x_a $ and its point of inflexion be $ x_b. \\\\ $Show that $ \lim_{n \rightarrow \infty} \left (\frac{x_b}{x_a} \right ) = 1 \\\\ $Interpret this result geometrically.$

bleakarcher · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Dont you mean the limit as n approaches infinity?

SpiralFlex · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

$\frac{x+y}{1-xy}=1$

$1-xy=x+y$

$y(x+1)=1-x$

$\therefore y=\frac{1-x}{1+x}$

Edit: Damn BOS NEVER LINKS ME TO THE LAST PAGE. >_>

Carrotsticks · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
Dont you mean the limit as n approaches infinity?

Thanks for that correction. It's become habit for me to type x -> infinity now haha.

nightweaver066 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Here's a bit of a tricky question, similar to that in the HSC (forgot which year, I think 2007?)

$Consider the curve $ y=x^{n}ln(x^{n}) $ for $ x>0 \\\\ $Let its stationary point be $ x_a $ and its point of inflexion be $ x_b. \\\\ $Show that $ \lim_{n \rightarrow \infty} \left (\frac{x_b}{x_a} \right ) = 1 \\\\ $Interpret this result geometrically.$

After tedious working out,

$\frac{x_b}{x_a} = e^\frac{1 - 2n^2}{n^2(n-1)}$

As n -> infinity, $\frac{1 - 2n^2}{n^2(n-1)} \rightarrow 0$

$\therefore e^\frac{1 - 2n^2}{n^2(n-1)} \rightarrow 1$

$\therefore \lim_{n \rightarrow \infty} \left (\frac{x_b}{x_a} \right ) = 1$

This means that as n approaches an infinitely large number, the stationary point and inflexion point will coincide (ignoring x = 0) and there will be a horizontal inflexion point. (is this right? lol)

bleakarcher · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Geometrically it means that as n approaches positive infinity the limiting ratio of the x-coordinates of the inflexion point to the stationary point of the curve y=x^n*log[e][x^n] is one?

bleakarcher · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Disregard my answer, this.

Carrotsticks · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
This means that as n approaches an infinitely large number, the stationary point and inflexion point will coincide (ignoring x = 0) and there will be a horizontal inflexion point. (is this right? lol)

Good argument. Indeed they tend to the same x value, and therefore the same y value as well.

There won't strictly be a horizontal inflexion point, though I can understand where you're coming from (point of inflexion + stationary point = HPOI).

bleakarcher said:
Geometrically it means that as n approaches positive infinity the limiting ratio of the x-coordinates of the inflexion point to the stationary point of the curve y=x^n*log[e][x^n] is one?

You just repeated the question basically haha. A stronger argument is required.

nightweaver066 · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Good argument. Indeed they tend to the same x value, and therefore the same y value as well.

There won't be a horizontal inflexion point, though I can understand where you're coming from (point of inflexion + stationary point = HPOI).

You just repeated the question basically haha. A stronger argument is required.

So just saying that they will tend to the same (x, y) values would be enough?

bleakarcher · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

I realised lol.

Carrotsticks · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
So just saying that they will tend to the same (x, y) values would be enough?

Yes.

Somebody post a new question (or answer seanieg's question).

largarithmic · Jan 23, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
An easy polynomials question:

$$Let $x_1,\ldots,x_{n+1}$ be $n+1$ distinct real numbers. Let $P(x)$ and $Q(x)$ be real polynomials of degree $n$ such that $P(x_j)=Q(x_j)$ for $j=1,\ldots,n+1.$ Prove that $P(x)=Q(x)$ for all real $x.$

Consider R(x) = P(x) - Q(x): note that it is of degree at most n.

Now suppose R(x) is not the zero polynomial; in this case it has at most n roots (from unique factorisation, which is provable from the division algorithm). However it clearly has n+1 roots x1, x2, ..., xn+1. Therefore R(x) is the zero polynomial, so P(x) is identically equal to Q(x).

Here's a decently nice question: by considering its complex roots, find all polynomials P(x) such that:
$P(x)^2 = P(x^2)$ for all complex x.

seanieg89 · Jan 24, 2012

Re: 2012 HSC MX2 Marathon

$Assume that $P(z)$ is not the zero polynomial.\\Suppose $\alpha\neq 0$ is a root of $P(z)$ with minimal argument in the range $0<\arg(\alpha)<2\pi$.\\ Then taking the principal square root of $\alpha$, we have $P(\alpha^{1/2})^2=P(\alpha)=0$ and $P(\alpha^{1/2})=0$. Since $\arg(\alpha^{1/2})=\frac{1}{2}\arg(\alpha)$, this contradicts our choice of $\alpha$. Hence every nonzero root of $P(z)$ lies on the positive real axis.\\ But any positive real root $x$ can be written as $(-r)^2$ for some positive real $r$. Whence $P(-r)^2=P(x)=0$ and $-r$ is a negative root of $P$, a contradiction.\\ So $P(z)=cz^n$ for some constant $c$ and some $n>0$ (or is constant and has no roots at all!). Running through the cases we arrive at the answer:\\$P(z)=z^n,1,0 \quad{n>0}.$$

Carrotsticks · Feb 6, 2012

Re: 2012 HSC MX2 Marathon

What happened to this thread?

$\\ $Show that the two curves $ x^2-y^2=c^2 $ and $ xy=c $ intersect at right angles.$$

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