HSC 2012 MX2 Marathon (archive) (1 Viewer)

Carrotsticks · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
Here's an interesting question I just did:

Generate a formula for the nth term of the Fibonacci series (1,1,2,3...)
Hence, find a formula for the nth term of the Lucas series (1,3,4,7...)given that $L_n=F_n+2F_{n-1}$

Very nice question.

SpiralFlex · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

$F_{n+2}=F_{n+1}+F_{n}$

$$Let F_{n}=a^n$

Forming a polynomial

$a^{n+2}=a^{n+1}+a^n$

$a^2-a-1=0$

$a=\psi, \varphi$

$\therefore F_{n}=\varphi^n$

System of linear equations,

$F_{n}=A \varphi^n +B\psi^n=1$

Since we know,

$F_{0}=A+B=1$

$F_{1}=A \varphi +B\psi=1$

Solve and arrange.

$F_{n}=\frac{\varphi^n - \psi^n}{\sqrt{5}}$

Now you can deduce Lucas my neighbour.

bleakarcher · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
$F_{n+2}=F_{n+1}+F_{n}$

$$Let F_{n}=a^n$

Forming a polynomial

$a^{n+2}=a^{n+1}+a^n$

$a^2-a-1=0$

$a=\psi, \varphi$

$\therefore F_{n}=\varphi^n$

System of linear equations,

$F_{n}=A \varphi^n +B\psi^n=1$

Since we know,

$F_{0}=A+B=1$

$F_{1}=A \varphi +B\psi=1$

Solve and arrange.

$F_{n}=\frac{\varphi^n - \psi^n}{\sqrt{5}}$

Now you can deduce Lucas my neighbour.

Spiral, you have used the symbol phi which represents the golden ratio. But what does the other symbol represent?

SpiralFlex · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Kekekekekeke

$\psi=\frac{1-\sqrt{5}}{2}=1-\varphi=-\frac{1}{\varphi}$

bleakarcher · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Kekekekekeke

$\psi=\frac{1-\sqrt{5}}{2}=1-\varphi=-\frac{1}{\varphi}$

Interesting, thanks.

largarithmic · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

A beautiful question for those who have an idea what modular arithmetic is (basically you reduce every number to its remainder when divided by a number, e.g. 13 becomes 3 modulo 5, 17 becomes 5 modulo 6. Its easy to check that if say A reduces to A', and B reduces to B', then A+B reduces to A'+B' or perhaps whatever that reduces to):

Prove that for every positive integer, there exists a Fibonacci number which is divisible by that integer.

Carrotsticks · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
A beautiful question for those who have an idea what modular arithmetic is (basically you reduce every number to its remainder when divided by a number, e.g. 13 becomes 3 modulo 5, 17 becomes 5 modulo 6. Its easy to check that if say A reduces to A', and B reduces to B', then A+B reduces to A'+B' or perhaps whatever that reduces to):

Prove that for every positive integer, there exists a Fibonacci number which is divisible by that integer.

Larg, why must u post interesting questions.just as i am about to sleep? Will try.tmr.during.work.

kingkong123 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\\textup{A relation is defined implicitly by: }x^{2}@plus;xy-2y^{2}=0.\textup{ Show that }\frac{dy}{dx}\textup{ has only two possible values: 1 and -}\frac{1}{2}. \\\textup{ Hence or otherwise, sketch the relation.}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\\textup{A relation is defined implicitly by: }x^{2}+xy-2y^{2}=0.\textup{ Show that }\frac{dy}{dx}\textup{ has only two possible values: 1 and -}\frac{1}{2}. \\\textup{ Hence or otherwise, sketch the relation.}" title="\\\textup{A relation is defined implicitly by: }x^{2}+xy-2y^{2}=0.\textup{ Show that }\frac{dy}{dx}\textup{ has only two possible values: 1 and -}\frac{1}{2}. \\\textup{ Hence or otherwise, sketch the relation.}" /></a>

bleakarcher · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
$F_{n+2}=F_{n+1}+F_{n}$

$$Let F_{n}=a^n$

Forming a polynomial

$a^{n+2}=a^{n+1}+a^n$

$a^2-a-1=0$

$a=\psi, \varphi$

$\therefore F_{n}=\varphi^n$

System of linear equations,

$F_{n}=A \varphi^n +B\psi^n=1$

Since we know,

$F_{0}=A+B=1$

$F_{1}=A \varphi +B\psi=1$

Solve and arrange.

$F_{n}=\frac{\varphi^n - \psi^n}{\sqrt{5}}$

Now you can deduce Lucas my neighbour.

Spiral, where did you get the expression after you wrote system of linear equations from?

Drongoski · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
$\\\textup{A relation is defined implicitly by: }x^{2}+xy-2y^{2}=0.\textup{ Show that }\frac{dy}{dx}\textup{ has only two possible values: 1 and -}\frac{1}{2}. \\\textup{ Hence or otherwise, sketch the relation.}$

$x^2+xy-2y^2 = 0 \implies 2x + [xy'+y] - 2\times2yy' = 0 \implies y' = \frac {dy}{dx} = \frac {2x+y}{4y-x}$

$x^2+xy-2y^2 = 0 \implies (x+2y)(x-y) = 0 \implies y=x,-\frac{1}{2}x$

By substituting each case for y, we get y' = 1 and -0.5 resp.

Nooblet94 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ x^2@plus;xy-2y^2=0\\ (x@plus;\frac{1}{2}y)^2-\frac{9}{4}y^2=0\\ (x@plus;2y)(x-y)=0~~~~\Rightarrow y=x$ or $y=-\frac{1}{2}x\\ ~\\ $Differentiating both functions we get$\\ \frac{dy}{dx}=1$ or $\frac{dy}{dx}=-\frac{1}{2}\\ ~\\ $Hence, the graph of the function is two straight lines intersecting at the origin$" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ x^2+xy-2y^2=0\\ (x+\frac{1}{2}y)^2-\frac{9}{4}y^2=0\\ (x+2y)(x-y)=0~~~~\Rightarrow y=x$ or $y=-\frac{1}{2}x\\ ~\\ $Differentiating both functions we get$\\ \frac{dy}{dx}=1$ or $\frac{dy}{dx}=-\frac{1}{2}\\ ~\\ $Hence, the graph of the function is two straight lines intersecting at the origin$" title="\\ x^2+xy-2y^2=0\\ (x+\frac{1}{2}y)^2-\frac{9}{4}y^2=0\\ (x+2y)(x-y)=0~~~~\Rightarrow y=x$ or $y=-\frac{1}{2}x\\ ~\\ $Differentiating both functions we get$\\ \frac{dy}{dx}=1$ or $\frac{dy}{dx}=-\frac{1}{2}\\ ~\\ $Hence, the graph of the function is two straight lines intersecting at the origin$" /></a>

kingkong123 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\\textup{Sketch the following locus of z on an Argand diagram. }\\\left | z @plus;2i \right |=Re(z)@plus;2i" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\\textup{Sketch the following locus of z on an Argand diagram. }\\\left | z +2i \right |=Re(z)+2i" title="\\\textup{Sketch the following locus of z on an Argand diagram. }\\\left | z +2i \right |=Re(z)+2i" /></a>

nightweaver066 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\\textup{Sketch the following locus of z on an Argand diagram. }\\\left | z @plus;2i \right |=Re(z)@plus;2i" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\\textup{Sketch the following locus of z on an Argand diagram. }\\\left | z +2i \right |=Re(z)+2i" title="\\\textup{Sketch the following locus of z on an Argand diagram. }\\\left | z +2i \right |=Re(z)+2i" /></a>

How can a modulus result in an imaginary number? lol..

SpiralFlex · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Wut?^

Think twas a typo.

bleakarcher said:
Spiral, where did you get the expression after you wrote system of linear equations from?

Substituted the F0, F1

bleakarcher · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
$F_{n+2}=F_{n+1}+F_{n}$

$$Let F_{n}=a^n$

Forming a polynomial

$a^{n+2}=a^{n+1}+a^n$

$a^2-a-1=0$

$a=\psi, \varphi$

$\therefore F_{n}=\varphi^n$

System of linear equations,

$F_{n}=A \varphi^n +B\psi^n=1$

Since we know,

$F_{0}=A+B=1$

$F_{1}=A \varphi +B\psi=1$

Solve and arrange.

$F_{n}=\frac{\varphi^n - \psi^n}{\sqrt{5}}$

Now you can deduce Lucas my neighbour.

Why is it that when I sub n=0 into F(n) I get 1/sqrt(5) when it is meant to be one?

kingkong123 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
How can a modulus result in an imaginary number? lol..

trueeee. there was a typo in the exercise. i think it should be Re(z) + 2 not 2i. In that case it will be a parabola, focal length 1, vertex (-1,-2) , concave up. right?

cutemouse · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

A bit of several variables calculus...

(i) Use the method of Lagrange multipliers, or otherwise, to find the maximum value of f(x,y)=4xy on the ellipse x^2 + 4y^2=4.

(ii) Give a geometrical interpretation of the result in (i).

Aesytic · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
trueeee. there was a typo in the exercise. i think it should be Re(z) + 2 not 2i. In that case it will be a parabola, focal length 1, vertex (-1,-2) , concave up. right?

mine ended up being a sideways parabola, so i had one that was "concave right"

D94 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Aesytic said:
mine ended up being a sideways parabola, so i had one that was "concave right"

This is right.

Trebla · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Another reminder that this thread is intended for HSC students. If you want to post some uni maths questions, go to the extracurricular topics forum.

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