HSC 2012 MX2 Marathon (archive) (1 Viewer)

Trebla · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

For those doing the HSC course, here is another question for you (non-HSC ppl please try to resist the temptation of answering it first):

There are three unknown numbers. It is known that the sum of these numbers is 4, the sum of the squares of these numbers is 10 and the sum of the cubes of these numbers is 16. Find the product of these three numbers.

Carrotsticks · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
What maths is that taught in? Haven't come across it yet...

Like gurmies said, it is like DEs. Recursive relationships can be homogenous or inhomogenous etc. The method of solving them is usually taught in discrete maths, despite the method of auxiliary eqns.being so similar to that of integral calc for DEs.

SpiralFlex · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

This question Deswa1 posted was in our 3U first assessment haha. (Side note: Slept in for 12 hours!)

SpiralFlex · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Thank you for that.

There is also a way to prove it using matrices, but I don't know how to type (nor can I be bothered lol) matrices.

I wanna see,

Just use

"\begin{bmatrix}
& & \\
& & \\
& &
\end{bmatrix}"

SpiralFlex · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

$\alpha+\beta+\gamma=4...(1)$

$\alpha^2+\beta^2+\gamma^2=(\sum \alpha)^2-2 \sum \alpha \beta= 10...(2)$

$\alpha^3+\beta^3+\gamma^3=16$

$(1) into (2)$

$16-2\sum \alpha \beta= 10$

$\sum \alpha \beta=3$

$x^3-4x^2+3x+c=0$

Now,

Since alpha, beta and gamma are roots

$\alpha^3=4\alpha^2-3\alpha+c$

Similarly,

$\beta^3=4\beta^2-3\beta+c$

$\gamma^3=4\gamma^2-3\gamma+c$

$16=3c+4(10)-3(4)$

$\therefore c=-4$

Hence the equation is

$x^3-4x^2+3x-4=0$

$Produck =4$

Nooblet94 · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:
fun question give it a go so I can compare your solutions to mine

[/URL][/IMG]

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $Prove true for n=1,2 assume true for $a_n$ and $a_{n-1}$ try to prove for $a_{n@plus;1}$ (Can't be bothered doing a proper proof at the moment)$\\ ~\\ \begin{align*} a_{n@plus;1}&=a_n@plus;6a_{n-1}\\ &=3^n-(-2)^n@plus;6(3^{n-1}-(-2)^{n-1})\\ &=3^{n-1}(3@plus;6)-(-2)^{n-1}(-2@plus;6)\\ &=3^{n-1}\cdot 3^2-(-2)^{n-1}\cdot (-2)^2\\ &=3^{n@plus;1}-(-2)^{n@plus;1} \end{align*}\\ $By the principle of mathematical induction blah blah blah it's true.$" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $Prove true for n=1,2 assume true for $a_n$ and $a_{n-1}$ try to prove for $a_{n+1}$ (Can't be bothered doing a proper proof at the moment)$\\ ~\\ \begin{align*} a_{n+1}&=a_n+6a_{n-1}\\ &=3^n-(-2)^n+6(3^{n-1}-(-2)^{n-1})\\ &=3^{n-1}(3+6)-(-2)^{n-1}(-2+6)\\ &=3^{n-1}\cdot 3^2-(-2)^{n-1}\cdot (-2)^2\\ &=3^{n+1}-(-2)^{n+1} \end{align*}\\ $By the principle of mathematical induction blah blah blah it's true.$" title="\\ $Prove true for n=1,2 assume true for $a_n$ and $a_{n-1}$ try to prove for $a_{n+1}$ (Can't be bothered doing a proper proof at the moment)$\\ ~\\ \begin{align*} a_{n+1}&=a_n+6a_{n-1}\\ &=3^n-(-2)^n+6(3^{n-1}-(-2)^{n-1})\\ &=3^{n-1}(3+6)-(-2)^{n-1}(-2+6)\\ &=3^{n-1}\cdot 3^2-(-2)^{n-1}\cdot (-2)^2\\ &=3^{n+1}-(-2)^{n+1} \end{align*}\\ $By the principle of mathematical induction blah blah blah it's true.$" /></a>

Also tried it Carrot's way, which I'll post in a second.

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $Rearranging the given equation we get $a_n=a_{n-1}@plus;6a_{n-2}\\ $The characteristic polynomial is $t^2-t-6=(t@plus;2)(t-3)\Rightarrow t=3,-2\\ ~\\ $Hence, the solution is in the form$\\ a_n=\alpha 3^n@plus;\beta(-2)^n\\ ~\\ $We're given$\\ a_1=3\alpha-2\beta=5~~~~(1)\\ a_2=9\alpha@plus;4\beta=5~~~~(2)\\ ~\\ $Rearranging (1) we get $ 2\beta=3\alpha-5 $ and upon subbing this into (2) we get $ 9\alpha@plus;2(3\alpha -5)=5\\ 15\alpha=15\\ \therefore \alpha=1\\ 2\beta=3-5\\ \therefore \beta=-1\\ ~\\ $Hence, the solution is $a_n=3^n-(-2)^n$" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $Rearranging the given equation we get $a_n=a_{n-1}+6a_{n-2}\\ $The characteristic polynomial is $t^2-t-6=(t+2)(t-3)\Rightarrow t=3,-2\\ ~\\ $Hence, the solution is in the form$\\ a_n=\alpha 3^n+\beta(-2)^n\\ ~\\ $We're given$\\ a_1=3\alpha-2\beta=5~~~~(1)\\ a_2=9\alpha+4\beta=5~~~~(2)\\ ~\\ $Rearranging (1) we get $ 2\beta=3\alpha-5 $ and upon subbing this into (2) we get $ 9\alpha+2(3\alpha -5)=5\\ 15\alpha=15\\ \therefore \alpha=1\\ 2\beta=3-5\\ \therefore \beta=-1\\ ~\\ $Hence, the solution is $a_n=3^n-(-2)^n$" title="\\ $Rearranging the given equation we get $a_n=a_{n-1}+6a_{n-2}\\ $The characteristic polynomial is $t^2-t-6=(t+2)(t-3)\Rightarrow t=3,-2\\ ~\\ $Hence, the solution is in the form$\\ a_n=\alpha 3^n+\beta(-2)^n\\ ~\\ $We're given$\\ a_1=3\alpha-2\beta=5~~~~(1)\\ a_2=9\alpha+4\beta=5~~~~(2)\\ ~\\ $Rearranging (1) we get $ 2\beta=3\alpha-5 $ and upon subbing this into (2) we get $ 9\alpha+2(3\alpha -5)=5\\ 15\alpha=15\\ \therefore \alpha=1\\ 2\beta=3-5\\ \therefore \beta=-1\\ ~\\ $Hence, the solution is $a_n=3^n-(-2)^n$" /></a>

Took a little longer than using induction, but I'm sure after I've used it a few more times it'll be much faster.

Nooblet94 · Mar 2, 2012

Re: 2012 HSC MX2 Marathon

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 $L_n=F_n+2F_{n-1}$

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..

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