HSC 2013 MX2 Marathon (archive) (1 Viewer)

cutemouse · Dec 23, 2012

Re: HSC 2013 4U Marathon

I don't think Ellipses with centre other than O are in the syllabus...

RealiseNothing · Dec 23, 2012

Re: HSC 2013 4U Marathon

cutemouse said:
I don't think Ellipses with centre other than O are in the syllabus...

I'm pretty sure they are? Unless that's only certain questions like find the centre of the ellipse given this equation etc.

HeroicPandas · Dec 23, 2012

Re: HSC 2013 4U Marathon

$A sequence of whole numbers \left \{ u_0, u_1, u_2,...,u_n,... \right \}$ $is generated by u_{r+2}=3r_{r+1}-2u_r$ $ where r = 0, 1, 2,... Defining u_{r+1} - u_r = d_r, (r\geq 0)$ $ prove that \\ i) d_{r+1} = 2d_r \\$ $ii) d_{n-1} = 2^{n-1}$ $d_0$

$Difficulty: Moderate-hard$

RealiseNothing · Dec 23, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$A sequence of whole numbers \left \{ u_0, u_1, u_2,...,u_n,... \right \}$ $is generated by u_{r+2}=3r_{r+1}-2u_r$ $ where r = 0, 1, 2,... Defining u_{r+1} - u_r = d_r, (r\geq 0)$ $ prove that \\ i) d_{r+1} = 2d_r \\$ $ii) d_{n-1} = 2^{n-1}$ $d_0$

$Difficulty: Moderate-hard$

Is the sequence defined correctly?

SpiralFlex · Dec 23, 2012

Re: HSC 2013 4U Marathon

cutemouse said:
I don't think Ellipses with centre other than O are in the syllabus...

Technically you see them more often in Trials than the actual HSC. They would fall under graph transformations anyway.

HeroicPandas · Dec 23, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Is the sequence defined correctly?

http://4unitmaths.com/SBHS1986.pdf

Last question (8, b)

seanieg89 · Dec 23, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
$Prove by induction or otherwise that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is $n$-times differentiable at $a\in\mathbb{R}$ if and only if there exists a polynomial $P$ of degree $n$ with: \\ \\ $\lim_{x\rightarrow a}\frac{f(x)-P(x-a)}{|x-a|^n}=0 \textrm{ and } f(a)=P(0).$ \\ (So $P$ is a good polynomial approximation to $f$ near $a$.) \\ \\Find an expression for this polynomial in terms of the derivatives of $f(x)$ at $a$.$

Perhaps this was a little difficult for a marathon question, although the needed tools are entirely within the HSC syllabus. I will post a solution a little later to illustrate how some different ideas in calculus interact.

Sy123 · Dec 23, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$A sequence of whole numbers \left \{ u_0, u_1, u_2,...,u_n,... \right \}$ $is generated by u_{r+2}=3r_{r+1}-2u_r$ $ where r = 0, 1, 2,... Defining u_{r+1} - u_r = d_r, (r\geq 0)$ $ prove that \\ i) d_{r+1} = 2d_r \\$ $ii) d_{n-1} = 2^{n-1}$ $d_0$

$Difficulty: Moderate-hard$

$$i) $d_{r+1}=u_{r+2}-u_{r+1}$

$d_r=u_{r+1}-u_r \\ 2d_r=2u_{r+1}-2u_r \\ 2d_r+u_{r+1}=3u_{r+1}-2u_r = u_{r+2} \\ \therefore 2d_r=u_{r+2}-u_{r+1}= d_{r+1}$

$$ii) $ d_{n-1}=2d_{n-2}=2(2d_{n-3})=2(2(2d_{n-4}))= 2(2...(d_0)..) = 2^{n-1}d_0$

RealiseNothing · Dec 23, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
http://4unitmaths.com/SBHS1986.pdf

Last question (8, b)

ok so it is defined wrong.

HeroicPandas · Dec 23, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) $d_{r+1}=u_{r+2}-u_{r+1}$

$d_r=u_{r+1}-u_r \\ 2d_r=2u_{r+1}-2u_r \\ 2d_r+u_{r+1}=3u_{r+1}-2u_r = u_{r+2} \\ \therefore 2d_r=u_{r+2}-u_{r+1}= d_{r+1}$

$$ii) $ d_{n-1}=2d_{n-2}=2(2d_{n-3})=2(2(2d_{n-4}))= 2(2...(d_0)..) = 2^{n-1}d_0$

Good work! (i couldn't do part ii lol)

Sy123 · Dec 26, 2012

Re: HSC 2013 4U Marathon

$$Consider the polynomial$ \\ \\ S(x)=Ax^3+Bx^2+Bx+A$

$$Show that if all roots are real, then $ \\ \\ \frac{B}{A} \geq 3 \ \ \ \ \frac{B}{A} \leq -1$

(keep in mind, that the astroid question is still left unanswered, it can be answered with only knowledge of implicit differentiation which you learn in Graphs.)

RealiseNothing · Dec 26, 2012

Re: HSC 2013 4U Marathon

I keep getting A=1 as the only possible value of A for the polynomial to have real roots.

Either something is seriously messed up with my working or it's surprisingly right.

Sy123 · Dec 26, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I keep getting A=1 as the only possible value of A for the polynomial to have real roots.

Either something is seriously messed up with my working or it's surprisingly right.

Setting A=2 B=7
Gives an equation with 3 real roots according to Geogebra.

What are the steps of your working?

RealiseNothing · Dec 26, 2012

Re: HSC 2013 4U Marathon

Wait, I just realised, I did something stupid in my first line of working :/

I went to type it up and was like "ohhhhhhhhhhhhh... facepalm".

RealiseNothing · Dec 26, 2012

Re: HSC 2013 4U Marathon

By inspection -1 is a root, therefore:

$S(x)=(x+1)($some quadratic$)$

Equating co-efficients gives:

$(x+1)(Ax^2+(B-A)x+A)$

Solving the quadratic for real roots gives:

$(B-A)^2 - 4A^2 \geq 0$

$(B-A)^2 \geq 4A^2$

$B-A \geq 2A$ or $B-A \leq -2A$

Simplifying each one gives:

$\frac{B}{A} \geq 3$ or $\frac{B}{A} \leq -1$

Sy123 · Dec 26, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
By inspection -1 is a root, therefore:

$S(x)=(x+1)($some quadratic$)$

Equating co-efficients gives:

$(x+1)(Ax^2+(B-A)x+A)$

Solving the quadratic for real roots gives:

$(B-A)^2 - 4A^2 \geq 0$

$(B-A)^2 \geq 4A^2$

$B-A \geq 2A$ or $B-A \leq -2A$

Simplifying each one gives:

$\frac{B}{A} \geq 3$ or $\frac{B}{A} \leq -1$

Nice work, its easier than my method as well, I let the roots be -1, alpha, 1/alpha then used sum of roots, found the quadratic in alpha and discriminant.

RealiseNothing · Dec 26, 2012

Re: HSC 2013 4U Marathon

Your question gave me an idea for a question, let me try it out and I'll post it if it works.

Sy123 · Dec 26, 2012

Re: HSC 2013 4U Marathon

$$Consider the Triangular Numbers, defined by the recurrence sequence $ \\ \\ T_n=n+T_{n-1} \ \ \ \ T_1=1$

$$i) By considering the identity $ \\ \\ j^2(j+1)^2-j^2(j-1)^2 = 4j^3 \\ \\ $Show that$ \ \ \ \ \sum_{k=1}^n k^3=\frac{n^2}{4}(n+1)^2$

$$ii) Hence show that $ \ \ \ \ T_n^2 = \sum_{k=1}^n k^3$

(Just thought it was a cool result)

RealiseNothing · Dec 26, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Consider the Triangular Numbers, defined by the recurrence sequence $ \\ \\ T_n=n+T_{n-1} \ \ \ \ T_1=1$

$$i) By considering the identity $ \\ \\ j^2(j+1)^2-j^2(j-1)^2 = 4j^3 \\ \\ $Show that$ \ \ \ \ \sum_{k=1}^n k^3=\frac{n^2}{4}(n+1)^2$

$$ii) Hence show that $ \ \ \ \ T_n^2 = \sum_{k=1}^n k^3$

(Just thought it was a cool result)

I've actually played around with this result a bit before. I'll post up a question linking heaps of things together, lemme type it up as a continue on from your question if you don't mind.

Sy123 · Dec 26, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I've actually played around with this result a bit before. I'll post up a question linking heaps of things together, lemme type it up as a continue on from your question if you don't mind.

Sure go ahead.

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