HSC 2013 MX2 Marathon (archive) (7 Viewers)

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.

HSC 2013 MX2 Marathon (archive) (7 Viewers)

Account Closed

what is that?It is Cowpea

Heroic!

what is that?It is Cowpea

Well-Known Member

Heroic!

Well-Known Member

This too shall pass

what is that?It is Cowpea

Heroic!

This too shall pass

what is that?It is Cowpea

This too shall pass

what is that?It is Cowpea

what is that?It is Cowpea

This too shall pass

what is that?It is Cowpea

This too shall pass

what is that?It is Cowpea

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 7)