HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Nov 4, 2012

Here, post questions that you have made/found for Extension 2 Maths, so we can all benefit from new experience to different questions

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The 2013 Question Compilation

Note the compilation is not in its final form, it is currently in the form of 5 PDFs, each of them containing questions, each PDF has a Difficulty level, from 1-5. This is NOT the FINAL version.

The difficulty is scaled so that Q15-Q16 is around Difficulty 2.7 or so. You may feel that some questions belong in a higher or lower difficulty category, keep in mind the difficulty recommendations are only subjective

I do plan on compiling it properly into a professional document, also sorted by topic. Currently some questions have been omitted.

HSC 4U 2013 Marathon Questions

The 2014 Advanced Questions have been moved to this thread
http://community.boredofstudies.org...2/hsc-2014-mx2-marathon-advanced-archive.html

Sy123 · Nov 4, 2012

Re: HSC 2013 4U Marathon

Stick to Complex Numbers/Polynomials at first since not many people have done other topics yet.

I will start off:

$Take$ \ \ \ \ \ z \ \ \ \text{as a complex number. And n as any integer greater than 2. Factorise } \\ \\ z^n-1=0 \ \ \ $ Using $z-1$ as a factor$\\ \\ $Hence find, the number of values $1+z+z^2+...+z^{n-1}$ can take$

EDIT: I made the question a little easier.
EDIT 2: Made a correction in the restriction of n.

barbernator · Nov 4, 2012

Re: HSC 2013 4U Marathon

good stuff sy, I challenge you to keep this running for the whole year.

Sy123 · Nov 4, 2012

Re: HSC 2013 4U Marathon

barbernator said:
good stuff sy, I challenge you to keep this running for the whole year.

Heh, Ill try man hopefully it will be more active than the 2012 3U one.

HeroicPandas · Nov 4, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Stick to Complex Numbers/Polynomials at first since not many people have done other topics yet.

I will start off:

$Take$ \ \ \ \ \ z \ \ \ \text{as a complex number. And n as any integer greater than 2. Factorise } \\ \\ z^n-1=0 \ \ \ $ Using $z-1$ as a factor$\\ \\ $Hence find, the number of values $1+z+z^2+...+z^{n-1}$ can take$

EDIT: I made the question a little easier.

z^n -1= (z-1)[1+z^2+z^3+......+z^(n-1)]

The number of values $$1+z+z^2+...+z^{n-1}$ can take is n>3?

EDIT: BOLD STUFF

distinction · Nov 4, 2012

Re: HSC 2013 4U Marathon

From Patel - Exercise 4O Q6

P is a variable point on the line x=4 and OPQR is a rectangle in which the length of OP is twice the length of OR. Find the locus of S in Cartesian form, where S is the point of intersection of the diagonals.

Sy123 · Nov 4, 2012

Re: HSC 2013 4U Marathon

Call the complex number P: z

Hence, since OP is perpendicular to OR (and half its size)

$OR:\frac{iz}{2}$

To find the midpoint S, we add and divide by 2:

$S:\frac{z}{4}(2+i)$

Define $z=4+iy$

$S=\frac{4+iy}{4}(2+i) \\ \\ S=\frac{8-y}{4}+i(\frac{y+2}{2})$

Finding in cartesian form:

$x=\frac{8-y}{4}$
$y=8-4x$

$\square$

InfinityBlade · Nov 5, 2012

Re: HSC 2013 4U Marathon

Hello guys and girls I'm new. May I ask how I enable LaTeX on this forum?

InfinityBlade · Nov 5, 2012

Re: HSC 2013 4U Marathon

Edit: Found a user called 'Jet' who made a wonderful guide on how to implement LaTeX onto the forum.

$$Let$\;P(x)=x^4+a_{1}x^3+a_{2}x^2+a_{3}x+a_{4}\;$where for$\;a_{k}, k=1...4 \in \mathbb{R}. \;$Suppose$\; z_{1}=\cos \theta_{1}+i\sin \theta_{1} \;$and$\;z_{2}=\cos \theta_{2}+i\sin \theta_{2} \;$are two roots of$\; P(x)=0,\;$where$\; 0<\theta_{1}<\theta_{2}<\pi.$

$(a) Show that:$

$$(I.)$\; P(x)=(x^2-2x\cos \theta_{1}+1)(x^2-2x\cos \theta_{2}+1)$

$$(II.)$\; P'(x)=2P(x)[\frac{x-\cos \theta_{1}}{x^2-2x\cos \theta_{1}+1}+\frac{x-\cos \theta_{2}}{x^2-2x\cos \theta_{2}+1}]$

$(b)$\; $Suppose w is a root of$\;P(x).\;$Show that:$

$$(I.)$\; \frac{P(x)}{x-w}=x^3+(w+a_{1})x^2+(w^2+a_{1}w+a_{2})x+(w^3+a_{1}w^2+a_{2}w+a_{3})$

$$(c) Define$\; S_{n}=z_{1}^n+(\overline{z_{1}})^n + z_{2}^n+(\overline{z_{2}})^n$

$(I.)$\; $Deduce that,$

$P'(x)=4x^3+(S_{1}+4a_{1})x^2+(S_{2}+a_{1}S_{1}+4a_{2})x+(S_{3}+S_{2}a_{1}+S_{1}a_{2}+4a_{3})$

$(II.) Hence finally show that,$

$S_{n}+a_{1}S_{n-1}+...+a_{n-1}S_{1}+na_{n}=0, \;$for$\; n=1,2,3,4$

$$Footnotes: Hint for C(I.):$\; \frac{2(x-\cos \theta_{r})}{x^2-2x\cos \theta_{r}+1} \equiv \frac{1}{x-z_{r}}+\frac{1}{x-\overline{z}_{r}}\;$for$\;r=1,2.$

Sy123 · Nov 5, 2012

Re: HSC 2013 4U Marathon

Alright I figured it out, some of them I arrived at constructively (not any of the deduce ones)
And I will not get into the working out in depth, some of it is quite tedious

Now:

$a) \ \ i) z_1, z_2, \bar{z_1}, \bar{z_2} \ \ \rightarrow $roots$$

$\therefore P(x)=(x-z_1)(x-z_2)(x-\bar{z_1})(x-\bar{z_2})$

$$Sub in the mod-arg form for each, and expand two brackets at a time, with $ z_1 $ and its conjugate $ z_2 $ and its conjugate, the arrived expression is$ \ \ \ \ P(x)=(x^2-2x\cos \theta_1+1)(x^2-2x\cos \theta_2+1)$

$ii) \ \ P'(x)=(2x-2\cos \theta_1)(x^2-2x\cos \theta_2+1)+(2x-2\cos \theta_2)(x^2-2x\cos \theta_1 +1) \ \ \ \ $by product rule$$

$P'(x)=2[\frac{(x-\cos \theta_1)(x^2-2x \cos \theta_2+1)(x^2-2x \cos \theta_1+1)}{(x^2-2x\cos \theta_1+1)}+\frac{(x-\cos \theta_2)(x^2-2x\cos \theta_1+1)(x^2-2x\cos \theta_2+1)}{x^2-2x\cos \theta_2+1}]$

$P'(x)=2P(x)[\frac{x-\cos \theta_1}{x^2-2x\cos \theta_1+1}+\frac{x-\cos \theta_2}{x^2-2x\cos \theta_2+1}]$

For b, I could not think of a good way of doing it, I just showed that expanding (x-w) and right hand side is indeed P(x)

$b) (x-w)(x^3+(w+a_1)x^2+(w^2+a_1w+a_2)x+(w^3+a_1w^2+a_2w+a_3) \\ \\ = x^4+wx^3+a_1x^3+w^2 x^2+a_1wx^2+a_2 x^2+w^3 x+a_1w^2 x+a_2 wx+a_3x - wx^3-w^2x^2-wa_1x^2-w^3x-a_1w^2x-wa_2x-w^4-a_1w^3-a_2w^2-a_3w$

$=x^4+a_1x^3+a_2x^2+a_3x-(w^4+a_1w^3+a_2w^2+a_3w) \ \ \ \ \ [1]$

$P(w)=w^4+a_1w^3+a_2w^2+a_3w+a_4$
$P(w)=0$
$\therefore -a_4=w^4+a_1w^3+a_2w^2+a_3w$

$\therefore $into eqn 1 $ =x^4+a_1x^3+a_2x^2+a_3x+a_4=P(x)$

$\therefore \frac{P(x)}{x-w} = x^3+(w+a_1)x^2+(w^2+a_1w+a_2)x+(w^3+a_1w^2+a_2w+a_3)$

$P'(x)=P(x)[\frac{1}{x-z_1}+\frac{1}{x-\bar{z_1}}+\frac{1}{x-z_2}+\frac{1}{x-\bar{z_2}}]$

Make a common denominator, the denominator then happens to be P(x) itself, it then cancels out with the P(x) at the front, then we expand everything on the numerator, we should get (after various manipulations)

$P'(x)=4x^3+(S_1+4a_1)x^2+(S_2+a_1S+1+4a^2)x+(S_3+S_2a+1+S_1a_2+4_3)$

My original method was showing that

$S_1=-a_1 $sum of roots$ \\ \\ S_1+4a_1=3a_1$
So I constructed each co-efficient by using sum sort of sum of roots rule, but I thought it was too constructive.

$ii)n=1 \rightarrow $co-efficient of x squared in P'(x)$ \\ \\ S_1+4a_1=3a_1 \\ \therefore S_1+a_1=0$

$n=2 \rightarrow $co-efficient of x in P'(x)$ \\ \\ S_2+a_1S_1+4a_2=2a_2 \\ \therefore S_2+a_1S_1+2a_2=0$

$n=3 \rightarrow $constant of P'(x)$ \\ \\ S_3+S_2a_1+S_1a_2+4a_3=a_3 \\ \therefore S_3+S_2a_1+S_1a_2+3a_3=0$

$n=4 \rightarrow $Following the pattern from above S_4+S_3a_1+S_2a_2+S_1a_3+4a_4=0 \ \ \ $equals zero within P'(x)$$

Not sure if I am allowed to assume pattern for n=4 however. However that is my attempt at the question. Will post one soon enough

SpiralFlex · Nov 5, 2012

Re: HSC 2013 4U Marathon

No product rule is required for II. You can use logarithmic differentiation by proving P'(x)/P(x)=Blah blah and shifting it to one side.

For the next one, use P(x)-P(w)=0

Sy123 · Nov 6, 2012

Re: HSC 2013 4U Marathon

An easier one:

$\text{If } z $ is a complex number on the Argand diagram$ \\ \\ \text{Describe the locus } \frac{1}{z}+\frac{1}{\bar{z}}=1$

(remember to post a question after you answer)

SpiralFlex · Nov 6, 2012

Re: HSC 2013 4U Marathon

Good question Sy this may separate people that just do mindless algebra.

Sy123 · Nov 6, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Stick to Complex Numbers/Polynomials at first since not many people have done other topics yet.

I will start off:

$Take$ \ \ \ \ \ z \ \ \ \text{as a complex number. And n as any integer greater than 2. Factorise } \\ \\ z^n-1=0 \ \ \ $ Using $z-1$ as a factor$\\ \\ $Hence find, the number of values $1+z+z^2+...+z^{n-1}$ can take$

EDIT: I made the question a little easier.
EDIT 2: Made a correction in the restriction of n.

Factorise:

$z^n-1=0 \\ \\ (z-1)(1+z+z^2+...+z^{n-1})=0$

By null factor law:

1+z+z^2+...+z^(n-1)=0

But in order for z^n-1=0, z=1 as well. Hence the other solution is 1+1+1^2+...+1^(n-1)=n

Hence two values that it can take, 0 and n

Carrotsticks · Nov 7, 2012

Sy123 said:
Factorise:

$z^n-1=0 \\ \\ (z-1)(1+z+z^2+...+z^{n-1})=0$

By null factor law:

1+z+z^2+...+z^(n-1)=0

But in order for z^n-1=0, z=1 as well. Hence the other solution is 1+1+1^2+...+1^(n-1)=n

Hence two values that it can take, 0 and n

Perhaps an explanation regarding the complex values of Z would make your solution a bit more complete. Z=1 is the trivial real solution, but it may be worth considering the non-trivial roots too.

Sy123 · Nov 7, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
Perhaps an explanation regarding the complex values of Z would make your solution a bit more complete. Z=1 is the trivial real solution, but it may be worth considering the non-trivial roots too.

Ah yes this is true. Is this explanation good enough?

All roots of z can be represented with:

$z_k=\cos \frac{2\pi}{k}+i\sin \frac{2\pi}{k} \ \ \ \ (=cis\frac{2\pi}{k})$

Now when any of these values undertake z (EXCEPT the trivial root when k=0, which is covered above (the original solution))

$z-1 \neq 0$
$1+z+z^2+...+z^{n-1}=0$

This is for ANY integer k, (excluding zero). Hence for any integer k excluding zero, the geometric series in z is always 0
In the trivial case where k=0, hence z=1m then the series initiates a value of n.

Hence for any complex z, there always is 2 solutions , 0 and n (although there are (n-1) different values of z that will make the series in z equal 0)

(keep in mind guys there is still a question unanswered)

hscwav2012 · Nov 8, 2012

Re: HSC 2013 4U Marathon

Inequality: Prove a^2+b^2>2ab

Carrotsticks · Nov 8, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Ah yes this is true. Is this explanation good enough?

All roots of z can be represented with:

$z_k=\cos \frac{2\pi}{k}+i\sin \frac{2\pi}{k} \ \ \ \ (=cis\frac{2\pi}{k})$

Now when any of these values undertake z (EXCEPT the trivial root when k=0, which is covered above (the original solution))

$z-1 \neq 0$
$1+z+z^2+...+z^{n-1}=0$

This is for ANY integer k, (excluding zero). Hence for any integer k excluding zero, the geometric series in z is always 0
In the trivial case where k=0, hence z=1m then the series initiates a value of n.

Hence for any complex z, there always is 2 solutions , 0 and n (although there are (n-1) different values of z that will make the series in z equal 0)

(keep in mind guys there is still a question unanswered)

That is more complete =) Nice.

InfinityBlade · Nov 9, 2012

Re: HSC 2013 4U Marathon

hscwav2012 said:
Inequality: Prove a^2+b^2>2ab

The inequality isn't correct for all cases.

Carrotsticks · Nov 9, 2012

Re: HSC 2013 4U Marathon

It is true for all real a and b (assuming a =/= b since there is a strict inequality there), since (a-b)^2 > 0 for all real a and b where a=/=b.

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