HSC 2013 MX2 Marathon (archive) (8 Viewers)

Lieutenant_21 · Feb 6, 2013

Re: HSC 2013 4U Marathon

asianese said:
$\text{Find }\int \sin^{-1} \left(\sqrt{x}\right) \; dx.$

$Let$ $u=\sqrt{x} \therefore dx=2udu \therefore I=2\int usin^{-1}udu$ $now do IBP (CBF doing it :P )$

Lieutenant_21 · Feb 6, 2013

Re: HSC 2013 4U Marathon

Do you guys like conics? I find it extremely boring

asianese · Feb 6, 2013

Re: HSC 2013 4U Marathon

Conics is quite bland I find... The proof of certain properties are memory based (some only, don't jump on me). Also there aren't that many applications nowadays and it isn't explored in depth in university so it has a lot less motivation than perhaps complex numbers or polynomials (and calculus

)

Lieutenant_21 · Feb 6, 2013

Re: HSC 2013 4U Marathon

$Solve the equation$ $x^{2}+(4x^{3}-3x)^{2}=1$ $in real numbers.$ $\ \ \ \fbox{2}$

SpiralFlex · Feb 6, 2013

Re: HSC 2013 4U Marathon

Conic sections is a beautiful topic

Lieutenant_21 · Feb 6, 2013

Re: HSC 2013 4U Marathon

asianese said:
Conics is quite bland I find... The proof of certain properties are memory based (some only, don't jump on me). Also there aren't that many applications nowadays and it isn't explored in depth in university so it has a lot less motivation than perhaps complex numbers or polynomials (and calculus )

I wish they put Real Analysis as a topic and remove conics

RealiseNothing · Feb 6, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
Do you guys like conics? I find it extremely boring

It's boring because of the questions, not because of the topic. They could make more interesting questions and it would be a very nice topic.

hayabusaboston · Feb 6, 2013

Re: HSC 2013 4U Marathon

wtf guys, conics is awesome! Probably one of the easiest topics in MX2 imo, and a lot of fun!

I find polynomials pree boring, complex numbers was very fun, and will be doing integration then curve sketching soon.

Lieutenant_21 · Feb 6, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
wtf guys, conics is awesome! Probably one of the easiest topics in MX2 imo, and a lot of fun!
I find polynomials pree boring, complex numbers was very fun, and will be doing integration then curve sketching soon.

Conics is a cool topic but in my opinion it is the most boring topic compared to the rest of MX2 topics.
Polynomials is considered to be one of the simplest topics by most school teachers but the questions can get very difficult and interesting. Most school teachers go over relation between roots and roots of multiplicity which are very simple and leave out the hard end...

Sy123 · Feb 7, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
$Solve the equation$ $x^{2}+(4x^{3}-3x)^{2}=1$ $in real numbers.$ $\ \ \ \fbox{2}$

$$let$ \ \ x=\cos \theta$

$\cos^2 \theta + \cos^2 3\theta = 1$

$\cos 3\theta = \pm \sin \theta$

Now notice that the function: $f(x)=x^2+(4x^3-3x)^2 = f(-x)$
So its an even function, hence the roots are opposites of each other.

So I will only take one of the above cases, namely:

$\cos 3\theta = \sin \theta$

$\cos 3\theta = \cos (\pi/2 - \theta)$

$3\theta = \pi/2 + \theta$

$\theta = \pi/4 \ \ \rightarrow x=\cos \pi/4 = \frac{1}{\sqrt{2}}$

$x=\frac{-1}{\sqrt{2}}$ is hence a solution

$3\theta = \pi/2 - \theta \ \ \rightarrow \theta = \pi/8 \ \ x=\pm \cos \frac{\pi}{8}$ (pm there for the extra solution)

$3\theta = 2\pi - (\pi/2 + \theta)$

$\theta = \frac{3\pi}{8} \ \ \ x=\pm \cos\frac{3\pi}{8}$

Hence those are the 6 solutions. Note how I skipped some cases for the pm and stuff only because they will just give the same answers that I have before.
There is probably a much easier way to do this question, and that I did it the long way, but oh well.

asianese · Feb 7, 2013

Re: HSC 2013 4U Marathon

Good observation on the cos3x expansion Sy!

I would definitely like to see it 'solved in 1-2 lines'...but the 'long way' would be expanding it all and factorising l0l

Sy123 · Feb 7, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$A sequence is defined recursively as follows$

$F_0 = 0 \ \ F_1=1$

$F_n=F_{n-1}+F_{n-2} \ \ \ n \geq 2$

$$i) Write down the first 7 numbers of this sequence$ \ \ \ \fbox{1}$

$ii) Consider a power series in$ \ z \ $whereby the co-efficients are the numbers of the defined sequence. That is$

$F(z)=\sum_{k=0}^{\infty} F_k z^k \equiv z+z^2+2z^3+3z^4+5z^5+8z^6+ \dots$

$$Show that$ \ \ \ F(z)=z+zF(z)+z^2F(z) \ \ \ \fbox{2}$

$$iii) Hence show that$ \ \ \ F(z) = \frac{-z}{z^2+z-1} \ \ \ \ \fbox{1}$

$iv) Find$ \ \ A, \ B, \ \alpha, \ \beta \ \ $within the below expression$

$F(z)=\frac{A}{1-\frac{z}{\alpha}} + \frac{B}{1-\frac{z}{\beta}}$

$$Where$ \ \ A \ \ $and$ \ \ B \ \ $are independent of$ \ \ z \\ \\ $ and$ \ \ \alpha, \ \beta \ \ $are roots of$ \ \ \ z^2+z-1 \ \ \ \ \fbox{2}$

$v) Hence prove that$

$F_k = \frac{1}{\sqrt{5}} \left ( \left(\frac{1+\sqrt{5}}{2} \right )^k - \left (\frac{1-\sqrt{5}}{2} \right )^k \right ) \ \ \ \fbox{2}$

i) 0, 1, 1, 2, 3, 5, 8

ii) and iii)

$F(z)= 0z^0 + 1z + \sum_{k=2}^{\infty} F_k z^k = z + z\sum_{k=2}^{\infty} F_{k-1}z^{k-1} + z^2 \sum_{k=2}^{\infty} F_{k-2}z^{k-2}$

$F(z) = z+z(F(z)-0z^0) + z^2(F(z))$

$-z=F(z)(z^2+z-1)$

$\therefore F(z)=\frac{-z}{z^2+z-1}$

iv) To find alpha and beta just solve the quadratic, to get A and B, just do common denominator and simultaneously solve for A and B. It is a long process, the end yields something hidden within my notes and I can't be bothered doing it again:

v)

Note how:

$F(z) = A \left(1+\frac{z}{\alpha}+\frac{z^2+}{\alpha^2})+ \dots \right) + B\left(1+\frac{z}{\beta}+\frac{z^2}{\beta^2}+ \dots \right)$

To find F_k we must equate the co-efficient of z^k on both sides

$F_k = \frac{A}{\alpha^k} + \frac{B}{\beta^k}$

Now, just sub in alpha, beta, A and B, rationalise denominators and some arithmetic, it should yield the final result.

======================================================

As for conics, I find it a little dry considering there are no advanced conics and the questions are relatively simple. i.e. no rotated ellipses hyperbola etc.
But its a pretty good topic in my opinion (and in spirit I made a question for it)

$The hyperbola$ \ \ \Omega \ \ \ \ \ \ \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1 \ \ \ $is known as a rectangular hyperbola$

$$i) Show that the focii are$ \ \ \ S(\pm a\sqrt{2} , 0) \\ \\ $and the directrices are$ \ \ \ x=\pm \frac{a}{\sqrt{2}} \ \ \ \fbox{1}$

$$A hyperbola is defined as the locus whereby the distance between the variable point and fixed point divided by the distance between the variable point and a fixed line is a constant$ \ \ e \ \ \ \ e > 1$

$ii) Considering the circles$ \ \ \ x^2+y^2 = 2a^2 \ \ \ \ x^2+y^2 = \frac{a^2}{2} \ \ \ $or otherwise$

$$And using the above definition of a hyperbola, prove that the equation of the curve created by rotating$ \ \ \Omega$

$by$ \ \ \frac{\pi}{4} \ \ $radians about the origin, is given by$

$xy=\frac{a^2}{2} \ \ \ \ \ \fbox{4}$

cutemouse · Feb 7, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$The hyperbola$ \ \ \Omega \ \ \ \ \ \ \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1 \ \ \ $is known as a rectangular hyperbola$

$$i) Show that the focii are$ \ \ \ S(\pm a\sqrt{2} , 0) \\ \\ $and the directrices are$ \ \ \ x=\pm \frac{a}{\sqrt{2}} \ \ \ \fbox{1}$

$$A hyperbola is defined as the locus whereby the distance between the variable point and fixed point divided by the distance between the variable point and a fixed line is a constant$ \ \ e \ \ \ \ e > 1$

$ii) Considering the circles$ \ \ \ x^2+y^2 = 2a^2 \ \ \ \ x^2+y^2 = \frac{a^2}{2} \ \ \ $or otherwise$

$$And using the above definition of a hyperbola, prove that the equation of the curve created by rotating$ \ \ \Omega$

$by$ \ \ \frac{\pi}{4} \ \ $radians about the origin, is given by$

$xy=\frac{a^2}{2} \ \ \ \ \ \fbox{4}$

This can also be proved using mappings in complex numbers/analysis.

Sy123 · Feb 7, 2013

Re: HSC 2013 4U Marathon

cutemouse said:
This can also be proved using mappings in complex numbers/analysis.

This is true (as had been demonstrated previous pages back) but I'm pretty sure the parametric equations of a hyperbola are out of syllabus (hyperbolic sine and cosine).

EDIT: I did it and it is still applicable to use complex numbers by simply using x+iy (disregard the above part)
But the question still stands lol

Sy123 · Feb 9, 2013

Re: HSC 2013 4U Marathon

$Sum the series$

$\cos \alpha + \cos(\alpha+2\beta)+ \dots + \cos (\alpha +2r\beta)+\dots + \cos(\alpha+2n\beta)$

HeroicPandas · Feb 11, 2013

Re: HSC 2013 4U Marathon

$$ The equation z^n - 1 = 0$ $ has roots: 1, z_1, z_2, \dots , z_{n-1}. $ $ Show that the constant term of the polynomial with roots: (1 - z_1), (1 - z_2), \dots , (1 - z_{n-1})$ $ is n ($ $ where n is even)$

RealiseNothing · Feb 11, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$$ The equation z^n - 1 = 0$ $ has roots: 1, z_1, z_2, \dots , z_{n-1}. $ $ Show that the constant term of the polynomial with roots: (1 - z_1), (1 - z_2), \dots , (1 - z_{n-1})$ $ is n ($ $ where n is even)$

The roots occur in conjugate pairs, that is:

$z_1 = \bar{z}_{n-1}$

Note that:

$z\bar{z} = |z|^2$

In this case $|z|=1$

The constant term is:

$1 + z_1z_{n-1} + z_2z_{n-2} + ...$

$1 + 1 + 1 + 1 +....$

$n$

asianese · Feb 12, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
The roots occur in conjugate pairs, that is:

$z_1 = \bar{z}_{n-1}$

Note that:

$z\bar{z} = |z|^2$

In this case $|z|=1$

The constant term is:

$1 + z_1z_{n-1} + z_2z_{n-2} + ...$

$1 + 1 + 1 + 1 +....$

$n$

Wouldn't there be n/2 pairs of roots...?

Sy123 · Feb 12, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$$ The equation z^n - 1 = 0$ $ has roots: 1, z_1, z_2, \dots , z_{n-1}. $ $ Show that the constant term of the polynomial with roots: (1 - z_1), (1 - z_2), \dots , (1 - z_{n-1})$ $ is n ($ $ where n is even)$

let $\alpha$ be a root of z^n-1=0
Lets form a polynomial with roots, 1-1, 1-z_1, 1-z_2, ....

$z=1-\alpha \ \ \alpha = 1-z$

Therefore the polynomial with roots:

$0, (1-z_1), (1-z_2) ... , (1-z_{n-1})$

is

$(1-z)^n - 1 = 0$

$z^n - nz^{n-1}+ \dots + (-1)^{n-1} n z = 0$

Now we will eliminate one solution, namely 0, or 1-1

$z^{n-1} - nz^{n-1} + \dots - n = 0$

The above equation has roots (1-z_1), (1-z_2), .... , (1-z_(n-1))

The constant term is -n ??

deadpatch · Feb 12, 2013

Re: HSC 2013 4U Marathon

William Sidis said:
Do you guys like conics? I find it extremely boring

I personally enjoy conics :L

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