HSC 2012 MX2 Marathon (archive) (3 Viewers)

IamBread · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
Prove that:

$3|z-1|^2 = |z+1|^2$ if and only if

$|z-2|^2 = 3$

$$Let $ z = x+ iy, \\\\ 3|z-1|^2 = |z+1|^2$

$3(x-1)^2 + 3y^2 = (x+1)^2 + y^2 \\\\ 3x^2-6x+3+3y^2=x^2+2x+1+y^2 \\\\ 2x^2 -8x +2 +2y^2 = 0 \\\\ 2(x-2)^2+2y^2=6 \\\\ (x-2)^2+y^2 = 3 \\\\ |x +iy-2|^2=3 \\\\ $As $ x+iy =z, |z-2|^2=3$

$New question... $ p(x) = 2x^3 -19x^2+112x+d $ has root $5+6i.\\\\ $Find the other roots and the value of d.$

nightweaver066 · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Let $ z = x+ iy, \\\\ 3|z-1|^2 = |z+1|^2$

$3(x-1)^2 + 3y^2 = (x+1)^2 + y^2 \\\\ 3x^2-6x+3+3y^2=x^2+2x+1+y^2 \\\\ 2x^2 -8x +2 +2y^2 = 0 \\\\ 2(x-2)^2+2y^2=6 \\\\ (x-2)^2+y^2 = 3 \\\\ |x +iy-2|^2=3 \\\\ $As $ x+iy =z, |z-2|^2=3$

$New question... $ p(x) = 2x^3 -19x^2+112x+d $ has root $5+6i.\\\\ $Find the other roots and the value of d.$

$As all the coefficients of p(x) are real, by conjugate root theorem, x = 5 - 6i is also a root$

$p(x) = (x - 5 + 6i)(x - 5 - 6i)(ax + b)$

$= (x^2 - 10x + 61)(ax + b) \equiv 2x^3 - 19x^2 + 112x + d$

$$Comparing the terms of x^3, $ a = 2$

$$Comparing the terms of x^2, $ b - 20 = - 19$

$b = 1$

$$Comparing the constants, $ d = 61$

$\therefore p(x) = (x^2 - 10x + 61)(2x + 1)$

$$The roots are x = 5 - 6i, 5 + 6i or $ -\frac{1}{2} $ and d = 61$

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
$As all the coefficients of p(x) are real, by conjugate root theorem, x = 5 - 6i is also a root$

$p(x) = (x - 5 + 6i)(x - 5 - 6i)(ax + b)$

$= (x^2 - 10x + 61)(ax + b) \equiv 2x^3 - 19x^2 + 112x + d$

$$Comparing the terms of x^3, $ a = 2$

$$Comparing the terms of x^2, $ b - 20 = - 19$

$b = 1$

$$Comparing the constants, $ d = 61$

$\therefore p(x) = (x^2 - 10x + 61)(2x + 1)$

$$The roots are x = 5 - 6i, 5 + 6i or $ -\frac{1}{2} $ and d = 61$

Possibly a simpler way to do this would be to use the sum of roots to find the final root, and the product of roots to find d.

IamBread · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
Possibly a simpler way to do this would be to use the sum of roots to find the final root, and the product of roots to find d.

I used sum of roots to find the final root, then subbed in x=-1/2 to find d. But either way works.

$P(a\cos\theta,b\sin\theta) $ and $ Q(a\cos\phi,b\sin\phi) $ lie on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \\\\ $Find the equation of line $PQ$

mnmaa · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:

I suspect that it's sufficient to solve Z=Z^3 and discard -1 as a solution. My reasoning is that there clearly can't be an imaginary part present in Z. With this in mind, we can proceed noting that |Z| = Z, Z > 0 and |Z| = -Z, Z < 0. My method yields Z = 0 or/ 1. Could be wrong here though.

cutemouse · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
Prove that:

$3|z-1|^2 = |z+1|^2$ if and only if

$|z-2|^2 = 3$

I asked that question in 2009, and you answered it oO.

cutemouse · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Let $ z = x+ iy, \\\\ 3|z-1|^2 = |z+1|^2$

$3(x-1)^2 + 3y^2 = (x+1)^2 + y^2 \\\\ 3x^2-6x+3+3y^2=x^2+2x+1+y^2 \\\\ 2x^2 -8x +2 +2y^2 = 0 \\\\ 2(x-2)^2+2y^2=6 \\\\ (x-2)^2+y^2 = 3 \\\\ |x +iy-2|^2=3 \\\\ $As $ x+iy =z, |z-2|^2=3$

From memory, I think there's an easier way to do this question. Ie. using z z (bar) = |z|^2

IamBread · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:

$z = 0, 1$

math man · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

cause this is a proof question i started with LHS:
$\!\!\!\!\!\!\!\!\!\!\!\!\!\! LHS= 3|z-1|^{2} \\ = 3(x-1)^{2} + 3y^{2} \\ = 3(x^{2}-2x+1) + 3y^{2} \\ = 3((x-2)^{2} +2x -3) +3y^{2} \\ = (x^{2}-4x+4 +6x -9)+y^{2} +2((x-2)^{2}+y^{2}) \\ = x^{2}+2x -5 +y^{2} +6 \\ =(x+1)^{2} + y^{2} \\ =|z+1|^{2}$

gurmies · Feb 13, 2012

Re: 2012 HSC MX2 Marathon

$|z-2|^{2}=3 \\\\ (z-2)\overline{(z-2)}=3 \\\\ (z-2)(\overline{z}-2)=3 \\\\ z\overline{z}-2z-2\overline{z}+4=3 \\\\ z\overline{z}-2z-2\overline{z}+1=0 \\\\ 2z\overline{z}-4z-4\overline{z}+2=0 \\\\ 3z\overline{z}-3z-3\overline{z}+3=z\overline{z}+z+\overline{z}+1 \\\\ 3(z\overline{z}-z-\overline{z}+1)=z\overline{z}+z+\overline{z}+1 \\\\ 3(z-1)(\overline{z}-1)=(z+1)(\overline{z}+1) \\\\ 3(z-1)\overline{(z-1)}=(z+1)\overline{(z+1)} \\\\ 3|z-1|^{2}=|z+1|^{2}$

cutemouse · Feb 14, 2012

Re: 2012 HSC MX2 Marathon

gurmies said:
$|z-2|^{2}=3 \\\\ (z-2)\overline{(z-2)}=3 \\\\ (z-2)(\overline{z}-2)=3 \\\\ z\overline{z}-2z-2\overline{z}+4=3 \\\\ z\overline{z}-2z-2\overline{z}+1=0 \\\\ 2z\overline{z}-4z-4\overline{z}+2=0 \\\\ 3z\overline{z}-3z-3\overline{z}+3=z\overline{z}+z+\overline{z}+1 \\\\ 3(z\overline{z}-z-\overline{z}+1)=z\overline{z}+z+\overline{z}+1 \\\\ 3(z-1)(\overline{z}-1)=(z+1)(\overline{z}+1) \\\\ 3(z-1)\overline{(z-1)}=(z+1)\overline{(z+1)} \\\\ 3|z-1|^{2}=|z+1|^{2}$

Pro (Y).

largarithmic · Feb 14, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
From memory, I think there's an easier way to do this question. Ie. using z z (bar) = |z|^2

It's also true because of

http://en.wikipedia.org/wiki/Circles_of_Apollonius

Nooblet94 · Feb 14, 2012

Re: 2012 HSC MX2 Marathon

Haven't seen many graphing questions, so here we go.

Graph the following parametric equation:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" title="\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" /></a>

rolpsy · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

lol

question:
graph
$x^3 = 1 + xy^2$
(perhaps a tad more accessible than that ^)

Carrotsticks · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

Rolpsy, what program was used to sketch that curve?

rolpsy · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Rolpsy, what program was used to sketch that curve?

texshop (i.e. latex) with pstricks (and pst-plot)

math man · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
Haven't seen many graphing questions, so here we go.

Graph the following parametric equation:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" title="\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" /></a>

nooblet what are you trying to imply here?

deswa1 · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

I haven't seen many integration questions... This is from someone at school (you know who you are

)

<a href="http://www.codecogs.com/eqnedit.php?latex=\int sin(lnx)dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int sin(lnx)dx" title="\int sin(lnx)dx" /></a>

nightweaver066 · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
I haven't seen many integration questions... This is from someone at school (you know who you are )

$\int sin(lnx)dx$

$\int sin(lnx)dx$

$Let u = lnx$

$e^u = x$

$du = \frac{1}{x}dx$

$\therefore \int sin(lnx) dx = \int e^usinudu$

$Applying I.B.P.$

$= e^usinu - \int e^ucosudu$

$= e^usinu - (e^ucosu + \int e^usinudu)$

$\int e^u sinu du= e^usinu - e^ucosu - \int e^usinudu ..... (1)$

$Applying I.B.P. the other way to the original integral$

$= -e^ucosu + \int e^ucosudu$

$\int e^u sinu du = -e^ucosu + e^usinu - \int e^usinudu ..... (2)$

$(1) + (2): 2\int e^u sinu du = 2e^u(sinu - cosu) - 2\int e^u sinu du$

$4\int e^u sinu du = 2e^u(sinu - cosu)$

$\int e^u sinu du = \frac{1}{2}e^u (sinu - cosu)$

$= \frac{x}{2}(sinlnx - coslnx) + c$

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