HSC 2015 MX2 Marathon (archive) (7 Viewers)

braintic · Feb 7, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Not sure if this is right but :
$1+i = \sqrt{2}cis(\frac{\pi}{4}) 1-i = \sqrt{2}cis(\frac{-\pi}{4})$

$\therefore $range of argz$ = -\frac{\pi}{4} \grq arg(z) \grq \frac{\pi}{4} ?$

Lower boundary is correct, but not the upper boundary.

You should find the cartesian equation of the locus defined by that condition, sketch it, then use 2 unit work to finish it off.

VBN2470 · Feb 7, 2015

Re: HSC 2015 4U Marathon

braintic said:
Lower boundary is correct, but not the upper boundary.

You should find the cartesian equation of the locus defined by that condition, sketch it, then use 2 unit work to finish it off.

$Hmm, I am not able to see how the lower bound is $-\frac{\pi}{4}$ since the line $y=-x$ is not tangent to the locus I got to be as $(x-\frac{3}{2})^2+(y-1)^2=\frac{9}{4}$. Upper bound for $\arg(z)$ should be $\frac{\pi}{2}.$

Please correct me if I am wrong on anything

Ekman · Feb 7, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Hmm, I am not able to see how the lower bound is $-\frac{\pi}{4}$ since the line $y=-x$ is not tangent to the locus I got to be as $(x-\frac{3}{2})^2+(y-1)^2=\frac{9}{4}$. Upper bound for $\arg(z)$ should be $\frac{\pi}{2}.$

Please correct me if I am wrong on anything

$\\* $Sub in z=x+iy into the equation$$

$\sqrt{(x-1)^2 + (y-1)^2} = x+1$

$\\* \therefore x^2-2x+1 + y^2-2y+1=x^2+2x+1$

$(y-1)^2 = 4x $ should be the locus$$

$$Sub in y=mx for the tangent, since z always goes through the origin $ (mx)^2+(-2(m+2))x+1=0$

$4(m+2)^2 - 4m^2 =0, $ thus $ m=-1$

$\therfore $At y=-1 and x=1, there is tangent from the origin, thus the gradient is -1, thus lowest angle is $ -\frac{\pi}{4}$

sirable1 · Feb 7, 2015

Re: HSC 2015 4U Marathon

$\mathrm{Let\, \mathit {p(x)= 3z-z^3+5z^4+z^6} and \,suppose\, that\, the\, six\, roots\, of\, p\,, are\, distinct. Prove\, that\, at\, least\, TWO\,of\,these\,roots\,are\,real.}$

Edit: Reference from MATH1131 book

FrankXie · Feb 7, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\mathrm{Let\, \mathit {p(x)= 3z-z^3+5z^4+z^6} and \,suppose\, that\, the\, six\, roots\, of\, p\,, are\, distinct. Prove\, that\, at\, least\, one\,of\,these\,roots\,are\,real.}$

lol, isn't z=0 a root, which is real?

sirable1 · Feb 7, 2015

Re: HSC 2015 4U Marathon

$\mathrm{Use\, the\, properties\, of\, the\, complex\, conjugates\, to\, show\, that\, if\, \mathit{a}\, and \mathit{b}\, are\, real\, and \, |z|\, =1, then\, show\, |\mathit{a}+\mathit{bz|}=|\mathit{az+b|}}$

Edit: Reference from MATH1131 booklet

Ekman · Feb 7, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
lol, isn't z=0 a root, which is real?

That's exactly what I was thinking as well. Im just scared that we are underestimating the question

VBN2470 · Feb 7, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\mathrm{Let\, \mathit {p(x)= 3z-z^3+5z^4+z^6} and \,suppose\, that\, the\, six\, roots\, of\, p\,, are\, distinct. Prove\, that\, at\, least\, one\,of\,these\,roots\,are\,real.}$

$$ We can see that $z=0$ is a root to the polynomial, and since the coefficients of $P(z)$ are real, then by the Complex conjugate root theorem, either two cases are possible for the remaining $5$ roots: \\ 1. Two pairs of complex conjugate roots + 1 real root \\ 2. All the other roots are real. \\ Thus, at least TWO of these roots are real for $P(z)$.$

EDIT. I think the question should be 'Prove that at least TWO of these distincts roots are real'.

Ekman · Feb 7, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\mathrm{Use\, the\, properties\, of\, the\, complex\, conjugates\, to\, show\, that\, if\, \mathit{a}\, and \mathit{b}\, are\, real\, and \, |z|\, =1, then\, show\, |\mathit{a}+\mathit{bz|}=|\mathit{az+b|}}$

$\frac{|a+bz|}{|az+b|} = 1$

$z \times \bar{z} = 1 $ (given in the question)$

$$Let $w=a+bz \therefore w \times \bar{w} = |w|^2$

$\therefore |w| = \sqrt{a^2 +ab(z+\bar{z}) +b^2}$

$$Let $y=az+b \therefore y \times \bar{y} = |y|^2$

$|y|=\sqrt{a^2 +ab(z+\bar{z})+b^2}$

$\frac{|w|}{|y|} = 1$

$\frac{|a+bz|}{|az+b|} = 1 \therefore |a+bz|=|az+b|$

sirable1 · Feb 7, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$$ We can see that $z=0$ is a root to the polynomial, and since the coefficients of $P(z)$ are real, then by the Complex conjugate root theorem, either two cases are possible for the remaining $5$ roots: \\ 1. Two pairs of complex conjugate roots with 1 real root \\ 2. All the other roots are real. \\ Thus, at least TWO of these roots are real for $P(z)$.$

EDIT. I think the question should be 'Prove that at least TWO of these distincts roots are real'.

Yeah sorry, it was prove at least TWO roots, which I have worded it incorrectly.

I was just looking over my old work and was wondering about this Q, hence I asked on this thread.

FrankXie · Feb 7, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\mathrm{Let\, \mathit {p(x)= 3z-z^3+5z^4+z^6} and \,suppose\, that\, the\, six\, roots\, of\, p\,, are\, distinct. Prove\, that\, at\, least\, TWO\,of\,these\,roots\,are\,real.}$

Edit: Reference from MATH1131 book

what is MATH1131, in which is Roll's theorem going to be covered?

sirable1 · Feb 7, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
what is MATH1131, in which is Roll's theorem going to be covered?

No Rolle's theorem in MATH1131, only the Mean Value theorem is covered. MATH1131 is the first year maths course at UNSW.

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

$$Find$ \prod_{r=0}^\infty (cos{\frac{\pi}{2^{r}}} +isin\frac{\pi}{2^r})$

braintic · Feb 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$Find$ \prod_{r=0}^\infty (cos{\frac{\pi}{2^{r}}} +isin\frac{\pi}{2^r})$

= cis pi . cis pi/2 . cis pi/4 . cis pi/8 . ...
= cis (pi + pi/2 + pi/4 + pi/8 + ...)
= cis pi/(1-1/2) .... [limiting sum of GP]
= cis 2pi
= 1

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

braintic said:
= cis pi . cis pi/2 . cis pi/4 . cis pi/8 . ...
= cis (pi + pi/2 + pi/4 + pi/8 + ...)
= cis pi/(1-1/2) .... [limiting sum of GP]
= cis 2pi
= 1

is that a rule?

Ekman · Feb 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
is that a rule?

When you multiply complex numbers together, the modulus gets multiplied and the arguments get added. Since the modulus is =1 and the arguments are decreasing by a factor of 1/2, thus you can make a limiting sum of the arguments.

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

Can we start posting permutations and combination questions from extension 2

sirable1 · Feb 8, 2015

Re: HSC 2015 4U Marathon

$\text{This problem was taken from a University Calculus (Spivak) textbook.} \\ \text{Use a substitution to find that} \int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\text{This problem was taken from a University Calculus (Spivak) textbook.} \\ \text{Use a substitution to find that} \int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$

Many of the 2015ers haven't studied extension 2 integration... But there is a separate thread for integration.

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

Anyways new question:
$arg(z-2) = arg(z+2) + \frac{\pi}{3} $Find the center, radius and area of this circle$$
I added the area bit in lets see how this goes

HSC 2015 MX2 Marathon (archive) (7 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Active Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Active Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

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