MX2 Marathon (2 Viewers)

dan964 · Oct 10, 2017

Welcome to the 2018 Maths Ext 2 Marathon

Post any questions within the scope and level of Mathematics Extension 2. Once a question is posted, it needs to be answered before the next question is raised.
This thread is mainly targeting Q1-15 difficulty in the HSC.

Q16/Q16+ material to be posted here:
http://community.boredofstudies.org...0463/hsc-2017-2018-mx2-marathon-advanced.html

I encourage all current HSC students in particular to participate in this marathon.

Have fun ^_^

Note: For quick access to previous year's marathons:
http://community.boredofstudies.org/14/mathematics-extension-2/361256/maths-ext-2-resource-list.html

Unanswered question from 2017 marathon:

Sy123 said:
$\\ $Suppose you are analysing the decay of particles from a radioactive source, suppose you discover that the probability that the source emits $ \ k \ $ particles from your source in an hour is$ \\\\ p_k = \frac{e^{-\lambda} \lambda^k}{k!}, \ k \geq 0 \\\\ $Where$ \ \lambda \ $is a constant and a positive integer, the rate of emission for your source.$

$\\ $(a) By considering the ratio $ \frac{p_{k+1}}{p_k} $ or otherwise, find the most likely number of particles to be emitted in an hour$$

$\\ $(b) Suppose you have a friend and she is analysing the decay from her own radioactive source, and that in fact the probability that $ \ n \ $ particles are emitted from her source in an hour is$ \\\\ q_n = \frac{e^{-\mu} \mu^n}{n!}, \ n \geq 0 \\\\ $Where$ \ \mu \ $is a constant and a positive integer, the rate of emission in her source.$$

$\\ $Show that the probability that the sum of your and her observations is $ \ m \ $ is given by, $ \\\\ r_m = \frac{e^{-(\lambda + \mu)} (\lambda + \mu)^m}{m!}, \ m \geq 0$

Sy123 · Oct 17, 2017

Re: HSC 2018 MX2 Marathon

$\\ $Consider the function in the complex plane, $ \ f(z) = z + i\text{Im}(z). \\\\ $i) Find a locus in the complex plane, where for every$ \ z \ $that lies on that locus, then$ \ |f(z)| = 1 \\\\ $ii) Find the locus in the complex plane of$ \ f(z) \ $for all$ \ |z| = 1 \ $, sketch this locus or describe its shape$$

TheZhangarang · Oct 17, 2017

Re: HSC 2018 MX2 Marathon

i) x^2+y^2=1

ii) x^2+(y-im(z))^2=1

I've probably misinterpreted the question given it does seem overly simple

Sy123 · Oct 18, 2017

Re: HSC 2018 MX2 Marathon

TheZhangarang said:
i) x^2+y^2=1

ii) x^2+(y-im(z))^2=1

I've probably misinterpreted the question given it does seem overly simple

I made a typo in writing the first question, it should be fixed now. Your second answer however has a 'z' there but this is not a proper cartesian equation for the locus, you want only 'x's and 'y's

dan964 · Oct 19, 2017

Re: HSC 2018 MX2 Marathon

si2136 said:
Are there answers to Sydney Boys 2002 4u trial? Thanks

https://thsconline.github.io/s/?view=5348&id=0ByEFYhkkDQBKSHRZUFRPZU9FM1U&n=Sydney Boys 2002 w. sol

sssona09 · Oct 28, 2017

Re: HSC 2018 MX2 Marathon

if (x+iy)(a+ib) = b+ia, express x,y in terms of a,b
thank you!!

dan964 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

sssona09 said:
if (x+iy)(a+ib) = b+ia, express x,y in terms of a,b
thank you!!

expand
(xa-by) +(ay+bx)i=b+ia
equate real and complex and solve simulataneously for x,y

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

dan964 said:
expand
(xa-by) +(ay+bx)i=b+ia
equate real and complex and solve simulataneously for x,y

ohh thank you

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

find x,y if
2z/(1+i) - 2z/i = 5/(2+i)

can't seem to get it,, do I sub x+iy later

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

si2136 said:
You could if you want to. Rationalise denominator and then solve for it by equating real and imaginary parts

oh okay thanks

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

E+ni is a root for ax2+bx+c=0
where a and b and c are real
show that an^2 =aE^2 + bE +c

so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/

1729 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

sssona09 said:
E+ni is a root for ax2+bx+c=0
where a and b and c are real
show that an^2 =aE^2 + bE +c

so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/

$$\noindent By the product of roots, $E^2 + n^2 = \frac{c}{a} \implies an^2 = -aE^2 + c \ (\star)$ and by the sum of roots, $2E = -\frac{b}{a} \implies 2aE^2 = -bE \implies -aE^2 = aE^2 + bE \ (\star \star)$ Substitute $(\star \star)$ in $(\star)$: $an^2 = aE^2 + bE + c$

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

1729 said:
$$\noindent By the product of roots, $E^2 + n^2 = \frac{c}{a} \implies an^2 = -aE^2 + c \ (\star)$ and by the sum of roots, $2E = -\frac{b}{a} \implies 2aE^2 = -bE \implies -aE^2 = aE^2 + bE \ (\star \star)$ Substitute $(\star \star)$ in $(\star)$: $an^2 = aE^2 + bE + c$

Ahh thank you!!!!!!!!!!

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

how do i divide x^3-2-2i by x+1-i

Sp3ctre · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

sssona09 said:
how do i divide x^3-2-2i by x+1-i

It's the same as any long division, but just treat -2-2i and 1-i each as one term.

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

if cube root (x+iy) =X+iY show that 4 (X^2-Y^2) = x/X + y/Y

1729 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

sssona09 said:
if cube root (x+iy) =X+iY show that 4 (X^2-Y^2) = x/X + y/Y

$$\noindent Cube both sides and equate real and imaginary parts, \\ \begin{align*}\quad x &= X^3 - 3XY^2 \implies \frac{x}{X} = X^2 - 3Y^2 \ (\star) \\ y &= 3X^2Y-Y^3 \implies \frac{y}{Y} = 3X^2 -Y^2 \ (\star \star) \\ (\star) + (\star \star) &: 4\left(X^2-Y^2\right) = \frac{x}{X}+\frac{y}{Y} \end{align*}$

sssona09 · Oct 29, 2017

Re: HSC 2018 MX2 Marathon

thank you so much!!

sssona09 · Nov 2, 2017

Re: HSC 2018 MX2 Marathon

rove by Induction that z1+z2+zn = z1 +z2 +zn

where the LHS has a big conjugate line on the top and RHS has small conjugate lines

edit: solved

dan964 · Nov 2, 2017

Re: HSC 2018 MX2 Marathon

sssona09 said:
rove by Induction that z1+z2+zn = z1 +z2 +zn

where the LHS has a big conjugate line on the top and RHS has small conjugate lines

Prove by Induction that $\overline{z_1+z_2+\dots + z_n} = \overline{z}_1 +\overline{z}_2 + \dots \overline{z}_n$
is this the question?

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