# MX2 Marathon (1 Viewer)

#### dan964

##### what
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:

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

Have fun ^_^

http://community.boredofstudies.org/14/mathematics-extension-2/361256/maths-ext-2-resource-list.html

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

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

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

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

#### Sy123

##### This too shall pass
Re: HSC 2018 MX2 Marathon

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

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#### TheZhangarang

##### New Member
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

##### This too shall pass
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
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

#### sssona09

##### Member
Re: HSC 2018 MX2 Marathon

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

#### dan964

##### what
Re: HSC 2018 MX2 Marathon

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

##### Member
Re: HSC 2018 MX2 Marathon

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

#### sssona09

##### Member
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

##### Member
Re: HSC 2018 MX2 Marathon

You could if you want to. Rationalise denominator and then solve for it by equating real and imaginary parts
oh okay thanks

#### sssona09

##### Member
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

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

#### sssona09

##### Member
Re: HSC 2018 MX2 Marathon

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

#### sssona09

##### Member
Re: HSC 2018 MX2 Marathon

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

#### Sp3ctre

##### Active Member
Re: HSC 2018 MX2 Marathon

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

##### Member
Re: HSC 2018 MX2 Marathon

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

#### 1729

##### Active Member
Re: HSC 2018 MX2 Marathon

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

##### Member
Re: HSC 2018 MX2 Marathon

thank you so much!!

#### sssona09

##### Member
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

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#### dan964

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