# Thread: HSC 2018-2019 MX2 Marathon

1. ## HSC 2018-2019 MX2 Marathon

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/...urce-list.html

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

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

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

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

2. ## Re: HSC 2018 MX2 Marathon

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

3. ## 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

4. ## Re: HSC 2018 MX2 Marathon

Originally Posted by TheZhangarang
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

5. ## Re: HSC 2018 MX2 Marathon

Originally Posted by si2136
Are there answers to Sydney Boys 2002 4u trial? Thanks
https://thsconline.github.io/s/?view...002%20w.%20sol

6. ## Re: HSC 2018 MX2 Marathon

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

7. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
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

8. ## Re: HSC 2018 MX2 Marathon

Originally Posted by dan964
expand
(xa-by) +(ay+bx)i=b+ia
equate real and complex and solve simulataneously for x,y
ohh thank you

9. ## 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

10. ## Re: HSC 2018 MX2 Marathon

Originally Posted by si2136
You could if you want to. Rationalise denominator and then solve for it by equating real and imaginary parts
oh okay thanks

11. ## 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 :/

12. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
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 :/
$\noindent By the product of roots, E^2 + n^2 = \frac{c}{a} \implies an^2 = -aE^2 + c \ (\star) and by the sum of roots, 2E = -\frac{b}{a} \implies 2aE^2 = -bE \implies -aE^2 = aE^2 + bE \ (\star \star) Substitute (\star \star) in (\star): an^2 = aE^2 + bE + c$

13. ## Re: HSC 2018 MX2 Marathon

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

14. ## Re: HSC 2018 MX2 Marathon

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

15. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
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.

16. ## Re: HSC 2018 MX2 Marathon

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

17. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
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*}

18. ## Re: HSC 2018 MX2 Marathon

thank you so much!!

19. ## 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

20. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
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?

21. ## Re: HSC 2018 MX2 Marathon

Originally Posted by dan964
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?
yes

22. ## Re: HSC 2018 MX2 Marathon

find z suvh that (Im)z=2 and z^2 is real

thank you

23. ## Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
find z suvh that (Im)z=2 and z^2 is real

thank you
So let z = x + 2i

(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0

Hence z = 2i

24. ## Re: HSC 2018 MX2 Marathon

Originally Posted by pikachu975
So let z = x + 2i

(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0

Hence z = 2i
thank you Pikachu!!

25. ## Re: HSC 2018 MX2 Marathon

Originally Posted by pikachu975
So let z = x + 2i

(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0

Hence z = 2i
how about if re(z) is 2Im(z) and z^2-4i is real

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