HSC 2014 MX2 Marathon (archive) (4 Viewers)

hit patel · Apr 10, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
If you mean I_(n-1) + I_n = 1/(2n-1) then this is fine since if we just make n -> n+1 we get

I_((n+1) - 1) + I_(n+1) = 1/(2(n+1)-1)

I_n + I_(n+1) = 1/(2n+1)

What do the other answers use? harder ext 1?

Sy123 · Apr 10, 2014

Re: HSC 2014 4U Marathon

hit patel said:
What do the other answers use? harder ext 1?

Yep

hit patel · Apr 10, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
Yep

Sorry that one remains to be learnt.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

Sketch the graph $y=\dfrac{x^3-a}{x^2-a}$ for:

i) $a>1$ ii) $0<a<1$

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon

$\\ $i) Explain why for any complex$ \ z \ $then$ \ |Re(z)| \leq |z| \\ $ii) Hence show that$ \ \ |ax-by| \leq \sqrt{a^2+b^2}\sqrt{x^2+y^2}$

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
Sketch the graph $y=\dfrac{x^3-a}{x^2-a}$ for:

i) $a>1$ ii) $0<a<1$

Doesnt seem correctg.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

You would be close if a=1. Hint: You should have two vertical asymtotes and one oblique.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $i) Explain why for any complex$ \ z \ $then$ \ |Re(z)| \leq |z| \\ $ii) Hence show that$ \ \ |ax-by| \leq \sqrt{a^2+b^2}\sqrt{x^2+y^2}$

I) $\vert z \vert =\vert Re(z)+iIm(z) \vert =\sqrt{Re^2(z)+Im^2(z)} \geq\sqrt{Re^2(z)}=\vert Re(z) \vert$

II) Let $z=(x+iy)(a+ib) \longrightarrow \vert z \vert= \sqrt{a^2+b^2}\sqrt{x^2+y^2}$ and $\vert Re(z) \vert=\vert ax-by \vert$

Hence using part (i) $|ax-by| \leq \sqrt{a^2+b^2}\sqrt{x^2+y^2}$

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon

(leave this marathon for the 2014ers please)

$\\ $i) Expand the complex number and find it in mod-arg form$ \\ (1+i)(1+2i)(1+3i) \\ $ii) Hence show that by finding each individual product in mod-arg form$ \\ \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1}3 = \pi$

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
You would be close if a=1. Hint: You should have two vertical asymtotes and one oblique.

Yes this is where I lose my marks. Sorry- This is 3 u Right? Damn. almost forgot about the original equation. yes x=+_a and y=x right? Sorry but why do the conditions change the graph?

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
(leave this marathon for the 2014ers please)

$\\ $i) Expand the complex number and find it in mod-arg form$ \\ (1+i)(1+2i)(1+3i) \\ $ii) Hence show that by finding each individual product in mod-arg form$ \\ \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1}3 = \pi$

wait why am i getting arc tan (3) x arctan (2) = 4?????

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
wait why am i getting arc tan (3) x arctan (2) = 4?????

10cis (pi/4 x arc tan 2 x arc tan 3)= -10 (equating answer from the comp number given)
cis (pi/4 x arctan 2 x arc tan 3)= -1
Therefore pi/4 x arc tan 2 x arctan3 = pi
therefore arc tan 2 x arctan2 = 4? ? ?

$arg(xy) = arg(x) + arg(y) \ $not$ \ arg(x) \cdot arg(y)$

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$arg(xy) = arg(x) + arg(y) \ $not$ \ arg(x) \cdot arg(y)$

argggggghhhhhhhhhh..... what am i thinking??????????? Its all because of the 1 hr english I studieed today and the caffeine I filled myself with to keep going through the one hour... sorry

10cis (pi/4+ arctan 3 + arc tan 2)= -10 by equating the expansions
cis (pi/4+ arctan 3 + arctan2) = -1
Therefore pi/4+arctan3 + arctan 2 = pi
Forgive my stupidity.

dunjaaa · Apr 11, 2014

Re: HSC 2014 4U Marathon

Asymptotes are both y=x (forgot to label in ss) by poly division

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

hit patel said:
Yes this is where I lose my marks. Sorry- This is 3 u Right? Damn. almost forgot about the original equation. yes x=+_a and y=x right? Sorry but why do the conditions change the graph?

Well consider the graph $y=x^{1/2}$ and $y=x^{1/3}$ , can you see how for $x>1$ , $y=x^{1/2}$ is greater, while for $0<x<1$ , $y=x^{1/3}$ is actually greater. What it means for the graph is where the x-intercept lies, between the asymtotes $(x=\pm \sqrt{a})$ or towards the right of the right asymtote. So can you see how it actually changes the graph significantly?

There are a few more details (which i will get to soon) but I think that would be the most significant change.

dunjaaa · Apr 11, 2014

Re: HSC 2014 4U Marathon

(ii) arg[(1+i)(1+2i)(1+3i)]=arg(-10)
Using arg(xy)=arg(x)+arg(y),
u get the required result noting that the argument of any negative real number is pi and also since the arguments of each individual complex numbers are acute

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
Well consider the graph $y=x^{1/2}$ and $y=x^{1/3}$ , can you see how for $x>1$ , $y=x^{1/2}$ is greater, while for $0<x<1$ , $y=x^{1/3}$ is actually greater. What it means for the graph is where the x-intercept lies, between the asymtotes $(x=\pm \sqrt{a})$ or towards the right of the right asymtote. So can you see how it actually changes the graph significantly?

Ah oh. Sorry today brain is not in the correct frame of thought. I swear I can match dunja's writing with someone I know.

dunjaaa · Apr 11, 2014

Re: HSC 2014 4U Marathon

do i know u?

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
(ii) arg[(1+i)(1+2i)(1+3i)]=arg(-10)
Using arg(xy)=arg(x)+arg(y),
u get the required result noting that the argument of any negative real number is pi and also since the arguments of each individual complex numbers are acute

Yep thats what I finally got after getting corrected by sy.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
View attachment 30264 Asymptotes are both y=x (forgot to label in ss) by poly division

Very good but there is one detail in part ii you missed, The graph (on the right branch) actually intercepts the asymtote at (1,1) and approaches y=x from the top side. Other than that well done.

And also you accidentally labeled asymtotes as $x=\pm a$ rather than $x=\pm \sqrt{a}$

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