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HSC 2014 MX2 Marathon (archive) (1 Viewer)

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hit patel

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Re: HSC 2014 4U Marathon

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?
 

Davo_01

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Re: HSC 2014 4U Marathon

Sketch the graph for:

i) ii)
 
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Davo_01

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Re: HSC 2014 4U Marathon

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

Sy123

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Re: HSC 2014 4U Marathon

(leave this marathon for the 2014ers please)

 

hit patel

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Re: HSC 2014 4U Marathon

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?
 

Sy123

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Re: HSC 2014 4U Marathon

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? ? ?
 

hit patel

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Re: HSC 2014 4U Marathon

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.
 
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dunjaaa

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Re: HSC 2014 4U Marathon

IMG_20140411_220230.jpg Asymptotes are both y=x (forgot to label in ss) by poly division
 

Davo_01

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Re: HSC 2014 4U Marathon

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 and , can you see how for , is greater, while for , is actually greater. What it means for the graph is where the x-intercept lies, between the asymtotes 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.
 
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dunjaaa

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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
 
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hit patel

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Re: HSC 2014 4U Marathon

Well consider the graph and , can you see how for , is greater, while for , is actually greater. What it means for the graph is where the x-intercept lies, between the asymtotes 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.
 

hit patel

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

Davo_01

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Re: HSC 2014 4U Marathon

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 rather than
 
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