HSC 2013 Maths Marathon (archive) (3 Viewers)

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Smile12345

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Re: HSC 2013 2U Marathon

This is probably an 'easy' question but still good revision for 2U's and a question I'm not sure about... (Thanks in advance

Solve 1/ x^6 - 5 / x^2 + 4 = 0

Trust the above is clear! :)
 
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Re: HSC 2013 2U Marathon

Ahh, damn! I just did it for 'x^2-6x... = 1' before I saw Sy123's post. Would my answer be right if that was = 1? IMG_0544.jpg
 
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Re: HSC 2013 2U Marathon

IMG_0545.jpg Wow... Also, Sorry for the blurry image :/ This one here is much better! :D
 
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HSC2014

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Re: HSC 2013 2U Marathon

what are these dots above x's
 

HSC2014

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Re: HSC 2013 2U Marathon

How did those accents come about? I've never even seen them :'( Only like the different level derivatives: y', y"
 

Menomaths

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Re: HSC 2013 2U Marathon

Guys don't change my comment OKAY? >:[
 

SpiralFlex

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Re: HSC 2013 2U Marathon

X dotdotdot is the rate of change of acceleration also know as a jerk
 

Sy123

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Re: HSC 2013 2U Marathon

i) point R(p,q) sub into y= 1/x ---> q=1/p
ii) dy/dx = -1/x^2 (sub x= p to get gradient) M=-1/p^2
y-y1= m(x-x1) ---> y-q= -1/p^2 (x-p) ---> y- 1/p = -x/p^2 +1/p
y= -x/p^2 +2/p (times by p^2 ) --> yp^2 = -x+ 2p ----> yp^2 +x = 2p
iii) when x = 0 y= 2/p at point Q when y=0 x = 2p at point P
A of triangle 0PQ= 1/2 x base x height = 1/2 x 2/p x 2p = 2 therefore constant
proof is complete [ ]
sy you need to teach me latex this is wasting too much time typing it this way :( put a harder question up fag :p
No latex 2/10

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