• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

HSC 2013 Maths Marathon (archive) (1 Viewer)

Status
Not open for further replies.

Smile12345

Active Member
Joined
May 30, 2013
Messages
827
Gender
Undisclosed
HSC
2014
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! :)
 
Joined
Mar 10, 2013
Messages
105
Gender
Male
HSC
2014
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
 
Joined
Mar 10, 2013
Messages
105
Gender
Male
HSC
2014
Re: HSC 2013 2U Marathon

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

HSC2014

Member
Joined
Jul 30, 2012
Messages
399
Gender
Male
HSC
N/A
Re: HSC 2013 2U Marathon

what are these dots above x's
 

HSC2014

Member
Joined
Jul 30, 2012
Messages
399
Gender
Male
HSC
N/A
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

Exaı̸̸̸̸̸̸̸̸lted Member
Joined
Jul 9, 2013
Messages
2,373
Gender
Male
HSC
2013
Re: HSC 2013 2U Marathon

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

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Re: HSC 2013 2U Marathon

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

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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

-----



 
Last edited:
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top