K kubekoo New Member Joined Aug 30, 2004 Messages 22 Gender Undisclosed HSC N/A Jul 22, 2005 #1 by doing JR past trial papers, i found some of their locus question asked to restriction of locus?! How do I obtion those restriction? cause they seems like coming out from no where to me T_T
by doing JR past trial papers, i found some of their locus question asked to restriction of locus?! How do I obtion those restriction? cause they seems like coming out from no where to me T_T
A acmilan I'll stab ya Joined May 24, 2004 Messages 3,989 Location Jumanji Gender Male HSC N/A Jul 23, 2005 #2 any particular examples?
K kubekoo New Member Joined Aug 30, 2004 Messages 22 Gender Undisclosed HSC N/A Jul 23, 2005 #3 the answer to part iii) is N[-pg(p+q) , p^2+q^2+pq+2 ] and part iv) has locus is y = x/2m + 4m^2 +2 ; it says it has restriction of y>3m^2 +2 and x > -2m^3 but i have no idea why....
the answer to part iii) is N[-pg(p+q) , p^2+q^2+pq+2 ] and part iv) has locus is y = x/2m + 4m^2 +2 ; it says it has restriction of y>3m^2 +2 and x > -2m^3 but i have no idea why....
D David_O Member Joined Jun 24, 2005 Messages 95 Location Auburn Gender Undisclosed HSC N/A Jul 23, 2005 #4 One of the more stupid JR q's. p+q = 2m :. rt(pq) <=m [because a^2+b^2>=2ab if you want to be pedantic] pq<=m^2 and you sub that into y = (p+q)^2 - pq + 2 Similarly, pq = -x/2m and you get x result.
One of the more stupid JR q's. p+q = 2m :. rt(pq) <=m [because a^2+b^2>=2ab if you want to be pedantic] pq<=m^2 and you sub that into y = (p+q)^2 - pq + 2 Similarly, pq = -x/2m and you get x result.
K kubekoo New Member Joined Aug 30, 2004 Messages 22 Gender Undisclosed HSC N/A Aug 1, 2005 #5 OMG, that is mad....... thanks so much!!!
