MedVision ad

BOS Trials 2013 2 Unit Discussion Thread (1 Viewer)

Joined
Apr 1, 2011
Messages
1,012
Location
District 12
Gender
Male
HSC
2013
14)b iv) y=mx passes through the origin. If you want to enclose an area you have to have 2 intersection points between the line y=mx and the curve

You can do this two ways, either find the gradient of the tangent to f(x) at x=0 then let m < gradient or solve the equations simultaneously and let the discriminant be positive for 2 solutions (i.e. 2 intersections). If you do the latter you will get a cubic, but note x=0 is an obvious solution so divide by x and get a quadratic.
lol I got the values out for m by inspection, not sure if it counts oops
 

JT145

ON is my homeboy
Joined
Mar 4, 2012
Messages
1,678
Gender
Female
HSC
2016
not gonna post here anymore until I get worked answers

I will be able to prove nothing except my own stupidity
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
well I knew that the gradient at the origin was 1 so clearly m has to be between 0 and 1 for there to be two intersections?
14)b iv) y=mx passes through the origin. If you want to enclose an area you have to have 2 intersection points between the line y=mx and the curve

You can do this two ways, either find the gradient of the tangent to f(x) at x=0 then let m < gradient or solve the equations simultaneously and let the discriminant be positive for 2 solutions (i.e. 2 intersections). If you do the latter you will get a cubic, but note x=0 is an obvious solution so divide by x and get a quadratic.
?
 

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Though may I suggest it would be easier instead of differentiating:



To put it in the form:



Then differentiate.
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,384
Gender
Male
HSC
2006
Official solutions (along with the results) will be released very soon!
 

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

Top