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BOS Trials 2013 2 Unit Discussion Thread (1 Viewer)

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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

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not gonna post here anymore until I get worked answers

I will be able to prove nothing except my own stupidity
 

RealiseNothing

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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

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Though may I suggest it would be easier instead of differentiating:



To put it in the form:



Then differentiate.
 

Trebla

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Official solutions (along with the results) will be released very soon!
 

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