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01-11-2008: 1st November 2008 14:21
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#16
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Exalted Member
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Join Date: Jul 2008
Location: Sofia, Bulgaria
Posts: 268
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Re: the 2006 oxford admissions test
Originally Posted by adie_raz
I was not sure with 3)ii).
To find k there needs to be one positive and one negative solution to f(x). I thought that for a cubic to have 2 solutions one must be repeated therefore if the graph was moved upwards, the minimum turning point would sit on the x axis.
Another idea was if the graph was shifted to the left then it would cross at
(0,0) and this meant that one solution was positive and one was negative whilst still having three solutions.
To be honest I went with the first idea, but this could mean that k couldhave several values as the repeated root could touch at various points on the x axis.
I am probably thinking too much into it and ended up confusing myself on this question. But I would like to know what other think.
I was thinking exactly the same - the first way is for k to be smth like 0.4. The other way is to shift it 1 unit to the left. But what about iii and iv? On iii) i get x1=1 and g(x1)=1/4. Strangely, iv seems the same to me
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