HSC 2013 MX2 Marathon (archive) (3 Viewers)

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RealiseNothing

what is that?It is Cowpea
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Re: HSC 2013 4U Marathon

bump.

Difficulty: 3.14159/5

That should be enough to keep 2013'ers busy for a while.
This is where I got up to last time, I'll try again though:

Dividing by gives:



Grouping appropriate terms:





Now using the quadratic formula:



Now let

We want to find the values of U such that it has atleast one real root:







So

Therefore going back to our quadratic equation and the roots:



 

RealiseNothing

what is that?It is Cowpea
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Re: HSC 2013 4U Marathon

This is where I got up to last time, I'll try again though:

Dividing by gives:



Grouping appropriate terms:





Now using the quadratic formula:



Now let

We want to find the values of U such that it has atleast one real root:







So

Therefore going back to our quadratic equation and the roots:



Ok now if we take the positive solution we get:



Make the square root the subject and squaring gives:







Now we want to find the minimum value of so we will make the the subject:





Hence we can deduce that:



The minimum value of is thus the minimum value of the RHS, so lets consider that side:





Upon completing the square:



Hence the minimum value of this is when the term with goes to 0:







Hence the minimum possible value of for the polynomial to have atleast one real root is
 
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Sy123

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Re: HSC 2013 4U Marathon



















 
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seanieg89

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Re: HSC 2013 4U Marathon








yes I am, thank you.

==========

Sean, is saying that in the interval [a,b] it is differentiable sufficient?
To be on the safe side I would say continuously differentiable on (a,b) and continuous on [a,b]. Continuity of the derivative implies that the integral exists, which we would not get from differentiability alone.
 
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