HSC 2013 MX2 Marathon (archive) (2 Viewers)

Status
Not open for further replies.

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
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
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
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
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2013 4U Marathon



















 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
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.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top