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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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Sy123

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Re: HSC 2013 4U Marathon






ii) It is obvious that one simply needs to pick the right function f(x), it is clear then that we need to find some function f so that:



This may not always give the correct answer, because it is possible the sum inside the limit may be manipulated or simplified further, but we can simply verify our result once we find a good guess as to what the function is.





We can verify this by simply putting 1+k/n back in.

So f(x) = 1/(1+x^2), the answer then is
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

Find, for some positive integer :

 
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Sy123

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Re: HSC 2013 4U Marathon

Find, for some positive integer :

Using De'Moivers Theorem:











With roots:





Sum of roots:












....

Finishing later.


EDIT: Is this sum even possible?
 
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braintic

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Re: HSC 2013 4U Marathon

Something easier than usual:



Simplifies to (x^2 - 1)^2 = x^3.
A graph of y = (x^2 - 1)^2 and y=x^3 on the same axes, and a substitution of x=2 into each equation, shows the result.
 

btx3

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Re: HSC 2013 4U Marathon

Simplifies to (x^2 - 1)^2 = x^3.
A graph of y = (x^2 - 1)^2 and y=x^3 on the same axes, and a substitution of x=2 into each equation, shows the result.
good to know brah but leave it to a student you know
 

Sy123

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Re: HSC 2013 4U Marathon

good to know brah but leave it to a student you know
A whole day and no one answered it, its up to anyone by then.

An alternative approach is simply letting it be f(x), finding f'(x) factorising x out and solving the quadratic.
======


 
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RealiseNothing

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Re: HSC 2013 4U Marathon

A whole day and no one answered it, its up to anyone by then.

An alternative approach is simply letting it be f(x), finding f'(x) factorising x out and solving the quadratic.
======


Unless the triangle is degenerate,

Now by triangle inequality:





Hence:

 
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Re: HSC 2013 4U Marathon

Where's that question with P Q and f(sinx)?? It was a nice one using the classic fact that

 

RealiseNothing

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Re: HSC 2013 4U Marathon

The upper bound is correct, but d can be negative.
Oh right, I was thinking that the side with length 1 was the shortest side. Also I just realised d=0 is possible, because you know, what even is an equilateral triangle?
 

RealiseNothing

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Re: HSC 2013 4U Marathon

A whole day and no one answered it, its up to anyone by then.

An alternative approach is simply letting it be f(x), finding f'(x) factorising x out and solving the quadratic.
======




Using triangle inequality twice.
 

Sy123

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Re: HSC 2013 4U Marathon

First find the function f(t) such that the equality holds:





For F the primitive of f:




















I have a feeling this argument is not valid but I cant see it right now, its a bit late
 
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