HSC 2012 MX2 Marathon (archive) (1 Viewer)

[Sindivyn]

Re: 2012 HSC MX2 Marathon

Partial f / Partial X means you differentiate f(x,y) purely with respect to x, and you treat Y as a constant.

Similarly for Partial f / Partial Y.

Ah, I see, thanks.

 

[math man]

Re: 2012 HSC MX2 Marathon

He explained his method to me through PM and I don't think it is valid, but we will see if someone can post a valid solution using MX2 only techniques (I don't think this is possible).

to prove my method i had to construct a quarter circle contour to show the integrals were the same, so the method works, but yeh it does need proof why it works using contour integration.

 

[math man]

Re: 2012 HSC MX2 Marathon

Here is a 4u question from cambridge, but done in a different way:

 

[asianese]

Re: 2012 HSC MX2 Marathon

Here is a 4u question from cambridge, but done in a different way:

View attachment 26157

Was that from an induction about lines intersecting and all?

PS: do we need to know how to do those induction questions relating to geometry?

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.

 

[asianese]

Re: 2012 HSC MX2 Marathon

I'm assuming a, b, c are real?

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

I'm assuming a, b, c are real?

"complex polynomial" = polynomial with complex coefficients.

 

[asianese]

Re: 2012 HSC MX2 Marathon

Oo didn't read lolty

 

[twinklegal19]

Re: 2012 HSC MX2 Marathon

 

[lolcakes52]

Re: 2012 HSC MX2 Marathon

Second part is done with Newton's method for x=k to find the approximation. I'm having trouble with the first part, possibly because I'm not sure about whether the next root is between k=1 and 1, greater than k, or less than zero.

 

[barbernator]

Re: 2012 HSC MX2 Marathon

Second part is done with Newton's method for x=k to find the approximation. I'm having trouble with the first part, possibly because I'm not sure about whether the next root is between k=1 and 1, greater than k, or less than zero.

Well the sum of the roots is k, so if one root is between 0 and 1, one is between k-1 and k, then the next must be between 0 and 1 as well to satisfy k

 

[math man]

Re: 2012 HSC MX2 Marathon

the third root lies between -1 and 0

 

[math man]

Re: 2012 HSC MX2 Marathon

a better question would be to deduce the smallest positive root plus the smallest negative root is positve

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.

.

 

[RealiseNothing]

Re: 2012 HSC MX2 Marathon

Here is a decent question:

The three roots of the complex polynomial:



all lie on the unit circle in the complex plane. Prove that the three roots of:



also lie on the unit circle.

So far I've got:



y/n?

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

So far I've got:



y/n?

I don't think q can possibly be that, that would imply that the product of roots of p has modulus 3.

 

[RealiseNothing]

Re: 2012 HSC MX2 Marathon

I don't think q can possibly be that, that would imply that the product of roots of p has modulus 3.

Then what have I done wrong? Have I assumed something that isn't true maybe?

Let the roots of p(x) be alpha, beta, and gamma. Then:









Since the roots lie on the unit circle and hence .

I then do this for 'b' and 'c' as well to obtain my expression for q(x).

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

From your second line to your third line of displayed equations.

 

[RealiseNothing]

Re: 2012 HSC MX2 Marathon

From your second line to your third line of displayed equations.

oh shit.

triangular inequality fail.