HSC 2012 MX2 Marathon (archive) (2 Viewers)

[Trebla]

Re: 2012 HSC MX2 Marathon

Another question:

 

[math man]

Re: 2012 HSC MX2 Marathon

bump

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

Let f(x)=x*arcsin(x). As the product of two odd functions f is even, so it suffices to prove the assertion for 0<=x<1.

f is the product of two positive and increasing functions on the interval 0<=x<=1. Hence 0=f(0)<=f(x) < f(1)=pi/2 for all -1 < x < 1.

 

[math man]

Re: 2012 HSC MX2 Marathon

Using the Gaussian integral:

and Euler's identity:

deduce the following special integrals:

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

The obvious method that comes to mind is looking at the sector contour with one side being part of the positive real axis, and an angle of pi/4 radians. The integral along the circular arc tends to zero which implies that half of the Gaussian integral is equal to the negative integral of e^(-z^2) over the diagonal ray. Which after applying Euler's identity gives us what we want.

Clearly this is beyond the scope of mx2 so perhaps you had another solution in mind.

 

[Johnstan]

Re: 2012 HSC MX2 Marathon

The obvious method that comes to mind is looking at the sector contour with one side being part of the positive real axis, and an angle of pi/4 radians. The integral along the circular arc tends to zero which implies that half of the Gaussian integral is equal to the negative integral of e^(-z^2) over the diagonal ray. Which after applying Euler's identity gives us what we want.

Clearly this is beyond the scope of mx2 so perhaps you had another solution in mind.

Couldn't have said it better myself.

 

[math man]

Re: 2012 HSC MX2 Marathon

yes i have a simple solution in mind, involving only 4u properties applying those 2 formulaes

 

[cutemouse]

Re: 2012 HSC MX2 Marathon

Probably using e^(i x^2) = cos (x^2) + i sin (x^2) and treat i as a constant and take the real and imaginary parts.

But yes, the complex analysis way is quite elegant I think.

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

Probably using e^(i x^2) = cos (x^2) + i sin (x^2) and treat i as a constant and take the real and imaginary parts.

But yes, the complex analysis way is quite elegant I think.

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).

 

[Sindivyn]

Re: 2012 HSC MX2 Marathon

Another question:

Could this be done geometrically?

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Could this be done geometrically?

An algebraic proof would probably be more convincing.

 

[Sindivyn]

Re: 2012 HSC MX2 Marathon

Could have generalised the problem by using the equation xy = k^2 instead =p

Oh and there is actually a nice and cheap way of calculating (or checking) the Implicit Derivative.




So for the equation xy= k^2, we have:



I know it doesn't seem worth it for this particular example, but it really helps when it comes to implicitly differentiating things like:

Would you mind going over the second example? Didn't really understand the method.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Would you mind going over the second example? Didn't really understand the method.

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.

 

[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.