HSC 2012 MX2 Marathon (archive) (4 Viewers)

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Second Question

Observe the diagram. The two loci share the same y value for their centres. The green circle and the blue circle are the largest possible values of k for it to intersect with the red circle whilst having two solutions (well, strictly 1 solution, but we will be using an inequality).

Looking at the radius of the red circle (which is obviously 2 as defined by the locus), the green and blue circle can only intersect on either side of the red circle, in order to have 1 solution.

This occurs when x=1 and x=5.

However since the centre of the circle lies on the line x=0, we can safely say that these x co-ordinates will also be the radius of the blue/green circle.

But the radius of the blue/green circle is the value of k.

Hence

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

for q2, thanks heaps man; perfect diagram understood everything you're a king. i was drawing |z-2i| = k as a circle with centre (0,-2) instead of (0,2) ie |z+2i|=k. i think it makes the question a lot harder? true?

anyway thanks a lot!

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

for the first question, i did it the same way;
i solved -npi/3 = 2kpi. therefore n =-6k where (k C Z) let k = -1 then n =6 but the answer in the solutions say n=7 :S

for q2, thanks heaps man; perfect diagram understood everything you're a king. i was drawing |z-2i| = k as a circle with centre (0,-2) instead of (0,2) ie |z+2i|=k. i think it makes the question a lot harder? true?

anyway thanks a lot!

Suppose the locus was |z+2i|=k.

You will need to use an algebraic proof as opposed to a geometric proof like I did. Here is the method if you were to receive such a question:

1. Find the Cartesian equation of both circles.

2. Attempt to solve them simultaneously. You will acquire a quadratic in terms of constants, x and k.

3. Let the discriminant of this quadratic be greater than 0 (for 2 roots)

4. Simplify and write down the inequality.

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

The answers must be incorrect then. If we were to have n=7, this means that we are rotating backwards 7 times. If we do so, we will not go back to our original vector (again, ignoring modulus). We will in fact overshoot it.

Suppose the locus was |z+2i|=k.

You will need to use an algebraic proof as opposed to a geometric proof like I did. Here is the method if you were to receive such a question:

1. Find the Cartesian equation of both circles.

2. Attempt to solve them simultaneously. You will acquire a quadratic in terms of constants, x and k.

3. Let the discriminant of this quadratic be greater than 0 (for 2 roots)

4. Simplify and write down the inequality.

oh okay, thanks for the method! ill use it if that type of question comes up

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

lol "You must spread some Reputation around before giving it to Carrotsticks again."

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

Answer to the argument question IS 7.



for some integer k.

Setting k=-1 gives us the simplest nontrivial solution of n=7.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Oh dear, I seem to have forgotten that is the identity vector. I did my working out on the incorrect basis that is the identity vector.

My solution for B stays, but take seanieg's solution for A.

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

Answer to the argument question IS 7.



for some integer k.

Setting k=-1 gives us the simplest nontrivial solution of n=7.

wait, i know when you divide complex numbers you subtract the arguments, does it work by the same token if you subtract two complex numbers you divide the arguments? is that what you did in the first step or am i missing something??

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

wait, i know when you divide complex numbers you subtract the arguments, does it work by the same token if you subtract two complex numbers you divide the arguments? is that what you did in the first step or am i missing something??

He was utilising the property:



Only works if you subtract the arguments, not the actual complex number itself.

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

He was utilising the property:



Only works if you subtract the arguments, not the actual complex number itself.

-.- oh yea :/

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Lol you are way too obsessed with complex analysis

Haha yeah probably, gonna be good when I get to uni and do it .

Would post other questions, just not sure what topics this years MX2 students have done.. lol.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Haha yeah probably, gonna be good when I get to uni and do it .

Would post other questions, just not sure what topics this years MX2 students have done.. lol.

At this stage, usually curve sketching, complex numbers and maybe polynomials too.

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

At this stage, usually curve sketching, complex numbers and maybe polynomials too.

Rweally? I haven't started Complex numbers yet. What's a complex number?

Many schools have done bits and pieces of topics.

Mainly, integration, inequalities, induction, conics, bit of probability. Depending on the school.

I think STHS usually does complex numbers and a bit of conics.

SBHS I think complex numbers, polynomials, integration and they include some questions on inverse trig and probability in their first task.

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

At this stage, usually curve sketching, complex numbers and maybe polynomials too.

Many schools have done bits and pieces of topics.

Mainly, integration, inequalities, induction, conics, bit of probability. Depending on the school.

I think STHS usually does complex numbers and a bit of conics.

SBHS I think complex numbers, polynomials, integration and they include some questions on inverse trig and probability in their first task.

Shit. My school's only finished complex numbers.

EDIT: Although, we've finished a fair amount of 2/3 unit (induction, series, locus, probability, perms & combs, geometry of the derivative, circle geo, binomials)

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Shit. My school's only finished complex numbers.

No. Most schools will do Complex Numbers and start one other topic.

It's not a race to see who finishes the course first. It's who knows it the best that counts in the end.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

So I'm guessing no school has done mechanics yet?

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

So I'm guessing no school has done mechanics yet?

NSGHS students claim to have started circular motion from what I have heard. Not sure if this is true though. Other than that no school I know has.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Might try posting a few up that I find then

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Might try posting a few up that I find then

You are best wait until close to HYs to post mechanics questions and get a fair amount of attention from the forum.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

You are best wait until close to HYs to post mechanics questions and get a fair amount of attention from the forum.

Yeah that's probably true.. I just really liked mechanics