HSC 2013 MX2 Marathon (archive) (5 Viewers)

Sy123 · Mar 6, 2013

Re: HSC 2013 4U Marathon

Fixing faulty question

seanieg89 · Mar 9, 2013

Re: HSC 2013 4U Marathon

$Find the minimum possible value of $a^2+b^2$ where $a$ and $b$ are two real numbers such that the polynomial:\\ \\ $x^4+ax^3+bx^2+ax+1$\\ \\ has at least one real root.$

Sy123 · Mar 9, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Find the minimum possible value of $a^2+b^2$ where $a$ and $b$ are two real numbers such that the polynomial:\\ \\ $x^4+ax^3+bx^2+ax+1$\\ \\ has at least one real root.$

That is from the IMO.

seanieg89 · Mar 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
That is from the IMO.

From the early 70s. Those problems are often doable by good mx2 students with no prior olympiad training and this one is no exception.

Sy123 · Mar 11, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
From the early 70s. Those problems are often doable by good mx2 students with no prior olympiad training and this one is no exception.

So would you say that most of the 1970/prior problems are good practise for problem solving skills for mx2 students?

=============

$Find an expression for$

$\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta)$

$Hence prove that$

$\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2}$

seanieg89 · Mar 11, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
So would you say that most of the 1970/prior problems are good practise for problem solving skills for mx2 students?

=============

$Find an expression for$

$\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta)$

$Hence prove that$

$\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2}$

Yep although not directly relevant to the HSC syllabus they are good practice for problem solving and rigorous argument, especially for those who want to pursue mathematics further. You certainly don't have to be of modern olympiad standard to solve most of them.

Exodia · Mar 15, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Find the minimum possible value of $a^2+b^2$ where $a$ and $b$ are two real numbers such that the polynomial:\\ \\ $x^4+ax^3+bx^2+ax+1$\\ \\ has at least one real root.$

$I used a very dodgy method lol$
$I wanted to find the discriminant of$ $x^{4}+ax^{3}+bx^{2}+ax+1$ $to be$ $\geq 0$ $so I made a=0$ $\Rightarrow x^{4}+bx^{2}+1\therefore b^{2}\geq 4\therefore a^{2}+b^{2}\geq 4$

Exodia · Mar 15, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Find the minimum possible value of $a^2+b^2$ where $a$ and $b$ are two real numbers such that the polynomial:\\ \\ $x^4+ax^3+bx^2+ax+1$\\ \\ has at least one real root.$

If it has at least 1 real root doesn't that mean it should have 2 real roots due to the conjugate complex root theorem and the fact that it is a quartic? Or are we not counting multiplicities here? I am sure my solution is wrong because when I sub a=1 and b=0 I get x=-1 as a real solution and 1<4.
What is the proper way to doing this question?

seanieg89 · Mar 15, 2013

Re: HSC 2013 4U Marathon

I won't spoil the question yet for those who are working on it, but there are likely several approaches.

A hint to doing it the way that I did is dividing the poly by x^2, (observing first that 0 cannot be a root of the poly).

Having at least one real root does not mean it must have two real roots, it could just as well have all four roots real.

And your first solution is wishful thinking really, we cannot arbitrarily set a=0, it just changes the problem you are solving.

Exodia · Mar 15, 2013

Re: HSC 2013 4U Marathon

$I get this:$ $\left ( x^{2}+\frac{1}{x^{2}} \right )+a\left ( x+\frac{1}{x} \right )+b=0$
$x+\frac{1}{x}=\frac{-a\pm \sqrt{a^{2}-4b}}{2}$

Lieutenant_21 · Mar 17, 2013

Re: HSC 2013 4U Marathon

@Sy123: I can do $\int\frac{\sin(2n)x}{\sin x}\ dx$ but couldn't do your one

give me a hint please!
By the way, does it have to do with expressing the sum in terms of euler's identity? i.e. the real part of: e^2ix+e^4ix + ... + e^4nix (which I don't think is within the MX2 syllabus).

Lieutenant_21 · Mar 19, 2013

Re: HSC 2013 4U Marathon

I got this: $\int_{0}^{\frac{\pi }{2}}\frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}}dx$ which doesn't look easier then the original lol

seanieg89 · Mar 19, 2013

Re: HSC 2013 4U Marathon

$To sum the cosine series, take the real part of a suitable geometric series. You definitely don't need anything like Euler's formula. I will let you do this yourself, but I shall provide an evaluation of the integral that does not require it.\\ \\To evaluate the integral:\\ $I_n=\int_0^{\pi/2}\frac{\sin ((2n+1)x)}{\sin(x)}\,dx$,\\ \\observe that $\sin((2n+1)x)-\sin((2n-1)x)=2\sin(x)\cos(2nx)$ by differences to products or whatever.\\ \\Hence $I_n=\int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin(x)}\,dx=\int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin(x)}\,dx+2\int_0^{\pi/2}\cos(2nx)\,dx.$\\ The cosine integral is zero, so we have that $I_n=I_{n-1}$ for every positive integer $n$. By induction this implies that $I_n=I_0=\pi/2$ for every positive integer $n$.$

Lieutenant_21 · Mar 19, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$To sum the cosine series, take the real part of a suitable geometric series. You definitely don't need anything like Euler's formula. I will let you do this yourself, but I shall provide an evaluation of the integral that does not require it.\\ \\To evaluate the integral:\\ $I_n=\int_0^{\pi/2}\frac{\sin ((2n+1)x)}{\sin(x)}\,dx$,\\ \\observe that $\sin((2n+1)x)-\sin((2n-1)x)=2\sin(x)\cos(2nx)$ by differences to products or whatever.\\ \\Hence $I_n=\int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin(x)}\,dx=\int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin(x)}\,dx+2\int_0^{\pi/2}\cos(2nx)\,dx.$\\ The cosine integral is zero, so we have that $I_n=I_{n-1}$ for every integer $n$. By induction this implies that $I_n=I_0=\pi/2$ for every integer $n$.$

The only geometric series I can think of are:
$(e^{2ix})^1 + (e^{2ix})^2 + \cdots + (e^{2ix})^n$
and (cis2x)+(cis2x)^2+...+(cis2x)^n

By the way, do we have to memorise the sums to products formulas? I have seen many questions that require them but I think they give the formula or ask to prove that i.e. 2sinacosb=(sin(a+b)+sin(a-b))

I am very interested in knowing the solution to your palindromic polynomial question. So far I am stuck but I found that the roots have to be 1 and/or -1.

seanieg89 · Mar 19, 2013

Re: HSC 2013 4U Marathon

1. You are on the right track. Forget about Euler's formula though. From the perspective of pretty much any MX2 level question it is just an alternative way of writing cis, and one that will cost you marks / potentially confuse you.

2. I didn't but you may find it helpful. What I did memorise was all the ideas though, if I needed the sums to product formula I would be able to figure it out very quickly. I think this is a more efficient use of memory.

3. I will not ruin it for others yet, but the roots certainly don't have to be 1 and/or -1.

HeroicPandas · Mar 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
So would you say that most of the 1970/prior problems are good practise for problem solving skills for mx2 students?

=============

$Find an expression for$

$\cos 2\theta + \cos 4\theta + \cos 6\theta + \dots + \cos (2n\theta)$

$Hence prove that$

$\int_0^{\pi/2} \frac{\sin(2n+1)x}{\sin x} \ dx = \frac{\pi}{2}$

I got ur integral if and only if n is even :O

Also is $cos 2\theta + \dots + cos 2n \theta = \frac{-1}{2} + \frac{sin((2n+1) \theta)}{2 sin \theta}$ ?

Sy123 · Mar 19, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
The only geometric series I can think of are:
$(e^{2ix})^1 + (e^{2ix})^2 + \cdots + (e^{2ix})^n$
and (cis2x)+(cis2x)^2+...+(cis2x)^n

By the way, do we have to memorise the sums to products formulas? I have seen many questions that require them but I think they give the formula or ask to prove that i.e. 2sinacosb=(sin(a+b)+sin(a-b))

I am very interested in knowing the solution to your palindromic polynomial question. So far I am stuck but I found that the roots have to be 1 and/or -1.

You are on the right track with geometric series. Make some strong manipulations so you can simplify your sum much more easily (or just grind out the result)

HeroicPandas said:
I got ur integral if and only if n is even :O

Also is $cos 2\theta + \dots + cos 2n \theta = \frac{-1}{2} + \frac{sin((2n+1) \theta)}{2 sin \theta}$ ?

That result is correct, it should be straightforward from there, I'm not sure how it would prove it for only n even though

HeroicPandas · Mar 19, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
That result is correct, it should be straightforward from there, I'm not sure how it would prove it for only n even though

nevermind... I did a dirty mistake lol

Lieutenant_21 · Mar 19, 2013

Re: HSC 2013 4U Marathon

What is the fastest way to prove the AM:GM inequality? My teacher uses induction + calculus which is a tedious method.

Sy123 · Mar 19, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
What is the fastest way to prove the AM:GM inequality? My teacher uses induction + calculus which is a tedious method.

In my opinion this is one of the few problems where induction isn't "cheap". Its not like we are getting the inequality out of nowhere since we can observe that its true for n=2 n=3 and n=4 rather easily. So its feasible to 'guess' the general case and prove by induction.

i.e. I dislike proofs of induction for identities where you can't possibly come guess the result.

HSC 2013 MX2 Marathon (archive) (5 Viewers)

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

This too shall pass

Well-Known Member

New Member

New Member

Well-Known Member

New Member

Member

Member

Well-Known Member

Member

Well-Known Member

Heroic!

This too shall pass

Heroic!

Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 5)