HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Dec 28, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
That feel when questions you made (legitimately) have already been asked in the HSC ='(

2009 Q7

Awww man :/

Heh, I guess I'll try make another one

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
I made this question just now:

$$If $ \ \ \ z= \cos \theta + i \sin \theta$

$$Consider $ \ \ \ (z+\frac{1}{z})^{2n}$

$$Show that$ \int_{0}^{\frac{\pi}{2}} \cos^{2n} \theta \ d\theta = \frac{\pi (2n)!}{2^{2n+1}(n!)^2}$

Noting that:

$z^n + z^{-n} = 2cos(n\theta)$

By expansion:

$(z+\frac{1}{z})^{2n} = z^{2n} + \binom{2n}{1}z^{2n-2} + \binom{2n}{2}z^{2n-4} + ... + \binom{2n}{2n-1}z^{-2n+2} + \binom{2n}{2n}z^{-2n}$

Also using the first identity:

$(z+\frac{1}{z})^{2n} = (2cos\theta)^{2n}$

Collecting similar powers but opposite in sign:

$(2cos\theta)^{2n} = (z^{2n} + z^{-2n}) + (\binom{2n}{1}z^{2n-2}+\binom{2n}{2n-1}z^{-2n+2}) + ... + (\binom{2n}{n-1}z^{2} + \binom{2n}{n+1}z^{-2}) + \binom{2n}{n}$

Using the first identity again and applying the symmetry property of combinatorics:

$(2cos\theta)^{2n} = 2cos(2n\theta}) + 2\binom{2n}{1}cos(2n-2)\theta + ... + 2\binom{2n}{n-1}cos2\theta + \binom{2n}{n}$

Now we integrate both sides:

$2^{2n} \int_{0}^{\frac{\pi}{2}} cos^{2n}\theta d\theta = \int_{0}^{\frac{\pi}{2}}2cos(2n\theta}) + 2\binom{2n}{1}cos(2n-2)\theta + ... + 2\binom{2n}{n-1}cos2\theta + \binom{2n}{n}$

$2^{2n} \int_{0}^{\frac{\pi}{2}} cos^{2n}\theta d\theta = [\frac{2sin2n\theta}{2n} + \binom{2n}{1}\frac{2sin(2n-2)\theta}{2n-2} + ... + \binom{2n}{n-1}\frac{2sin2\theta}{2} + \binom{2n}{n}\theta]_0^{\frac{\pi}{2}}$

Now when we substitute in the limits, we end up with many:

$sin(2k\pi)$

Where $k$ is an integer. They are all equal to 0, so it all cancels out besides the last term, which leaves us with:

$2^{2n} \int_{0}^{\frac{\pi}{2}} cos^{2n}\theta d\theta = \binom{2n}{n}\frac{\pi}{2}$

Now changing the combinatorics into factorial notion, we obtain:

$2^{2n} \int_{0}^{\frac{\pi}{2}} cos^{2n}\theta d\theta = \frac{\pi}{2} \times \frac{(2n)!}{n!n!}$

Taking the power of 2 to the other side gives us the required result:

$\int_{0}^{\frac{\pi}{2}} cos^{2n}\theta d\theta = \frac{\pi(2n)!}{2^{2n+1}(n!)^2}$

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

ok let me fix all them typos up lol.

kazemagic · Dec 28, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
That feel when questions you made (legitimately) have already been asked in the HSC ='(

2009 Q7

please tell me how the hell do you rmbr these stuff

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

Find and justify the value of the following in terms of $n$ :

$tan^2\theta(\frac{1-sin^{2n}\theta}{1-cos^{2n}\theta})$

When $\theta = 0$ and $n$ is an integer.

Hence find:

$\lim_{n \to \infty} tan^2\theta(\frac{1-sin^{2n}\theta}{1-cos^{2n}\theta})$

When $\theta = 0$

Note this is more of a research kinda thing rather than an actual problem. I found an answer to this but it makes no sense at all besides all my working seeming legit. Curious to see what people get.

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

^^^ this question is trippin' me out. I was playing around with something and found the above result. But the result doesnt make that much sense.

seanieg89 · Dec 28, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Find and justify the value of the following in terms of $n$ :

$tan^2\theta(\frac{1-sin^{2n}\theta}{1-cos^{2n}\theta})$

When $\theta = 0$ and $n$ is an integer.

Hence find:

$\lim_{n \to \infty} tan^2\theta(\frac{1-sin^{2n}\theta}{1-cos^{2n}\theta})$

When $\theta = 0$

Note this is more of a research kinda thing rather than an actual problem. I found an answer to this but it makes no sense at all besides all my working seeming legit. Curious to see what people get.

Isn't the first thing undefined at theta=0?

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Isn't the first thing undefined at theta=0?

You end up with $\frac{0}{0}$

So you have to manipulate it so it doesn't end up like this (well that's what I'm intending). You could always L'Hopital's it but that's no fun.

Unless you can't do this? (because upon manipulation you get a value). I'm going to take a guess and say you can't.

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

What if I was to make it such that $\theta \to 0$

seanieg89 · Dec 28, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
You end up with $\frac{0}{0}$

So you have to manipulate it so it doesn't end up like this (well that's what I'm intending). You could always L'Hopital's it but that's no fun.

Unless you can't do this? (because upon manipulation you get a value). I'm going to take a guess and say you can't.

Well firstly you asked for its value at theta=0 rather than its limit as theta->0. But assuming you meant the latter then your second question might not work out the way you want, as the two limiting processes (theta->0,n->inf) don't have to commute. I will think about it a little more properly now.

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Well firstly you asked for its value at theta=0 rather than its limit as theta->0. But assuming you meant the latter then your second question might not work out the way you want, as the two limiting processes (theta->0,n->inf) don't have to commute. I will think about it a little more properly now.

Alright thanks. I was thinking more along the lines of $\theta \to 0$ but didn't really put that to paper.

seanieg89 · Dec 28, 2012

Re: HSC 2013 4U Marathon

Yep, so then the first question is fine and pretty straightforward using mx2 methods: 1/n.

The latter question turns out to be true taking limits in either order, ie:

$\lim_{n\rightarrow \infty}\lim_{\theta\rightarrow 0} \frac{\tan^2(\theta)(1-\sin^{2n}(\theta))}{1-\cos^{2n}(\theta)}=\lim_{\theta\rightarrow 0}\lim_{n\rightarrow\infty}\frac{\tan^2(\theta)(1-\sin^{2n}(\theta))}{1-\cos^{2n}(\theta)}=0$

seanieg89 · Dec 28, 2012

Re: HSC 2013 4U Marathon

My solution to 1.

$Let $u=\sin^2\theta,v=\cos^2\theta.$\\Then $\lim_{\theta\rightarrow 0} \frac{u}{v}\cdot\frac{1-u^n}{1-v^n}=\lim_{\theta\rightarrow 0}\frac{u}{v}\cdot\frac{1-u}{1-v}\cdot\frac{1+u+\ldots+u^{n-1}}{1+v+\ldots+v^{n-1}}\\=\lim_{\theta\rightarrow 0}\frac{u}{v}\cdot\frac{v}{u}\cdot\frac{1+u+\ldots+u^{n-1}}{1+v+\ldots+v^{n-1}}=\frac{1+0+\ldots+0}{1+1+\ldots+1}=\frac{1}{n}.$$

Sy123 · Dec 28, 2012

Re: HSC 2013 4U Marathon

$The nth Fermat number $ \ \ F_n \ \ $is defined by$

$F_n= 2^{(2^n)} + 1$

$$Prove by Mathematical Induction or otherwise$ \\ \\ F_0 F_1 F_2 ... F_{k-1} \equiv \prod_{m=0}^{k-1}F_{k-1} = F_{k} - 2$

$Hence deduce that no two Fermat numbers have a common factor greater than 1$

$$Hence show that there are infinitely many prime numbers$^{(*)}$

$(*) \ \ \rightarrow \ \ \ $A prime number is defined as a natural number with no common factor besides 1 and itself. 1 is not itself a prime$$

RealiseNothing · Dec 28, 2012

Re: HSC 2013 4U Marathon

You can use a really simple proof by contradiction to get the same result:

Assume that there is a finite amount of primes, and let there be $n$ amount, denoted by $p_k$ for the k'th prime.

Then we do:

$p_1p_2p_3...p_n+1$

This makes a prime which has not be accounted for, and hence there is more than $n$ amount of primes. Thus it is a contradiction, and the result follows.

I'll attempt your way though.

Sy123 · Dec 28, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
You can use a really simple proof by contradiction to get the same result:

Assume that there is a finite amount of primes, and let there be $n$ amount, denoted by $p_k$ for the k'th prime.

Then we do:

$p_1p_2p_3...p_n+1$

This makes a prime which has not be accounted for, and hence there is more than $n$ amount of primes. Thus it is a contradiction, and the result follows.

I'll attempt your way though.

Wow, very good. This is just a 'filler' question while I find/make something else to ask, but I want to stick with Complex Numbers/Polynomials so others can contribute.

Also, this question is from another STEP paper but I forget which one.

seanieg89 · Dec 28, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
You can use a really simple proof by contradiction to get the same result:

Assume that there is a finite amount of primes, and let there be $n$ amount, denoted by $p_k$ for the k'th prime.

Then we do:

$p_1p_2p_3...p_n+1$

This makes a prime which has not be accounted for, and hence there is more than $n$ amount of primes. Thus it is a contradiction, and the result follows.

I'll attempt your way though.

Lies! p1p2...pn doesn't have to be prime!

But it does have to contain a prime not previously listed as a factor.

Sy123 · Dec 30, 2012

Re: HSC 2013 4U Marathon

$$Question Pell - 15 marks$

$Define the Pell numbers by the recursive relation$

$P_n=\begin{cases} 0 \ \ \ \ \ $if$ \ \ n = 0 \\ 1 \ \ \ \ \ $if$ \ \ n = 1 \\ 2P_{n-1}+P_{n-2} \ \ \ \ \ n \geq 2 \end{cases}$

$$i) Find the first 5 Pell numbers$ \ \ \ \ \ \fbox{1}$

$$ii) Prove using Mathematical Induction that$ \\ \\ P_{n+1}P_{n-1}-P_n^2=(-1)^n \ \ \ \ \ $for$ \ \ n \geq 1 \ \ \ \ \ \fbox{3}$

$$iii) Hence prove that $ \ \ \ x=P_{2n-1}+P_{2n} \ \ $and$ \ \ y= P_{2n} \ \ \ \ $satisfy the equation for all positive integers n$ \\ \\ x^2-2y^2=1 \ \ \ \fbox{3}$

$$The kth triangular number can be found by$ \ \ \ t_k = \frac{k(k+1)}{2} \ \ \ $and the mth square number can be found by$ \ \ \ s_m = m^2$
$Let $ \ \ N_p \ \ $denote the pth number that is both triangular and square$
$$Hence there exists integers, k and m such that$ \ \ m^2=\frac{k(k+1)}{2} = N_p$

$$iv) Show that$ \ \ \ x=2k+1 \ \ \ y=2m \\ $satisfy the equation$ \ \ \ x^2-2y^2=1 \ \ \ \fbox{3}$

$$v) Hence using part (iii) and (iv), show that N_p = \left ( \frac{P_{2p}}{2} \right ) ^2 \ \ \ \fbox{2}$

$$vi) Considering the equality$ \ \ \ \ \lim_{j \to \infty} \frac{P_{j+1}}{P_j} = \lim_{j \to \infty} \frac{P_{j}}{P_{j-1}}$

$$Find$ \ \ \lim_{p\to \infty} \frac{N_{p+1}}{N_p} \ \ \ \ \fbox{3}$

RivalryofTroll · Dec 30, 2012

Re: HSC 2013 4U Marathon

Just double checking my answers for a question:

If (1+i) is a root of the equation:

x^3 - ax^2 + bx - 4 = 0

Find values a & b. Hence, solve the equation.

I got a = 4 and b = 6.

Then roots are (1+i), (1-i), 2?

Sy123 · Dec 30, 2012

Re: HSC 2013 4U Marathon

RivalryofTroll said:
Just double checking my answers for a question:

If (1+i) is a root of the equation:

x^3 - ax^2 + bx - 4 = 0

Find values a & b. Hence, solve the equation.

I got a = 4 and b = 6.

Then roots are (1+i), (1-i), 2?

You can find all roots without finding a and b by considering conjugate root theorem and product of roots:

$4=(1+i)(1-i) \alpha \ \ \ \ \rightarrow \alpha=2$

And then considering sum of roots 2 at a time and sum of roots we can then find a and b. You are correct.

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