HSC 2016 MX2 Marathon (archive) (3 Viewers)

Trebla · Mar 26, 2016

Re: HSC 2016 4U Marathon

Just a reminder to please stick to questions related to the HSC syllabus.

Paradoxica · Mar 27, 2016

Re: HSC 2016 4U Marathon

$$\noindent The following system of equations holds true:$ \\ x^2 + xy + y^2 = 9 \\ y^2 + yz + z^2 = 16 \\ z^2 + zx + x^2 = 25 \\ $Find the value of $(x+y+z)^2$, leaving your answer in the form $a+b\sqrt{c}$,$\\$where $c$ is a square free integer, and $a,b$, are integers.$

Drsoccerball · Mar 27, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent The following system of equations holds true:$ \\ x^2 + xy + y^2 = 9 \\ y^2 + yz + z^2 = 16 \\ z^2 + zx + x^2 = 25 \\ $Find the value of $(x+y+z)^2$, leaving your answer in the form $a+b\sqrt{c}$,$\\$where $a,b,c,$ are square free integers.$

Trebla said:
Just a reminder to please stick to questions related to the HSC syllabus.

.

Paradoxica · Mar 27, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
.

You're one to talk, you didn't even do that in the 2014 and 2015 marathons.

porcupinetree · Mar 27, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent The Beta Function B$(x,y)$ is defined as B$(x,y) =\int_0^1 t^{x-1} (1-t)^{y-1} $d$t \\ \indent $Consider the Beta Function for natural numbers $x,y.\\ $a) Show the following: $\\ \indent $i) B$(x,y) = $ B$(y,x) \\ \indent $ii) B$(x,y)=$ B$(x,y+1)+$ B$(x+1,y)\\ \indent $iii) B$(x+1,y)= \frac{x\,\text{B}(x,y)}{x+y}\\$b) Evaluate B$(1,1)\\$c) Prove the following statement using Induction:$\\ \indent $B$(x,y) = \frac{(x-1)!(y-1)!}{(x+y-1)!} $ for all natural numbers $x,y.$

Leaked question 11(a) of 2016 4U HSC exam

DatAtarLyfe · Mar 27, 2016

Re: HSC 2016 4U Marathon

porcupinetree said:
Leaked question 11(a) of 2016 4U HSC exam

what

Paradoxica · Mar 27, 2016

Re: HSC 2016 4U Marathon

porcupinetree said:
Leaked question 11(a) of 2016 4U HSC exam

actually though

nekminnit

InteGrand · Mar 28, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
A game of chance uses a simple harmonic oscillator. Its amplitude of oscillation is 1 but its frequency is unknown to you.

To play the game, you have to pay $1, and you get to choose a finite number of closed subintervals of [-1,1].

You win if after a random large amount of time, the oscillating particle lies in one of your subintervals, and you win $(1/L) where L is the sum of the lengths of your subintervals. So for example, if you were to cover the whole interval [-1,1] with a single subinterval, you will always "win", but as you only get $0.50 of your $1 back, this is not a good strategy!

Q/ How large an expected profit can you obtain with a well chosen bet? Provide proof.

I believe you can obtain an arbitrarily large expected profit by choosing your subintervals to be arbitrarily small and at the endpoints of the whole interval, e.g. choose two intervals like this: [1, 1 – L/2] and [-1, -1 + L/2], and make L arbitrarily small (the total length of these subintervals is L). The motivation for placing them at the ends is that a simple harmonic oscillator spends more time per unit length at the ends, as it is slower there, so any length of subinterval we use should be chosen as close to the ends as possible. We will show that this allows for an arbitrarily high expected value.

$\noindent We can assume wlog that the particle starts at $1$. Let the angular frequency be $\omega >0$. Then the position as a function of time is $x(t) = \cos \omega t$. Let's just consider $0\leq t \leq \frac{\pi}{2\omega}$ (so the first quarter-cycle), as after this everything repeats symmetrically. We will find the fraction of time the particle spends in our chosen subintervals. The particle reaches the end of our subinterval when $x(t) = \cos \omega t= 1-\frac{L}{2}\Longleftrightarrow t=\frac{\cos ^{-1}\left(1-\frac{L}{2}\right)}{\omega}$. Hence the proportion of time $\varpi$ in the first quarter (and by symmetry overall) that the particle spends in our subintervals is $\varpi = \frac{\frac{\cos ^{-1}\left(1-\frac{L}{2}\right)}{\omega}}{\frac{\pi}{2\omega}}=\frac{2}{\pi}\cos ^{-1}\left(1-\frac{L}{2}\right)$.$

$\noindent Hence the probability that the particle lies in our chosen subintervals after an arbitrarily large time is $\frac{2}{\pi}\cos ^{-1}\left(1-\frac{L}{2}\right)$. (It is always a good idea to check if extreme cases work. If $L=2$, i.e. we covered the whole interval, the probability should be 1, and substituting $L=2$ into our formula indeed gives $1$. If $L=0$, the probability should be $0$, and indeed it is if we substitute $L=0$.) So our expected profit is $\frac{1}{L}\cdot \frac{2}{\pi}\cos ^{-1}\left(1-\frac{L}{2}\right) -1$ (subtracting the 1 as this is the game's fee), and as $L\to 0^{+}$, $\frac{\cos ^{-1}\left(1-\frac{L}{2}\right)}{L}$ has limit$

$\begin{align*} \lim_{L\to 0^{+}}\frac{\cos ^{-1} \left(1-\frac{L}{2}\right)}{L} &= \lim_{L \to 0^{+}} \frac{ -\frac{1}{2}\cdot \frac{-1}{\sqrt{1-\left(1-\frac{L}{2}\right)^2}}}{1} \quad (\text{using L'Hospital's rule, as both numerator and denominator go to }0) \\ &= \lim_{L\to 0^{+}} \frac{1}{2\sqrt{1-\left(1-\frac{L}{2}\right)^2}} \\ &= \infty.\end{align*}$

Hence we can make the expected profit arbitrarily large.

Paradoxica · Mar 28, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
I believe you can obtain an arbitrarily large expected profit by choosing your subintervals to be arbitrarily small and at the endpoints of the whole interval, e.g. choose two intervals like this: [1, 1 – L/2] and [-1, -1 + L/2], and make L arbitrarily small (the total length of these subintervals is L). The motivation for placing them at the ends is that a simple harmonic oscillator spends more time per unit length at the ends, as it is slower there, so any length of subinterval we use should be chosen as close to the ends as possible. We will show that this allows for an arbitrarily high expected value.

$\noindent We can assume wlog that the particle starts at $1$. Let the angular frequency be $\omega >0$. Then the position as a function of time is $x(t) = \cos \omega t$. Let's just consider $0\leq t \leq \frac{\pi}{2\omega}$ (so the first quarter-cycle), as after this everything repeats symmetrically. We will find the fraction of time the particle spends in our chosen subintervals. The particle reaches the end of our subinterval when $x(t) = \cos \omega t= 1-\frac{L}{2}\Longleftrightarrow t=\frac{\cos ^{-1}\left(1-\frac{L}{2}\right)}{\omega}$. Hence the proportion of time $\varpi$ in the first quarter (and by symmetry overall) that the particle spends in our subintervals is $\varpi = \frac{\frac{\cos ^{-1}\left(1-\frac{L}{2}\right)}{\omega}}{\frac{\pi}{2\omega}}=\frac{2}{\pi}\cos ^{-1}\left(1-\frac{L}{2}\right)$.$

$\noindent Hence the probability that the particle lies in our chosen subintervals after an arbitrarily large time is $\frac{2}{\pi}\cos ^{-1}\left(1-\frac{L}{2}\right)$. (It is always a good idea to check if extreme cases work. If $L=2$, i.e. we covered the whole interval, the probability should be 1, and substituting $L=2$ into our formula indeed gives $1$. If $L=0$, the probability should be $0$, and indeed it is if we substitute $L=0$.) So our expected profit is $\frac{1}{L}\cdot \frac{2}{\pi}\cos ^{-1}\left(1-\frac{L}{2}\right) -1$ (subtracting the 1 as this is the game's fee), and as $L\to 0^{+}$, $\frac{\cos ^{-1}\left(1-\frac{L}{2}\right)}{L}$ has limit$

$\begin{align*} \lim_{L\to 0^{+}}\frac{\cos ^{-1} \left(1-\frac{L}{2}\right)}{L} &= \lim_{L \to 0^{+}} \frac{ -\frac{1}{2}\cdot \frac{-1}{\sqrt{1-\left(1-\frac{L}{2}\right)^2}}}{1} \quad (\text{using L'Hospital's rule, as both numerator and denominator go to }0) \\ &= \lim_{L\to 0^{+}} \frac{1}{2\sqrt{1-\left(1-\frac{L}{2}\right)^2}} \\ &= \infty.\end{align*}$

Hence we can make the expected profit arbitrarily large.

What would happen if it didn't have a simple frequency? e.g. cos(t^2) or something like that.

seanieg89 · Mar 28, 2016

Re: HSC 2016 4U Marathon

Good stuff Integrand, that is exactly correct.

Of course all this question really amounts to is the statement that the PDF of the particles position at a random large time tends to infinity as you approach +-1.

seanieg89 · Mar 28, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
What would happen if it didn't have a simple frequency? e.g. cos(t^2) or something like that.

For simple things like power functions, you would still be able to get arbitrarily large EV, but the calculations are messier.

Try to show this yourself.

To find a g where sin(g(t)) (with g(t) increasing) did not have this property, you could cook up g so that the particle slows down heaps every time it is near zero for instance.

Paradoxica · Mar 28, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
For simple things like power functions, you would still be able to get arbitrarily large EV, but the calculations are messier.

Try to show this yourself.

To find a g where sin(g(t)) (with g(t) increasing) did not have this property, you could cook up g so that the particle slows down heaps every time it is near zero for instance.

$g(t) = t+\sin{t}$

I can't seem to find a simple function where the particle spends most of it's time near the origin...

Time to go on a function hunt.

Hmm... interesting... I have found a function where the particle spends significantly more time at -1 than 1...

Paradoxica · Mar 28, 2016

Re: HSC 2016 4U Marathon

$\noindent$g(t) \equiv t+\sin{t}+\sin{\left(t+\sin{t}\right)}\\ \sin(g(t)) $ results in a particle that spends a highly disproportionate amount of time near the origin. Now, I presume ''simple'' harmonic oscillation is always a linear input of time?$

seanieg89 · Mar 28, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent$g(t) \equiv t+\sin{t}+\sin{\left(t+\sin{t}\right)}\\ \sin(g(t)) $ results in a particle that spends a highly disproportionate amount of time near the origin. Now, I presume ''simple'' harmonic oscillation is always a linear input of time?$

Yes simple harmonic motion is just the case where g is a linear polynomial in time, ie the case treated by Integrand.

seanieg89 · Mar 31, 2016

Re: HSC 2016 4U Marathon

A classic:

1. Use the trapezoidal approximation to show that there exists a positive constant A such that $An^{n+1/2}e^{-n}\leq n!$ for all positive integers n.

2. Using another inequality coming from integration, show there exists a positive constant B such that $n! \leq Bn^{n+1/2}e^{-n}.$

3. By considering the binomial coefficient $\binom{2n}{n}$ , find $\lim_{n\rightarrow \infty} \frac{n!}{n^{n+1/2}e^{-n}}.$ (You may assume without proof that this limit exists.)

Hint for 3: You might find it useful to think about powers of trig functions.

E. (Extension) It is a basic lemma in real analysis that if a sequence of real numbers is monotonic and bounded, it is convergent. Using this lemma or otherwise, fill in the gap in Q3 and prove that $\lim_{n\rightarrow \infty} \frac{n!}{n^{n+1/2}e^{-n}}$ exists.

Nailgun · Apr 1, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent The following system of equations holds true:$ \\ x^2 + xy + y^2 = 9 \\ y^2 + yz + z^2 = 16 \\ z^2 + zx + x^2 = 25 \\ $Find the value of $(x+y+z)^2$, leaving your answer in the form $a+b\sqrt{c}$,$\\$where $c$ is a square free integer, and $a,b$, are integers.$

$Observe that each equation follows the cosine rule i.e.$\\ { a }^{ 2 }={ b }^{ 2 }+{ c }^{ 2 }-2bc\cos { A } \\ $Hence each equation can be used to construct a triangle with an angle of 120 degrees as $ -2bc\cos { A } =bc,\quad \therefore \quad A=120°$

$$ You should have three triangles, T1. with sides x, y, and 3 (opp. to 120 degree angle) T2. with sides z, y, and 4 (opp. to 120 degree angle) and T3. with sides z, x, 5 (opp. to 120 degree angle). Thus since we have three 120 degree angles and 3 pairs of matching sides (x,y,z), we can place the triangles together and construct a triangle (120*3 = 360) forming a right angled triangle with sides 3,4,5.$

$A_{ TOTAL }=\frac { 1 }{ 2 } \times 3\times 4=6\quad units^{ 2 }\\ A_{ 1 }=\frac { 1 }{ 2 } xy\sin { 120 } =\frac { \sqrt { 3 } }{ 4 } xy\quad units^{ 2 }\\ \\ A_{ 2 }=\frac { 1 }{ 2 } yz\sin { 120 } =\frac { \sqrt { 3 } }{ 4 } yz\quad units^{ 2 }\\ \\ A_{ 3 }=\frac { 1 }{ 2 } xz\sin { 120 } =\frac { \sqrt { 3 } }{ 4 } xz\quad units^{ 2 }$

$A_{ TOTAL }=A_{ 1 }+A_{ 2 }+A_{ 3 }\\ 6=\frac { \sqrt { 3 } }{ 4 } \left( xy+yz+xz \right) \\ xy+yz+xz=\frac { 24 }{ \sqrt { 3 } }$

$$Adding the original equations together$\\{ 2x }^{ 2 }+{ 2y }^{ 2 }+{ 2z }^{ 2 }+xy+yz+xz=50\\ \\ { 2x }^{ 2 }+{ 2y }^{ 2 }+{ 2z }^{ 2 }=50-\frac { 24 }{ \sqrt { 3 } } \\ \\ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=\frac { 25\sqrt { 3 } -12 }{ \sqrt { 3 } }$

$$Expanding $ (x+y+z)^{ 2 }={ x }^{ 2 }+y^{ 2 }+{ z }^{ 2 }+2xy+2yz+2zx\\ (x+y+z)^{ 2 }=\frac { 25\sqrt { 3 } -12 }{ \sqrt { 3 } } +2\left( \frac { 24 }{ \sqrt { 3 } } \right) \\ \\ (x+y+z)^{ 2 }=25+12\sqrt { 3 } \\ \\ \therefore \quad a=25,\quad b=12,\quad c=3$

leehuan · Apr 3, 2016

Re: HSC 2016 4U Marathon

I had to help a student do this maths question just now. I feel it would be good to leave it on the 4U marathon as an instructive exercise for any E3 or above aiming student to attempt.

$Suppose $\frac{z-1+i}{z+1-i}=ki$ where $k$ is a real number. \\ \\ By drawing a diagram or otherwise, find the value of $|z|$. Show all reasoning.$

Drsoccerball · Apr 3, 2016

Re: HSC 2016 4U Marathon

leehuan said:
I had to help a student do this maths question just now. I feel it would be good to leave it on the 4U marathon as an instructive exercise for any E3 or above aiming student to attempt.

$Suppose $\frac{z-1+i}{z+1-i}=ki$ where $k$ is a real number. \\ \\ By drawing a diagram or otherwise, find the value of $|z|$. Show all reasoning.$

$|z| = \sqrt{2}$ ?

leehuan · Apr 3, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
$|z| = \sqrt{2}$ ?

Yes lol

But I want the 2016ers to attempt the question with full working. If they don't want to upload a diagram they can just briefly explain it

Drsoccerball · Apr 3, 2016

Re: HSC 2016 4U Marathon

leehuan said:
Yes lol

But I want the 2016ers to attempt the question with full working. If they don't want to upload a diagram they can just briefly explain it

I didn't use a diagram

you don't need to

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