HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

Sy123 · Feb 22, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
$0 < ^nC_k (p)^k(1-p)^{n-k} < 1$

Why is this a condition for k tails being the most probable outcome?

InteGrand · Feb 22, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $A biased coin is tossed$ \ n \ $times, and the probability that the coin lands tails is$ \ p $. The most probable outcome of the coin tosses is that tails comes up$ \ k \ $times. Find all possible values of$ \ p.$

$\text{Let }p_j \text{ be the probability of getting }j \text{ tails } (j\in \{0,1,2,\cdots,n\}).$

$\text{Then }p_j=\binom{n}{j}p^j q^{n-j},\text{ where } q \equiv 1-p.$ .

$\text{We are given that the largest value of }p_j \text{ is }p_k. \text{ So }k\text{ satisfies}$

$\frac{p_k}{p_{k-1}}>1 \Leftrightarrow \frac{\binom{n}{k}p^k q^{n-k}}{\binom{n}{k-1}p^{k-1} q^{n-k+1}}>1$ $(1)$

$\text{and}$

$\frac{p_k}{p_{k+1}}>1 \Leftrightarrow \frac{\binom{n}{k}p^k q^{n-k}}{\binom{n}{k+1}p^{k+1} q^{n-k-1}}>1.$ $(2)$

$\text{Solving } (1)\text{ yields }p<\frac{k+1}{n+1}. \text{ Solving } (2) \text{ yields } p >\frac{k}{n+1}. \text{ Hence the range of values } p \text{ can take is } \frac{k}{n+1}<p<\frac{k+1}{n+1}.$

Drsoccerball · Feb 22, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
Why is this a condition for k tails being the most probable outcome?

Thats just the probabilty of getting tails

RealiseNothing · Feb 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
Thats just the probabilty of getting tails

yer but what Sy is asking is why is that important?

Sy123 · Feb 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Show that$ \ y^2 = x^2 + x + 1 \ $has no integer solutions$$

Sy123 · Feb 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Determine the integer solutions to$ \ x^2 - y^2 = p \ $for primes$ \ p \ $and positive integers$ \ x,y$

InteGrand · Feb 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Show that$ \ y^2 = x^2 + x + 1 \ $has no integer solutions$$

What about $(x,y) = (0, \pm 1)$ .

seanieg89 · Feb 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Show that$ \ y^2 = x^2 + x + 1 \ $has no integer solutions$$

It has no POSITIVE integer solutions because x^2 < x^2+x+1 < (x+1)^2.

Sy123 · Feb 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
What about $(x,y) = (0, \pm 1)$ .

seanieg89 said:
It has no POSITIVE integer solutions because x^2 < x^2+x+1 < (x+1)^2.

Right, my bad

seanieg89 · Feb 23, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Determine the integer solutions to$ \ x^2 - y^2 = p \ $for primes$ \ p \ $and positive integers$ \ x,y$

By factorising the LHS we are forced to have x-y=1 and x+y=p.

So x=y+1 and p=2y+1.

So there are only solutions for odd primes, and for each odd prime p we have the unique solution pair ((p+1)/2,(p-1)/2).

Sy123 · Feb 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

(Doesn't fit in the other thread)

$\\ $Show that there does not exist a polynomial$ \ f(x) \ $such that$ \\ xf(x-1) = (x+1)f(x)$

InteGrand · Feb 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
(Doesn't fit in the other thread)

$\\ $Show that there does not exist a polynomial$ \ f(x) \ $such that$ \\ xf(x-1) = (x+1)f(x)$

Well, there is the zero polynomial.

seanieg89 · Feb 25, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
(Doesn't fit in the other thread)

$\\ $Show that there does not exist a polynomial$ \ f(x) \ $such that$ \\ xf(x-1) = (x+1)f(x)$

x=0 tells you 0 is a root.

x=k for positive integral k tells you

k-1 is a root => k is a root.

The only poly with infinitely many roots is the zero poly.

FrankXie · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

next question

$f(x) $ is an even function defined for all real values of $ x. $ The graph of $ y=f(x) $ is symmetrical about the line $ x=1. $ For every $ 0\leq x_1, x_2\leq\frac{1}{2}, $ there holds $ f(x_1+x_2)=f(x_1)\cdot f(x_2). $ Let $ a_n=f\left(2n+\frac{1}{2n}\right), $ for $ n=1, 2, 3, \cdots. $ Find $ \lim_{n\rightarrow\infty}a_n.$

seanieg89 · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

FrankXie said:
next question

$f(x) $ is an even function defined for all real values of $ x. $ The graph of $ y=f(x) $ is symmetrical about the line $ x=1. $ For every $ 0\leq x_1, x_2\leq\frac{1}{2}, $ there holds $ f(x_1+x_2)=f(x_1)\cdot f(x_2). $ Let $ a_n=f\left(2n+\frac{1}{2n}\right), $ for $ n=1, 2, 3, \cdots. $ Find $ \lim_{n\rightarrow\infty}a_n.$

Note that composing a reflection about 0 and a reflection about 1 (in that order) gives us the translation $x\mapsto x+2$ .

This means that $f(x)=f(x+2)$ for all real x.

In particular, $a_n=f(1/2n)$ .

Now, by a straightforward induction we have $f(nx)=f(x)^n$ for $x\in [0,1/2n].$ (The constraint on x here ensures that in multiple applications of the $f(x+y)=f(x)f(y)$ rule we stay inside the interval in which this rule is valid.)

This implies $f(1/2n)=f(1/2)^{1/n}.$

So $a_n\rightarrow 1$ unless $f(1/2)=0$ , in which case $a_n\rightarrow 0$ .

Both situations are possible because the constant functions 1 and 0 satisfy the desired properties.

Sy123 · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Show that$ \ \left (\frac{n+1}{2} \right)^n \geq n! \ $for integers$ \ n \geq 1$

SilentWaters · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Apply the AP-GP inequality for integers 1 to n:

$\frac{1}{n}\cdot\frac{n(n+1)}{2}\geq\sqrt[n]{n!}$

Making the cancellation and moving the root to the other side, we obtain the result.

Drsoccerball · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

SilentWaters said:
Apply the AP-GP inequality for integers 1 to n:

$\frac{1}{n}\cdot\frac{n(n+1)}{2}\geq\sqrt[n]{n!}$

Making the cancellation and moving the root to the other side, we obtain the result.

Or use induction

seanieg89 · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Find all functions $f:\mathbb{Q}\rightarrow \mathbb{Q}$ such that:

$f(xy)=f(x)f(y)-f(x+y)+1.$

Sy123 · Feb 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
Find all functions $f:\mathbb{Q}\rightarrow \mathbb{Q}$ such that:

$f(xy)=f(x)f(y)-f(x+y)+1.$

$f(xy) = f(x) f(y) - f(x+y) + 1$

$\\ $Let$ \ x=y=0 \ \Rightarrow \ f(0) = f(0)^2 - f(0) + 1 \Rightarrow \ f(0) = 1$

$\\ $Let$ \ x = 1, y = -1 \ \Rightarrow \ f(-1)(f(1) - 1) = 0 \\ \\ \therefore \ f(-1) = 0 \ $or$ \ f(1) = 1$

$\\ $If$ \ f(1) = 1 \ $then let$ \ y =1 \ \Rightarrow \ f(x) + f(x+1) = f(x) + 1 \Rightarrow \ f(x) = 1 \ $is a solution$$

$\\ $If$ \ f(-1) = 0 \ $then let$ \ y = -1 \ \Rightarrow \ f(-x) + f(x-1) = 1$

$x \mapsto x + \frac{1}{2} \ \Rightarrow \ f\left(-x - \frac{1}{2} \right) + f\left(x - \frac{1}{2} \right) = 1$

$\\ $Let$ \ g(x) = f\left(x - \frac{1}{2} \right)$

$\\ \therefore \ g(x) + g(-x) = 1$

$\\ \therefore \ g(x) = \frac{1}{2} + O(x) \ $for some odd function$ \ O: \mathbb{Q} \rightarrow \mathbb{Q}$

$\\ \therefore \ f(x) = \frac{1}{2} + O\left(x + \frac{1}{2} \right) \ $for any odd function$ \ O: \mathbb{Q} \rightarrow \mathbb{Q}$

$\\ \therefore \ $The functions$ \ f: \mathbb{Q} \rightarrow \mathbb{Q} \ $are$ \\ \\ f(x) = 1 \ $(if f(1) = 1)$ \\ \\ f(x) = \frac{1}{2} + O\left(x+ \frac{1}{2} \right) \ $(otherwise)$$

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I don't think I made any mistakes, is that the solution you're looking for?

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