HSC 2013 MX2 Marathon (archive) (5 Viewers)

gustavo28 · Oct 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove$

$x^x + y^y > x^y +y^x$

$$For distinct positive reals$ \ \ x,y$

edit: my bad

Assume that x>y since the problem is symmetric in x and y.
We can manipulate the inequality into wanting to prove that

x^y(x^(x-y)-1)/y^y >y^(x-y)-1. Now, x^a>y^a if a,x,y are positive since x^a is increasing for positive x (f'(x)=ax^(a-1)). So, x^(x-y)-1>y^(x-y)-1. Moreover x^y/y^y>1. So multiplying these two inequalities together we get the desired result. (Note y^(x-y)-1 is positive since x-y>0)

HeroicPandas · Oct 17, 2013

Re: HSC 2013 4U Marathon

nvm lol

Sy123 · Oct 17, 2013

Re: HSC 2013 4U Marathon

$Find$

$\prod_{m=1}^{n-1} \sin \frac{m\pi}{n}$

gustavo28 · Oct 17, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
So, first of all suppose there is a monic rational quadratic with the root X+1/X with X as the cube root of a non-cube rational. From now on this root will be known as T.
Let this quadratic be x^2+ax+b. So, (x-a)(x^2+ax+b)=x^3+ax^2+bx-ax^2-a^2x-ab=x^3+x(b-a^2)-ab=Q.

So, using the P that has been referred to previously, we get P-Q=some linear rational polynomial which also must have T as a root. But this cant have the irrational root T.

Now, rational polynomials of degree 1 and 2 cant have T as a root then. Suppose some rational polynomial S(x) has T as a root. Note it must be of degree greater than or equal to 2. So, S(x)=M(x)P(x)+R(x) where deg R is 2 or less. Now, R has T as a root. But R can't be a constant, linear or quadratic rational polynomial. So R=0 and P has to divide S.

I think there's a fairly subtantial problem with my proof in that I assume that X+1/X is irrational where X is the cube root of a non-cube rational. However, I'm not entirely sure whether or not this is provable within the syllabus.

hayabusaboston · Oct 17, 2013

Re: HSC 2013 4U Marathon

Looks like Sy123 and RealiseNothing have a new challenger............

lol

RayRay · Oct 17, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
Looks like Sy123 and RealiseNothing have a new challenger............

lol

And im still in cardboard 7 elo

Sy123 · Oct 17, 2013

Re: HSC 2013 4U Marathon

$$Approximate without the use of a calculator$ \ \ \tan^{-1} (2)$

psychotropic · Oct 17, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
I think there's a fairly subtantial problem with my proof in that I assume that X+1/X is irrational where X is the cube root of a non-cube rational. However, I'm not entirely sure whether or not this is provable within the syllabus.

I think there is a much simpler way to do this- btw who are you?
A proof would be along these lines
x^(1/3)+x^(-1/3) is rational =>
(x^(1/3)+x^(-1/3))^2=x^(2/3)+x^(-2/3)+2 is rational=>
x^(2/3)+x^(-2/3) is rational=>
x(x^(1/3)+x^(-1/3))=x^(2/3)+x^(4/3) is rational=>
(x^(2/3)+x^(-2/3))-(x^(2/3)+x^(4/3))=x^(-2/3)+x^(4/3)=(x+x^(-1))x^(1/3) is rational=>
x^(1/3) is rational.
therefore x has to be a rational cube?

ColdLipstick · Oct 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Approximate without the use of a calculator$ \ \ \tan^{-1} (2)$

Sy123 · Oct 17, 2013

Re: HSC 2013 4U Marathon

ColdLipstick said:

Yea well done

(someone post a question please)

psychotropic · Oct 17, 2013

Re: HSC 2013 4U Marathon

Given four concyclic points. For each subset of three points take the incenter. Show that the four incenters from a rectangle.

twinklegal19 · Oct 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yea well done

(someone post a question please)

$A ring of mass 0.6kg is attached to a point P on a string AB of length 1.4m, where AP is 0.8m. The ends A and B are attached to two points 1.0 m apart in a vertical line, A being above B. The ring is made to travel in a horizontal circle with speed v m/s.$

$(i) What is the smallest possible value of v if neither portion of the string is slack?$

$(ii) If v=4.2 m/s, calculate the tension in the portion AP of the string$

RealiseNothing · Oct 17, 2013

Re: HSC 2013 4U Marathon

you're kidding twink

twinklegal19 · Oct 17, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
you're kidding twink

hey, well he asked lol

integral95 · Oct 17, 2013

Re: HSC 2013 4U Marathon

https://mail-attachment.googleuserc...551&sads=Kk6_lPAkNilhsmdT9pXI21JQhms&sadssc=1

Little help pls

bottleofyarn · Oct 17, 2013

Re: HSC 2013 4U Marathon

Here's a doable one, four lines of working and one of my favourites.

$$if$ \, \frac{sin(\pi x)}{\pi x} = \prod_{n = 1}^{\infty}(1 - \frac{x^2}{n^2})$
$$show that$ \; \pi \cot(\pi x) = \frac{1}{x} + \sum_{n = 1}^{\infty}\left(\frac{2x}{x^2 - n^2}\right)$
$where \prod_{n=1}^{\infty}=1\times 2\times 3\times ...$

Sy123 · Oct 17, 2013

Re: HSC 2013 4U Marathon

MethewYan said:
https://mail-attachment.googleuserc...551&sads=Kk6_lPAkNilhsmdT9pXI21JQhms&sadssc=1

Little help pls

Link doesn't work

bottleofyarn said:
Here's a doable one, four lines of working and one of my favourites.

$$if$ \, \frac{sin(\pi x)}{\pi x} = \prod_{n = 1}^{\infty}(1 - \frac{x^2}{n^2})$
$$show that$ \; \pi \cot(\pi x) = \frac{1}{x} + \sum_{n = 1}^{\infty}\left(\frac{2x}{x^2 - n^2}\right)$
$where \prod_{n=1}^{\infty}=1\times 2\times 3\times ...$

1. Take the natural logarithm of both sides
2. Split it up using log laws
3. Differentiate
4. Simplify

integral95 · Oct 17, 2013

Re: HSC 2013 4U Marathon

Screen shot 2013-10-10 at 11.11.40 PM.png

lol ok here it is again

Sy123 · Oct 17, 2013

Re: HSC 2013 4U Marathon

MethewYan said:
View attachment 28799

lol ok here it is again

$i) Re-arranging the property of the polynomial$

$P(k) = \frac{k}{k+1} \Rightarrow \ \ (k+1)P(k) - k = 0 \ \ \ $for$ \ \ k=0,1, \dots , n$

We are being asked for the roots of the polynomial

$(x+1)P(x) - x = 0$

So, basically, for what values of x will that equation hold, well it is given from the above property, if:

$x=0,1, \dots, n$

then the equation holds hence the roots of the polynomial (x+1) P(x) - x

$$ii)$ \ \ (x+1)P(x) - x = Ax(x-1)(x-2)(x-3)\dots (x-n)$

$x = -1 \Rightarrow \ \ 1 = A(-1)(-2)(-3) \dots (-(n+1))$

$A = \frac{(-1)^{n+1}}{(n+1)!}$

$$iii)$ \ \ (n+2)P(n+1) - (n+1) = \frac{(-1)^{n+1}}{(n+1)!} (n+1)(n)\dots (1)$

Then just re-arrange

seanieg89 · Oct 17, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
I think there's a fairly subtantial problem with my proof in that I assume that X+1/X is irrational where X is the cube root of a non-cube rational. However, I'm not entirely sure whether or not this is provable within the syllabus.

This is easy enough:

If X+1/X=r rational, then X is a nonzero irrational root of a rational quadratic. So we can write X=a+b*sqrt(c), with a,b,c rational, c not a perfect square (from the definition of X). By cubing this implies that we must have 3a^2b+cb^3=0. As c is irrational, this forces a=b=0, contradiction.

HSC 2013 MX2 Marathon (archive) (5 Viewers)

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