HSC 2012 MX2 Marathon (archive) (1 Viewer)

deswa1 · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

Do it geometrically:
<img src="http://latex.codecogs.com/gif.latex?arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" title="arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" />

Try going from here. Tell us how you go (I haven't tried the question yet so I don't know the answer.

Carrotsticks · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

Cyberbully said:
How would you approach this algebraically (or any general explanation for that matter):

find the locus of arg(z(z-(sqrt(3)+i))) = pi/6

$\\ \arg(z(z-(\sqrt{3}+i))) = \arg(z) + \arg(z-(\sqrt{3}+i)) \qquad $using a property of the argument.$$

Try it from here.

And beaten to it.

deswa1 · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

Yes, I finally beat Carrotsticks to something

.

Cyberbully · Apr 17, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Do it geometrically:
<img src="http://latex.codecogs.com/gif.latex?arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" title="arg(z(z-(\sqrt{3}+i))=\frac{\pi}{6}\\ argz+arg(z-(\sqrt{3}+i))=\frac{\pi}{6}" />

Try going from here. Tell us how you go (I haven't tried the question yet so I don't know the answer.

tried that.

didn't get very far.

the "answer" is a straight line through origin and (sqrt(3) + i), excluding the point (sqrt(3) + i) on an argand diagram if it helps...

math man · Apr 29, 2012

Re: 2012 HSC MX2 Marathon

New question: Prove by induction that every natural number greater than 2 has a prime divisor.

Inference · May 1, 2012

Re: 2012 HSC MX2 Marathon

math man said:
New question: Prove by induction that every natural number greater than 2 has a prime divisor.

There is a stronger result, that every natural number greater than 1 has a prime divisor. Anyhow, several methods, first as per the question states:

Assume a set $Q = \{2, 3, \cdots, k-1\}$ where $\forall \ n \in Q$ , then n > 1 and n has a prime divisor. Now to show that k exists in Q is quite straightforward with the following partition:

If k has no other factors than k or 1, then k is trivially in Q.

If k is nonprime, then k will have a factor w such that $1<w<k$ , thus $w \in Q$ by the inductive hypothesis.

hence $k \in Q$

Next method is to show a even stronger result, in other words, the fundmental theorem of arithmetic.

Assume that there are numbers which can not be expressed as a product of primes. Let the smallest possible number of this kind be $m$ .

$m$ can not be $1$ since $1$ is neither composite nor prime. $m$ can not be prime since the PPF of a prime number is just itself. Thus $m$ must be a composite number.

Let the composition of $m = ab$ where $\{a,b\} < m$

Since $m$ was the smallest number that can not be expressed as a product of primes, this means $a$ and $b$ can be expressed as a product of primes and consequently we get $m = ab$ where $a$ and $b$ can be both expressed as primes. Contradiction!

Thus $m$ can also be expressed as a product of primes.

Lemma 1: If $p$ is a prime and $p|a_1a_2a_3 \cdots a_n$ then $p|a_i$ for some $i$ .

Lemma 2: If $p$ and $q$ are primes and $e$ is a natural number and $p|q^e$ then $p=q$ .

Let's assume that for some number $k$ that there are $2$ (at least) ways of expressing its PPF.

$\therefore k = p_1^{f_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = q_1^{e_1}q_2^{e_2}q_3^{e_3} \cdots q_m^{e_m}$

Clearly for all $\mbox{n}$ , $p_n|k$

$\implies p_n| q_1^{e_1}q_2^{e_2}q_3^{e_3} \cdots q_m^{e_m}$

By Lemma 1 $p_n|q_m^{e_m}$ for any $m$ .

By Lemma 2 $p_n=q_m$

This means that for all $\mbox{n}$ and all $m$ there are values of $p$ which equals to those of $q$ . For example, $p_3$ could equal to $q_5$ , or $p_6=q_3$ etc. This also means we have created a bijection between $p$ and $q$ such that $n = m$ .

Therefore if the number $k$ has $2$ PPF's then the prime number 'base' will be exactly the same, the only different would be in the powers, namely $f_n$ and $e_m$ .

Now since each $p_n$ has a corespondent equivalent $q_m$ we can rewrite $k$ as:

$p_1^{f_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = p_1^{e_1}p_2^{e_2}p_3^{e_3} \cdots p_n^{e_n}$

$p_1^{f_1-e_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = p_2^{e_2}p_3^{e_3} \cdots p_n^{e_n}$

$\mbox{LHS}|p_1$ however $\mbox{RHS}$ can not be divided by $p_1$ unless for some $2 \le j \le n$ such that $p_j = p_1$

But since $p_1 \neq p_2 \neq p_3 \neq \cdots \neq p_n$ we have a contradiction.

lolcakes52 · May 5, 2012

Re: 2012 HSC MX2 Marathon

Question 1: ß and ∂ are complex roots of the equation x^3 + 5x + 1=0. Show that the other root is -1/|ß|^2
Question 2: Show that ß∂ is a root of the equation x^3 -5x^2 -1=0.

Aesytic · May 5, 2012

Re: 2012 HSC MX2 Marathon

1.since the equation is a cubic, there must be at least 1 real root, so therefore the last root must be real since alpha and beta are complex roots
also, since the coefficients of the polynomial are all real, the 2 complex roots alpha and beta must be conjugates of one another because the other root is real, and the conjugate of a real number is the same, leaving alpha and beta to form a conjugate pair
using the product of roots, and letting the last root be gamma, alpha*beta*gamma = -1
gamma = -1/alpha*beta
when a complex number and its conjugate is multiplied together, the result is the modulus squared
.'. gamma = -1/|beta|^2

2. subbing gamma into the equation x^3 + 5x + 1,
-1/(|beta|^2)^3 - 5/|beta|^2 + 1 = 0 since -1/|beta|^2 is a root of this equation
mutiplying everything by |beta|^6,
-1 - 5|beta|^4 + |beta|^6 = 0
(|beta|^2)^3 - 5(|beta|^2)^2 - 1 = 0
looking at this equation, we can therefore see that |beta|^2 is a root of the equation x^3 - 5x^2 - 1 = 0, and from before it was established that alpha*beta = |beta|^2
.'. alpha*beta is a root of the equation x^3 -5x^2 - 1 =0

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

good stuff, someone revised this thread

AAEldar · May 5, 2012

Re: 2012 HSC MX2 Marathon

I'll add a question into the loop...

$Find without using a trigonometric substitution:$

$\int \frac{1}{\sqrt{1+x^2}}\;dx$

Want to see the working and I don't want to see any hyperbolic functions being thrown in there!

Also I doubt you'd ever get something like that, you'd probably be given the first step or a substitution but it's still fun!

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

AAEldar said:
I'll add a question into the loop...

$Find without using a trigonometric substitution:$

$\int \frac{1}{\sqrt{1+x^2}}\;dx$

Want to see the working and I don't want to see any hyperbolic functions being thrown in there!

Also I doubt you'd ever get something like that, you'd probably be given the first step or a substitution but it's still fun!

this is on the standard integrals sheet lol

AAEldar · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
this is on the standard integrals sheet lol

Yea but a good exercise to show it

lolcakes52 · May 5, 2012

Re: 2012 HSC MX2 Marathon

Okay.
Question 1: The ellipse (x^2)/2 +y^2 = 1 is intercepted at two distinct points P(x1,y1) and Q(x2,y2) by the line y=mx+3. Show that points x1 and x2 are the solutions to the equation (1+2m^2)x^2 +12mx +16 =0
Question 2: Find the values of m for which x1 and x2 are real distinct points. (sorry for the poor phrasing)
Question 3: Find the equations of the tangents to the ellipse from the point (0,3).

Aesytic · May 5, 2012

Re: 2012 HSC MX2 Marathon

1. solving simultaneously with the ellipse and the line,
(x^2)/2 + (mx+3)^2 = 1
x^2 + 2(mx+3)^2 = 2
x^2 + 2m^2x^2 + 12mx + 18 = 2
(1+2m^2)x^2 + 12mx + 16 = 2
since the line and the ellipse intersect when x = x1 and when x=x2, then x1 and x2 are solutions to this equation
2. in order for x1 and x2 to be real distinct points, the discriminant of the equation must be >0
.'. 144m^2 - 4*16(1+2m^2) > 0
144m^2 - 64 - 128m^2 > 0
16m^2 - 64 > 0
m^2 - 4 > 0
(m-2)(m+2)>0
.'. m>2 , m<-2
3. the line y=mx + 3 is a tangent to the ellipse that passes through (0,3) when the points x1 and x2 are the same
x1 and x2 are the same when the discriminant equals 0
replacing the inequality in question 2 with an equality sign,
m=2 or -2
.'. y = 2x + 3 or y = -2x + 3

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

prove that the area subtended by the tangent P to the curve xy=32, the X axis, and the Y axis is constant. Find the value of this constant

karnbmx · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
prove that the area subtended by the tangent P to the curve xy=32, the X axis, and the Y axis is constant. Find the value of this constant

basically, you find the equation of the tangent at P using parametric coordinates (root(32)t,root(32)/t), then using X and Y intercepts, calculate the area (the parameters should cancel out and give a constant value).

I'll type up the full solution soon.

Just out of interest Barbernator, where did you get this question from?

EDIT:

here you go:

x = 4root(2)t y = 4root(2)/t, where t is a parameter

m = -1/t^2

therefore, y -4root(2)/t = -1(x-4root(2)/t)/t^2

Therefore x int: (8root(2)t, 0) y-int: (0,8root(2)/t)

Hence Area of Triangle = 8root(2)t * 8root(2)/t * 1/2 = 32root2 units squared

As there are no parameters present, the area must be a constant, with a value of 32root(2).

I sort of did this in a rush, so sorry if the answer is wrong.

nightweaver066 · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
prove that the area subtended by the tangent P to the curve xy=32, the X axis, and the Y axis is constant. Find the value of this constant

Diagram:
http://i.imgur.com/kEmRg.jpg

$h: xy = 32$

$Let~P(ct, \frac{c}{t})~be~a~point~on~h~where~c^2 = 32$

$$Differentiating both sides w.r.t. x$

$x\frac{dy}{dx} + y = 0$

$\frac{dy}{dx} = \frac{-y}{x}$

$= -\frac{1}{t^2}~at~P(ct, \frac{c}{t})$

$\therefore $ Equation of tangent at P:$$

$y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$

$$At Y, x = 0$, \therefore y - \frac{c}{t} = \frac{c}{t}$

$y = \frac{2c}{t} => Y(0, \frac{2c}{t})$

$$At X, y = 0$, \therefore \frac{c}{t} = \frac{1}{t^2}(x - ct)$

$x - ct = ct$

$x = 2ct => X(2ct, 0)$

$Area~of~\Delta XPY = \frac{1}{2} \times 2ct \times \frac{2c}{t}$

$= 2c^2$

$= 64$

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

karnbmx said:
basically, you find the equation of the tangent at P using parametric coordinates (root(32)t,root(32)/t), then using X and Y intercepts, calculate the area (the parameters should cancel out and give a constant value).

I'll type up the full solution soon.

Just out of interest Barbernator, where did you get this question from?

i made it up, but i have seen questions like this before. but there is the geometric property of the rectangular hyperbola that states that the area is constant.

Carrotsticks · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
i made it up, but i have seen questions like this before. but there is the geometric property of the rectangular hyperbola that states that the area is constant.

Could have generalised the problem by using the equation xy = k^2 instead =p

Oh and there is actually a nice and cheap way of calculating (or checking) the Implicit Derivative.

$\frac{dy}{dx} = - \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$

So for the equation xy= k^2, we have:

$\\ f(x,y) = xy - k^2 \\\\ \frac{\partial f}{\partial x} = y \\\\ \frac{\partial f}{\partial y} = x \\\\ \frac{dy}{dx} = - \frac{y}{x}$

I know it doesn't seem worth it for this particular example, but it really helps when it comes to implicitly differentiating things like:

$3x^2 + 2x^2y + 2xy^2 + y^2 + xy = 1$

juantheron · May 5, 2012

Re: 2012 HSC MX2 Marathon

$\hspace{-16}\bf{\int\frac{1}{\sqrt{1+x^2}}dx}$\\\\\\ Let $\bf{1+x^2=y^2\Leftrightarrow 2xdx=2ydy}$\\\\\\ $\bf{xdx=ydy\Leftrightarrow \frac{dx}{y}=\frac{dy}{x}=\frac{d(x+y)}{x+y}}$\\\\\\ Now Integral Convert into \\\\\\ $\bf{=\int \frac{dx}{y}=\int \frac{d(x+y)}{(x+y)}}$\\\\\\ $\bf{=\ln \mid x+y \mid+C}$\\\\\\ So $\bf{\int \frac{1}{\sqrt{1+x^2}}dx = \ln \mid x+\sqrt{1+x^2} \mid+C}$$

HSC 2012 MX2 Marathon (archive) (1 Viewer)

Well-Known Member

Retired

Well-Known Member

Where do you live?

Member

New Member

Member

Member

Active Member

Premium Member

Active Member

Premium Member

Member

Member

Active Member

Member

Well-Known Member

Active Member

Retired

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)