HSC 2013 MX2 Marathon (archive) (1 Viewer)

braintic · Mar 15, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
I stole this question from a math tutorial on usyd, it had that restriction. I'm guessing because if b=0 then the function is a straight line and the only case for its own inverse is the trivial y=x

I don't think y=x is possible for this question.
Just checking - is the answer c=1, no restriction on a,b?
In which case your trivial answer would be y=1-x

Sy123 · Mar 15, 2014

Re: HSC 2014 4U Marathon

braintic said:
I don't think y=x is possible for this question.
Just checking - is the answer c=1, no restriction on a,b?
In which case your trivial answer would be y=1-x

Just did the question, that is what I got

RealiseNothing · Mar 16, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Define the notation$ \ f : \mathbb{R} \rightarrow \mathbb{R} \ $to mean that the function$ \ f \ $has inputs in the real numbers and outputs in the real numbers$$

$Note$ \ \mathbb{C} = \ $complex numbers$

$\\ $Also define a function$ \ f \ $ to have property$ \ S \ $if the entire output domain of the function can be arrived at through some input$ \\ \\ $i.e.$ \ f: \mathbb{R} \rightarrow \mathbb{R} \ , \ \ f(x) = x^3 \ \ $has the property$ \ S$

$\\ $Which of these functions have the property$ \ S$

$\\ $i)$ \ f: \mathbb{R} \rightarrow \mathbb{R} \ , \ \ f(x) = x^2 \\ \\ $ii)$ \ f: \mathbb{C} \rightarrow \mathbb{C} \ , \ f(z) = z^2 \\ \\ $iii)$ \ f: \mathbb{C} \rightarrow \mathbb{C} \ , \ \ f(z) = z+ \bar{z} \\ \\ $iv)$ \ f: \mathbb{C} \rightarrow \mathbb{R} \ , \ \ f(z) = z + \bar{z}$

Practicing 1901 I see.

Sy123 · Mar 17, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Practicing 1901 I see.

To be fair they are very easy for MATH1901 lol

----------------------

$\\ $In the complex plane consider the points$ \ A,B \ $represented by the complex numbers$ \ z,w \ $respectively$ \\ \\ $Another point$ \ C \ $represented by the complex number$ \ v \ $divides$ \ AB \ $in the ratio of$ \ \alpha : \beta \\ \\ $Show that$ \\ \\ v = \frac{\beta z + \alpha w}{\alpha + \beta}$

Sy123 · Mar 17, 2014

Re: HSC 2014 4U Marathon

$\\ $Prove that for the real polynomial$ \\ \\ a_0 + a_1x+ \dots +a_nx^n \\ \\ $The expression can only be equal to 0 across all$ \ x \ $if and only if$ \\ \\ a_0=a_1= \dots = a_n = 0$

HeroicPandas · Mar 18, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
To be fair they are very easy for MATH1901 lol

----------------------

$\\ $In the complex plane consider the points$ \ A,B \ $represented by the complex numbers$ \ z,w \ $respectively$ \\ \\ $Another point$ \ C \ $represented by the complex number$ \ v \ $divides$ \ AB \ $in the ratio of$ \ \alpha : \beta \\ \\ $Show that$ \\ \\ v = \frac{\beta z + \alpha w}{\alpha + \beta}$

So we draw a diagram

Look at vector AB

Now

vec(AC)/vec(BC) = alpha/beta

(v-z)/(w-v) = alpha/beta

v(beta) - z(beta) = (w)alpha - (v)alpha

v(alpha + beta) = (w)alpha + z(beta)

v = (bz+aw)/(a+b) as required!!!

Resource · Mar 19, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
To be fair they are very easy for MATH1901 lol

----------------------

$\\ $In the complex plane consider the points$ \ A,B \ $represented by the complex numbers$ \ z,w \ $respectively$ \\ \\ $Another point$ \ C \ $represented by the complex number$ \ v \ $divides$ \ AB \ $in the ratio of$ \ \alpha : \beta \\ \\ $Show that$ \\ \\ v = \frac{\beta z + \alpha w}{\alpha + \beta}$

z = A(x1=Re(z),y1=Im(z)), w=B(x2=Re(w),y2=(Im(w)), v= C(x=Re(v),y=Im(v)), ratio of: a:b (m:n)

v=C( (ax2+bx1)/(a+b),(ay2+by1)/(a+b) ) (Formula for point, p (v), that divides an interval)
v=Re(v)+Im(v)
Re(v)=(a*Re(w)+b*Re(z))/(a+b)
Im(v)=(a*Im(w)+b*Im(z))/(a+b)
v=((a*Re(w)+b*Re(z)+a*Im(w)+b*Im(z))/(a+b)
v=(a*(Re(w)+Im(w)) +b*(Re(z)+Im(z)))/(a+b)
Therefore v=(aw+bz)/(a+b)

HSC 2013 MX2 Marathon (archive) (1 Viewer)

braintic

Well-Known Member

Sy123

This too shall pass

RealiseNothing

what is that?It is Cowpea

Sy123

This too shall pass

Sy123

This too shall pass

HeroicPandas

Heroic!

Resource

New Member

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