HSC 2013 MX2 Marathon (archive) (6 Viewers)

Carrotsticks · Mar 5, 2014

Re: HSC 2014 4U Marathon

dragerx said:
Graph y= x

I think this is better suited for the 5 Unit Mathematics Marathon Thread.

RealiseNothing · Mar 6, 2014

Re: HSC 2014 4U Marathon

By using a suitable transformation, solve:

$x^4-32x^3+387x^2-2096x+4290=0$

Sy123 · Mar 7, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
By using a suitable transformation, solve:

$x^4-32x^3+387x^2-2096x+4290=0$

Seeking to transform this equation into a reducible quadratic, we substitute, $x=t+q$

$(t+q)^4 -32(t+q)^3 +387(t+q)^2 - 2096(t+q) + 4290$

Now, we find the value q such that the co-efficient of t^3 and t are zero, by equating co-efficients, we find that if we let q=8, the co-efficient of t^3 is zero, and by co-incidence so is t (since the question is constructed in this way).

Applying the substitution, $x=t+8$

$\Rightarrow \ t^4 + 3t^2 + 2 = 0 \\ \\ t^2_1 = -1 \ \ t^2_2 = -2 \\ \\ t_{1,2} = \pm i \\ \\ t_{3,4} = \pm \sqrt{2} i$

$$Therefore the roots of the original equation is$ \\ \\ 8 + i , 8-i, 8 + \sqrt{2}i, 8 - \sqrt{2} i$

VBN2470 · Mar 8, 2014

Re: HSC 2014 4U Marathon

Prove the following statement (i) using mathematical induction (ii) without using mathematical induction for all positive integer values of $n$

Sy123 · Mar 8, 2014

Re: HSC 2014 4U Marathon

$\\ $Consider$ \ \ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \\ \\ Re(1+i)^n = \left(\binom{n}{0} + \binom{n}{4} + \binom{n}{8} \right) - \left(\binom{n}{2} + \binom{n}{6} + \dots \right) \ \ \ \ (1) \\ \\ $Also consider that since$ \ \binom{n}{0} + \binom{n}{2} + \dots = \binom{n}{1} + \binom{n}{3} + \dots \\ \\ $Then$ \ 2^{n-1} = \binom{n}{0} + \binom{n}{2} + \dots \ \ \ \ (2) \ \ \ \ $By adding the LHS to the RHS$ \\ \\ (1)+(2) \ \Rightarrow \ 2^{n-1} + Re(1+i)^n = 2 (S) \\ \\ S = \frac{1}{4}(2^n + 2Re(1+i)^n) \\ \\ (1+i) = \sqrt{2} \left(\cos \frac{\pi}{4} + \sin \frac{\pi}{4} \right) \\ \therefore (\sqrt{2})^n \cos \frac{n\pi}{4} = Re(1+i)^n \\ \\ $Therefore$ \ \ \sum \binom{n}{4k} = \frac{1}{4} (2^n + \sqrt{2^n} \cos \frac{n\pi}{4})$

braintic · Mar 10, 2014

Re: HSC 2014 4U Marathon

Define a 'composition' of a positive integer n to be an ordered list of positive integers whose sum is n.
For example, the compositions of 3 are:
1+1+1
1+2
2+1
3

Let c(n) be the number of compositions of n.
So c(3) = 4 (from the above example)

(i) Show that c(n) = c(n-1) + c(n-2) + ... + c(2) + c(1) + 1

(ii) Hence prove by mathematical induction that c(n) = 2^(n-1)

(iii) Now find a slick way to justify why c(n) = 2^(n-1) without using part (i) or induction.

Sy123 · Mar 13, 2014

Re: HSC 2014 4U Marathon

$\\ $Prove using complex numbers that for real$ \ x,y,u,v \\ \\ \sqrt{(x^2+u^2)(y^2+v^2)} \geq xu+yv$

Sy123 · Mar 13, 2014

Re: HSC 2014 4U Marathon

$\\ $Solve$ \ x^3-3x-1=0$

RealiseNothing · Mar 13, 2014

Re: HSC 2014 4U Marathon

Suppose that $m$ boys and $n$ girls are going to the movies such that $m>n$ .

If no girl wants to sit next to each other, find the amount of possible arrangements that they can sit in a row in the cinema.

RealiseNothing · Mar 13, 2014

Re: HSC 2014 4U Marathon

Let the above situation be A, and now consider the same scenario but instead they all sit around a circular dinner table. Call this B.

Define $C(x)$ to be the amount of arrangements of a situation $x$ .

Show that, despite the girls refusing to sit next to each other, the ratio $\frac{C(A)}{C(B)}$ is independent on the amount of girls that attend.

RealiseNothing · Mar 13, 2014

Re: HSC 2014 4U Marathon

And yes, me and Sy did go to the USYD computers during our break and sat down and posted questions together. Don't judge.

btx3 · Mar 13, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
And yes, me and Sy did go to the USYD computers during our break and sat down and posted questions together. Don't judge.

you cuties, where was my invite though?

VBN2470 · Mar 13, 2014

Re: HSC 2014 4U Marathon

Someone please answer the following question (stuck mainly on parts (b) - (d)):

braintic · Mar 14, 2014

Re: HSC 2014 4U Marathon

VBN2470 said:
Someone please answer the following question (stuck mainly on parts (b) - (d)):
View attachment 30123

See attached solution: View attachment Misc.pdf

Sy123 · Mar 14, 2014

Re: HSC 2014 4U Marathon

$\\ $For what values of$ \ f(x) = \frac{x-a}{bx-c} \ $constants$ \ a,b,c | b \neq 0$

$Is the function its own inverse?$

braintic · Mar 14, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $For what values of$ \ f(x) = \frac{x-a}{bx-c} \ $constants$ \ a,b,c | b \neq 0$

$Is the function its own inverse?$

Why the restriction that b can't be zero? Doesn't it still work?

Carrotsticks · Mar 15, 2014

Re: HSC 2014 4U Marathon

I'm guessing so we still have a hyperbola.

Sent from my GT-N8010 using Tapatalk

Sy123 · Mar 15, 2014

Re: HSC 2014 4U Marathon

braintic said:
Why the restriction that b can't be zero? Doesn't it still work?

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

seanieg89 · 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

(And y = -x + k, k being anything at all).

Sy123 · Mar 15, 2014

Re: HSC 2014 4U Marathon

$\\ $Define the notation$ \ 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}$

HSC 2013 MX2 Marathon (archive) (6 Viewers)

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