HSC 2016 MX2 Marathon (archive) (2 Viewers)

seanieg89 · Mar 1, 2016

Re: HSC 2016 4U Marathon

Yep. What Integrand said.

All you can deduce about the geometry of the set of four roots is that this set is closed under the map z->1/z.

(In addition of course to being closed under conjugation, if our coefficients are real.)

InteGrand · Mar 1, 2016

Re: HSC 2016 4U Marathon

These polynomials are called palindromic polynomials. The curious HSC 4U student may want to read about some of their properties here: https://en.wikipedia.org/wiki/Reciprocal_polynomial#Palindromic_and_antipalindromic_polynomials

Paradoxica · Mar 1, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent c) By adding terms in pairs and considering the parity of $n$, evaluate $S_n \\ $Parity refers to whether something is odd or even.$

$$\noindent We will use the following result: $\frac{1}{1-\alpha}+\frac{1}{1-\frac{1}{\alpha}} \equiv 1\\$Observe that for $0 \leq k \leq n$, $\frac{1}{1-r_{k+1}}+\frac{1}{1-r_{n-k}}=1\\ S_n = \frac{1}{1-r_1}+\frac{1}{1-r_2}+\cdots+\frac{1}{1-r_{n-1}}+\frac{1}{1-r_n}\\S_n = \frac{1}{1-r_n}+\frac{1}{1-r_{n-1}}+\cdots+\frac{1}{1-r_{2}}+\frac{1}{1-r_1}\\S_n+S_n = 1+1+1+1+\cdots+1+1+1+1\\2S_n = n \Leftrightarrow S_n = \frac{n}{2}$

Post-hoc, I realised there was no need to consider the parity of n, as the sum could be evaluated in a manner similar to little gauss did with his sum from 1 to 100

Paradoxica · Mar 1, 2016

Re: HSC 2016 4U Marathon

$$\noindent Show that the curve $y = \sqrt[3]{x-1}+\sqrt[3]{x}+\sqrt[3]{x+1} =0$ intersects the $x$-axis at the origin, and at no other point.$

KingOfActing · Mar 2, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent Show that the curve $y = \sqrt[3]{x-1}+\sqrt[3]{x}+\sqrt[3]{x+1} =0$ intersects the $x$-axis at the origin, and at no other point.$

Note that (0,0) is a point on the curve.
$\frac{dy}{dx} = \frac{1}{3\sqrt[3]{x-1}^2} + \frac{1}{3\sqrt[3]{x}^2} + \frac{1}{3\sqrt[3]{x+1}^2} > 0\\$Hence the function is increasing everywhere, and cannot pass through the x axis again.$$

Paradoxica · Mar 2, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent Show that the curve $y = \sqrt[3]{x-1}+\sqrt[3]{x}+\sqrt[3]{x+1} =0$ intersects the $x$-axis at the origin, and at no other point.$

KingOfActing said:
Note that (0,0) is a point on the curve.
$\frac{dy}{dx} = \frac{1}{3\sqrt[3]{x-1}^2} + \frac{1}{3\sqrt[3]{x}^2} + \frac{1}{3\sqrt[3]{x+1}^2} > 0\\$Hence the function is increasing everywhere, and cannot pass through the x axis again.$$

$\noindent Alternatively, using the identity$\\a^3+b^3+c^3-3abc \equiv (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\$We have $x-1+x+x+1=\sqrt[3]{x^3-x} \Leftrightarrow x^3 = x^3-x \Leftrightarrow x=0 \\ $As the resultant equation has no extraneous solutions, and the only solution is the one we already know, this is sufficient to show there is no other root.$

wu345 · Mar 2, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent We shouldn't do it like this (without first proving the roots lie on the unit circle, or unless we are told to assume it without proof). Rather, sub. $y = x+\frac{1}{x}$ and observe we will obtain a quadratic in $y$ $\Big{(}$using manipulations like $x^2 + \frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2 - 2 = y^2 -2\Big{)}$, which we can then solve for $y$, and hence for $x$ (this is a general method that'll always work for degree 4 symmetric polynomials). In this particular case, the roots of the original polynomial all have modulus 1, so your substitution would lead you to the right answer. But in general, a symmetric polynomial need not have all roots have unit modulus $\Big{(}$as a simple example, consider $f(x) = x^2 -4x + 1$. The roots are $\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm \sqrt{12}}{2}$, and these are not unit modulus$\Big{)}$.$

Yep thanks, that's how we were taught to do it but my teacher also showed the cis theta way though he wasnt sure why we could just assume the modulus is 1 (now i know you cant)

juantheron · Mar 3, 2016

Re: HSC 2016 4U Marathon

$Numbers of ordered pair,s $(z,\omega)$ of the complex number $z$ and $\omega$ satisfying$

$the system of equations $z^3+\bar{\omega}^7=0$ and $z^5\cdot \omega^{11} = 1$

Paradoxica · Mar 4, 2016

Re: HSC 2016 4U Marathon

juantheron said:
$Numbers of ordered pair,s $(z,\omega)$ of the complex number $z$ and $\omega$ satisfying$

$the system of equations $z^3+\bar{\omega}^7=0$ and $z^5\cdot \omega^{11} = 1$

There are at most 11 pairs that can satisfy these equations, but there are only 10 pairs.

The proof of this is unknown to me, if it is even elementary. (Although it probably is.)

Drsoccerball · Mar 4, 2016

Re: HSC 2016 4U Marathon

$$ Derive the formula for the chord of contact of an ellipse with the exterior point $ P(x_0,y_0)$

seanieg89 · Mar 4, 2016

Re: HSC 2016 4U Marathon

A short question to test student understanding of differentiability:

Suppose the function f is differentiable to all orders at the real number a.

Explain why $E(h):=\frac{f(a+h)-f(a)}{h}-f'(a)\rightarrow 0$ and $\tilde{E}(h):=\frac{f(a+h)-f(a-h)}{2h}- f'(a)\rightarrow 0$ as $h$ tends to zero.

What about the quantities $E(h)/h$ and $\tilde{E}(h)/h$ ? Do they tend to zero as well? Why/why not?

Recall that a function $f$ is said to be differentiable at $a$ precisely if $E(h)\rightarrow 0$ . Could we replace the $E$ with $\tilde{E}$ to obtain an equivalent definition? Why/why not?

agha · Mar 10, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
A short question to test student understanding of differentiability:

Suppose the function f is differentiable to all orders at the real number a.

Explain why $E(h):=\frac{f(a+h)-f(a)}{h}-f'(a)\rightarrow 0$ and $\tilde{E}(h):=\frac{f(a+h)-f(a-h)}{2h}- f'(a)\rightarrow 0$ as $h$ tends to zero.[/tex]

$\\ $By the very definition of differentiability$ \ f \ $is differentiable at$ \ a \ $if and only if the limit$ \ \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \ $exists$$

$\\ f'(a) \ $is also simply the name we give to this limit, and since$ \ f'(a) \ $is independent of$ \ h \ $then:$$

$\\ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

$\\ \Rightarrow \lim_{h \to 0} \left( \frac{f(a+h) - f(a)}{h} \right) - f'(a) = 0$

$\\ \Rightarrow \lim_{h \to 0} \left( \frac{f(a+h) - f(a)}{h} - f'(a) \right) = 0 \ (1)$

$\\ $Which is equivalent to the statement in question.$$

$\\ $To prove the second statement, simply substitute$ \ h = - j \ $and replace$ \ h \ $with$ \ j \ $after simplifying (essentially, map$ \ h \to -h $)$$

$\lim_{h \to 0} \left( \frac{f(a+h) - f(a)}{h} - f'(a) \right) = 0$

$\\ \Rightarrow \lim_{h \to 0} \left( \frac{f(a-h) - f(a)}{-h} - f'(a) \right) = 0$

$\Rightarrow \lim_{h \to 0} \left( f(a) - f(a-h)}{h} \right) - f'(a) \right) = 0 \ (2)$

$\\ $Adding equations,$ \ (1) \ $and$ \ (2):$

$\\ \Rightarrow \lim_{h \to 0} \left( \frac{f(a) - f(a-h)}{h} - f'(a) \right) + \lim_{h \to 0} \left( \frac{f(a+h) - f(a)}{h} - f'(a) \right) = 0$

$\\ $Collect and divide by$ \ 2:$

$\\ \Rightarrow \lim_{h \to 0} \left( \frac{f(a+h) - f(a-h)}{2h} - f'(a) \right) = 0$

$\\ $Which is equivalent to the second statement$$

What about the quantities $E(h)/h$ and $\tilde{E}(h)/h$ ? Do they tend to zero as well? Why/why not?

$\\ $Counter example for$ \ \frac{E(h)}{h} \ $is to let$ \ f(x) = x^2$

$\\ $Then$ \ \lim_{h \to 0} \frac{E(h)}{h} = \lim_{h \to 0} \frac{f(h)}{h^2} = 1$

$\\ $Counter example for$ \ \frac{\overline{E}(h)}{h} \ $is to let$ \ f(x) = \begin{cases} x^2 , x \geq 0 \\ -x^2, x < 0 \end{cases}$

$\\ $Then$ \ \lim_{h \to 0} \frac{\overline{E}(h)}{h} = 1$

$\\ $The heuristic explanation for this fact is that$ \ E(h) \ $and$ \ \overline{E}(h) \ $in some cases may go to zero linearly fast or slower than that$$

$\\ $More concretely, the approximation of a secant for the tangent for some functions have their error go to zero in a sufficiently slow manner$$

Recall that a function $f$ is said to be differentiable at $a$ precisely if $E(h)\rightarrow 0$ . Could we replace the $E$ with $\tilde{E}$ to obtain an equivalent definition? Why/why not?

$\\ $If$ \ \tilde{E} \ $was a sufficient definition, then for every case in which$ \ E(h) \rightarrow 0 \ $then$ \ \tilde{E}(h) \rightarrow 0 \ $(and conversely)$$

$\\ $However, in the case$ \ f(x) = \begin{cases} x, x \geq 0 \\ 0 , x < 0 \end{cases}$

$\\ \tilde{E}(h) \rightarrow \frac{1}{2} \ $whereas the limit for$ \ E \ $doesn't exist. The function is sharp at$ \ 0 \ $and so it should not be considered differentiable, yet$ \ \tilde{E} \ $implies that it is, therefore, it cannot be substituted in to replace$ \ E$

agha · Mar 10, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
$$ Derive the formula for the chord of contact of an ellipse with the exterior point $ P(x_0,y_0)$

$\\ $Let$ \ P(x_1,y_1) \ $and$ \ Q(x_2, y_2) \ $be points on ellipse$ \ b^2x^2 + a^2y^2 = a^2b^2$

$\\ $Let the tangents at$ \ P \ $and$ \ Q \ $meet at$ \ R(x_0, y_0)$

$\\ $Immediately the equation of$ \ PQ \ $is$ \ x \frac{y_2 - y_1}{x_1y_2 - x_2 y_1} - y \frac{x_2 - x_1}{x_1y_2 - x_2y_1} = 1 \ (*)$

$\\ $Using calculus, the equation of the tangent at$ \ P \ $is$ \ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$

$\\ $Simultaneously solving them leads to the following:$$

$\\x_0 = a^2 \frac{y_2-y_1}{x_1y_2 - x_2y_1} \ $and$ \ y_0 = -b^2 \frac{x_2-x_1}{x_1y_2 - x_2y_1}$

$\\ $Substitute these back into$ \ (*)$

$\\ $So the equation of the chord of contact is$ \ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$

InteGrand · Mar 10, 2016

Re: HSC 2016 4U Marathon

agha said:
$\\ $Let$ \ P(x_1,y_1) \ $and$ \ Q(x_2, y_2) \ $be points on ellipse$ \ b^2x^2 + a^2y^2 = a^2b^2$

$\\ $Let the tangents at$ \ P \ $and$ \ Q \ $meet at$ \ R(x_0, y_0)$

$\\ $Immediately the equation of$ \ PQ \ $is$ \ x \frac{y_2 - y_1}{x_1y_2 - x_2 y_1} - y \frac{x_2 - x_1}{x_1y_2 - x_2y_1} = 1 \ (*)$

$\\ $Using calculus, the equation of the tangent at$ \ P \ $is$ \ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$

$\\ $Simultaneously solving them leads to the following:$$

$\\x_0 = a^2 \frac{y_2-y_1}{x_1y_2 - x_2y_1} \ $and$ \ y_0 = -b^2 \frac{x_2-x_1}{x_1y_2 - x_2y_1}$

$\\ $Substitute these back into$ \ (*)$

$\\ $So the equation of the chord of contact is$ \ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$

$$\noindent (Best to avoid using $P$ as a point on the ellipse, as $P$ is given in the question to be an exterior point $(x_0,y_0)$.)$

$\noindent Alternative method is to note that as $P(x_0,y_0)$ lies on the tangents to both $(x_1,y_1)$ and $(x_2,y_2)$ on the ellipse (the two points whose tangents go through $P$), we have $\frac{x_0 x_i}{a^2} + \frac{y_0 y_i}{b^2} = 1$ for $i=1,2$ (using the equation of a tangent to a point $(x_i,y_i)$ on the ellipse). Hence both points on the ellipse satisfy the equation $\frac{xx_0}{a^2} +\frac{yy_0}{b^2} = 1$. Since two points uniquely determine a straight line, this is the equation of the chord of contact.$

agha · Mar 10, 2016

Re: HSC 2016 4U Marathon

juantheron said:
$Numbers of ordered pair,s $(z,\omega)$ of the complex number $z$ and $\omega$ satisfying$

$the system of equations $z^3+\bar{\omega}^7=0, \ (1)$ and $z^5\cdot \omega^{11} = 1, \ (2)$

$\\ z^3 = -\overline{w}^7 \ $(from (1))$$

$\\ z^5 w^11 = - z^2 w^7 \overline{w}^7 w^4$

$1 = - z^2 w^4 |w|^{14}$

$\\ \Rightarrow |z|^2 |w|^{18} = 1, \ (3)$

$\\ $(from (2))$ |z|^5 |w|^{11} = 1, \ (4)$

$\\ (3) - (4): \ |z|^2 |w|^{11}(|w|^7 - |z|^3) = 0$

$\\ |z| \neq 0 , |w| \neq 0, \therefore |z| = |w|$

$\\ $From (2) by equating moduli$ \ |z| = |w| = 1$

$\\ $So,$ \ z = \cos \alpha + i \sin \alpha$

$\\ w = \cos \beta + i \sin \beta$

$\\ $Using (2)$ \ 5\alpha + 11\beta = 0, \ (5)$

$\\ $Using (1)$ \ \cos 3\alpha + \cos 7\beta = 0, (6) \ $and$ \ \sin 3\alpha - \sin 7\beta = 0, \ (7)$

$\\ $From (7):$ \ \sin 3\alpha = \sin \frac{-35}{11}\alpha = \sin \left(2n\pi - \frac{35}{11}\alpha \right)$

$\\ $So,$ \ 3\alpha = 2n\pi - \frac{35}{11}\alpha \Rightarrow \ \alpha = \frac{11n}{34}\pi$

$\\ $From (6):$ \ \cos 3\alpha = - \cos \frac{-35}{11}\alpha$

$\\ $So,$ \ \alpha = \frac{11m}{68}\pi \ $or$ \ \alpha = \frac{-11q}{2} \pi$

$\\ $Values of$ \ \alpha \ $for which both equation$ \ (6) \ $and$ \ (7) \ $hold true imply both equations to be true$$

$\\ $Those values of$ \ \alpha \ $occur only when$ \ n=0,1,2, \dots, 6 \ $and$ \ m=2, 4, 6, \dots, 12$$

$\\ $So, there are (not including$ \ z=1 \ $) 6 values of$ \ z \ $(and for$ \ w\ $ as no solutions for it are in common with$ \ z\ $) such that both equations hold true$$

$\\ $So, there are in total$ \ 7 \ $different pairs of complex numbers satisfying the equations$$

Paradoxica · Mar 10, 2016

Re: HSC 2016 4U Marathon

$$\noindent Consider the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, and the hyperbola $xy=1$. The point $P(a\cos{\theta},b\sin{\theta})$, where $0<\theta<\frac{\pi}{2}$ has it's normal constructed and intersects the hyperbola in two distinct points on both branches. Let the point where the normal meets the ellipse again be $Q$ and the positive and negative hyperbola branch intersection points be $A$ and $B$ respectively. Let the mid-point of $PQ$ and $ST$ be $R$ and $C$ respectively. Find the values of $a,b,\theta$ such that $|PA|=|AR|=|RC|=|CQ|=|QB|$

agha · Mar 10, 2016

Re: HSC 2016 4U Marathon

$\\ $Let there be polynomials of the form$ \\ p(x) = x^n - nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} + a_{n-3}x^{n-3} + \dots + a_1x + a_0 \\ $the roots of which are$ \ \alpha_1, \alpha_2, \dots , \alpha_n \\\\ $i) Show that$ \ (\alpha_1 - 1)^2 + (\alpha_2 - 1)^2 + \dots + (\alpha_n - 1)^2 = 0 \\ \\ $ii) Given that all of the roots of$ \ p(x) \ $are real, explain why$ \ p(x) = (x-1)^n$

Paradoxica · Mar 15, 2016

Re: HSC 2016 4U Marathon

agha said:
$\\ $Let there be polynomials of the form$ \\ p(x) = x^n - nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} + a_{n-3}x^{n-3} + \dots + a_1x + a_0 \\ $the roots of which are$ \ \alpha_1, \alpha_2, \dots , \alpha_n \\\\ $i) Show that$ \ (\alpha_1 - 1)^2 + (\alpha_2 - 1)^2 + \dots + (\alpha_n - 1)^2 = 0 \\ \\ $ii) Given that all of the roots of$ \ p(x) \ $are real, explain why$ \ p(x) = (x-1)^n$

$$\noindent Sum of roots: $\sum_{k=1}^n \alpha_k = n \\ $Sum of roots taken two at a time: $\sum_{1\leq i,j \leq n}^{i \neq j} \alpha_{i}\alpha_{j} = \frac{n(n-1)}{2} \\ \left(\sum_{k=1}^n \alpha_k \right)^2 = \sum_{k=1}^n \alpha_k^2 + 2 \sum_{1\leq i,j \leq n}^{i \neq j} \alpha_{i}\alpha_{j}\\n^2 = \sum_{k=1}^n \alpha_k^2 + n^2 - n \Leftrightarrow \sum_{k=1}^n \alpha_k^2 = n\\ \sum_{k=1}^n \alpha_k = n = \sum_{k=1}^n 1 \\ \sum_{k=1}^n (\alpha_k^2 + 1) = 2\sum_{k=1}^n \alpha_k \\ \sum_{k=1}^n (\alpha_k^2-2\alpha_k +1)=0 \Leftrightarrow \sum_{k=1}^n (\alpha_k -1)^2 =0 \\$As required.$ \\ $Since all the roots are real, the only way the above statement can possibly be true is if every term is zero. But that means every root is equal to every other root, which is equal to 1. Thus, $p(x) \equiv (x-1)^n$

seanieg89 · Mar 15, 2016

Re: HSC 2016 4U Marathon

A game of chance uses a simple harmonic oscillator. Its amplitude of oscillation is 1 but its frequency is unknown to you.

To play the game, you have to pay $1, and you get to choose a finite number of closed subintervals of [-1,1].

You win if after a random large amount of time, the oscillating particle lies in one of your subintervals, and you win $(1/L) where L is the sum of the lengths of your subintervals. So for example, if you were to cover the whole interval [-1,1] with a single subinterval, you will always "win", but as you only get $0.50 of your $1 back, this is not a good strategy!

Q/ How large an expected profit can you obtain with a well chosen bet? Provide proof.

Paradoxica · Mar 26, 2016

Re: HSC 2016 4U Marathon

$$\noindent The Beta Function B$(x,y)$ is defined as B$(x,y) =\int_0^1 t^{x-1} (1-t)^{y-1} $d$t \\ \indent $Consider the Beta Function for natural numbers $x,y.\\ $a) Show the following: $\\ \indent $i) B$(x,y) = $ B$(y,x) \\ \indent $ii) B$(x,y)=$ B$(x,y+1)+$ B$(x+1,y)\\ \indent $iii) B$(x+1,y)= \frac{x\,\text{B}(x,y)}{x+y}\\$b) Evaluate B$(1,1)\\$c) Prove the following statement using Multi-Dimensional Induction:$\\ \indent $B$(x,y) = \frac{(x-1)!(y-1)!}{(x+y-1)!} $ for all natural numbers $x,y.$

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