HSC 2016 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Feb 14, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$ \noindent 1. Given $P(z) = z^{n+1}-1$ and $ S_n = \frac{1}{1-r_1}+\frac{1}{1-r_2}+\cdots+\frac{1}{1-r_{n-1}}+\frac{1}{1-r_n}\\$where $r_k$ is the $k^{\text{th}}$ complex root of $P(z) \\$Note that $r_0 = 1$, which does not appear anywhere in $S_n.\\ $a) By transforming $P(z)$, evaluate $S_n$

$$ a) $ x= \frac{1}{1-\alpha}$

$\alpha= \frac{x-1}{x}$

$0=(x+1)^{n+1} - x^{n+1}$

$\frac{-b}{a} = \frac{\binom{n+1}{2}}{\binom{n+1}{1}}= \frac{n}{2}=S_n$

wu345 · Feb 18, 2016

Re: HSC 2016 4U Marathon

Hey guys, can you please help me with this?

Please don't use Euler's formula, this was in the DeMoivre's chapter

Paradoxica · Feb 18, 2016

Re: HSC 2016 4U Marathon

wu345 said:
Hey guys, can you please help me with this?
View attachment 32922

Please don't use Euler's formula, this was in the DeMoivre's chapter

Divide down z^8 and make the substitution.

wu345 · Feb 18, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Divide down z^8 and make the substitution.

okay so
${ u }^{ 8 }=1,\quad $where$\quad u=\frac { 1 }{ z } +1,u\neq 1\\ let\quad u=cis\theta \\ cis8\theta =cis0\\ u=cis(\frac { n\pi }{ 4 } ),\quad where\quad n=\pm 1,\pm 2,\pm 3,4\\ 1+\frac { 1 }{ z } =cis(\frac { n\pi }{ 4 } )\\$
Now what? Sorry, I am bad

Paradoxica · Feb 18, 2016

Re: HSC 2016 4U Marathon

wu345 said:
okay so
${ u }^{ 8 }=1,\quad $where$\quad u=\frac { 1 }{ z } +1,u\neq 1\\ let\quad u=cis\theta \\ cis8\theta =cis0\\ u=cis(\frac { n\pi }{ 4 } ),\quad where\quad n=\pm 1,\pm 2,\pm 3,4\\ 1+\frac { 1 }{ z } =cis(\frac { n\pi }{ 4 } )\\$
Now what? Sorry, I am bad

Use your condition for "u" on the set of solutions you have now obtained, and make your conclusion.

InteGrand · Feb 18, 2016

Re: HSC 2016 4U Marathon

wu345 said:
okay so
${ u }^{ 8 }=1,\quad $where$\quad u=\frac { 1 }{ z } +1,u\neq 1\\ let\quad u=cis\theta \\ cis8\theta =cis0\\ u=cis(\frac { n\pi }{ 4 } ),\quad where\quad n=\pm 1,\pm 2,\pm 3,4\\ 1+\frac { 1 }{ z } =cis(\frac { n\pi }{ 4 } )\\$
Now what? Sorry, I am bad

$\noindent Note that $z =\frac{1}{u-1}$, $u\neq 1$ (so we lose the solution of $u=1$). Also, you may find it helpful to show that $\frac{1}{\mathrm{cis}\, \theta -1} = -\frac{1}{2} \left(1+i\cot \frac{\theta}{2}\right)$, where $\theta \in \mathbb{R}$ and $\sin \frac{\theta}{2}\neq 0$.$

Drsoccerball · Feb 19, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$b) By considering $\log(P(z))$, the fundamental theorem of algebra, differentiation, and using L'H$\hat{\text{o}}$pital's rule, evaluate $S_n \\$

I fail to see how this provides us any information to find $S_n$ There was nothing in the indeterminate forms that I saw. Can you provide a solution please?

Paradoxica · Feb 19, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
I fail to see how this provides us any information to find $S_n$ There was nothing in the indeterminate forms that I saw. Can you provide a solution please?

$$\noindent$ \log(P(x)) = \log(x-1) + \log(x-r_1) + \cdots + \log(x-r_{n-1}) + \log(x-r_n) \\ \frac{P'(x)}{P(x)} = \frac{1}{x-1}+\frac{1}{x-r_1} + \cdots + \frac{1}{x-r_{n-1}}+\frac{1}{x-r_n} \\ \frac{(n+1)x^n}{x^{n+1}-1} = \frac{1}{x-1}+\frac{1}{x-r_1} + \cdots + \frac{1}{x-r_{n-1}}+\frac{1}{x-r_n} \\ $The sum we are after is every term on the right hand side except the first term. Further observe that when you directly substitute $x=1$, neither expressions yield indeterminate forms that call for L'H$\hat{\text{o}}$pital's Rule. Thus we must manipulate them until they yield an indeterminate form upon direct substitution.$ \\ S_n =\lim_{x \to 1} \frac{(n+1)x^n}{x^{n+1}-1} - \frac{1}{x-1} = \frac{(x-1)(n+1)x^n - x^{n+1}+1}{(x^{n+1}-1)(x-1)} = \frac{nx^{n+1} -(n+1)x^n +1}{x^{n+2}-x^{n+1}-x+1} \\ $Substituting $x=1$ results in an indeterminate form. Applying L'H$\hat{\text{o}}$pital's Rule, we have $S_n = \lim_{x \to 1} \frac{(n+1)nx^n -(n+1)nx^{n-1}}{(n+2)x^{n+1}-(n+1)x^n-1}$.$

$$\noindent The expression still yields indeterminate form upon direct substitution. Applying L'H$\hat{\text{o}}$pital's Rule again, we get $S_n = \lim_{x \to 1} \frac{(n+1)n^2x^{n-1} -(n+1)n(n-1)x^{n-2}}{(n+2)(n+1)x^n-(n+1)nx^{n-1}}\\S_n = \frac{n^2 -n(n-1)}{n+2-n}=\frac{n^2 - n^2 + n}{2} = \frac{n}{2}$

davidgoes4wce · Feb 20, 2016

Re: HSC 2016 4U Marathon

I am studying the ellipse and was stuck on this question.

$$ If a=13 and the foci are $ (\pm{5},0) $ then the equation of the ellipse is $:$

$A) \frac{x^2}{100}+\frac{y^2}{169}=1$

$B) \frac{x^2}{144}+\frac{y^2}{25}=1$

$C) \frac{x^2}{169}+\frac{y^2}{25}=1$

$D \frac{x^2}{169}+\frac{y^2}{144}=1$

$$ I know the formula for the Foci is $ (\pm ae, 0), Directrices =\frac{\pm{a}}{e}$

I'm looking at the choices its either C or D but how do I go about solving for the 'b' constant?

Drsoccerball · Feb 20, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
I am studying the ellipse and was stuck on this question.

$$ If a=13 and the foci are $ (\pm{5},0) $ then the equation of the ellipse is $:$

$A) \frac{x^2}{100}+\frac{y^2}{169}=1$

$B) \frac{x^2}{144}+\frac{y^2}{25}=1$

$C) \frac{x^2}{169}+\frac{y^2}{25}=1$

$D \frac{x^2}{169}+\frac{y^2}{144}=1$

$$ I know the formula for the Foci is $ (\pm ae, 0), Directrices =\frac{\pm{a}}{e}$

I'm looking at the choices its either C or D but how do I go about solving for the 'b' constant?

$b^2=a^2(1-e^2) $ by first finding e $$

davidgoes4wce · Feb 20, 2016

Re: HSC 2016 4U Marathon

I am struggling a bit to visualise this question.

The maths behind it to get the answer is fine.

$$ Using cartersian coordinates, find the locus of a Point P in a plane such that is its distance from the point $ (0,3) is \frac{3}{5} $ of its distance from the line $ y=\frac{25}{3}$

InteGrand · Feb 20, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
I am struggling a bit to visualise this question.

The maths behind it to get the answer is fine.

$$ Using cartersian coordinates, find the locus of a Point P in a plane such that is its distance from the point $ (0,3) is \frac{3}{5} $ of its distance from the line $ y=\frac{25}{3}$

$\noindent This is an ellipse by the focus-directrix definition of ellipse $\Big{(}$its eccentricity is $\frac{3}{5}$ and a focus is $(0,3)$, with corresponding directrix $y=\frac{25}{3}$$\Big{)}$. So in terms of visualisation, observe that any point in an ellipse has its distance to a focus be a fixed ratio (less than 1) of its distance to a certain line (corresponding directrix).$

davidgoes4wce · Feb 22, 2016

Re: HSC 2016 4U Marathon

Going through the 2011 HSC exam paper,

$$ A bit confused as to how to obtain the angle. Now Im assuming that $ \angle{AOB}= \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}. $ The answer of the $ \theta $ value is $\frac{5 \pi}{6}.$

InteGrand · Feb 22, 2016

Re: HSC 2016 4U Marathon

davidgoes4wce said:
Going through the 2011 HSC exam paper,

$$ A bit confused as to how to obtain the angle. Now Im assuming that $ \angle{AOB}= \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}. $ The answer of the $ \theta $ value is $\frac{5 \pi}{6}.$

$\noindent Since adjacent angles of a rhombus add up to $180^\circ$, the value of $\theta $ is $\pi -\angle AOB = \pi - \frac{\pi}{6} =\frac{5\pi}{6}$, or in degrees, $\theta = 150^\circ$.$

davidgoes4wce · Feb 24, 2016

Re: HSC 2016 4U Marathon

I am not sure but is this a type in the solutions?

$From Fitzpatrick Exercise 7.1 Q 9a, At time t a particle of mass m is moving in a straight line under the action of a force given by $ F=\frac{m(3-5x)}{x^3}. $ The particle starts from rest at x $=\frac{1}{3}. $ Find its velocity in terms of x$

$$ I had an answer of $ v=\pm \sqrt{10x-3-3x^2}$

Paradoxica · Feb 28, 2016

Re: HSC 2016 4U Marathon

$$\noindent By considering the expansions of $\cos{k\theta}$ for $k \in \{2,3,4\} \\ $Solve for the zeroes of the polynomial $\\ P(x) \equiv 4x^4 - 2x^3 - 3x^2 + x$

Drsoccerball · Feb 28, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent By considering the expansions of $\cos{k\theta}$ for $k \in \{2,3,4\} \\ $Solve for the zeroes of the polynomial $\\ P(x) \equiv 4x^4 - 2x^3 - 3x^2 + x$

$$ Hint: $ P(x) = \frac{1}{2} Re(\sum_{k=1}^{4}((-1)^kcis{k\theta}))$

KingOfActing · Feb 28, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$$\noindent By considering the expansions of $\cos{k\theta}$ for $k \in \{2,3,4\} \\ $Solve for the zeroes of the polynomial $\\ P(x) \equiv 4x^4 - 2x^3 - 3x^2 + x$

$$Expanding $\cos{k\theta}$ in terms of $\cos\theta$, and then making the substitution $x = \cos\theta\\P(x) = 4\cos^4{\theta} - 2\cos^3{\theta} - 3\cos^2{\theta} + \cos\theta \\= \frac{1}{2}(\cos{4\theta} + \cos{2\theta} - \cos{3\theta} - \cos{\theta}) = 0\\$Using the sum to product formula, we get $\\\cos\theta = 0 $ or $ \sin{\frac52\theta} = 0 $ or $ \sin{\frac12\theta} = 0 \\$From which we deduce that $ x = 0, 1, \cos{\frac{2\pi}{5}}, \cos{\frac{4\pi}{5}}$

wu345 · Mar 1, 2016

Re: HSC 2016 4U Marathon

Just a quick question - when solving polynomials with symmetrical co-efficients such as 2x^4 +3x^3 +5x^2 +3x+2=0, if I were to divide by x^2 and use the substitution x=cis theta , how would I know that the roots definitely have modulus 1?

InteGrand · Mar 1, 2016

Re: HSC 2016 4U Marathon

wu345 said:
Just a quick question - when solving polynomials with symmetrical co-efficients such as 2x^4 +3x^3 +5x^2 +3x+2=0, if I were to divide by x^2 and use the substitution x=cis theta , how would I know that the roots definitely have modulus 1?

$\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$\Big{)}$.$

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