HSC 2015 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

Kaido said:
In the meantime, someone try solving this (just curious):
x^5+x=1, find real x

$\text{By estimation of roots, the root is approximately at .7545}$

Ekman · Feb 1, 2015

Re: HSC 2015 4U Marathon

$$The roots of the cubic equation $ z^3 -2z^2+pz+10=0 $ are $ \alpha, \beta, \gamma. \\ $It is given that $ \alpha^3 + \beta^3 + \gamma^3 = -4$

$$ (i) Write down the value of $ \alpha+\beta+\gamma$

$\\* $ (ii) (1) Explain why $ \alpha^3 - 2\alpha^2 +p\alpha +10=0 \\ $(2) Hence show that $ \alpha^2 + \beta^2 + \gamma^2 = p+13$

$$ (iii) Deduce that $ p=-3$

$$ (iv) Solve $ z^3 - 2z^2 - 3z +10 = 0$

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$The roots of the cubic equation $ z^3 -2z^2+pz+10=0 $ are $ \alpha, \beta, \gamma. \\ $It is given that $ \alpha^3 + \beta^3 + \gamma^3 = -4$

$$ (i) Write down the value of $ \alpha+\beta+\gamma$

$\\* $ (ii) (1) Explain why $ \alpha^3 - 2\alpha^2 +p\alpha +10=0 \\ $(2) Hence show that $ \alpha^2 + \beta^2 + \gamma^2 = p+13$

$$ (iii) Deduce that $ p=-3$

$$ (iv) Solve $ z^3 - 2z^2 - 3z +10 = 0$

$$ i) $ \alpha +\beta +\gamma =2$
$ii)$ \alpha^3 -2\alpha^2 +p\alpha +10=0 $ Since alpha is a root of the equation$
$$iii)$ \alpha^3 =2\alpha^2 -p\alpha -10 \\ \therefore -4=2(\alpha^2 +\beta^2 +\gamma^2) -p(2) -30 \\ \therefore \alpha^2 +\beta^2 + \gamma^2=13 +p$
$$iv)$ \alpha^2 +\beta^2 +\gamma^2 = (\alpha +\beta +\gamma)^2 -2(p) \\ \therefore p=-3$
$$v)$ -2 $is a root$ \therefore (z+2)(z^2 -4z +5) = P(z) \\ \therefore $roots are:$ -2, 2+i,2-i$

Ekman · Feb 1, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION:

$$ (i) Show that $ (\alpha^2 - 1)(\alpha^4 -14\alpha^2 +1) = \alpha^6 - 15\alpha^4 + 15\alpha^2 -1.$

$\\ $ (ii) Use de Moivre's theorem to show that $ \cot(6\theta) = \frac{(\alpha^2 -1)(\alpha^4 - 14\alpha^2 +1)}{2\alpha(3\alpha^4 - 10\alpha^2 +3)}$ where $\alpha = \cot(\theta) $(and $ 6\theta $ is not a multiple of $ \pi $).$$

$$ (iii) Hence show that $ \cot^2(\frac{\pi}{12}) + \cot^2(\frac{5\pi}{12}) = 14.$

$$ (iv) Deduce that $\cot(\frac{\pi}{12}) + \tan(\frac{\pi}{12}) = 4$

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

Ekman said:
NEXT QUESTION:

$$ (i) Show that $ (\alpha^2 - 1)(\alpha^4 -14\alpha^2 +1) = \alpha^6 - 15\alpha^4 + 15\alpha^2 -1.$

$\\ $ (ii) Use de Moivre's theorem to show that $ \cot(6\theta) = \frac{(\alpha^2 -1)(\alpha^4 - 14\alpha^2 +1)}{2\alpha(3\alpha^4 - 10\alpha^2 +3)}$ where $\alpha = \cot(\theta) $(and $ 6\theta $ is not a multiple of $ \pi $).$$

$$ (iii) Hence show that $ \cot^2(\frac{\pi}{12}) + \cot^2(\frac{5\pi}{12}) = 14.$

$$ (iv) Deduce that $\cot(\frac{\pi}{12}) + \tan(\frac{\pi}{12}) = 4$

$$For i) expand and you get the answer$ \\ $For ii expand$ (cos\theta +isin\theta)^6 $and use De Moivres thereom to equate$ sin^6\theta $and$ cos^6\theta $Then do$ \frac{cos6\theta}{sin6\theta} $After that use i) to factorise then the proof is complete$ \\ $For iii) find values for x which $ cos6\theta = 0 $ \sum\alpha\beta = 14 \\ \text{iv) was a little tricky but you need to realize that} cot(\frac{\pi}{12})= tan(\frac{5\pi}{12}) \text{and vise versa.} \\ \text{After you've done this make a quadratic equation with} a^4 -14a^2 +1 \text{and do sum of roots}$
Not sure if iv is right though a bit delirious

dan964 · Feb 1, 2015

Re: HSC 2015 4U Marathon

Part (iv) is the easiest part if you complete the square by noting that cot x * tan x = 1
And that cot 5pi/12 = tan pi/12
And you're done

Good effort drsoccerball.
You could do it that method it is just more painful trying to factorise that polynomial further.

dan964 · Feb 1, 2015

Re: HSC 2015 4U Marathon

Another question:

A man has a monkey
And k bananas.
Each day he gets k bananas and divides them amongst the amount of monkeys so that he has at least has
1 banana leftover for himself presumably.

Every day he also buys another monkey.
Show that the amount of bananas given to his original monkey over h days
Can be approximated as k* ln(1+h)

Hint: It involves an integral

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

dan964 said:
Part (iv) is the easiest part if you complete the square by noting that cot x * tan x = 1
And that cot 5pi/12 = tan pi/12
And you're done

Good effort drsoccerball.
You could do it that method it is just more painful trying to factorise that polynomial further.

Completing the square of a^4 -14a^2 +1 or?..

dan964 · Feb 1, 2015

Re: HSC 2015 4U Marathon

Complete the square using part (iii) only

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon

dan964 said:
Complete the square using part (iii) only

can you do it

? I dont see what you're talking about

FrankXie · Feb 1, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
can you do it ? I dont see what you're talking about

$\cot^2\frac{\pi}{12}+\cot^2\frac{5\pi}{12}=14 \Rightarrow\cot^2\frac{\pi}{12}+\tan^2\frac{\pi}{12}=14\Rightarrow\left(\tan\frac{\pi}{12}+\cot\frac{\pi}{12}\right)^2-2=14\Rightarrow\left(\tan\frac{\pi}{12}+\cot\frac{\pi}{12}\right)^2=16\Rightarrow\tan\frac{\pi}{12}+\cot\frac{\pi}{12}=4 $ because it is positive. $$

Ekman · Feb 2, 2015

Re: HSC 2015 4U Marathon

This thread has been quiet for some time, here is another question:

$\\$(i) Use De Moivre's Theorem to show that $ (\cot(\theta) + i)^n + (\cot(\theta) - i)^n = \frac{2\cos(n\theta)}{sin^n(\theta)}$

$\\ $(ii) Show that the equation $ (x+i)^5 + (x-i)^5 = 0 $ has roots 0, $ +-\cot(\frac{\pi}{10}), $ $ +-\cot(\frac{3\pi}{10}).$

$\\ $(iii) Hence show that the equation $ x^4 - 10x^2 + 5= 0 $ has roots $ +-\cot(\frac{\pi}{10}), $ $ +-\cot(\frac{3\pi}{10}).$

$$(iv) Hence show that $ \cot(\frac{\pi}{10}) = \sqrt{5+2\sqrt{5}}$

Drsoccerball · Feb 2, 2015

Re: HSC 2015 4U Marathon

Ekman said:
This thread has been quiet for some time, here is another question:

$\\$(i) Use De Moivre's Theorem to show that $ (\cot(\theta) + i)^n + (\cot(\theta) - i)^n = \frac{2\cos(n\theta)}{sin^n(\theta)}$

$\\ $(ii) Show that the equation $ (x+i)^5 + (x-i)^5 = 0 $ has roots 0, $ +-\cot(\frac{\pi}{10}), $ $ +-\cot(\frac{3\pi}{10}).$

$\\ $(iii) Hence show that the equation $ x^4 - 10x^2 + 5= 0 $ has roots $ +-\cot(\frac{\pi}{10}), $ $ +-\cot(\frac{3\pi}{10}).$

$$(iv) Hence show that $ \cot(\frac{\pi}{10}) = \sqrt{5+2\sqrt{5}}$

$$Question i) we have to factorise the LHS$ \\LHS= (cot\theta +i)^n +(cot\theta -i)^n \\ = \frac{cis(n\theta)+(cos\theta -isin\theta)^n}{sin^n\theta} $Make$ \theta= -\theta \text{Therefore once simplifying} \frac{2cosn\theta}{sin^n\theta}$

$$Question ii) let x=$ cot\theta $and let$ n=5, cos(5\theta)=0 \text{Therfore by solving that equation you get the roots}$

$$Question iv) \text{Solve simultaneous equations} \sum\alpha\beta= -10 \\ \therefore 10= cot^2 (\frac{\pi}{10}) + cot^2 (\frac{3\pi}{10}) \\ \sum\alpha\beta\gamma\delta =5 \\ \text{Let} cot(\frac{\pi}{10})=u \\ u^4 -10u^2 +5=0 \text{solutions are:} \sqrt{5 +2\sqrt{5}} \{And all the other values which i cant be bothered coding}$

Ekman · Feb 2, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$Question i) we have to factorise the LHS$ \\LHS= (cot\theta +i)^n +(cot\theta -i)^n \\ = \frac{cis(n\theta)+(cos\theta -isin\theta)^n}{sin^n\theta} $Make$ \theta= -\theta \text{Therefore once simplifying} \frac{2cosn\theta}{sin^n\theta}$

$$Question ii) let x=$ cot\theta $and let$ n=5, cos(5\theta)=0 \text{Therfore by solving that equation you get the roots}$

$$Question iv) \text{Solve simultaneous equations} \sum\alpha\beta= -10 \\ \therefore 10= cot^2 (\frac{\pi}{10}) + cot^2 (\frac{3\pi}{10}) \\ \sum\alpha\beta\gamma\delta =5 \\ \text{Let} cot(\frac{\pi}{10})=u \\ u^4 -10u^2 +5=0 \text{solutions are:} \sqrt{5 +2\sqrt{5}} \{And all the other values which i cant be bothered coding}$

Please Drsoccerball, improve your LaTeX coding because everything looks clustered...

LostInHSC · Feb 2, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Please Drsoccerball, improve your LaTeX coding because everything looks clustered...

Just looking at both of your last few posts, i think these would be helpful:

$\pm$ = \ pm (no space)

$$ And all the other values which i cant be bothered coding $ =$ $ And all the other values which i cant be bothered coding $

If you want space between your characters without moving onto the next line, place these ' \, ' (however many times you want) wherever you wish

Ekman · Feb 2, 2015

Re: HSC 2015 4U Marathon

LostInHSC said:
Just looking at both of your last few posts, i think these would be helpful:

$\pm$ = \ pm (no space)

$$ And all the other values which i cant be bothered coding $ =$ $ And all the other values which i cant be bothered coding $

If you want space between your characters without moving onto the next line, place these ' \, ' wherever you wish

Thanks, I was looking for the plus minus code thing

FrankXie · Feb 3, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$Find the values of $ k $ for which the equation $ 4x^3+4x^2+x+k=0 $ has more than one real root.$

VBN2470 · Feb 3, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$Find the values of $ k $ for which the equation $ 4x^3+4x^2+x+k=0 $ has more than one real root.$

$$ Consider the graph of $ P(x) = x(2x+1)^2 $ and observe where the line $ y=-k $ intersects the graph. For $ y>0 $, the line intersects the curve once (hence we exclude this region), for $ y\le0 $, the line then intersects the curve once or twice. To determine where exactly it cuts twice (or more), we need to find the stationary points of $ P(x) $, namely the local minimum (since this forms one bound for which our $ k $ will fall under). \\ $ P'(x) = 12x^2+8x+1 = (6x+1)(2x+1) $ and setting $ P'(x) = 0 $ we have our turning points at $ x=-\frac{1}{6} $ and $ x=-\frac{1}{2} $. Since $ x = -\frac{1}{2} $ is a double root (and hence a turning point at $ y=0 $) then we are only interested in $ x = -\frac{1}{6} $. Substituting this into $ P(x) $ yields $ P(-\frac{1}{6}) = -\frac{2}{27} $. \\ So the values for which $y=-k$ intersects the curve more than once occurs when $ -\frac{2}{27}\le{y}\le0 \Rightarrow $ {0}\,\le{k}\le\frac{2}{27} $.$

VBN2470 · Feb 3, 2015

Re: HSC 2015 4U Marathon

Next Question:

$Suppose that $ a $ and $ b $ are real numbers (not both zero) and $ w=\frac{az+bz^{-1}}{bz+az^{-1}}. $ $ \\ $ Show that if $ |z|=1 $, then $ |w|=1. $$

Ekman · Feb 3, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
Next Question:

$Suppose that $ a $ and $ b $ are real numbers (not both zero) and $ w=\frac{az+bz^{-1}}{bz+az^{-1}}. $ $ \\ $ Show that if $ |z|=1 $, then $ |w|=1. $$

Nice question:

$w\times \bar{w} = |w|^2$

$w=\frac{az^2+b}{bz^2+a}$

$\therefore \frac{az^2+b}{bz^2+a} \times \frac{a\bar{z}^2+b}{b\bar{z}^2 + a} = |w|^2$

$$We know that $ z\times \bar{z} = 1$

$\therefore \frac{a^2 + ab(z^2+\bar{z}^2)+ b^2}{b^2 + ab(z^2+ \bar{z}^2) + a^2} = |w|^2$

$\therefore 1=|w|^2 $ ,thus $ |w|=1 $ since $|w|$ is always positive. $$

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