HSC 2015 MX2 Marathon (archive) (3 Viewers)

Trebla · Dec 7, 2014

Post any questions within the scope and level of Mathematics Extension 2. Once a question is posted, it needs to be answered before the next question is raised.

This thread is mainly targeting Q1-15 difficulty in the HSC. Any questions of Q16 difficulty or harder should be posted in this thread:
http://community.boredofstudies.org...1666/hsc-2015-4u-marathon-advanced-level.html

I encourage all current students in particular to participate in this marathon.

To start off:

$\noindent$Suppose that $A, B, $ and $C$ represent complex numbers $\alpha, \beta, $ and $\gamma $ on the Argand diagram. Show that if $\Delta ABC$ is equilateral then \\\\$(\alpha+\beta+\gamma)^2=3(\alpha\beta+\beta \gamma +\alpha\gamma)$

Sy123 · Dec 7, 2014

Re: HSC 2015 4U Marathon

$\\ $If$ \ \alpha, \beta, \gamma \ $are an equilateral triangle, then$ \\ \\ (\beta - \alpha) (cis \frac{\pi}{3}) = (\gamma - \alpha) \\ \\ (\gamma - \beta) (cis \frac{\pi}{3}) = (\alpha - \beta) \\ \\ (\alpha - \gamma) (cis \frac{\pi}{3}) = (\beta - \gamma)$

$\\ $So,$ \ \frac{\gamma - \alpha}{\beta - \alpha} = \frac{\alpha - \beta}{\gamma - \beta} \ \Rightarrow \ \alpha^2 + \beta^2 + \gamma^2 = \alpha \beta + \alpha \gamma + \beta \gamma \\ \\ $So,$ \ (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \alpha \gamma + \beta \gamma) = \alpha^2 + \beta^2 + \gamma^2 = \alpha \beta + \alpha \gamma + \beta \gamma \\ \\ $So,$ \ (\alpha + \beta + \gamma)^2 = 3(\alpha \beta + \alpha \gamma + \beta \gamma)$

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$NEXT QUESTION$

$$a) Expand$ \ (1+i)(3+2i)$

$$b) Simplify$ \ \frac{2-i}{7 + 17i}$

Axio · Dec 7, 2014

Re: HSC 2015 4U Marathon

$$a)$\ (1+i)(3+2i)=3+2i+3i-2=1+5i$

$$b)$\ \frac{2-i}{7 + 17i} =\frac{(2-i)(7-17i)}{(7+17i)(7-17i)}=\frac{14-34i-7i-17}{49+289}=\frac{-3-41i}{338}$

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$$Show that if \ $x^3+px+r=0$ has a root of multiplicity 2, then \ 27r^2+4p^3=0$

Sy123 · Dec 7, 2014

Re: HSC 2015 4U Marathon

Axio said:
$$a)$\ (1+i)(3+2i)=3+2i+3i-2=1+5i$

$$b)$\ \frac{2-i}{7 + 17i} =\frac{(2-i)(7-17i)}{(7+17i)(7-17i)}=\frac{14-34i-7i-17}{49+289}=\frac{-3-41i}{338}$

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$$Show that if \ $x^3+px+r=0$ has a root of multiplicity 2, then \ 27r^2+4p^3=0$

$\\ $If$ \ p(x) = x^3 + px + r = 0 \ $has root multiplicity of 2, then$ \ p'(x) \ $has a root in common with$ \ p(x) \\ \\ p'(x) = 3x^2 + p = 0 \Rightarrow \ x = \pm \sqrt{\frac{-p}{3}} \\ \\ $If$ \ x = \sqrt{\frac{-p}{3}} \ $is a root, then$ \ p\left( \sqrt{\frac{-p}{3}} \right) = 0 \\ \\ \Rightarrow \ \frac{2p}{3} \sqrt{\frac{-p}{3}} + r = 0 \Rightarrow \ 4p^3 + 27r^2 = 0 \\ \\ $And, similarly if$ \ x = -\sqrt{\frac{-p}{3}} \ $was the double root$$

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$$NEXT QUESTION$ \\ \\ $a) Expand$ \ (1-i)(1+2i)(5+19i) \\ \\ $b) Simplify$ \ \frac{(6-i)(4+3i)}{(17+22i)(1+61i)} \cdot i^5$

Axio · Dec 7, 2014

Re: HSC 2015 4U Marathon

$$a)$ \ =(3+i)(5+19i)=-4+62i$

$$b)$ \ =\frac{i(27+14i)}{-1325+1059i}=\frac{(-14+27i)(-1325-1059i)}{(-1325+1059i)(-1325-1059i)}=\frac{47143-20949i}{2877106}$

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$$If \alpha , \beta , \gamma$ \ \ $are the roots of the equation x^3-px^2+qx-r=0,$

$$find \alpha ^3 + \beta ^3 + \gamma ^3$ \ \ $in terms of p, q and r$

Soulful · Dec 7, 2014

Re: HSC 2015 4U Marathon

Axio said:
$$a)$ \ =(3+i)(5+19i)=-4+62i$

$$b)$ \ =\frac{i(27+14i)}{-1325+1059i}=\frac{(-14+27i)(-1325-1059i)}{(-1325+1059i)(-1325-1059i)}=\frac{47143-20949i}{2877106}$

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$$If \alpha , \beta , \gamma$ \ \ $are the roots of the equation x^3-px^2+qx-r=0,$

$$find \alpha ^3 + \beta ^3 + \gamma ^3$ \ \ $in terms of p, q and r$

$\alpha + \beta + \gamma = p$

$\alpha\beta + \alpha\gamma + \beta\gamma = q$

$\begin{align*} \alpha^2 + \beta^2 + \gamma^2 &= (\alpha + \beta + \gamma)^2-2(\alpha\beta + \alpha\gamma + \beta\gamma) \\ &= p^2 - 2q \end{align*}$

$$As \alpha, \beta, \gamma$ \ \ $are roots of the equation x^3-px^2+qx-r=0, then$

$\alpha^3 - p\alpha^2+q\alpha - r=0 $ (factor theorem)$$

$$Similarly $ \beta^3 - p\beta^2+q\beta - r=0 \ \ $and $ \gamma^3 - p\gamma^2+q\gamma - r=0$

$Add them all up$

$\begin{align*} \alpha^3 + \beta^3 + \gamma^3 &= p(\alpha^2 + \beta^2 + \gamma^2) - q(\alpha + \beta + \gamma) + 3r \\ &= p(p^2-2q) - qp + 3r \\&= p^3 - 3qp + 3r \end{align*}$

ok wow 1st time latex user (it isnt as scary as I thought haha) so I spent more time working out how this thing actually works than the question itself so I might have added some superfluous steps :'(

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$$If $ \frac{|a|}{3} = \frac{|b|}{4} = \frac{|a-b|}{5} $ show that $ 16a^2+9b^2=0$

(this is a complex numbers question)

Axio · Dec 7, 2014

Re: HSC 2015 4U Marathon

Soulful said:
$\alpha + \beta + \gamma = p$

$\alpha\beta + \alpha\gamma + \beta\gamma = q$

$\begin{align*} \alpha^2 + \beta^2 + \gamma^2 &= (\alpha + \beta + \gamma)^2-2(\alpha\beta + \alpha\gamma + \beta\gamma) \\ &= p^2 - 2q \end{align*}$

$$As \alpha, \beta, \gamma$ \ \ $are roots of the equation x^3-px^2+qx-r=0, then$

$\alpha^3 - p\alpha^2+q\alpha - r=0 $ (factor theorem)$$

$$Similarly $ \beta^3 - p\beta^2+q\beta - r=0 \ \ $and $ \gamma^3 - p\gamma^2+q\gamma - r=0$

$Add them all up$

$\begin{align*} \alpha^3 + \beta^3 + \gamma^3 &= p(\alpha^2 + \beta^2 + \gamma^2) - q(\alpha + \beta + \gamma) + 3r \\ &= p(p^2-2q) - qp + 3r \\&= p^3 - 3qp + 3r \end{align*}$

ok wow 1st time latex user (it isnt as scary as I thought haha) so I spent more time working out how this thing actually works than the question itself so I might have added some superfluous steps :'(

---

$$If $ \frac{|a|}{3} = \frac{|b|}{4} = \frac{|a-b|}{5} $ show that $ 16a^2+9b^2=0$

(this is a complex numbers question)

$$ The statement means that the triangle with sides \left |a \right |, \left | b \right |, \left | a-b \right |$

$$is a right-angled triangle (ratio of sides).$

$\therefore $ \ bcis90^{\circ}=\frac{4a}{3}$

$\Rightarrow 16a^2+9b^2=0$

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$$Show that cos^5\alpha =\frac{1}{16}(cos5\alpha +5cos3\alpha +10cos\alpha )$

FrankXie · Dec 8, 2014

Re: HSC 2015 4U Marathon

$$ Let $ z=\cos\alpha+i\sin\alpha, $ then from DeMoivre's theorem, $ z+z^{-1}=2\cos\alpha, z^3+z^{-3}=2\cos3\alpha, z^5+z^{-5}=2\cos5\alpha.$

$$ From the Binomial Theorem, $ (z+z^{-1})^5=(z^5+z^{-5})+5(z^3+z^{-3})+10(z+z^{-1}), $ which gives $ 32\cos^5\alpha=2\cos5\alpha+10\cos3\alpha+20\cos \alpha.$

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NEXT QUESTION

$$ If $ |z+\sqrt3+i|\leq1, $ find the greatest and least values of (i) $ |z-2i|, $ (ii) $ \rm{arg}(z-2i)$

Joshmosh2 · Dec 9, 2014

Re: HSC 2015 4U Marathon

FrankXie said:
$$ Let $ z=\cos\alpha+i\sin\alpha, $ then from DeMoivre's theorem, $ z+z^{-1}=2\cos\alpha, z^3+z^{-3}=2\cos3\alpha, z^5+z^{-5}=2\cos5\alpha.$

$$ From the Binomial Theorem, $ (z+z^{-1})^5=(z^5+z^{-5})+5(z^3+z^{-3})+10(z+z^{-1}), $ which gives $ 32\cos^5\alpha=2\cos5\alpha+10\cos3\alpha+20\cos \alpha.$

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

NEXT QUESTION

$$ If $ |z+\sqrt3+i|\leq1, $ find the greatest and least values of (i) $ |z-2i|, $ (ii) $ \rm{arg}(z-2i)$

Here's my attempt:
http://imgur.com/OsVeIMC

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Find the number of real roots of the equation x^4 + 2x^3 + 8x + 15 = 0 and find the integral values of x between each

FrankXie · Dec 9, 2014

Re: HSC 2015 4U Marathon

Joshmosh2 said:
Here's my attempt:
http://imgur.com/sMtHOlv

part one is a well done. part two was a lot confusing: make sure you understand the definition of the argument of a complex number

Note: both part one and two can be solved from a geometric point of view

Joshmosh2 · Dec 9, 2014

Re: HSC 2015 4U Marathon

FrankXie said:
part one is a well done. part two was a lot confusing: make sure you understand the definition of the argument of a complex number

Note: both part one and two can be solved from a geometric point of view

I think i misread the locus on my 1st attempt
My 2nd attempt (not the most elegant approach..): http://imgur.com/NTfoHbn

turntaker · Dec 9, 2014

Re: HSC 2015 4U Marathon

Does anyone know how to find roots using Euler's

FrankXie · Dec 9, 2014

Re: HSC 2015 4U Marathon

Joshmosh2 said:
I think i misread the locus on my 1st attempt
My 2nd attempt (not the most elegant approach..): http://imgur.com/NTfoHbn

excellent!

Axio · Dec 9, 2014

Re: HSC 2015 4U Marathon

Joshmosh2 said:
Here's my attempt:
http://imgur.com/OsVeIMC

---
Find the number of real roots of the equation x^4 + 2x^3 + 8x + 15 = 0 and find the integral values of x between each

let equation = P(x)
P'(x)=4x^3+6x^2+8
P'(x)=(x+2)(4x^2-2x+4)
Therefore x=-2, y=-1 is the only stationary point and thus 2 real roots exist, with one root being between -3 and -2 and the other being between -2 and -1.

FrankXie · Dec 9, 2014

Re: HSC 2015 4U Marathon

NEXT QUESTION

$Show that $ x^2+y^2+xy=4 $ represents an ellipse by rewriting the equation into the form $ d_S=e\times d_D, $ where $ d_S $ denotes the distance of $ P(x,y) $ from some fixed point S and $ d_D $ denotes the distance of $ P(x,y) $ from some fixed straight line D and $ 0<e<1 $ is a constant.$

Sy123 · Dec 10, 2014

Re: HSC 2015 4U Marathon

$\\ $The polynomial$ \ x^3- 2x^2 + 3x - 1 \ $has roots$ \ \alpha, \beta, \gamma \\ \\ $Find the value of$ \ (\alpha + \beta - \gamma)(\alpha + \gamma - \beta)(\beta + \gamma - \alpha)$

authenticity · Dec 10, 2014

Re: HSC 2015 4U Marathon

Is it possible to do it this way?
http://i.imgur.com/E4R20ER.jpg

Sy123 · Dec 10, 2014

Re: HSC 2015 4U Marathon

authenticity said:
Is it possible to do it this way?
http://i.imgur.com/E4R20ER.jpg

That's almost how I would do it:

After showing

$(\alpha + \beta - \gamma)(\alpha + \gamma - \beta)(\beta + \gamma - \alpha) = 8(1- \alpha)(1-\beta)(1-\gamma)$

$p(x) = x^3 - 2x^2 + 3x - 1 = (x - \alpha)(x- \beta)(x-\gamma)$

$\therefore \ p(1) = 1 = (1-\alpha)(1-\beta)(1-\gamma)$

So no need to expand!

authenticity · Dec 11, 2014

Re: HSC 2015 4U Marathon

Sy123 said:
That's almost how I would do it:

After showing

$(\alpha + \beta - \gamma)(\alpha + \gamma - \beta)(\beta + \gamma - \alpha) = 8(1- \alpha)(1-\beta)(1-\gamma)$

$p(x) = x^3 - 2x^2 + 3x - 1 = (x - \alpha)(x- \beta)(x-\gamma)$

$\therefore \ p(1) = 1 = (1-\alpha)(1-\beta)(1-\gamma)$

So no need to expand!

that's so much smarter! It didn't even cross my mind at all lol

Sy123 · Dec 11, 2014

Re: HSC 2015 4U Marathon

$\\ $Describe the locus of$ \ z \ $in the Argand plane if$ \ \frac{1}{z} + \frac{1}{\bar{z}} = 1$

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