HSC 2016 MX2 Marathon (archive) (5 Viewers)

relativity1 · Nov 6, 2015

Re: HSC 2016 4U Marathon

Realise the fraction
and that gives you
(cos a)(cos b)-(cos a)(isin b)+(cos b)(isin a)+(sin a)(sin b) note: denominator = 1 as sin^2b+cos^2b=1
=cos(a-b)+isin(a-b) using trigonometric identities (cos expansion and sin expansion)

leehuan · Nov 6, 2015

Re: HSC 2016 4U Marathon

$$Describe the key features of exponentiating a given graph. \\ i.e. What happens when $y=f\left( x \right) \Rightarrow y={ e }^{ f\left( x \right) }$

(Too much effort giving a graph and making people sketch it)

Paradoxica · Nov 6, 2015

Re: HSC 2016 4U Marathon

leehuan said:
$$Describe the key features of exponentiating a given graph. \\ i.e. What happens when $y=f\left( x \right) \Rightarrow y={ e }^{ f\left( x \right) }$

(Too much effort giving a graph and making people sketch it)

All convexity and concavity is preserved. Vertical Asymptotes remain so, and horizontal asymptotes are now the exponentiation of their original value. A similar case occurs for oblique asymptotes. Critical points remain critical for the input values.

leehuan · Nov 6, 2015

Re: HSC 2016 4U Marathon

There's also the special case of x-intercepts always being shifted up one ordinate, and when graphing if the graph approaches the vertical asymptote in the negative direction, you should plot a discontinuity (hole in graph) at where an "x-intercept" would be on the exponentiated graph (however, this is only for when you're actually sketching).
-------------------------------
Too tired to think up a new question for 4u.

InteGrand · Nov 7, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
All convexity and concavity is preserved.

$This part is not true. If $y=\mathrm{e}^{f(x)}$ and $f$ is twice differentiable, then $y^\prime = f^\prime (x)\mathrm{e}^{f(x)}$, and so$

$\begin{align*}y'' &= f'' (x)\mathrm{e}^{f(x)} + f^\prime (x) \times f^\prime (x)\mathrm{e}^{f(x)} \\ &= f'' (x) \mathrm{e}^{f(x)} + \left( f^\prime (x)\right)^2 \mathrm{e}^{f(x)}. \end{align*}$

$So if $f''(x) \geq 0$ for all $x$ in the domain (so $f$ is convex (concave up)), then $y=\mathrm{e}^{f(x)}$ is also convex. However, if $f''(x) \leq 0$ for all $x$ in the domain (so $f$ is concave (concave down)), then we do not necessarily have $y''$ be negative/zero always ($y'' \leq 0$ if and only if $f''(x) \leq -\left( f^\prime (x)\right)^2$). As an example, look at the graph of $y=\mathrm{e}^{\sqrt{x}}$ (linked below). Clearly the square root function is concave (i.e. concave down), but $y=\mathrm{e}^{\sqrt{x}}$ changes concavity.$

Graph: https://graphsketch.com/?eqn1_color..._lines=1&line_width=4&image_w=850&image_h=525

InteGrand · Nov 7, 2015

Re: HSC 2016 4U Marathon

$Simple complex numbers Q:$

$Let $z$ be a non-zero complex number. Show that $\frac{1}{z}$ is real if and only if $z$ is real.$

leehuan · Nov 7, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$This part is not true. If $y=\mathrm{e}^{f(x)}$ and $f$ is twice differentiable, then $y^\prime = f^\prime (x)\mathrm{e}^{f(x)}$, and so$

$\begin{align*}y'' &= f'' (x)\mathrm{e}^{f(x)} + f^\prime (x) \times f^\prime (x)\mathrm{e}^{f(x)} \\ &= f'' (x) \mathrm{e}^{f(x)} + \left( f^\prime (x)\right)^2 \mathrm{e}^{f(x)}. \end{align*}$

$So if $f''(x) \geq 0$ for all $x$ in the domain (so $f$ is convex (concave up)), then $y=\mathrm{e}^{f(x)}$ is also convex. However, if $f''(x) \leq 0$ for all $x$ in the domain (so $f$ is concave (concave down)), then we do not necessarily have $y''$ be negative/zero always ($y'' \leq 0$ if and only if $f''(x) \leq -\left( f^\prime (x)\right)^2$). As an example, look at the graph of $y=\mathrm{e}^{\sqrt{x}}$ (linked below). Clearly the square root function is concave (i.e. concave down), but $y=\mathrm{e}^{\sqrt{x}}$ changes concavity.$

Graph: https://graphsketch.com/?eqn1_color..._lines=1&line_width=4&image_w=850&image_h=525

Huh, go figure, I was differentiating that in my head last night as well and was thinking something was weird. But I let it be because I was unsure of my theory.

qwert73 · Nov 7, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$Simple complex numbers Q:$

$Let $z$ be a non-zero complex number. Show that $\frac{1}{z}$ is real if and only if $z$ is real.$

I'm not too sure how to do these ones but here is what I did.

1/z = 1/(x+iy) = (x-iy)/(x^2 +y^2)
therefor if y does not equal zero it is a complex number, if y = 0 it is a real number

if y = 0 then z is a real number
therefor z must be real

leehuan · Nov 7, 2015

Re: HSC 2016 4U Marathon

$\\ $Prove that for any polynomial with real coefficients, if a root is not real, it's complex conjugate is also a root.$$
---------------------------------------
$\\ $Prove that for any monic polynomial with integer roots, the roots are always factors of the constant term.$

Paradoxica · Nov 7, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$This part is not true. If $y=\mathrm{e}^{f(x)}$ and $f$ is twice differentiable, then $y^\prime = f^\prime (x)\mathrm{e}^{f(x)}$, and so$

$\begin{align*}y'' &= f'' (x)\mathrm{e}^{f(x)} + f^\prime (x) \times f^\prime (x)\mathrm{e}^{f(x)} \\ &= f'' (x) \mathrm{e}^{f(x)} + \left( f^\prime (x)\right)^2 \mathrm{e}^{f(x)}. \end{align*}$

$So if $f''(x) \geq 0$ for all $x$ in the domain (so $f$ is convex (concave up)), then $y=\mathrm{e}^{f(x)}$ is also convex. However, if $f''(x) \leq 0$ for all $x$ in the domain (so $f$ is concave (concave down)), then we do not necessarily have $y''$ be negative/zero always ($y'' \leq 0$ if and only if $f''(x) \leq -\left( f^\prime (x)\right)^2$). As an example, look at the graph of $y=\mathrm{e}^{\sqrt{x}}$ (linked below). Clearly the square root function is concave (i.e. concave down), but $y=\mathrm{e}^{\sqrt{x}}$ changes concavity.$

Graph: https://graphsketch.com/?eqn1_color..._lines=1&line_width=4&image_w=850&image_h=525

I think a more blatantly obvious example is ln(x). e^ln(x) = x, which is not concave or convex at all. But thanks for pointing out my false assumption.

InteGrand · Nov 7, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
I think a more blatantly obvious example is ln(x). e^ln(x) = x, which is not concave or convex at all. But thanks for pointing out my false assumption.

$Technically $y=x$ is both a concave and convex function (which is why I didn't use it as the example). However, it is not a \textsl{strictly convex} function.$

Drsoccerball · Nov 7, 2015

Re: HSC 2016 4U Marathon

$(i) One day Integrand realised he needed to pee and decided to play a mathematical game. He stood at the toilet at a distance of R and fired his projectile at a speed of V and at an angle of $ \theta $ to the horizontal.$

$$ Given the height of the projectile is H and the height of the toilet is h, Prove that the time taken to reach the toilet is :$

$T= \frac{2Vsin\theta \pm \sqrt{2V^2sin^2\theta -8gh+8gH}}{2g}$

$$ where $ H \geq h$

$(ii) Integrand then realises that he has a limited supply of the projectile. Assuming that he maintains an angle of $ \theta $ to the horizontal and the rate at which the substance decreases is k, at what rate must he walk in order to always hit the toilet. For $ 0 \leq t \leq P +Q $ where Q is the time the last drop escapes and P is the time taken for the projectile to travel its journey.$

qwert73 · Nov 8, 2015

Re: HSC 2016 4U Marathon

So what happens when we don't know how to answer because we haven't learnt it at school yet, do we just wait for someone else to answer it?

Sy123 · Nov 8, 2015

Re: HSC 2016 4U Marathon

Needed: Only complex numbers

$\\ $Let$ \ \alpha \ $and$ \ \beta \ $be complex numbers so that:$ \\ \alpha - \beta = ik \ $and$ \ \frac{\alpha}{\beta} = im \ $where$ \ k,m \ $are real$$

$\\ $i) Show that if$ \ z = \alpha + \beta \ $then$ \ z = \frac{2mk}{m^2 + 1} + i \frac{m^2-1}{m^2+1}k$

$\\ $ii) Hence show that every complex number can be represented as the sum of two complex numbers, whose difference and quotient are purely imaginary$$

leehuan · Nov 8, 2015

Re: HSC 2016 4U Marathon

qwert73 said:
So what happens when we don't know how to answer because we haven't learnt it at school yet, do we just wait for someone else to answer it?

For now, that exception is acceptable and the question can just be ignored after 5 days without guilt. But later in the year, follow the rules.

Note that I wouldn't say the same thing for the advanced marathon however. That one usually involves ex-HSC students as well due to it's difficulty.

Drsoccerball · Nov 8, 2015

Re: HSC 2016 4U Marathon

leehuan said:
For now, that exception is acceptable and the question can just be ignored after 5 days without guilt. But later in the year, follow the rules.

Note that I wouldn't say the same thing for the advanced marathon however. That one usually involves ex-HSC students as well due to it's difficulty.

Mybad

I just thought of it and wanted to make it before I forgot.

leehuan · Nov 8, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
Mybad I just thought of it and wanted to make it before I forgot.

Nah your question's fine, I posted a polynomials question earlier as well and let it slip

omegadot · Nov 9, 2015

Re: HSC 2016 4U Marathon

Sy123 said:
$\\ $The polynomial$ \ ax^3 + bx^2 + bx + a \ $is such that$ \ a,b \ $are positive and all the roots are real, prove that$ \ b \geq 3a$

$$Not as slick as Paradoxica's solution, here I give another approach.$

$\noindent By inspection $-1$ is a root of the given polynomial (which is symmetric). Thus$

$\begin{align*}ax^3 + bx^2 + bx + a = (x + 1)(a x^2 + (b - a) x + a).\end{align*}$

$For the remaining two roots to be real we require$

$\begin{align*}\Delta^2 = (b - a)^2 - 4a^2 &\geqslant 0\\ \Rightarrow (b - a)^2 &\geqslant 4a^2\\ \Rightarrow |b - a| &\geqslant |2a| = 2a, \quad \mbox{since \textit{a} is positive.}\end{align*}$

$\noindent This gives $ b \geqslant - a$ which is always true as $a,b$ are both positive and $b \geqslant 3a,$ as reuired to show.$

RachelGreen · Nov 9, 2015

Re: HSC 2016 4U Marathon

oops

KingOfActing · Nov 10, 2015

Re: HSC 2016 4U Marathon

Sy123 said:
Needed: Only complex numbers

$\\ $Let$ \ \alpha \ $and$ \ \beta \ $be complex numbers so that:$ \\ \alpha - \beta = ik \ $and$ \ \frac{\alpha}{\beta} = im \ $where$ \ k,m \ $are real$$

$\\ $i) Show that if$ \ z = \alpha + \beta \ $then$ \ z = \frac{2mk}{m^2 + 1} + i \frac{m^2-1}{m^2+1}k$

$\\ $ii) Hence show that every complex number can be represented as the sum of two complex numbers, whose difference and quotient are purely imaginary$$

$i) Solving the equations simultaneously we obtain $ \alpha = \frac{ikm}{m+i}$, $ \beta = \frac{k}{m+i}$ and as such $ z = \alpha \ +\ \beta = \frac{ikm + k}{m + i} = \frac{(ikm+k)(m-i)}{m^2 + 1} = \frac{ikm^2 + km + km - ik}{m^2 + 1} = \frac{2mk}{m^2 + 1} + i\frac{m^2-1}{m^2 + 1}k $ as required.$\\\\$By making the substitution $m = \tan{\frac{\frac{\pi}{2} - \theta}{2}}$ and using the t-formulae, our expression for $z$ becomes $ k\sin{\left(\frac{\pi}{2} - \theta\right)} + ik\cos{\left (\frac{\pi}{2} - \theta\right)} = k(\cos{\theta} \ +\ i\sin{\theta}) $ which is the polar form of any complex number.$

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