HSC 2012 MX2 Marathon (archive) (2 Viewers)

bleakarcher · Jan 11, 2012

Re: 2012 HSC MX2 Marathon

true.

deswa1 · Jan 11, 2012

Re: 2012 HSC MX2 Marathon

Thanks guys. The way I did it was Carrotsticks way but I don't get how that can only be a one mark question...

seanieg89 · Jan 11, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
It's good to see new methods pop up. I never thought of this. Nice work!

However, It is a wee bit slow. Another problem I can see coming is that you will need to actually show that it is in fact max/min, which can get quite tedious.

It can be made pretty fast...observe that arctan is an increasing function of the positive reals and that $x^2+4=(x-2)^2+4x\geq 4x$ with equality attained at x=2.

So
$\arctan(\theta)\geq \arctan(1)\Rightarrow \arg(z)\geq \pi/4.$
with equality attained for some z.

Carrotsticks · Jan 11, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Thanks guys. The way I did it was Carrotsticks way but I don't get how that can only be a one mark question...

If it's only a 1 mark question, then a solution by simple inspection will suffice (seanieg89's way).

If it is more, then perhaps a more 'definite' proof may be required.

maths lover · Jan 11, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
The method I'm using below can be handy sometimes, because it gives me max AND min argument immediately. Of course there are other ways, but it's really just preference.:

$\text{The locus is defined to be } y=\frac{1}{4} (x^2+4) \\ \text{The line going through the origin and touching the curve, hence yielding the max/min argument, is in the form: }\\y=mx$

$\text{Substitute into one another to acquire a quadratic in terms of x and m: }\\ x^2-4mx+4=0$

$\text{Let the discriminant be equal to 0, since we are looking for tangents (which have 1 root with the curve) and solve for m:}$

$m=\pm 1\\ \\ \therefore tan\theta =\pm 1 \\ \\ \therefore \theta = \pm \frac{\pi}{4}$

$\text{Thus, minimum argument is } \frac{\pi}{4}$

EDIT:

How did you use calculus for this problem?

i am going to rep you for this very intuitive method(me thinks). avoids allot of the tedious trigonometry i usually use for these types of questions.

rolpsy · Jan 12, 2012

Re: 2012 HSC MX2 Marathon

$\\\textup{The exponential function can be defined as: } e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots,\\ \textup{and the sine function as: } \sin (x)= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots.$

$\begin{align*}\textup{(a) } &\textup{Show that:}\\\textit{i. }& \frac{\mathrm{d} }{\mathrm{d} x} \left ( e^x \right ) = e^x\\\textit{ii. }& \cos (x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \\\textup{(b) } &\textup{Show:}\\&e^{i \theta} = \cos (\theta) + i \sin(\theta)\\\textup{(c) } &\textup{Prove:}\\\textit{i. }& \cos(x) = \frac{e^{ix} + e^{-ix}}{2}\\\textit{ii. }& \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}\end{align*}$

SpiralFlex · Jan 12, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
$\\\textup{The exponential function can be defined as: } e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots,\\ \textup{and the sine function as: } \sin (x)= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots.$

$\begin{align*}\textup{(a) } &\textup{Show that:}\\&\textit{i. } \frac{\mathrm{d} }{\mathrm{d} x} \left ( e^x \right ) = e^x\\&\textit{ii. } \cos (x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \\\textup{(b) } &\textup{Show:}\\&e^{i \theta} = \cos (\theta) + i \sin(\theta)\end{align*}$

$i) \frac{d}{dx}(e^x)=0+1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...$ By differentiation.

$\therefore \frac{d}{dx}(e^x)=e^x$

$ii)\frac{d}{dx}\sin x = \cos x$

$\therefore \frac{d}{dx}(x-\frac{x^3}{3!}+\frac{x^5}{5!}-....)=1-\frac{x^2}{2!}+\frac{x^4}{4!}...=\cos x$

b)

Let $x=i\theta$

Solve from there. By using RHS then it will be in the form of LHS.

Laptop stuffing up so I can't type much.

For the last parts.

Use $e^{ix}=\cos x + i\sin x$

$e^{-ix}=\cos x- i\sin x$

Substitute and solve.

rolpsy · Jan 12, 2012

Re: 2012 HSC MX2 Marathon

$\textup{By writing } \cos^2(x) \textup{ as } \left ( \frac{e^{ix} + e^{-ix}}{2} \right )^2, \textup{ find } \int \cos^2(x) \, \mathrm{d} x. \textup{ (treat }i\textup{ as a constant)}$

SpiralFlex · Jan 12, 2012

Re: 2012 HSC MX2 Marathon

Rolpsy, you are really tempting me to not sleep and do questions. (I'll miss Scischool tomorrow.

)

$\int \frac{e^{2ix}+2+e^{-2ix}}{4} dx$

$=\frac{1}{4}[\frac{e^{2ix}}{2}+2x-\frac{e^{-2ix}}{2i}]+C$

By using the identity of

$Re(z) = \frac{e^{ix}-e^{-ix}}{2i}=\sin x$ Taking $z=e^{ix}$

But substitute $2x$ instead,

$=\frac{1}{4}(\sin 2x + 2x)+C$

rolpsy · Jan 12, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Rolpsy, you are really tempting me to not sleep and do questions. (I'll miss Scischool tomorrow. )

ok I'll stop lol – until tomorrow

$\begin{align*}&\textup{Let } I = \int^{\infty}_{0} \frac{\log(x)}{x^2 + a^2} \, \mathrm{d} x, \, a>0.\\&\textup{By using the substitution } x =\frac{a^2}{u} \textup{ (or otherwise), show that}\\&I=\frac{\pi}{2a}\log(a).\end{align*}$

IamBread · Jan 13, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
ok I'll stop lol – until tomorrow

$\begin{align*}&\textup{Let } I = \int^{\infty}_{0} \frac{\log(x)}{x^2 + a^2} \, \mathrm{d} x, \, a>0.\\&\textup{By using the substitution } x =\frac{a^2}{u} \textup{ (or otherwise), show that}\\&I=\frac{\pi}{2a}\log(a).\end{align*}$

Interesting question.

$\text{let } x = \frac{a^2}{u}$

$\frac{dx}{du}=-\frac{a^2}{u^2}$

$\text{when }x=\infty,u=0$

$\text{when }x=0,u=\infty$

$\therefore\text{I}=\int^{\infty}_{0} \frac{\log(\frac{a^2}{u})}{{\frac{a^4}{u^2} + a^2}}\times\frac{a^2}{u^2}du$

$=\int^{\infty}_{0} \frac{2\log(a)-\log(u)}{{u^2 + a^2}}du$

$=\int^{\infty}_{0} \frac{2\log(a)}{{u^2 + a^2}}du - \int^{\infty}_{0} \frac{\log(x)}{{x^2 + a^2}}dx$

$\therefore 2I=\int^{\infty}_{0} \frac{2\log(a)}{{u^2 + a^2}}du$

$=[2\log(a)\tan^{-1}\frac{u}{a}|^\infty_0$

$\therefore I=\frac{\pi}{2a}\log(a) \text{ as required}$

IamBread · Jan 13, 2012

Re: 2012 HSC MX2 Marathon

First time using latex, that took awhile...

rolpsy · Jan 13, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Interesting question.

$\text{let } x = \frac{a^2}{u}$

$\frac{dx}{du}=-\frac{a^2}{u^2}$

$\text{when }x=\infty,u=0$

$\text{when }x=0,u=\infty$

$\therefore\text{I}=\int^{\infty}_{0} \frac{\log(\frac{a^2}{u})}{{\frac{a^4}{u^2} + a^2}}\times\frac{a^2}{u^2}du$

$=\int^{\infty}_{0} \frac{2\log(a)-\log(u)}{{u^2 + a^2}}du$

$=\int^{\infty}_{0} \frac{2\log(a)}{{u^2 + a^2}}du - \int^{\infty}_{0} \frac{\log(x)}{{x^2 + a^2}}dx$

$\therefore 2I=\int^{\infty}_{0} \frac{2\log(a)}{{u^2 + a^2}}du$

$=[2\log(a)\tan^{-1}\frac{u}{a}|^\infty_0$

$\therefore I=\frac{\pi}{2a}\log(a) \text{ as required}$

cool (although you should probably include limit swap step: $-\int_{\infty}^{0} \frac{\log \left (\frac{a^2}{u} \right )}{{\frac{a^4}{u^2} + a^2}}\times\frac{a^2}{u^2} \, {\mathrm{d} u} = \int^{\infty}_{0} \frac{\log \left (\frac{a^2}{u} \right )}{{\frac{a^4}{u^2} + a^2}}\times\frac{a^2}{u^2} \, {\mathrm{d} u}$ )

$\begin{align*}\textup{(a)} &\textup{ Let } S_n = 1 - x + x^2 - x^3 + \cdots + (-x)^n.\\i.& \textup{ Find } S_{\infty} \textup{ for }|x|<1.\\ii.& \textup{ Show } \ln(1+x) =x-\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \textup{ for }|x| <1.\end{align*}$

Carrotsticks · Jan 13, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
$\begin{align*}\textup{(a)} &\textup{ Let } S_n = 1 - x + x^2 - x^3 + \cdots + (-x)^n.\\i.& \textup{ Find } S_{\infty} \textup{ for }|x|<1.\\ii.& \textup{ Show } \ln(1+x) =x-\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \textup{ for }|x| <1.\end{align*}$

The question is actually invalid because it does not take into account the fact that there is an error term.

IamBread · Jan 13, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
cool (although you should probably include limit swap step: $-\int_{\infty}^{0} \frac{\log \left (\frac{a^2}{u} \right )}{{\frac{a^4}{u^2} + a^2}}\times\frac{a^2}{u^2} \, {\mathrm{d} u} = \int^{\infty}_{0} \frac{\log \left (\frac{a^2}{u} \right )}{{\frac{a^4}{u^2} + a^2}}\times\frac{a^2}{u^2} \, {\mathrm{d} u}$ )

Yeah I did on paper, but didn't want to write more then was absolutely necessary with latex

IamBread · Jan 13, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
The question is actually invalid because it does not take into account the fact that there is an error term.

Error term?

IamBread · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Where do you guys get these questions from?

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Many 'difficult' or 'interesting' Extension 2 questions are proofs for cool identities.

These proofs can be found everywhere on the Internet, or any higher-level Maths textbooks.

ie: The question rolpsy just posted was actually a 'proof' (though technically flawed, the idea is correct) of the Taylor Series expansion for the function ln(x+1). It also leads to the fact that the Alternating Harmonic Series converges to ln(2).

rolpsy · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

oh.. well anyway:

$\begin{align*}S_{\infty}&=\frac{a}{1-r}\\&=\frac{1}{1-(-x)}\\&=\frac{1}{1+x}, \textup{ for } |x| <1.\end{align*}$

$\begin{align*}\therefore \frac{1}{1+x} &= 1 - x + x^2 - x^3 + x^4 - \cdots, \textup{ for } |x| <1\\\int \frac{1}{1+x} \, \mathrm{d}x &= \int 1 - x + x^2 - x^3 + x^4 - \cdots \, \mathrm{d} x \\\ln(1+x) + C &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\\\textup{let } x=0, \, \ln(1) + C &= 0 - \frac{0^2}{2} + \frac{0^3}{3} - \frac{0^4}{4} + \cdots\\\therefore C&=0\\\therefore \ln(1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots, \textup{ for } |x| <1.\\\end{align*}$

new question:
$\\\textup{If } \omega \textup{ is a cube root of unity, show } \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega^2.$

nightweaver066 · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

$\frac{a + bw + cw^2}{c + aw + bw^2}$

$= \frac{c + aw + bw^2}{w(c + aw + bw^2)}$

$= \frac{1}{w}$

$= \frac{w^3}{w^4}$

$= \frac{w^3}{w}$

$= w^2$

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