HSC 2012 MX2 Marathon (archive) (3 Viewers)

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Haha ok, I'll let you go to the toilet

Plus I'm out of standard integrals to post lol.

Well you could always do some Uni integrals.

$\int \frac{1}{a^2-x^2}\:dx=\frac{1}{2a}\log |\frac{a+x}{a-x}|+C$

$\int \frac{1}{x^2-a^2}\:dx=\frac{1}{2a}\log |\frac{x-a}{x+a}|+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

One last question then sleep for me.

$$Prove via mathematical induction $ (cosx + isinx)^n = cosnx + isinnx$

Goodnight.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Well you could always do some Uni integrals.

$\int \frac{1}{a^2-x^2}\:dx=\frac{1}{2a}\log |\frac{a+x}{a-x}|+C$

$\int \frac{1}{x^2-a^2}\:dx=\frac{1}{2a}\log |\frac{x-a}{x+a}|+C$

I was thinking of doing things like that, but wasn't too sure if people had done partial fractions.

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
One last question then sleep for me.

$$Prove via mathematical induction $ (cosx + isinx)^n = cosnx + isinnx$

Goodnight.

I will do for n>0. Where n is an integer. I can't be bothered.

Let n = 1,

$LHS=(\cos x+i\sin x)^1=\cos x+i\sin x$

$RHS=\cos (1)x+i\sin (1)x$

$\therefore LHS=RHS$

$The statement is true for n=1.$

$Assume true for n=k$

$(\cos x+i\sin x)^k=\cos kx+i\sin kx$

$$R.T.P for n=k+1$

$R.T.P \: (\cos x+i\sin x)^{k+1}=\cos {k+1}x+i\sin {k+1}x$

$LHS=(\cos x+i\sin x)^{k+1}$

$=(\cos x+i\sin x)^k (\cos x+i\sin x)$

$=(\cos kx+i\sin kx)(\cos x+i\sin x)$ (From the assumption.)

$=\cos kx cos x+i\sin x \cos kx +i\sin kx\cos x -\sin kx \sin x$

$=\cos(k+1)x+i\sin (k+1)x$

$=RHS$

$\therefore$ True for n=k+1$$

Insert conclusion here.

Parvee · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Why does this thread make me feel way behind in 4u?

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Parvee said:
Why does this thread make me feel way behind in 4u?

Don't feel like that. We just like to post random stuff. Many of the stuff you see here aren't even testable for 4U.

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
Hey your thing says HSC 2011, are you doing science / advanced maths this year and if so at which uni?

Also may as well post a question. Dunno how suitable thing one is but its neat:

Part (A). Trapezium ABCD is given with AB parallel to CD. Let diagonals AC and BD intersect at X. Prove that triangles ADX and BCX have the same area.

Part (B). A triangle and any point on one of its sides are given. Show how to construct, with straight edge and compass, a line through this given point that bisects the triangle into two regions of equal area.

(A) Using alternate angle theorem and vertically opposite angles, you can prove that the two triangles ADX and BCX are similar. Then use the expression $Area=\frac{1}{2} a b \sin C$ to prove equality of areas using identities from the ratios between the lengths of similar triangles.

(B) Using basic straight edge and compass techniques (the ones taught in Year 8 Maths), draw the median of the opposite side from given point. The median splits the triangle into two equal segments.

asianese · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Parvee said:
Why does this thread make me feel way behind in 4u?

THIS THIS THIS ^^^^

Getting very insecure..

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

asianese said:
THIS THIS THIS ^^^^

Getting very insecure..

Like Spiral said, there is little need for feeling bad about it. Here are my reasons.

1. The majority of the questions here won't actually appear in exams. They are simply questions that test our thinking abilities, not necessarily a reflection of what the HSC will be like.

2. The majority of the posts from this thread have been from 2012'ers who most likely have Mathematics as their strongest subject, hence why they are able to do the questions.

3. It is likely that the people posting there are being/have been tutored (perhaps even ahead of time) in Extension 2, hence why they are able to do other topics usually taught later.

4. Some of the questions here were posted by people who have either graduated or are currently studying at University. Many of these questions (like tywebb's proof for the Generalised Wallis Product) are very difficult and perhaps even beyond what is expected of even the best of Extension 2 students.

5. You will be able to do the majority of these questions by the time you have your Trials.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Haha, don't worry guys, most of questions are just random stuff we find/ think of, even if it is beyond the 4u course.

rolpsy · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

$\\\textup{Let } z = r\, \mathrm{cis}(\theta).\\\textup{Show that } |z+1|\times|z-1| = 1 \textup{ implies } r^2 = 2\cos(2\theta).$

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
$\\\textup{Let } z = r\, \mathrm{cis}(\theta).\\\textup{Show that } |z+1|\times|z-1| = 1 \textup{ implies } r^2 = 2\cos(2\theta).$

Rolpsy, I very much like your questions.

May I ask from where do you acquire them, unless you make them yourself?

tywebb · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Many of these questions (like tywebb's proof for the Generalised Wallis Product) are very difficult and perhaps even beyond what is expected of even the best of Extension 2 students.

tywebb said:
$\ \newline(a)\text{ If }y=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-n^2}\text{ for }0<x<1\text{ use the fact that }\pi^2=\sum_{n=1}^\infty{6\over n^2}\text{ to show }\newline\ \newline{\hskip0.25in} \pi^2+y'+y^2=0\newline\ \newline (b)\text{ Hence show }y=\pi\cot\pi x\text{ for }0<x<1\newline\ \newline{\hskip0.25in}\text{ and }\cot x=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-\pi^2n^2}\text{ for }0<x<\pi\newline\ \newline (c)\text{ Find }\int_0^x(\cot t-{1\over t})\ \!dt\text{ and hence show }\newline\ \newline{\hskip0.25in}\sin x=\textstyle x\prod_{n=1}^\infty(1-{x^2\over\pi^2n^2})\text{ for }0<x<\pi\newline\ \newline (d)\text{ Let }x=\textstyle{\pi\over k}\text{ to show }\prod_{n=1}^\infty{(kn)^2\over(kn)^2-1}={\pi/k\over\sin(\pi/k)}\text{ for }k>1\newline\ \newline (e)\text{ Let }k=2\text{ to prove Wallis' Product, }\prod_{n=1}^\infty{(2n)^2\over (2n)^2-1}={\pi\over2}$

$\textstyle\text{Letting }k=2,\ \therefore\prod_{n=1}^\infty{(2n)^2\over (2n)^2-1}={\pi/2\over\sin(\pi/2)}={\pi\over2}$

It wasn't really that hard.

Next question:

$\text{Sketch }y=e^{f(x)}$

Does it have inflections?

Carrotsticks · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

^^^^^ I had a very similar proof for that question. The main difficulty was finding the necessary substitution in order to acquire the telescoping sum. I also got stuck during the calculations for Part (A). I got the exact same as you for line 3, except I had 8x^2 instead of 4x^2, and I could not proceed (similarly to the way that $(a+b)^{2}=a^{2}+b^{2}+2ab$

However, I do not think many students will be able to recognise:

1. The substitution to be used.

2. The use of Partial fractions in your solutions line 3

Here were my solutions, which I had written a while ago, but never got round to uploading them. I included explanations for 2012'ers who may not understand all the concepts there.

seanieg89 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
(B) Using basic straight edge and compass techniques (the ones taught in Year 8 Maths), draw the median of the opposite side from given point. The median splits the triangle into two equal segments.

The point is an arbitrary point on one of the triangles sides, not a vertex.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

I don't think anyone did this question, so I'll repost it

$$Using $ sinx = \frac{e^{ix}-e^{-ix}}{2i} $, show sin^{-1}x = -i\ln(ix+\sqrt{1-x^2})$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Let $y=\sin x=\frac{e^{ix}-e^{-ix}}{2i}$

Noting: $\sin^{-1}y=x$

$\frac{e^{ix}-e^{-ix}}{2i}$

Let $z=cis x =e^{ix}$

$\frac{1}{z}=\frac{1}{cis x}=e^{-ix}$

$\frac{z-\frac{1}{z}}{2i}=y$

$z^2-2iyz-1=0$

$z=iy \pm \sqrt{1-y^2}$

$\therefore iy \pm \sqrt{1-y^2}=e^{ix}$

Solve for $x$

$\ln (iy \pm \sqrt{1-y^2})=ix$

$\therefore x=-i\ln (iy \pm \sqrt{1-y^2})$

Taking the positive solution,

$\therefore \sin^{-1}{y}=-i\ln (iy + \sqrt{1-y^2})$

Editing

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

I'm not sure why you added the $z=cisx$ part for, but that's right.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Show the following:

$1. |z+w|^2 = |z|^2 +2Re(z\overline{w}) +|w|^2$

$2. |z-w|^2 = |z|^2 -2Re(z\overline{w}) +|w|^2$

$3. |z+w|^2 +|z-w|^2= 2(|z|^2 +|w|^2)$

Drongoski · Jan 20, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Show the following:

$1. |z+w|^2 = |z|^2 +2Re(z\overline{w}) +|w|^2$

$2. |z-w|^2 = |z|^2 -2Re(z\overline{w}) +|w|^2$

$3. |z+w|^2 +|z-w|^2= 2(|z|^2 +|w|^2)$

$1) |z+w|^2 = (z+w) \overline {(z+w)} =(z+w)(\overline {z} +\overline {w} ) = z \overline{z} + z \overline {w} + \overline {z} w + w \overline {w} \\ \\ = |z|^2 + (z\overline {w} + \overline {z\overline{w}}) + |w|^2 = |z|^2 +2 Re (z \overline {w}) + |w|^2$

2) Similar to 1)

3) Add 1) and 2)

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