HSC 2013 MX2 Marathon (archive) (7 Viewers)

asianese · Dec 11, 2013

Re: HSC 2014 4U Marathon

anomalousdecay said:
The hardest thing I saw was in this year's paper, and that was still pretty easy (full marks).

Also, to do the multiple choice question, you kinda had to know L'Hopital's (I forgot to use this and lost the mark).

Why would you ever need L'Hopital's rule in the HSC? Are you referring to the sinx/x? If you are then I'm very sorry for you.

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

obliviousninja said:
Sy, how hard can stuff in graphing get? I never really saw anything hard.

Not sure actually, depends on what you mean by hard, sketching weird implicit graphs can be seen as hard, such as:

$$Sketch$ \ \ x^2 - y^3 +xy -2y = 0$

integral95 · Dec 11, 2013

Re: HSC 2014 4U Marathon

asianese said:
$$Using de Moivre's theorem$

$(\cos \theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta),$

$$show that for positive, even integers $ n$,$

$\cos(n\theta) = \sum_{k=0}^{n/2} \binom{n}{2k} (-1)^k\sin^{n-2k}\theta\cos^{2k}\theta $ \\and derive a similar expression for $ \sin(n\theta).$

(not sure if this is 100% right, just off the top of my head)

lol I dunno if 2014 guys learn binomial at this stage .

but anyways

$\sin(n\theta) = \sum_{k=1}^{n/2} \binom{n}{2k-1} (-1)^(k-1)\sin^{2k-1}\theta\cos^{n-(2k-1)}\theta $.$

I meant (-1)^(k-1) I'm just learning latex

integral95 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Not sure actually, depends on what you mean by hard, sketching weird implicit graphs can be seen as hard, such as:

$$Sketch$ \ \ x^2 - y^3 +xy -2y = 0$

I think oblivious is only asking about the topic graphing ( not combined with conics )

obliviousninja · Dec 11, 2013

I can't view any questions, I'm on my phone and there's no latex. will view later.

anomalousdecay said:
The hardest thing I saw was in this year's paper, and that was still pretty easy (full marks).

Also, to do the multiple choice question, you kinda had to know L'Hopital's (I forgot to use this and lost the mark).

Lol. The board doesn't approve that method. Did you get this from UNSW mx2 day?

integral95 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $i) Expand$ \ \ (1+i\sqrt{3})(1+i) \\ \\ $ii) Hence show that$ \ \ \cos \left(\frac{\pi}{24} \right) = \frac{1}{2} \sqrt{2 + \sqrt{2+\sqrt{3}}}$

If you want problems suited to a 2014er then I suggest posting some questions or answering the complex numbers that I have posted before

Wouldn't it be better to expand $(1+i\sqrt{3})(1-i)$

since that gives (something) cis pi/12 rather than cis 7pi/12

asianese · Dec 11, 2013

Re: HSC 2014 4U Marathon

MethewYan said:
lol I dunno if 2014 guys learn binomial at this stage .

but anyways

$\sin(n\theta) = \sum_{k=1}^{n/2} \binom{n}{2k-1} (-1)^{k-1} \sin^{2k-1}\theta\cos^{n-(2k-1)}\theta $.$

I meant (-1)^(k-1) I'm just learning latex

ftfy

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

MethewYan said:
Wouldn't it be better to expand $(1+i\sqrt{3})(1-i)$

since that gives (something) cis pi/12 rather than cis 7pi/12

Indeed but the initial cis 7pi/12 provides a small obstacle to getting cis pi/12, so yes it makes it slightly necessarily harder

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Maybe this is better

$\\ $Find in the form$ \ (a+ib) \\ \\ $i)$ \ (1+i)(2+i) \\ \\ $ii)$ \ \frac{1+i}{2+i} \\ \\ $iii)$ (5+6i)+ (7+i) \\ \\ $iv)$ \ \frac{3+36i}{2+15i} \\ \\ $v)$ \ (6+i) - (\pi + \pi i) \\ \\ $vi)$ \ \frac{1+i}{1+i} \\ \\ $vii)$ \ \frac{1+12i}{4+3i} \\ \\ $viii)$ \ (2+i)(8+5i) \\ \\ $ix)$ \ (1+4i) - (3+2i) \\ \\ $x)$ \ \frac{3+20i}{2+7i}$

anomalousdecay · Dec 11, 2013

Re: HSC 2014 4U Marathon

asianese said:
Why would you ever need L'Hopital's rule in the HSC? Are you referring to the sinx/x? If you are then I'm very sorry for you.

Yes I was. I just referred to that identity as a use of L'Hopital's rule, Hence my use of "kinda".

Oh well, at least by making that mistake, I made sure to remember to quickly check my answer in ext.1 two days after.

obliviousninja said:
I can't view any questions, I'm on my phone and there's no latex. will view later.

Lol. The board doesn't approve that method. Did you get this from UNSW mx2 day?

Yeah lol. You told me you were on the other side of the room with the guys from that school clapping and congratulating some guy who made a mistake.. Relax it was an m/c question (q9 I think).

MethewYan said:
lol I dunno if 2014 guys learn binomial at this stage .

Yeah first topic I did last year. Depends on the school I guess.

Sy123 said:
Maybe this is better

You should make a textbook.

RealiseNothing · Dec 11, 2013

Re: HSC 2014 4U Marathon

anomalousdecay said:
The hardest thing I saw was in this year's paper, and that was still pretty easy (full marks).

Also, to do the multiple choice question, you kinda had to know L'Hopital's (I forgot to use this and lost the mark).

I'm assuming this was the multiple choice question? That was 2U lol.

obliviousninja said:
Lol. The board doesn't approve that method. Did you get this from UNSW mx2 day?

L'Hopital's Rule is widely known, not everyone got it from the UNSW MX2 day...

seanieg89 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $The sum of the positive reals$ \ a_1, a_2 , \dots , a_n \ $is$ \ 1 \\ \\ $Show that$ \\ \\ \left(1 + \frac{1}{a_1}\right) \left(1 + \frac{1}{a_2} \right) \left(1 + \frac{1}{a_3} \right) \dots \left(1+ \frac{1}{a_n} \right) \geq (n+1)^n$

$Let $y_k=na_k$ and let $S_k$ be the $k$-th elementary symmetric sum of the set $\{y_k^{-1}\}_{k=1}^n.$\\ \\ Our constraint on the $a_j$ translates to $\sum_k y_k=n$, and so by AM-GM we have $\prod_k y_k\leq 1.$\\ \\ Applying AM-GM to the sum $S_k$ and using the above gives\\ \\ $S_k\geq {n \choose k}$.\\ \\ Expanding out the product in our question then gives:\\ \\ $\prod_{k=1}^n (1+n/y_j)=\sum_{k=0}^n n^kS_k\geq \sum_{k=0}^n n^k{n \choose k}=(1+n)^n.$$

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
$Let $y_k=na_k$ and let $S_k$ be the $k$-th elementary symmetric sum of the set $\{y_k^{-1}\}_{k=1}^n.$\\ \\ Our constraint on the $a_j$ translates to $\sum_k y_k=n$, and so by AM-GM we have $\prod_k y_k\leq 1.$\\ \\ Applying AM-GM to the sum $S_k$ and using the above gives\\ \\ $S_k\geq {n \choose k}$.\\ \\ Expanding out the product in our question then gives:\\ \\ $\prod_{k=1}^n (1+n/y_j)=\sum_{k=0}^n n^kS_k\geq \sum_{k=0}^n n^k{n \choose k}=(1+n)^n.$$

Nice solution, here is mine

$\\ $By a little manipulation we simply need to prove:$$

$\frac{1}{n+1} \geq \sqrt[n]{\prod_1^n \frac{a_k}{a_k+1}}$

$\\ $By AM-GM$ \\ \\ \sqrt[n]{\prod_1^n \frac{a_k}{1+a_k}} \leq \frac{1}{n} \sum_1^n \frac{a_k}{1+a_k} \\ \\ \Rightarrow \ \frac{1}{n} \sum_1^n \frac{a_k}{1+a_k} = \frac{1}{n} \left(n - \sum_1^n \frac{1}{1+a_k} \right) \\ \\ $From the inequality$ \ \ \left(\sum_1^n (1+a_k) \right) \left(\sum_1^n \frac{1}{1+a_k} \right) \geq n^2 \ \ $easily proven by AM-GM, we yield the required inequality$ \\ \\ \rightarrow \frac{1}{n} \left(n - \sum_1^n \frac{1}{a_k+1} \right) \leq \frac{1}{n} \left(n - \frac{n^2}{n+1} \right) = \frac{1}{n+1} \\ \\ $Therefore$ \ \ \frac{1}{n+1} \geq \sqrt[n]{\prod_1^n \frac{a_k}{a_k+1}}$

Fawun · Dec 11, 2013

Re: HSC 2014 4U Marathon

Okay you know i'm shit at maths so let me attempt this

Sy123 said:
$\\ $i) Expand$ \ \ (1+i\sqrt{3})(1+i) \\ \\ $ii) Hence show that$ \ \ \cos \left(\frac{\pi}{24} \right) = \frac{1}{2} \sqrt{2 + \sqrt{2+\sqrt{3}}}$

If you want problems suited to a 2014er then I suggest posting some questions or answering the complex numbers that I have posted before

$\begin{align*}\textup{i}) (1 + i\sqrt{3})(1+i) &= 1 + i + i\sqrt{3} + \sqrt{3}i^2 \\ &= (1-\sqrt{3} + i(1+\sqrt{3}) \\ &= \sqrt{8}\textup{ cis} (-\frac{5\pi}{12})\end{align*}$

smh i got -5 pi/12 and so i have no idea where you got pi/24 from

Fawun · Dec 11, 2013

Re: HSC 2014 4U Marathon

A for effort fawun

#fawunsroadtoanE4

anomalousdecay · Dec 11, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
I'm assuming this was the multiple choice question? That was 2U lol.

Yeah. I made the silliest mistake, which could possibly cost me e4. I was supposed to use L'Hopital's to check quickly but I didn't and lost the mark.
Also, the complex region graph and the asymptotes of the other graph are not the average type of questions from recent MX2 papers. Luckily I got all of those marks.

RealiseNothing said:
L'Hopital's Rule is widely known, not everyone got it from the UNSW MX2 day...

I did though lol. We were both in the same room, learnt L'Hopital's but didn't meet

Fawun said:
A for effort fawun

#fawunsroadtoanE4

Don't worry. I couldn't do this type of stuff last year and I might potentially get one. Its just practice.

Sy123 · Dec 11, 2013

Re: HSC 2014 4U Marathon

Fawun said:
Okay you know i'm shit at maths so let me attempt this

$\begin{align*}\textup{i}) (1 + i\sqrt{3})(1+i) &= 1 + i + i\sqrt{3} + \sqrt{3}i^2 \\ &= (1-\sqrt{3} + i(1+\sqrt{3}) \\ &= \sqrt{8}\textup{ cis} (-\frac{5\pi}{12})\end{align*}$

smh i got -5 pi/12 and so i have no idea where you got pi/24 from

Yes the version in the form (a+ib) is correct, how did you get the De Moivere's theorem version from that though?

As far as how to get cos(pi/24) is concerned, you need to find how to get from cos(7pi/12) to cos(pi/24)

RealiseNothing · Dec 11, 2013

Re: HSC 2014 4U Marathon

anomalousdecay said:
I did though lol. We were both in the same room, learnt L'Hopital's but didn't meet

Wait was this that 5 day thing in January?

anomalousdecay · Dec 11, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Wait was this that 5 day thing in January?

No. Open day. What 5 day thing?

RealiseNothing · Dec 11, 2013

Re: HSC 2014 4U Marathon

anomalousdecay said:
No. Open day. What 5 day thing?

ok nvm, I never went to open day

HSC 2013 MX2 Marathon (archive) (7 Viewers)

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