HSC 2014 MX2 Marathon (archive) (2 Viewers)

mreditor16 · Apr 16, 2014

Re: HSC 2014 4U Marathon

kurosaki gets part i)

good job.

Kurosaki · Apr 16, 2014

Re: HSC 2014 4U Marathon

mreditor16 said:
kurosaki gets part i) good job.

Yay! haha I realised I forgot to divide by 2 the first time around :L. Herp derp.

is part 2, is it 43/49? Not sure about this one though.
1081/1225 is the answer for part (ii) I think.
OK, working: so we want to divide 48 coins among 3 pirates. Naturally, dividers comes to mind, but we don't need to add any objects as they are already there, in the form of the two lead coins.
So we permute them to get $\frac{50!}{2!48!}=1225$
For the second part, imagine 48 gold coins lying like so: .........

Now, there are 49 places where we can insert the two lead coins. But we wish to guarantee that all the pirates get gold coins, so the lead coins cannot be on the ends. thus, there are 47 places to insert them, and since they are identical coins, we get 47C2 ways. Divide by 1225 to get $\frac{1081}{1225}$

awesome-0_4000 · Apr 16, 2014

Re: HSC 2014 4U Marathon

So to get 1225 for part (i) would you do:
50!/(48!x2!) ?

juantheron · Apr 16, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Find a root of the equation$ \ \ (x-2)(x-3)(x-4)(x-5) = 24$

Solution :: $$Here $ \ \ (x-2)(x-5)\cdot (x-3)(x-4) = 24\Rightarrow (x^2-7x+10)\cdot(x^2-7x+12) = 24$

$$Let$ \ \ (x^2-7x) = y\;, $Then \; equation \; is \;$ (y+10)\cdot(y+12) = 24$

$\Rightarrow y^2+22y+120 = 24\Rightarrow y^2+22y+96 = 0\Rightarrow y = -16\; y = -6$

$Now \; we \; will \; solve \ for$

$\Rightarrow x^2-7x = -16\Rightarrow x^2-7x+16 = 0\; $and$ \; x^2-7x = -6\Rightarrow x^2-7x+6 = 0$$

$\displaystyle $we\; get \;,$\; x = \frac{1}{2}\left(7\pm i\sqrt{15}\right)\;\; $and$\; x = 1\;,6$

Davo_01 · Apr 16, 2014

Re: HSC 2014 4U Marathon

$\int{\dfrac{x^2}{x^6-1} dx}$

Immortality · Apr 16, 2014

Re: HSC 2014 4U Marathon

mreditor16 said:
Three pirates are sharing out the contents of a treasure chest containing forty-eight
gold coins and two lead coins. The first pirate takes out coins one at a time until a lead
coin is taken. The second pirate then takes out coins one at a time until the second lead
coin is taken. The third pirate then takes all of the remaining coins.
(i) In how many ways can the coins be distributed? (1 mark)
(ii) What is the probability that all three pirates receive some gold coins? (2 mark)

i) 50!/48!2! = 1225
ii) 47C2/1225 = 1081/1225

Davo_01 · Apr 16, 2014

Re: HSC 2014 4U Marathon

Let $f_1(x)=ln(x)$

$f_2(x)=ln(ln(x))$

$f_3(x)=ln(ln(ln(x)))$

$f_n(x)=ln(ln(...ln(x)))$

Prove by Mathematical Induction for $n>1$ :

$\dfrac{d(f_{n})}{dx}=\dfrac{1}{x\cdot f_1\cdot f_2 \cdot ...\cdot f_{n-1}}$

Davo_01 · Apr 16, 2014

Re: HSC 2014 4U Marathon

(i) Show that:

$x^6+x^3+1=(x^2-2x\cos{\frac{2\pi}{9}}+1)(x^2-2x\cos{\frac{4\pi}{9}}+1)(x^2-2x\cos{\frac{8\pi}{9}}+1)$

(ii) Hence evaluate:

$\cos{\frac{2\pi}{9}}+\cos{\frac{4\pi}{9}}+$ $\cos{\frac{8\pi}{9}}$

$\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}+$ $\cos{\frac{4\pi}{9}} \cos{\frac{8\pi}{9}}+$ $\cos{\frac{8\pi}{9}}\cos{\frac{2\pi}{9}}$

$\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}\cos{\frac{8\pi}{9}}$

braintic · Apr 16, 2014

Re: HSC 2014 4U Marathon

A reminder of Trebla's intruction when he set up this thread 4 weeks ago:

"Once a question is posted, it needs to be answered before the next question is raised."

Immortality · Apr 16, 2014

Re: HSC 2014 4U Marathon

Davo_01 said:
$\int{\dfrac{x^2}{x^6-1} dx}$

$\int{\dfrac{x^2}{x^6-1} dx} \\ \\ $let$ k=x^3 \\ \\ \frac{dk}{dx}=3x^2 \\ \\ \frac{dk}{3} = x^3 dx \\ \\ = \frac{1}{3}\int{\dfrac{dk}{k^2-1}} \\ \\ = \frac{1}{3}\int{\dfrac{dk}{(k-1)(k+1)}} \\ \\ = \frac{1}{3}\int[{\frac{1}{2(k-1)}-\frac{1}{2(k+1)}]dk} \\ \\ = \frac{1}{6}[ln|k-1|-ln|k+1|]+c \\ \\ = \frac{1}{6}ln|\frac{x^3-1}{x^3+1}|+c$

dunjaaa · Apr 17, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-04-17 at 1.03.42 AM.png

:d

kden · Apr 17, 2014

Re: HSC 2014 4U Marathon

Curve sketching, don't hate <a href="http://www.codecogs.com/eqnedit.php?latex=Sketch\,\,&space;the\,\,&space;curve\,\,&space;x^2&space;-9xy+y^2&space;=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Sketch\,\,&space;the\,\,&space;curve\,\,&space;x^2&space;-9xy+y^2&space;=1" title="Sketch\,\, the\,\, curve\,\, x^2 -9xy+y^2 =1" /></a>

unnecessary use of latex

dunjaaa · Apr 17, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-04-17 at 1.25.07 AM.png

Nice question, took me while to figure the pattern

dunjaaa · Apr 17, 2014

Re: HSC 2014 4U Marathon

Not sure if rotation of axes of conic sections are within the constraints of the 4U syllabus

dunjaaa · Apr 17, 2014

Re: HSC 2014 4U Marathon

It is a hyperbola that has had its x-y axis rotated 45 degrees anti-clockwise. The equation of this hyperbola on the new X-Y axis is 11y^2-7x^2=2 (I think) and cuts the x-y axis at y=-1 and y=1. Learned something new today

Parvee · Apr 17, 2014

Re: HSC 2014 4U Marathon

kden said:
Curve sketching, don't hate <a href="http://www.codecogs.com/eqnedit.php?latex=Sketch\,\,&space;the\,\,&space;curve\,\,&space;x^2&space;-9xy+y^2&space;=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Sketch\,\,&space;the\,\,&space;curve\,\,&space;x^2&space;-9xy+y^2&space;=1" title="Sketch\,\, the\,\, curve\,\, x^2 -9xy+y^2 =1" /></a>

unnecessary use of latex

theres a neat way of doing these with eigenvalues/vectors and creating principles axes
shame you guys dont do that in the hsc :L

Davo_01 · Apr 17, 2014

Re: HSC 2014 4U Marathon

braintic said:
A reminder of Trebla's intruction when he set up this thread 4 weeks ago:

"Once a question is posted, it needs to be answered before the next question is raised."

Oh sorry about that

Davo_01 · Apr 17, 2014

Re: HSC 2014 4U Marathon

Immortality said:
$\int{\dfrac{x^2}{x^6-1} dx} \\ \\ $let$ k=x^3 \\ \\ \frac{dk}{dx}=3x^2 \\ \\ \frac{dk}{3} = x^3 dx \\ \\ = \frac{1}{3}\int{\dfrac{dk}{k^2-1}} \\ \\ = \frac{1}{3}\int{\dfrac{dk}{(k-1)(k+1)}} \\ \\ = \frac{1}{3}\int[{\frac{1}{2(k-1)}-\frac{1}{2(k+1)}]dk} \\ \\ = \frac{1}{6}[ln|k-1|-ln|k+1|]+c \\ \\ = \frac{1}{6}ln|\frac{x^3-1}{x^3+1}|+c$

Yep very good:

Davo_01 · Apr 17, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
View attachment 30298 Nice question, took me while to figure the pattern

Very nice, alternatively the first one could be done using sum of roots and the third one could be done by letting x=1:
$1+1+1=8(1-\cos{\frac{2\pi}{9}})(1-\cos{\frac{4\pi}{9}})(1-\cos{\frac{8\pi}{9}})$
$\frac{3}{8}=1-(\cos{\frac{2\pi}{9}}$ $+\cos{\frac{4\pi}{9}}+\cos{\frac{8\pi}{9}})$ $+(\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}+$ $\cos{\frac{4\pi}{9}}\cos{\frac{8\pi}{9}}$ $+\cos{\frac{8\pi}{9}}\cos{\frac{2\pi}{9}})-\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}\cos{\frac{8\pi}{9}}$
Hence $\cos{\frac{2\pi}{9}}\cos{\frac{4\pi}{9}}\cos{\frac{8\pi}{9}}=-\frac{1}{8}$

I found equating coefficients only necessary for the second one (unless someone has an alt solution) and its fairly tedious when dealing with 3 factors.

edit:Oh whoops replied to wrong post, my bad, but nice work on this question aswell

mreditor16 · Apr 17, 2014

Re: HSC 2014 4U Marathon

Kurosaki said:
Yay! haha I realised I forgot to divide by 2 the first time around :L. Herp derp.

is part 2, is it 43/49? Not sure about this one though.
1081/1225 is the answer for part (ii) I think.
OK, working: so we want to divide 48 coins among 3 pirates. Naturally, dividers comes to mind, but we don't need to add any objects as they are already there, in the form of the two lead coins.
So we permute them to get $\frac{50!}{2!48!}=1225$
For the second part, imagine 48 gold coins lying like so: .........

Now, there are 49 places where we can insert the two lead coins. But we wish to guarantee that all the pirates get gold coins, so the lead coins cannot be on the ends. thus, there are 47 places to insert them, and since they are identical coins, we get 47C2 ways. Divide by 1225 to get $\frac{1081}{1225}$

nice jon kurosaki!

that's exactly how i did it.

HSC 2014 MX2 Marathon (archive) (2 Viewers)

Well-Known Member

True Fail Kid

New Member

Active Member

Active Member

Rekt by Adv. English

Active Member

Active Member

Well-Known Member

Rekt by Adv. English

Active Member

New Member

Active Member

Active Member

Active Member

(╯°□°）╯︵ ┻━┻

Active Member

Active Member

Active Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)