HSC 2013 MX2 Marathon (archive) (3 Viewers)

funnytomato · Apr 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Find$

$\int \frac{dx}{x\sqrt{2x-1}}$

$Let $ u^2=2x-1$$

$\int \frac{dx}{x\sqrt{2x-1}} =2\int \frac{du}{1+u^2}=2\tan^{-1}u+C=2\tan^{-1}\sqrt{2x-1}+C$

Sy123 · Apr 9, 2013

Re: HSC 2013 4U Marathon

Nice job.

$Find$

$\int \sin(\ln(x)) + \cos (\ln(x)) \ dx$

funnytomato · Apr 9, 2013

Re: HSC 2013 4U Marathon

$20 oranges are to be distributed among 15 students. The number of oranges given to any student can be any integer from 0 to 20. Find the probability that at least 1 student will have at least 3 oranges$

RealiseNothing · Apr 9, 2013

Re: HSC 2013 4U Marathon

funnytomato said:
$20 oranges are to be distributed among 15 students. The number of oranges given to any student can be any integer from 0 to 20. Find the probability that at least 1 student will have at least 3 oranges$

I got:

$1-\frac{\binom{30}{20}}{15^{20}}$

Which is basically 1.

RealiseNothing · Apr 9, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I got:

$1-\frac{\binom{30}{20}}{15^{20}}$

Which is basically 1.

Basically we want to find the probability that all students have either 0, 1, or 2 oranges, and subtract that from 1 (complement etc).

Now the total arrangement we can make is $15^{20}$ as we give 20 oranges to 15 distinct people, so multiply by 15 per orange.

Now we want to find the amount of arrangements such that all students get either 0, 1, or 2 oranges, consider the following:

$1(O,O)$

$2(O,O)$

$3(O,O)$

$4(O,O)$

$5(O,O)$

$6(O,O)$

$7(O,O)$

$8(O,O)$

$9(O,O)$

$10(O,O)$

$11(O,O)$

$12(O,O)$

$13(O,O)$

$14(O,O)$

$15(O,O)$

The numbers represent the person, and the brackets represent how many oranges they can be given. Hence we have to place the 20 oranges into the brackets, with each (O) representing a spot for an orange to go. Hence there are 30 places to put 20 oranges, so we can arrange this in the following amount of ways:

$\binom{30}{20}$

Putting all this together yields the solution of:

$1 - \frac{\binom{30}{20}}{15^{20}}$

GoldyOrNugget · Apr 9, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Hence there are 30 places to put 20 oranges, so we can arrange this in the following amount of ways:

If I understand your solution correctly, you're counting (*, O) as different to (O, *) where * is an orange, whereas these both correspond to a single orange for a single student (i.e. you'd be double-counting)

RealiseNothing · Apr 10, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
If I understand your solution correctly, you're counting (*, O) as different to (O, *) where * is an orange, whereas these both correspond to a single orange for a single student (i.e. you'd be double-counting)

Right, I didn't take that into account.

RealiseNothing · Apr 10, 2013

Re: HSC 2013 4U Marathon

My new answer is now:

$1 - \frac{\binom{30}{20} - 3\binom{15}{2} - 15\binom{15}{4} - 63\binom{15}{6} - 255\binom{15}{8} - 1023\binom{15}{10}}{15^{20}}$

Have no idea as to whether or not this is correct. I'm going to bed now though, I'll post my working out tomorrow. Basically if a person has either 0 or 2 oranges, they have not been double counted. So we consider the cases when a person has only been given 1 orange, and we use powers of 2 subtract 1, multipled by the amount of ways we choose this to get how many cases we double counted, then subtract from our total.

This probably makes no sense, I'm way too tired.

GoldyOrNugget · Apr 10, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
My new answer is now:

$1 - \frac{\binom{30}{20} - 3\binom{15}{2} - 15\binom{15}{4} - 63\binom{15}{6} - 255\binom{15}{8} - 1023\binom{15}{10}}{15^{20}}$

Have no idea as to whether or not this is correct. I'm going to bed now though, I'll post my working out tomorrow. Basically if a person has either 0 or 2 oranges, they have not been double counted. So we consider the cases when a person has only been given 1 orange, and we use powers of 2 subtract 1, multipled by the amount of ways we choose this to get how many cases we double counted, then subtract from our total.

This probably makes no sense, I'm way too tired.

Sorry bro... 15^20 is also double counting! If you have 2 oranges and 15 people, the answer isn't 2^15, it's 16. I can see what you were trying to do, assign oranges to students, but your method thinks that assigning the second orange to the first student and the first orange to the second student is different from assigning the first orange to the first student and the second orange to the second student.

Anyway, after messing around with recurrence equations in Excel, I get an answer of 198493615/198853663. I'd love to know if this is correct.

Mathematica · Apr 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Nice job.

$Find$

$\int \sin(\ln(x)) + \cos (\ln(x)) \ dx$

$Let$ $I = \int sin(ln(x))dx$

$Let$ $lnx=y \Rightarrow x=e^{y} \therefore dx=e^{y}dy$

$Sub into the original integral:$ $I=\int sin(y)e^{y}dy$

$Now using IBP twice,$

$I=e^{y}sin(y)-\left ( e^{y}cos(y))+\int sin(y)e^{y}dy \right )$

$=e^{y}sin(y)-e^{y}cos(y))-I$

$\therefore I=\frac{ln(x)}{2}\left ( sin(ln(x))-cos(ln(x)) \right )$

$Do the same for the cosine then add them.$

Mathematica · Apr 10, 2013

Re: HSC 2013 4U Marathon

$Prove the following inequality:$

$\sum_{n=1}^{\infty }\frac{1}{(n+1)\sqrt{n}}< 2$

GoldyOrNugget · Apr 10, 2013

Re: HSC 2013 4U Marathon

Mathematica said:
$Prove the following inequality:$

$\sum_{n=1}^{\infty }\frac{1}{(n+1)\sqrt{n}}< 2$

$\int_0^\infty \negthickspace \frac{1}{(n+1)\sqrt{n}}\;\text{d}n > \sum_{n=1}^{\infty }\frac{1}{(n+1)\sqrt{n}} $ by a sum-of-rectangles argument$$

Substitute $u=x+1$ into the integral, which evaluates into $2\tan^{-1}(\sqrt{x})\]_0^\infty \rightarrow 2^-$ . Therefore since the integral is smaller than 2 and the sum is smaller than the integral, then the sum is smaller than 2.

Sy123 · Apr 10, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
$\int_0^\infty \negthickspace \frac{1}{(n+1)\sqrt{n}}\;\text{d}n > \sum_{n=1}^{\infty }\frac{1}{(n+1)\sqrt{n}} $ by a sum-of-rectangles argument$$

Substitute $u=x+1$ into the integral, which evaluates into $2\tan^{-1}(\sqrt{x})\]_0^\infty \rightarrow 2^-$ . Therefore since the integral is smaller than 2 and the sum is smaller than the integral, then the sum is smaller than 2.

Unfortunately that integral evaluates to $\pi$ , not 2.

What needs to be done is:

$\int_{3}^{\infty} \frac{dx}{\sqrt{x} (x+1)} > \sum_{k=4}^{\infty} \frac{1}{(k+1)\sqrt{k}}$

$\frac{\pi}{3} > \sum_{k=4}^{\infty} \frac{1}{(k+1)\sqrt{k}}$

Now lets add to both sides the first 3 terms of the summation.

$\frac{\pi}{3} + \frac{1}{2} + \frac{1}{3\sqrt{2}} + \frac{1}{4\sqrt{3}} > \sum_{n=1}^{\infty} \frac{1}{(n+1)\sqrt{n}}$

Computation yields that on the LHS we have, 1.92723...

$\therefore 2 > 1.9272.... > \sum$

I guess there is a more formal approach to it by first bounding Pi and square root functions but yeah.

Sy123 · Apr 10, 2013

Re: HSC 2013 4U Marathon

$i) Prove that$

$\tan^{-1} \left (\frac{x}{x+1} \right ) - \tan^{-1} \left (\frac{x-1}{x} \right ) = \tan^{-1} \left (\frac{1}{2x^2}\right)$

$ii) Hence prove that$

$\frac{\pi}{4} = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{18} + \tan^{-1} \frac{1}{32} + \tan^{-1} \frac{1}{50} + \dots \dots$

Argaroth · Apr 12, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$i) Prove that$

$\tan^{-1} \left (\frac{x}{x+1} \right ) - \tan^{-1} \left (\frac{x-1}{x} \right ) = \tan^{-1} \left (\frac{1}{2x^2}\right)$

$ii) Hence prove that$

$\frac{\pi}{4} = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{8} + \tan^{-1} \frac{1}{18} + \tan^{-1} \frac{1}{32} + \tan^{-1} \frac{1}{50} + \dots \dots$

i)

ii)

Sy123 · Apr 12, 2013

Re: HSC 2013 4U Marathon

Argaroth said:
i)

ii)

Nice work.

================================

$Evaluate$

$\int_{1}^{\sqrt{3}} \sqrt{\frac{1-x}{3+x}} \ dx$

HeroicPandas · Apr 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work.

================================

$Evaluate$

$\int_{1}^{\sqrt{3}} \sqrt{\frac{1-x}{3+x}} \ dx$

$\int_{1}^{\sqrt{3}} \sqrt{\frac{1-x}{3+x}} \ dx . \frac{\sqrt{1-x}}{\sqrt{1-x}} \\ = \int_{1}^{\sqrt{3}} \frac{1-x}{\sqrt{-(x^2 + 2x - 3)}} \ dx = \int_{1}^{\sqrt{3}} \frac{1-x}{\sqrt{4 - (x+1)^2}} \\ = \int_{1}^{\sqrt{3}} \frac{1}{\sqrt{4-(x+1)^2}} + \frac{1}{2}[\frac{-2x -2}{\sqrt{4-(x+1)^2}} + \frac{2}{\sqrt{4-(x+1)^2}}] \\ = \int_{1}^{\sqrt{3}} \frac{2}{\sqrt{4 - (x+1)^2}} + \frac{1}{2} \int_{1}^{\sqrt{3}} \frac{-2x -2}{\sqrt{4 - (x+1)^2}} \\ = \left [ 2sin^{-1} \frac{x+1}{4} + \sqrt{4 - (x+1)^2} \right ]^{\sqrt3} _1$

?

Sy123 · Apr 13, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$\int_{1}^{\sqrt{3}} \sqrt{\frac{1-x}{3+x}} \ dx . \frac{\sqrt{1-x}}{\sqrt{1-x}} \\ = \int_{1}^{\sqrt{3}} \frac{1-x}{\sqrt{-(x^2 + 2x - 3)}} \ dx = \int_{1}^{\sqrt{3}} \frac{1-x}{\sqrt{4 - (x+1)^2}} \\ = \int_{1}^{\sqrt{3}} \frac{1}{\sqrt{4-(x+1)^2}} + \frac{1}{2}[\frac{-2x -2}{\sqrt{4-(x+1)^2}} + \frac{2}{\sqrt{4-(x+1)^2}}] \\ = \int_{1}^{\sqrt{3}} \frac{2}{\sqrt{4 - (x+1)^2}} + \frac{1}{2} \int_{1}^{\sqrt{3}} \frac{-2x -2}{\sqrt{4 - (x+1)^2}} \\ = \left [ 2sin^{-1} \frac{x+1}{4} + \sqrt{4 - (x+1)^2} \right ]^{\sqrt3} _1$

?

That looks right and it was a little different to my method, nice work, I think the sqrt 3 was supposed to be 3 sorry about that.

=============

This is quite hard:

$Find$

$\int \frac{1-t^2}{1+10t^2+t^4} \ dt$

twinklegal19 · Apr 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
That looks right and it was a little different to my method, nice work, I think the sqrt 3 was supposed to be 3 sorry about that.

=============

This is quite hard:

$Find$

$\int \frac{1-t^2}{1+10t^2+t^4} \ dt$

not sure if I'm on the right track, but does the answer happen to have a lot of $\sqrt{24}$ ?

Sy123 · Apr 13, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
not sure if I'm on the right track, but does the answer happen to have a lot of $\sqrt{24}$ ?

I don't think so, but I think its a bit unreasonable of me to ask it like that with no hint whatsoever.

Redone:

$$Find, using the substitution$ \ \ t=\tan \frac{x}{2}$

$\int \frac{1-t^2}{1+10t^2+t^4} \ dt$

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