HSC 2013 MX2 Marathon (archive) (3 Viewers)

HeroicPandas · Apr 20, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Note the binomial co-efficient at the front =)

NOOOO! damn...................

i think i got it...NOT!

HeroicPandas · Apr 20, 2013

Re: HSC 2013 4U Marathon

$$ Consider (1+x)^n = $\sum_{r=0}^{n} \binom{n}{r} x^r \\ $Let x = cis2\beta \\ \therefore (1+cis2\beta)^n = $\sum_{r=0}^{n} \binom{n}{r}cis(2r\beta) \\ $Using double angles of cosine and sine and factorising something... \\ \therefore 2^n cos^n \beta (cosn\beta + isinn\beta) = $\sum_{r=0}^{n} \binom{n}{r} cos2r\beta + isin2r\beta \\ $Equate reals and imaginaries to create 2 equations \\ 2^n cos^n \beta cosn\beta = $\sum_{r=0}^{n} \binom{n}{r} cos2r\beta \dots [1] \\ 2^n cos^n \beta sinn\beta = \sum_{r=0}^{n} \binom{n}{r}sin2r\beta \dots [2] \\ $Multiply [1] and [2] $ $by $cos\alpha$ and sin\alpha $ $respectively$

$$Now, \\ 2^n cos^n \beta cosn\beta cos\alpha = $\sum_{r=0}^{n} \binom{n}{r} cos2r\beta cos\alpha \dots [3] \\ 2^n cos^n \beta sinn\beta sin\alpha = \sum_{r=0}^{n} \binom{n}{r}sin2r\beta sin\alpha \dots [4] \\ $Now, [3] - [4] \\ \therefore $\sum_{r=0}^{n}\binom{n}{r} cos(\alpha + 2r\beta) = 2^n cos^n \beta [cos(n\beta + \alpha)]$

Can u see any mistakes?

Sy123 · Apr 20, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$$ Consider (1+x)^n = $\sum_{r=0}^{n} \binom{n}{r} x^r \\ $Let x = cis2\beta \\ \therefore (1+cis2\beta)^n = $\sum_{r=0}^{n} \binom{n}{r}cis(2r\beta) \\ $Using double angles of cosine and sine and factorising something... \\ \therefore 2^n cos^n \beta (cosn\beta + isinn\beta) = $\sum_{r=0}^{n} \binom{n}{r} cos2r\beta + isin2r\beta \\ $Equate reals and imaginaries to create 2 equations \\ 2^n cos^n \beta cosn\beta = $\sum_{r=0}^{n} \binom{n}{r} cos2r\beta \dots [1] \\ 2^n cos^n \beta sinn\beta = \sum_{r=0}^{n} \binom{n}{r}sin2r\beta \dots [2] \\ $Multiply [1] and [2] $ $by $cos\alpha$ and sin\alpha $ $respectively$

$$Now, \\ 2^n cos^n \beta cosn\beta cos\alpha = $\sum_{r=0}^{n} \binom{n}{r} cos2r\beta cos\alpha \dots [3] \\ 2^n cos^n \beta sinn\beta sin\alpha = \sum_{r=0}^{n} \binom{n}{r}sin2r\beta sin\alpha \dots [4] \\ $Now, [3] - [4] \\ \therefore $\sum_{r=0}^{n}\binom{n}{r} cos(\alpha + 2r\beta) = 2^n cos^n \beta [cos(n\beta + \alpha)]$

Can u see any mistakes?

Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)

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

$Find$

$I=\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \ dx$

HeroicPandas · Apr 20, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)

YESSSSSSSS!!!!!!

You have made my day Sy!

RealiseNothing · Apr 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$x^2=u$

$du= 2x dx$

$I = 2 \int \frac{x}{x+1}$

$I = 2(x+1) - 2\ln(x+1) + c$

Isn't it meant to be 2x not 2(x+1)?

bleakarcher · Apr 21, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Isn't it meant to be 2x not 2(x+1)?

Go to sleep bro, you're off your ass.

Sy123 · Apr 21, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Isn't it meant to be 2x not 2(x+1)?

Ah yes, I'm making too many mistakes at the moment :/
However the only differing thing between this and the real solution is constant, so very technically it can be part of the +c right?

bleakarcher · Apr 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ah yes, I'm making too many mistakes at the moment :/
However the only differing thing between this and the real solution is constant, so very technically it can be part of the +c right?

Yeah, both are correct expressions for the primitive.

HeroicPandas · Apr 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)

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

$Find$

$I=\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \ dx$

$I=\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \ dx \\ $Let u = sin^2 x, du = 2sinxcosx.dx \\ I = \frac{1}{2}$\int \frac{du}{2sin^4 x - 2sin^2 x + 1} = \frac{1}{2} \int \frac{du}{2u^2 - 2u + 1} = \frac{1}{4}\int \frac{du}{(u-\frac{1}{2})^2 + \frac{1}{4}} = \frac{1}{2}[tan^{-1} \frac{u-\frac{1}{2}}{\frac{1}{2}}] = \frac{1}{2} tan^{-1} \frac{2sin^2 x - 1}{1} =- \frac{1}{2} tan^{-1} \frac{cos2x}{1} + c$

Question:

$$ Let I_n = $\int_{0}^{x} (1+t^2)^n dt, n = 1,2,3,... \\$Show that I_n = $\frac{1}{2n+1}(1+x^2)^nx + \frac{2n}{2n+1}I_{n-1}$

RealiseNothing · Apr 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
However the only differing thing between this and the real solution is constant, so very technically it can be part of the +c right?

True.

Drongoski · Apr 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep I was just about to post that I made a mistake when I was doing it, and somehow arrived at that, but I was able to arrive at what you just got there, well done =)

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

$Find$

$I=\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \ dx$

Just trying!

$x^4+y^4 = x^4+y^4+2x^2y^2 - 2x^2y^2 = (x^2+y^2)^2 - 2x^2y^2$

$\therefore I = \int \frac {\frac {1}{2} sin2x}{(sin^2 x+cos^2x)^2 - 2sin^2 x cos^2x} dx$

$= \int \frac { - \frac {1}{4}}{(1 - \frac {1}{2}sin^2 2x)} dcos2x$

$= - \int \frac {\frac {1}{4}}{\frac {1}{2} (2 - sin^2 2x)} dcos2x$

$= - \frac {1}{2}\int \frac {1}{1+cos^2 2x}dcos 2x$

$= - \frac {1}{2} tan^{-1}cos2x + C$

Sy123 · Apr 21, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$I=\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} \ dx \\ $Let u = sin^2 x, du = 2sinxcosx.dx \\ I = \frac{1}{2}$\int \frac{du}{2sin^4 x - 2sin^2 x + 1} = \frac{1}{2} \int \frac{du}{2u^2 - 2u + 1} = \frac{1}{4}\int \frac{du}{(u-\frac{1}{2})^2 + \frac{1}{4}} = \frac{1}{2}[tan^{-1} \frac{u-\frac{1}{2}}{\frac{1}{2}}] = \frac{1}{2} tan^{-1} \frac{2sin^2 x - 1}{1} =- \frac{1}{2} tan^{-1} \frac{cos2x}{1} + c$

Question:

$$ Let I_n = $\int_{0}^{x} (1+t^2)^n dt, n = 1,2,3,... \\$Show that I_n = $\frac{1}{2n+1}(1+x^2)^nx + \frac{2n}{2n+1}I_{n-1}$

$I_n = t(1+t^2)^n |_0^x - 2n \int_0^x t^2(1+t)^{n-1} \ dt$

$(1+t^2)^n-(1+t^2)^{n-1} = t^2(1+t^2)^{n-1}$

$\therefore \ \ I_n = x(1+x^2)^n - 2n(I_n - I_{n-1})$

$(2n+1)I_n = (1+x^2)^n x +2nI_{n-1}$

$\therefore \ \ I_n = \frac{1}{2n+1} (1+x^2)^n x + \frac{2n}{2n+1} I_{n-1}$

Sy123 · Apr 21, 2013

Re: HSC 2013 4U Marathon

$$i) Explain why, for$ \ \ x \geq 3$

$\frac{1}{x(x+1)(x+2)} < \frac{1}{x^3} < \frac{1}{x(x-1)(x-2)}$

$ii) Hence prove that$

$\frac{7}{6} < \sum_{n=1}^{\infty} \frac{1}{n^3} < \frac{11}{8}$

HeroicPandas · Apr 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$I_n = t(1+t^2)^n |_0^x - 2n \int_0^x t^2(1+t)^{n-1} \ dt$

$(1+t^2)^n-(1+t^2)^{n-1} = t^2(1+t^2)^{n-1}$

$\therefore \ \ I_n = x(1+x^2)^n - 2n(I_n - I_{n-1})$

$(2n+1)I_n = (1+x^2)^n x +2nI_{n-1}$

$\therefore \ \ I_n = \frac{1}{2n+1} (1+x^2)^n x + \frac{2n}{2n+1} I_{n-1}$

very nice!

Sy123 · Apr 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Explain why, for$ \ \ x \geq 3$

$\frac{1}{x(x+1)(x+2)} < \frac{1}{x^3} < \frac{1}{x(x-1)(x-2)}$

$ii) Show that$

$\frac{1}{x(x+1)(x+2)} = \frac{\frac{1}{2}}{x} - \frac{1}{x+1} + \frac{\frac{1}{2}}{x+2}$

$iii) Hence prove that$

$\frac{7}{6} < \sum_{n=1}^{\infty} \frac{1}{n^3} < \frac{11}{8}$

Redone

Sy123 · Apr 22, 2013

Re: HSC 2013 4U Marathon

$$A cylinder has a height of$ \ 2r \ \ $and radius$ \ \ r$

$Inside the cylinder a cone with height$ \ \ 2r \ \ $and radius$ \ \ r \ \ $is fit inside$

$Also a sphere of radius$ \ \ r \ \ $is fit inside, such that all 3 solids share a certain volume$

$$Show that this volume that is shared between all 3 solids is$ \ \ \ \frac{12 \pi r^3}{25}$

seanieg89 · Apr 24, 2013

Re: HSC 2013 4U Marathon

A man walks into a casino with $10 and the only bet he knows how to make is on a colour in roulette (the probability of winning this bet is 18/37, and you win as much as you bet).

Rather than spending all his money at once with one bet, he decides to "play it safe" and bet $1 at a time until he either reaches $20 or goes broke.

a) What is the probability that he will walk away with $20 and what is the probability that he will go broke?

b) Is this better or worse than he would do if he just made a single bet with all of his money?

anomalousdecay · Apr 24, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
A man walks into a casino with $10 and the only bet he knows how to make is on a colour in roulette (the probability of winning this bet is 18/37, and you win as much as you bet).

Rather than spending all his money at once with one bet, he decides to "play it safe" and bet $1 at a time until he either reaches $20 or goes broke.

a) What is the probability that he will walk away with $20 and what is the probability that he will go broke?

b) Is this better or worse than he would do if he just made a single bet with all of his money?

a) P($20) = (18/37)^10
P($0) = (19/37)^10

b) If he makes a single bet with his money, he will have a higher probability of winning.
However, he has a higher probability of losing as well.
If he only wants to make a few dollars, it is better for him to bet $1 at a time.
If he only wants double-or-nothing then he must do it all in one bet.

P.S. The Probability of me getting this question wrong is 40%

as I thought that it may be possible to use another method to work it out, but I thought this method is most correct.
Please correct me if I am wrong. We all learn from our mistakes.

Sy123 · Apr 24, 2013

Re: HSC 2013 4U Marathon

anomalousdecay said:
a) P($20) = (18/37)^10
P($0) = (19/37)^10

b) If he makes a single bet with his money, he will have a higher probability of winning.
However, he has a higher probability of losing as well.
If he only wants to make a few dollars, it is better for him to bet $1 at a time.
If he only wants double-or-nothing then he must do it all in one bet.

P.S. The Probability of me getting this question wrong is 40% as I thought that it may be possible to use another method to work it out, but I thought this method is most correct.
Please correct me if I am wrong. We all learn from our mistakes.

But he can lose say 9$ then win 19 to win, your probability is for when he wins 10 times in a row, it is merely one case.

anomalousdecay · Apr 24, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
But he can lose say 9$ then win 19 to win, your probability is for when he wins 10 times in a row, it is merely one case.

Yeah that makes more sense.
So then would it be......

a)
P($20) = 10C0 (18/37)^10(19/37) + 10C1(18/37)^11(19/37) + 10C2(18/37)^12(19/37)^2 + 10C3(18/37)^13(19/37)^3 +10C4(18/37)^14(19/37)^4 + 10C5(18/37)^15(19/37)^5 + 10C6(18/37)^16(19/37)^6 +10C7(18/37)^17(19/37)^7 + 10C8(18/37)^18(19/37)^8 + 10C9(18/37)^19(19/37)^9

P($0) = 1- P($20)

b) If he makes a single bet with his money, he will have a higher probability of winning.
However, he has a higher probability of losing as well.
If he only wants to make a few dollars, it is better for him to bet $1 at a time.
If he only wants double-or-nothing then he must do it all in one bet.

?????? For some reason I am thinking that I missed something again

.

HSC 2013 MX2 Marathon (archive) (3 Viewers)

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