HSC 2013 MX2 Marathon (archive) (2 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) (2 Viewers)

Heroic!

Heroic!

This too shall pass

Heroic!

what is that?It is Cowpea

Active Member

This too shall pass

Active Member

Heroic!

what is that?It is Cowpea

Well-Known Member

This too shall pass

This too shall pass

Heroic!

This too shall pass

This too shall pass

Well-Known Member

Premium Member

This too shall pass

Premium Member

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