HSC 2013-14 MX1 Marathon (archive) (3 Viewers)

asianese · Jan 25, 2013

Re: HSC 2013 3U Marathon Thread

Very un-Sy123, that question...

Sy123 · Jan 25, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
Very un-Sy123, that question...

How so? :s
I know I can't go all out like I can in the 4U Marathon.

asianese · Jan 25, 2013

Re: HSC 2013 3U Marathon Thread

You usually post much more challenging stuff haha.

Sometimes 3U can be harder than 4U so don't be afraid to construct a killer question.

Trebla · Jan 25, 2013

Re: HSC 2013 3U Marathon Thread

Prove that log₂3 is irrational.

Sy123 · Jan 25, 2013

Re: HSC 2013 3U Marathon Thread

Trebla said:
Prove that log₂3 is irrational.

Good question, I will post my solution if no-one has solved in by tomorrow.

Exia_ · Jan 26, 2013

Re: HSC 2013 3U Marathon Thread

Prove by mathematical induction that if p is a positive integer then (1+p)^n -1 is divisible by p for all values of n > or equal to 1.
Hence deduce that 12^n - 4^n -3^n +1 is divisible by 6 for all positive integers n> or equal to 1.

omgiloverice · Jan 26, 2013

Re: HSC 2013 3U Marathon Thread

Trebla said:
Prove that log₂3 is irrational.

$assume \log_2 3 = \frac {x}{y}$ \; $with x and y an rational integer$
$since \log_2 3 > 0$
$therefore x,y is greater than 0$
$$when we rearrange the equation we get 2^x = 3^y$
$$since x > 0$
$$This will imply that 3^y \; \text{is an even number, which is not true}$
$therefore \; \log_2 3 $is irrational$$

Sy123 · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

$$Prove that$ \ \ \ \ \int_0^{\pi/2}\cos x \sin^{n-1} x \ dx = \frac{1}{n}$

$$Hence prove$ \ \ \ \sum_{k=1}^{n} \frac{1}{k} = \int_{0}^{\pi/2} \cos x \cdot \frac{1-\sin^n x}{1-\sin x} \ dx$

HeroicPandas · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Prove that$ \ \ \ \ \int_0^{\pi/2}\cos x \sin^{n-1} x \ dx = \frac{1}{n}$

$$Hence prove$ \ \ \ \sum_{k=1}^{n} \frac{1}{k} = \int_{0}^{\pi/2} \cos x \cdot \frac{1-\sin^n x}{1-\sin x} \ dx$

ok:

First part-

$\int_{0}^{\pi/2} cosx sin^{n-1}x.dx =\frac{1}{n}\\ \\ $Let u = sinx \\ \\ $ $du = cos.dx, when x=0, u=0, when x =\pi/2, u =1 \\ \\ \therefore $ $ I = \int_{0}^{1}u^{n-1}.du = \frac{1}{n} \\ \\ \therefore \int_{0}^{\pi/2} cosx sin^{n-1}x.dx =\frac{1}{n}$

Second part-
$$ RTP: \sum_{k=1}^{n} \frac{1}{k} = \int_{0}^{\frac{\pi}{2}} cosx.\frac{1-sin^n x}{1-sinx}.dx \\ \\ = \int_{0}^{\frac{\pi}{2}}cosx. \frac{sin^n x -1}{sinx -1}.dx \\ \\ \int_{0}^{\frac{\pi}{2}}cosx. \frac{(sinx - 1)(1 + sinx+sin^2x+...+sin^{n-1}x)}{sinx - 1} \\ \\ = \int_{0}^{\frac{\pi}{2}} cosx (1 + sinx + sin^2x+...+sin^{n-1}x).dx \\ \\ = \int_{0}^{\frac{\pi}{2}}cosx + cosxsinx +...+cosxsin^{n-1}x.dx = 1 + 1/2 + 1/3 +...+1/n \\ \\ = \sum_{k=1}^{n} \frac{1}{k}$

Sy123 · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
ok:

First part-

$\int_{0}^{\pi/2} cosx sin^{n-1}x.dx =\frac{1}{n}\\ \\ $Let u = sinx \\ \\ $ $du = cos.dx, when x=0, u=0, when x =\pi/2, u =1 \\ \\ \therefore $ $ I = \int_{0}^{1}u^{n-1}.du = \frac{1}{n} \\ \\ \therefore \int_{0}^{\pi/2} cosx sin^{n-1}x.dx =\frac{1}{n}$

Second part-
$$ RTP: \sum_{k=1}^{n} \frac{1}{k} = \int_{0}^{\frac{\pi}{2}} cosx.\frac{1-sin^n x}{1-sinx}.dx \\ \\ = \int_{0}^{\frac{\pi}{2}}cosx. \frac{sin^n x -1}{sinx -1}.dx \\ \\ \int_{0}^{\frac{\pi}{2}}cosx. \frac{(sinx - 1)(1 + sinx+sin^2x+...+sin^{n-1}x)}{sinx - 1} \\ \\ = \int_{0}^{\frac{\pi}{2}} cosx (1 + sinx + sin^2x+...+sin^{n-1}x).dx \\ \\ = \int_{0}^{\frac{\pi}{2}}cosx + cosxsinx +...+cosxsin^{n-1}x.dx = 1 + 1/2 + 1/3 +...+1/n \\ \\ = \sum_{k=1}^{n} \frac{1}{k}$

Correct, though for the first part, it only comes down to simple inspection to notice that:

$d/dx \sin^n x = n\cos x \sin^{n-1} x$

Then we take constant to other side and etc.

HeroicPandas · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

$\int \frac{1 + cosx}{1 - cosx}.dx =$ ?$

HeroicPandas · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Correct, though for the first part, it only comes down to simple inspection to notice that:

$d/dx \sin^n x = n\cos x \sin^{n-1} x$

Then we take constant to other side and etc.

oh ok

braintic · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$\int \frac{1 + cosx}{1 - cosx}.dx =$ ?$

Not a 3U question.

nightweaver066 · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

braintic said:
Not a 3U question.

It's doable.

SpiralFlex · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

braintic said:
Not a 3U question.

This sort of manipulation is 3U

Sy123 · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
$\frac{3x}{2}+ 2sin (x) +\frac{sin(2x)}{4}$
aha finally it works. is this wrong? not sure

Incorrect.

Sy123 · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
how about now?I edited the post

if you forgot the +C I would of pointed it out lol, its still incorrect.

Did you use t-substitution (Weiserstrass)?

Kurosaki · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
if you forgot the +C I would of pointed it out lol, its still incorrect.

Did you use t-substitution (Weiserstrass)?

? I used double angle, cos(2x) thingy if thats what u mean?Oh wait fack nvm. i think i see where i went wrong now. I misread the question...

.
Thanks for checking my answer Sy =)

Drongoski · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

$\int \frac {1+cos x}{1-cos x} dx = \int \frac {2cos^2 {\frac {x}{2}}}{2sin^2 \frac{x}{2}} dx = \int (cosec^2 \frac {x}{2} - 1) dx = - 2cot \frac {x}{2} - x + C$

So, indeed, it is doable in 3U.

Someone please post another question.

Edit:

Corrected correctly hopefully.

SpiralFlex · Jan 27, 2013

Re: HSC 2013 3U Marathon Thread

Sure?

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