HSC 2013 MX2 Marathon (archive) (1 Viewer)

jyu · Apr 28, 2013

Re: HSC 2013 4U Marathon

I was misled by the diagram.

jyu · Apr 28, 2013

Re: HSC 2013 4U Marathon

Capt Rifle · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$\int \frac{1-x}{1-\sqrt{x}} \ dx = \int \frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}} = \int 1+\sqrt{x} \ dx = x + \frac{2}{3}x^{\frac{3}{2}} + c$

==========

$\int \frac{dx}{e^x + e^{-x}}$

Nice... took me a while to figure out the 2 squares lol

asianese · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$\int \frac{dx}{e^x + e^{-x}}$

$\frac{1}{2}\ln(e^x+1)+C$

asianese · Apr 28, 2013

Re: HSC 2013 4U Marathon

$\tan^{-1}(e^x)+k$

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

asianese said:
$\tan^{-1}(e^x)+k$

How?

HeroicPandas · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
(Another early IMO question).

$Solve the equation$

$\cos^2 x + \cos^2(2x) + \cos^2(3x) = 1$

$\cos^2 x + \cos^2 2x + \cos^2 3x = 1 \\$Easiest way to do this is to convert all these into cosx using the double and triple angle formulae \\ \\ \cos^2 x + (2\cos^2 x - 1)^2 + (4\cos^3 x - 3\cosx)^2 = 1 \\ \\ $ $Expand and let a = $cosx \\ 16a^6 - 20a^4 + 6a^2 = 0 \\ a^2 (8a^4 - 10a^2 + 3) = 0 \\ a^2 (4a^2 -3)(2a^2 - 1) = 0 \\ \\ blah blahblah$

HeroicPandas · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
How?

I think he let x = lnu and used int by substitution

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$\cos^2 x + \cos^2 2x + \cos^2 3x = 1 \\$Easiest way to do this is to convert all these into cosx using the double and triple angle formulae \\ \\ \cos^2 x + (2\cos^2 x - 1)^2 + (4\cos^3 x - 3\cosx)^2 = 1 \\ \\ $ $Expand and let a = $cosx \\ 16a^6 - 20a^4 + 6a^2 = 0 \\ a^2 (8a^4 - 10a^2 + 3) = 0 \\ a^2 (4a^2 -3)(2a^2 - 1) = 0 \\ \\ blah blahblah$

The constant term does not cancel out. I think you have made a silly mistake.

asianese · Apr 28, 2013

Re: HSC 2013 4U Marathon

Wtf lol

$\int \frac{dx}{e^x+\frac{1}{e^x}}=\int \frac{e^x}{e^{2x}+1} dx= \tan^{-1}(e^x)+k$

Trebla · Apr 28, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
I think he let x = lnu and used int by substitution

Isn't it pretty obvious once you multiply numerator and denominator by e^x?

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
I think he let x = lnu and used int by substitution

I would have accepted the answer with only the above sentence ^^

===========

$$Find for all values of$ \ \ \alpha$

$\int \frac{dx}{x^{\alpha + 1} + x}$

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

asianese said:
Wtf lol

$\int \frac{dx}{e^x+\frac{1}{e^x}}=\int \frac{e^x}{e^{2x}+1} dx= \tan^{-1}(e^x)+k$

Thank you =)
That way people who browse this thread who don't know how to do it at least get a solution.

HeroicPandas · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
The constant term does not cancel out. I think you have made a silly mistake.

made a typo:

$\cos^2 x + \cos^2 2x + \cos^2 3x = 1 \\ \cos^2 x + (2\cos^2 x -1)^2 + (4cos^3 x - 3cosx)^2 = 1$

$16cos^6 x - 20cos^4 x + 6cos^2 x =0$
and so the 1's disappear

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
made a typo:

$\cos^2 x + \cos^2 2x + \cos^2 3x = 1 \\ \cos^2 x + (2\cos^2 x -1)^2 + (4cos^3 x - 3cosx)^2 = 1$

and so the 1's disappear

Ah yes ok.

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

$Find$

$\int_{-9001}^{9001} \sin(x^3) e^{-\pi^{77} x^{14}} \tan^{-1}(\ln(x^2)) (x^{191919923134}-1) \ dx$

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Find$

$\int_{-9001}^{9001} \sin(x^3) e^{-\pi^{77} x^{14}} \tan^{-1}(\ln(x^2)) (x^{191919923134}-1) \ dx$

hahahahhahaha it's just a tad obvious

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
hahahahhahaha it's just a tad obvious

(;

seanieg89 · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
hahahahhahaha it's just a tad obvious

Well not entirely, he should probably change the log to something else. Logs explode at 0 so its not at all clear that the integral is well defined

.

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Well not entirely, he should probably change the log to something else. Logs explode at 0 so its not at all clear that the integral is well defined .

lol I got owned

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