HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

seanieg89 · Mar 22, 2013

Re: HSC 2013 3U Marathon Thread

Oh ye of little faith.

asianese · Mar 22, 2013

Re: HSC 2013 3U Marathon Thread

Ok ok proof coming.

$\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{1}{3^{m+n}}=\\ \sum_{m=0}^\infty \left[ \sum_{n=0}^\infty \frac{1}{3^m} \cdot \frac{1}{3^n} \right]\\ = \sum_{m=0}^\infty \left( \frac{1}{3^m} \left[ \sum_{n=0}^\infty \frac{1}{3^n} \right] \right)\\= \sum_{m=0}^\infty \frac{1}{3^m} \left[ \frac{1}{1}+\frac{1}{3}+\frac{1}{9}+\cdots+\frac{1}{3^n}+\cdots \right] \\ = \sum_{m=0}^\infty \frac{1}{3^m} \left[ \frac{1}{1-\frac{1}{3}} \right] \\ = \sum_{m=0}^\infty \frac{1}{3^m} \left[ \frac{3}{2}\right] \\ = \frac{3}{2}\times\frac{3}{2}\\ = \frac{9}{4}.$

The 'trick' is to recognise that you can split the terms, and you can take out whatever does not depend on the summation. Then it is a matter of recognising an infinite sum. Since the result is true for the 'n' sum, then it is the same for the 'm' sum and the result is (3/2)^2 = 9/4.

You bereive me nao?

asianese · Mar 22, 2013

Re: HSC 2013 3U Marathon Thread

Capt Rifle said:
<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" title="Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" /></a>

Edit: This is at a 3U difficulty, right guys?

This is more leaning toward a 4U question.

Capt Rifle · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
This is more leaning toward a 4U question.

Alrighty then what about this:

<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int \frac{x^2dx}{\sqrt{x-1}}\: \: \: using\: the\: substitution\: x=u^2@plus;1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int \frac{x^2dx}{\sqrt{x-1}}\: \: \: using\: the\: substitution\: x=u^2+1" title="Find \int \frac{x^2dx}{\sqrt{x-1}}\: \: \: using\: the\: substitution\: x=u^2+1" /></a>

HeroicPandas · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

Capt Rifle said:
<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" title="Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" /></a>

Edit: This is at a 3U difficulty, right guys?

A 3U human can do it...

asianese · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

Actually sorry - I didn't see that the substitution was given. It is doable by a 3U student.

Capt Rifle · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
A 3U human can do it...
View attachment 27884

That is correct.

Capt Rifle · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
Actually sorry - I didn't see that the substitution was given. It is doable by a 3U student.

I was going to say lol, I thought of it as a 3u question

Sy123 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

$$Prove that$ \ \ \ \lim_{\alpha \to 0}\int_{\alpha}^{x} \ln t \ dt = x\ln x -x$

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Prove that$ \ \ \ \lim_{\alpha \to 0}\int_{\alpha}^{x} \ln t \ dt = x\ln x -x$

Is this doable without IBP?

Mr Slick · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

Lieutenant_21 said:
Is this doable without IBP?

doubt it

seanieg89 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

By just considering areas you wouldn't need parts.

Sy123 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
By just considering areas you wouldn't need parts.

This.

I'm not silly enough to post a 4U question in a 3U thread.
In fact that integral is 2U.

Sy123 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

$Utilising a suitable substitution, find$

$\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \ dx$

seanieg89 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

$More generally, if $f(x)$ is an increasing function on $\mathbb{R}$ with an anti-derivative $F(x)$, explain why $f$ has an inverse and find a closed-form expression for: \\ \\ $\int_a^b f^{-1}(t)\,dt$\\ \\where $a$ and $b$ are arbitrary real numbers.\\ \\ Hence find anti-derivatives of the logarithm and arcsine functions.$

Lieutenant_21 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Utilising a suitable substitution, find$

$\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \ dx$

$Let$ $I=\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \ dx$

$Let$ $u=x+\sqrt{x}$

$\therefore dx=\frac{2\sqrt{x}dx}{2\sqrt{x}+1}du$

$I=\int \frac{2\sqrt{x}}{2\sqrt{x}(u)}du=\int\frac{du}{u}=lnu+c=ln(x+\sqrt{x})+c$

Sy123 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
$More generally, if $f(x)$ is an increasing function on $\mathbb{R}$ with an anti-derivative $F(x)$, explain why $f$ has an inverse and find a closed-form expression for: \\ \\ $\int_a^b f^{-1}(t)\,dt$\\ \\where $a$ and $b$ are arbitrary real numbers.\\ \\ Hence find anti-derivatives of the logarithm and arcsine functions.$

Great question.

Lieutenant_21 said:
$Let$ $I=\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \ dx$

$Let$ $u=x+\sqrt{x}$

$\therefore dx=\frac{2\sqrt{x}dx}{2\sqrt{x}+1}du$

$I=\int \frac{2\sqrt{x}}{2\sqrt{x}(u)}du=\int\frac{du}{u}=lnu+c=ln(x+\sqrt{x})+c$

Nice work.

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

$Find$

$\int \frac{1}{x^2+10x+50} \ dx$

HeroicPandas · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Great question.

Nice work.

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

$Find$

$\int \frac{1}{x^2+10x+50} \ dx$

complete the square

int 1/[(x+5)^2 + 25] = 0.2 arctan(x+5/5) +C

woops thought this was 4U marathon...

$\lim_{x\rightarrow -5}\frac{\sqrt {20-x} -5}{x+5} = ?$

hayabusaboston · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Great question.

Nice work.

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

$Find$

$\int \frac{1}{x^2+10x+50} \ dx$

1/5 . tan^-1 (x+5)/5?

Oh right hero posted alread y haha.

Sy123 · Mar 23, 2013

Re: HSC 2013 3U Marathon Thread

$Prove using Binomial Theorem or Combinatorics or otherwise$

$\binom{k}{k} + \binom{k+1}{k} + \binom{k+2}{k} + \dots + \binom{n}{k} = \binom{n+1}{k+1}$

EDIT: Have I already posted this question? I have a feeling I already have.

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