2011 hsc mx1 marathon (1 Viewer)

taeyang · Oct 14, 2011

nightweaver066 said:
$\int \pi (1+tany)^2 dy$
$= \pi \int 1 + 2tany + tan^2y dy$
$= \pi \int 1 + 2tany + sec^2y - 1 dy$
$= \pi (-2lncosy + tany) + c$
$= \pi (tany - 2lncosy) + c$

New question:

$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.....+\frac{n}{(n+1)!} = \frac{(n+1)!-1}{(n+1)!}\\\\ $Going to do this really non-professionally to save time$\\\\ $Prove true for n = 1 $\\\\ \frac{1}{(1+1)!} = \frac{(1+1)!-1}{(1+1)!}\\ \therefore 1 = 1\\\\ $Assume statement true for n = k\\\\ $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.....+\frac{k}{(k+1)!} = \frac{(k+1)!-1}{(k+1)!}\\\\$\\ Prove statement true for n = k + 1\\\\ $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.....+\frac{k+1}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}$\\\\$ $$By applying the series up to n = k from the assumption to the above thingy$\\\\ \frac{(k+1)!-1}{(k+1)!} + \frac{k+1}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!((k+1)!-1)+(k+1)!(k+1)}{(k+1)!(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!(k+1)!-(k+2)!+(k+1)!(k+1)}{(k+1)!(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!(k+1)!-(k+2)(k+1)!+(k+1)!(k+1)}{(k+1)!(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!-(k+2)+(k+1)}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!-1}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}$

nightweaver066 · Oct 14, 2011

hup said:
there is only one variable x and the constant so just let x = 0

definite integrals wont work if you hav to integrate twice

Thanks for that, had slipped my mind after i couldn't find a question where i had actually used it..

taeyang · Oct 14, 2011

Woops, Next Question.... Uhh let me just make one up..

$\int \frac{9x+7}{3x^2 + 3}dx$

May require some -------------------lateral thinking---------------------

bleakarcher · Oct 15, 2011

(3/2)log[e][3x^2+3]+(7/3)tan^(-1)x+C

brettymaccc · Oct 15, 2011

taeyang said:
Woops, Next Question.... Uhh let me just make one up..

$\int \frac{9x+7}{3x^2 + 3}dx$

May require some -------------------lateral thinking---------------------

$\int \frac{9x}{3x^2+3} + \int \frac{7}{3x^2+3}$

$\frac{3}{2}\int \frac{2x}{x^2+1} + \frac{7}{3}\int \frac{1}{x^2+1}$

$\frac{3}{2}ln(x^2+1) + \frac{7}{3}tan^{-1}x + c$

artosis · Oct 15, 2011

taeyang said:
Woops, Next Question.... Uhh let me just make one up..

$\int \frac{9x+7}{3x^2 + 3}dx$

May require some -------------------lateral thinking---------------------

thats a 4u q1 question.
but nonetheless its still adequate for 3u students

artosis · Oct 15, 2011

since no one posted a question heres a fun one:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{-1}^{1}\frac{1}{x^2} dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{-1}^{1}\frac{1}{x^2} dx" title="\int_{-1}^{1}\frac{1}{x^2} dx" /></a>

hup · Oct 15, 2011

artosis said:
since no one posted a question heres a fun one:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{-1}^{1}\frac{1}{x^2} dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{-1}^{1}\frac{1}{x^2} dx" title="\int_{-1}^{1}\frac{1}{x^2} dx" /></a>

lol
invalid but still works out

apollo1 · Oct 15, 2011

heres a question:

Screen shot 2011-10-15 at 11.42.51 AM.png

3PiCFail · Oct 15, 2011

apollo1 said:
heres a question:

View attachment 23278

i) 7x7!
ii) 2 x 6!
? probably wrong here... I haven't revised probability etc. yet...

apollo1 · Oct 15, 2011

3PiCFail said:
i) 7x7!
ii) 2 x 6!
? probably wrong here... I haven't revised probability etc. yet...

yeh ur wrong.

3PiCFail · Oct 15, 2011

apollo1 said:
yeh ur wrong.

i fail so hard... ill wait for someone to answer it, i actually have no idea. or send me the solution? =)

blackops23 · Oct 15, 2011

taeyang said:
$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.....+\frac{n}{(n+1)!} = \frac{(n+1)!-1}{(n+1)!}\\\\ $Going to do this really non-professionally to save time$\\\\ $Prove true for n = 1 $\\\\ \frac{1}{(1+1)!} = \frac{(1+1)!-1}{(1+1)!}\\ \therefore 1 = 1\\\\ $Assume statement true for n = k\\\\ $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.....+\frac{k}{(k+1)!} = \frac{(k+1)!-1}{(k+1)!}\\\\$\\ Prove statement true for n = k + 1\\\\ $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+.....+\frac{k+1}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}$\\\\$ $$By applying the series up to n = k from the assumption to the above thingy$\\\\ \frac{(k+1)!-1}{(k+1)!} + \frac{k+1}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!((k+1)!-1)+(k+1)!(k+1)}{(k+1)!(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!(k+1)!-(k+2)!+(k+1)!(k+1)}{(k+1)!(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!(k+1)!-(k+2)(k+1)!+(k+1)!(k+1)}{(k+1)!(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!-(k+2)+(k+1)}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}\\\\ \frac{(k+2)!-1}{(k+2)!} = \frac{(k+2)!-1}{(k+2)!}$

Woah.... proving statement for n = k+1

$\textup{Prove for n = k+1}\\\\S_{k}\, +\: T_{k+1}\\\\\therefore \frac{(k+1)! - 1}{(k+1)!} \; \; \; + \; \; \; \frac{k+1}{(k+2)!}\\\\=\, \frac{(k+2)(k+1)! - (k+2)\: +\: (k+1)}{(k+2)!}\\\\=\frac{(k+2)! - 2 + 1}{(k+2)!}\\\\= \frac{(k+2)! - 1}{(k+2)!}\\\\= S_{k+1}$

Therefore, true for n=1 and true for n=k+1 when n=k, k integer, therefore true for n=2, n=3 and true for all positive integers n.

Much quicker

bleakarcher · Oct 15, 2011

^ mine didnt take as long either lol.

bleakarcher · Oct 15, 2011

though as he said he did unprofessionally lol

acemusic415 · Oct 15, 2011

Find the probability that of 5 cards chosen from a 52-card deck that 2 will be of the same suit.

apollo1 · Oct 15, 2011

Screen shot 2011-10-14 at 4.06.56 AM.png

i got the number of objects as n = 7. can someone confirm my answer.

Comeeatmebro · Oct 15, 2011

Don't you reckon you guys should be doing a bit of English??? :S :S :S

someth1ng · Oct 15, 2011

acemusic415 said:
Find the probability that of 5 cards chosen from a 52-card deck that 2 will be of the same suit.

P(pair)=1

apollo1 said:
heres a question:

View attachment 23278

Okay,
Ways to pick the loner=7
Ways to create 3 pairs= 6C2 * 4C2
Total ways=7 * 6C2 * 4C2 = 630 ways

Ways that not together = 630 - Ways together
Ways together = 5 * 4C2 = 30 ways
.`. Ways that not together = 600 - 30 = 600 ways.

someth1ng · Oct 15, 2011

apollo1 said:
heres a question:

View attachment 23278

(i) $7 * 6C2 * 4C2 = 630 ways$

(ii) $7 * 6C2 * 4C2 - 5 * 4C2 = 600 ways$

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