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

Make a Difference – Donate to Bored of Studies!

2011 hsc mx1 marathon (1 Viewer)

Member

Well-Known Member

Member

Active Member

Member

Member

Member

Member

Banned

Member

Banned

Member

Member

Active Member

Active Member

Member

Banned

New Member

Retired Nov '14

Retired Nov '14

Users Who Are Viewing This Thread (Users: 0, Guests: 1)