HSC 2013 Maths Marathon (archive) (3 Viewers)

Sy123 · May 12, 2013

Re: HSC 2013 2U Marathon

Kostie100 said:
Can you explain his to me. I have probably just forgotten something really simple lol

Simply logarithm laws:

$a^{\log_a(b)} = b$

omgiloverice · May 12, 2013

Re: HSC 2013 2U Marathon

nightweaver066 · May 12, 2013

Re: HSC 2013 2U Marathon

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

what's wrong with my code ._.

shongaponga · May 12, 2013

Re: HSC 2013 2U Marathon

nightweaver066 said:
$\int \frac{x^3}{x^2+1} dx$

what's wrong with my code ._.

Fixed: $\int \frac{x^3}{x^2+1} dx$

I put a space between the "tex" wraps and the code itself and it worked.

Makematics · May 12, 2013

Re: HSC 2013 2U Marathon

nightweaver066 said:
$\int \frac{x^3}{x^2+1} dx$

what's wrong with my code ._.

shit that one scared me for a sec. btw if you do 4 unit or have finished 4 unit i think that we should leave the questions to 2U/3U people to attempt. if noone is posting up a solution for a while then a solution should be posted.

HeroicPandas · May 12, 2013

Re: HSC 2013 2U Marathon

Makematics said:
shit that one scared me for a sec. btw if you do 4 unit or have finished 4 unit i think that we should leave the questions to 2U/3U people to attempt. if noone is posting up a solution for a while then a solution should be posted.

i agree

nightweaver066 · May 12, 2013

Re: HSC 2013 2U Marathon

$\int \frac{x}{\sqrt{x-1}}dx$ (no substitution.)

fishrushed · May 12, 2013

Re: HSC 2013 2U Marathon

Registered User said:
$$Solve$

$\cos(x)=\cos(3x)$

Makematics said:
General solutions aren't in the 2 unit course?

Yeah, they aren't. Since it's out of 2U I'll do it for fun and try out LaTeX

$cos(x)=\cos(3x)\\ x = \pm 3x+k360 \: $where k is an integer$ \\\\ $Consider$ \: x =-3x+k360\\ 4x =k360\\ x =k90 \\\\ $Consider$\: x =3x+k360\\ -2x =k360\\ x =k180\\\\ $Combining both solutions$ \\ x =k90 $(where k is an integer)$$

EDIT: align environment didn't seem to work at first so i removed it

asianese · May 12, 2013

Re: HSC 2013 2U Marathon

fishrushed said:
Yeah, they aren't. Since it's out of 2U I'll do it for fun and try out LaTeX

$cos(x)=\cos(3x)\\ x = \pm 3x+k360 \: $where k is an integer$ \\\\ $Consider$ \: x =-3x+k360\\ 4x =k360\\ x =k90 \\\\ $Consider$\: x =3x+k360\\ -2x =k360\\ x =k180\\\\ $Combining both solutions$ \\ x =k90 $(where k is an integer)$$

EDIT: align environment didn't seem to work

You didn't use it?

fishrushed · May 12, 2013

Re: HSC 2013 2U Marathon

asianese said:
You didn't use it?

I did before, but for some reason it completely removed the whole tex and so I removed it.

HeroicPandas · May 13, 2013

Re: HSC 2013 2U Marathon

nightweaver066 said:
$\int \frac{x^3}{x^2+1} dx$

$\int \frac{x^3}{x^2 + 1}dx = \int \frac{x^3 +x - x}{x^2+1}dx$

nightweaver066 said:
$\int \frac{x}{\sqrt{x-1}}dx$ (no substitution.)

$\int \frac{x}{\sqrt{x-1}}dx = \int \frac{x-1+1}{\sqrt{x-1}}dx$

Here are some bad hints so we can increase the activity in this 2u marathon

HeroicPandas · May 15, 2013

Re: HSC 2013 2U Marathon

Another hint...for these 2 integration questions

$\frac{a+b+c+ \dots + y}{z} = \frac{a}{z}+ \frac{b}{z}+ \frac{c}{z} + \dots + \frac{y}{z}$

Makematics · May 15, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
$\int \frac{x^3}{x^2 + 1}dx = \int \frac{x^3 +x - x}{x^2+1}dx$

$\int \frac{x}{\sqrt{x-1}}dx = \int \frac{x-1+1}{\sqrt{x-1}}dx$

Here are some bad hints so we can increase the activity in this 2u marathon

2u people y u no motivated

more ppl do 2u than 3u or 4u

omgiloverice · May 15, 2013

Re: HSC 2013 2U Marathon

$\int \frac{{x^3}}{x^2 +1}dx =\int x - \frac{x}{x^2 +1}dx$

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

$= \ \frac{x^2}{2} \ - \ \frac{1}{2} ln(x^2 +1)+c$

tbh your hint gave me an even harder time solving those integration questions :L
I reckon some 2U people just feel intimidated on these marathons, so if your 2U and your reading this just give it a try~

I can't seem to solve the square root one without substitution, haha I rely on that too much

HeroicPandas · May 15, 2013

Re: HSC 2013 2U Marathon

omgiloverice said:
tbh your hint gave me an even harder time solving those integration questions :L
I reckon some 2U people just feel intimidated on these marathons, so if your 2U and your reading this just give it a try~

I can't seem to solve the square root one without substitution, haha I rely on that too much

very nice! u got it, yeh the hint didnt look that helpful

yes i encourage all 2U to participate in this marathon - it'll help u greatly

nightweaver066 · May 15, 2013

Re: HSC 2013 2U Marathon

omgiloverice said:
$\int \frac{{x^3}}{x^2 +1}dx =\int x - \frac{x}{x^2 +1}dx$

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

$= \ \frac{x^2}{2} \ - \ \frac{1}{2} ln(x^2 +1)+c$

tbh your hint gave me an even harder time solving those integration questions :L
I reckon some 2U people just feel intimidated on these marathons, so if your 2U and your reading this just give it a try~

I can't seem to solve the square root one without substitution, haha I rely on that too much

I hope you didn't perform polynomial division

If you didn't, it's the same technique applied to the 2nd question.

omgiloverice · May 17, 2013

Re: HSC 2013 2U Marathon

It is about time I give the answer to my question...
$Since it is known that the distance from point P(x,y) to point A(p,q) is equal to p(x,y) and to line y = 1$

$\therefore \ $using perpendicular formula \sqrt {(x-p)^2 +(y-q)^2} = \frac{|0+y+1|}{\sqrt{0^2 + 1^2 }}$

$$ After solving the whole thing you'll get$ \ y=\frac{x^2-2px +p^2 -q^2 -1}{2q-2}$

anomalousdecay · May 17, 2013

Re: HSC 2013 2U Marathon

omgiloverice said:
I can't seem to solve the square root one without substitution, haha I rely on that too much

I also rely way too much on substitution, that my teacher took marks from me by using it in 2-unit exams.
Now I learnt the quick way to do those questions as I was forced to.

Sy123 · May 21, 2013

Re: HSC 2013 2U Marathon

$\int \frac{dx}{x\ln x}$

HeroicPandas · May 21, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$\int \frac{dx}{x\ln x}$

is this possible in 2U maths? (they dont learn integration by substitution)

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