HSC Mathematics Marathon (2 Viewers)

hscishard · Feb 14, 2011

That was lame...
Anyway how would you solve it? This looks like natural decay...but it isn't

slyhunter · Feb 14, 2011

jm01 you genius.

Trebla · Feb 14, 2011

hscishard said:
Why are uni students dominating this thread?

Uni students are still on holidays while HSC students have already started school. Also, I have tried not to answer any of the questions posed in this thread. I only contribute to asking the questions if the need arises, with the expectation that hopefully a HSC student can answer them (though obviously this doesn't always happen).

cutemouse · Feb 14, 2011

hscishard said:
That was lame...
Anyway how would you solve it? This looks like natural decay...but it isn't

lol it's been deleted now.

It's a question similiar to one in the USyd MATH1013 notes. Except that one has Romeo and Juliet's fate of love in it... Trebla posted it once I think.

Drongoski · Feb 14, 2011

Have not posted a marathon question. Here's one from an old algebra book (1st published 1887)

Factorise: a(b-c)³ + b(c-a)³ + c(a-b)³

cutemouse · Feb 14, 2011

Bored Of Fail said:
you have put in alot of effort to try and find out shit about me

yeah right.

ohexploitable · Feb 14, 2011

Drongoski said:
Have not posted a marathon question. Here's one from an old algebra book (1st published 1887)

Factorise: a(b-c)³ + b(c-a)³ + c(a-b)³

$\\a(b-c)^3+b(c-a)^3+c(a-b)^3\\=ab^3-3ab^2c+3abc^2-ac^3+bc^3-3abc^2+3a^2bc-a^3b+a^3c-3a^2bc+3ab^2c-b^3c\\=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\\=a^3(c-b)-a(c^3-b^3)+bc(c^2-b^2)\\=a^3(c-b)-a(c-b)(c^2+cb+b^2)+bc(c-b)(c+b)\\=(c-b)[a^3-a(c^2+cb+b^2)+bc(c+b)]\\=(c-b)[a(a^2-b^2)-c^2(a-b)-bc(a-b)]\\=(c-b)(a-b)[a(a+b)-c^2-bc]\\=(c-b)(a-b)[(a^2-c^2)+b(a-c)]\\=(c-b)(a-b)(a-c)(a+b+c)$

ohexploitable · Feb 14, 2011

Is there a quicker way of doing that?

Drongoski · Feb 14, 2011

Don't know. Too hard for me.

Edit

On 2nd thought maybe.

$Let f(c) = a(b-c)^3 + b(c-a)^3 + c(a-b)^3$

$\therefore f(b) = a(b-b)^3 + b(b-a)^3 + b(b-a)^3 = 0$

$\therefore$ (c-b) or (b-c) is a factor

In the same way (or by symmetry) (c-a) and (a-b) are also factors.

As the terms are all uniformly of degree 4, $\exists$ one remaining factor involving a, b & c of deg 1.

By inspection &/or trial-and-error you get 4th factor (a+b+c)

jyu · Feb 14, 2011

Drongoski said:
Have not posted a marathon question. Here's one from an old algebra book (1st published 1887)

Factorise: a(b-c)³ + b(c-a)³ + c(a-b)³

=(b-c)(c-a)(a-b)(a+b+c) by factor theorem

e.g. in the original exp, replace b by c gives zero, .: b-c is a factor. The original expression is cyclic, .: c-a and a-b are also factors, .....

Drongoski · Feb 14, 2011

Brilliant effort ohexploitable and Mr jyu.

jyu · Feb 15, 2011

Drongoski said:
Brilliant effort ohexploitable and Mr jyu.

You are very polite Mr Goski.

FriedRice1 · Feb 16, 2011

QUESTION:

Integrate (6x+1)/(x^2+2x+3) dx

Drongoski · Feb 16, 2011

FriedRice1 said:
QUESTION:

Integrate (6x+1)/(x^2+2x+3) dx

integral

$= \int \frac{6x+1}{x^2 + 2x+3} dx = \int \frac {6x+6 - 5 }{x^2 + 2x + 3}dx$

$= \int \frac{6x+6}{x^2+2x+3} - \frac{5}{(\sqrt 2)^2 + (x+1)^2} dx$

$= 3ln(x^2+2x+3) - \frac {5}{\sqrt 2} tan^{-1} \frac {x+1}{\sqrt 2} + C$

Drongoski · Feb 16, 2011

UnrealAnchovies · Feb 16, 2011

Drongoski said:
$\int \frac {cot x}{ln sin x} dx$

Let u = ln(sinx)

$\frac {du}{dx} = \frac {sinx}{cosx}$

$ducosx = dxsinx$

:. $ducotx = dx$

so $\int \frac {cot x}{ln sin x} dx = \int \frac {1}{u} du$

= ln(u)

= ln[ln(sinx)]

First ever Latex post. :~~~~~)

--------------------------------

New question:

$\int \frac {x^1/2}{(16-x)^1/2} dx$

I can't seem to put the sq. root sign in. But the WHOLE SUM IS UNDER A SQ. ROOT.

khfreakau · Feb 16, 2011

UnrealAnchovies said:
Let u = ln(sinx)
$\frac {du}{dx} = \frac {sinx}{cosx}$
$ducosx = dxsinx$
:. $ducotx = dx$

so $\int \frac {cot x}{ln sin x} dx = \int \frac {1}{u} du$
= log(u)
= log(sinx)

You mean ln(ln(sinx))

Also I would have simply done it by derivative on function, derivative of ln(sinx) is cotx, therefore integral is ln(ln(sinx)) + C.

UnrealAnchovies · Feb 16, 2011

khfreakau said:
You mean ln(ln(sinx))
Also I would have simply done it by derivative on function, derivative of ln(sinx) is cotx, therefore integral is ln(ln(sinx)) + C.

Ugh... Latex...

You're right about the method you mentioned. Can't believe I didn't see that.

deterministic · Feb 16, 2011

You mean:
$\int \sqrt{\frac{x}{16-x}} dx$

UnrealAnchovies · Feb 16, 2011

deterministic said:
You mean:
$\int \sqrt{\frac{x}{16-x}} dx$

Yep.

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