HSC Mathematics Marathon (4 Viewers)

[hscishard]

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

 

[slyhunter]

jm01 you genius.

 

[Trebla]

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]

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]

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]

you have put in alot of effort to try and find out shit about me

yeah right.

 

[ohexploitable]

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

 

[ohexploitable]

Is there a quicker way of doing that?

 

[Drongoski]

Don't know. Too hard for me.

Edit

On 2nd thought maybe.





(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, one remaining factor involving a, b & c of deg 1.

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

 

[jyu]

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]

Brilliant effort ohexploitable and Mr jyu.

 

[jyu]

Brilliant effort ohexploitable and Mr jyu.

You are very polite Mr Goski.

 

[FriedRice1]

QUESTION:

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

 

[Drongoski]

QUESTION:

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

integral

 

[Drongoski]

 

[UnrealAnchovies]

Let u = ln(sinx)





:.

so

= ln(u)

= ln[ln(sinx)]

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

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

New question:



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

 

[khfreakau]

Let u = ln(sinx)


:.

so
= 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]

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]

You mean:

 

[UnrealAnchovies]

You mean:

Yep.