HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

[Aesytic]

Re: HSC 2012 Marathon

2^x = e^[log(2^x)]
= e^[xlog(2)]
= (e^x)*(e^[x(log2 - 1)])
.'. integral of 2^x dx = integral (e^x)*(e^[x(log2 - 1)]) dx
let u = e^x
du = (e^x)*dx
when x = 1, u = e
when x= 0, u = 1
.'. integral 2^x dx = integral u^(log2 - 1) du
= [(u^log2)/log2] from 1 to e
= 2/log2 - 1/log2
= 1/log2

might be wrong :/

 

[barbernator]

Re: HSC 2012 Marathon

nah its not

other then that last step where i missed the fraction, can you show where my substitution working has gone wrong deswa or timske? also, wolfram alpha confirmed the result as 1/ln(2) as well.

 

[deswa1]

Re: HSC 2012 Marathon

Oh crap- I know why I said you guys had a answer different to what I thought the answer should me, I mistyped the question- MASSIVE SORRY. The real question is:

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}2^{lnx}dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{1}2^{lnx}dx" title="\int_{0}^{1}2^{lnx}dx" /></a>

All of you guys are right with your original working.

 

[nightweaver066]

Re: HSC 2012 Marathon

other then that last step where i missed the fraction, can you show where my substitution working has gone wrong deswa or timske? also, wolfram alpha confirmed the result as 1/ln(2) as well.

You wrote xln(2), not x/ln(2).

Edit: lol didn't see your 'correction'

 

[Timske]

Re: HSC 2012 Marathon

other then that last step where i missed the fraction, can you show where my substitution working has gone wrong deswa or timske? also, wolfram alpha confirmed the result as 1/ln(2) as well.

My bad didnt see ur edit there

 

[Timske]

Re: HSC 2012 Marathon

Its gotta be this -4 -3sqrt(3)

EDIT: <a href="http://www.codecogs.com/eqnedit.php?latex=tan(\theta ) = -4 - 3\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(\theta ) = -4 - 3\sqrt{3}" title="tan(\theta ) = -4 - 3\sqrt{3}" /></a>

 

[barbernator]

Re: HSC 2012 Marathon

Oh crap- I know why I said you guys had the wrong answer- the question was 2 to the power of log x. Sorry, I mistyped it- this question is much better.

All of you guys are right with your original working.

2^log(x) diverges as x-->0, so the integral cannot be evaluated by HSC means, this is just going out on a limb, im not really sure lol

edit: shit it converges ahhhhhhhh

 

[barbernator]

Re: HSC 2012 Marathon

Its gotta be this -4 -3sqrt(3)

EDIT: <a href="http://www.codecogs.com/eqnedit.php?latex=tan(\theta ) = -4 - 3\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(\theta ) = -4 - 3\sqrt{3}" title="tan(\theta ) = -4 - 3\sqrt{3}" /></a>

so so so close, just check one of your signs

 

[deswa1]

Re: HSC 2012 Marathon

2^log(x) diverges as x-->0, so the integral cannot be evaluated by HSC means

I might be wrong but I think you can- this is a question from Camrbdige and wolfram gives a definite answer.

Try this:
We know that e^ln(x)=x. Now try expressing 2^x in a similar form and then integrate.

 

[Timske]

Re: HSC 2012 Marathon

so so so close, just check one of your signs

I keep getting 8 + 5sqrt(3) now, ill post my solution so u can check it

 

[barbernator]

Re: HSC 2012 Marathon

i think I have it,

is it this?

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{log_{2}e}{1@plus;log_{2}e}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{log_{2}e}{1+log_{2}e}" title="\frac{log_{2}e}{1+log_{2}e}" /></a>

 

[deswa1]

Re: HSC 2012 Marathon

i think I have it,

is it this?

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{log_{2}e}{1@plus;log_{2}e}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{log_{2}e}{1+log_{2}e}" title="\frac{log_{2}e}{1+log_{2}e}" /></a>

Yep- My answer was different (1/1+ln2) but they are the same thing. Well done

 

[barbernator]

Re: HSC 2012 Marathon

Yep- My answer was different (1/1+ln2) but they are the same thing. Well done

fuark that is a crazy question

ok next one, find,<a href="http://www.codecogs.com/eqnedit.php?latex=\lim_{x\rightarrow -5}\frac{\sqrt{20-x}-5}{5@plus;x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow -5}\frac{\sqrt{20-x}-5}{5+x}" title="\lim_{x\rightarrow -5}\frac{\sqrt{20-x}-5}{5+x}" /></a>

 

[deswa1]

Re: HSC 2012 Marathon

fuark that is a crazy question

How did you do it- there's a relatively simple way (note the word relatively...)

 

[Timske]

Re: HSC 2012 Marathon



EDIT: Barb any easier way to do this lol

 

[barbernator]

Re: HSC 2012 Marathon

How did you do it- there's a relatively simple way (note the word relatively...)

ahah maybe your form of simple is my form of hard lol. I mean more conceptually difficult then actually process difficult, and more because of the fact that i have never seen anything like it and that in a test I would have tried substitution for 10 minutes and gave up. I used change of base formula for loge(x)=log2(x)/log2(e) and then got x to the power of 1/log2(e) and went from there. Is that the simplest way?

 

[barbernator]

Re: HSC 2012 Marathon

EDIT: Barb any easier way to do this lol

oh no you copied the question wrong originally but that is how you do it, except if you can, dont switch radians for degrees

 

[deswa1]

Re: HSC 2012 Marathon

ahah maybe your form of simple is my form of hard lol. I mean more conceptually difficult then actually process difficult, and more because of the fact that i have never seen anything like it and that in a test I would have tried substitution for 10 minutes and gave up. I used change of base formula for loge(x)=log2(x)/log2(e) and then got x to the power of 1/log2(e) and went from there. Is that the simplest way?

That's not a bad method. I did this:
<a href="http://www.codecogs.com/eqnedit.php?latex=2^x=e^{ln2^{lnx}}\\ =e^{ln2lnx}\\ =(e^{lnx})^{ln2}\\ =x^{ln2}\\ \therefore \int_{0}^{1}2^xdx=\int_{0}^{1} x^{ln2}=\left [ \frac{x^{ln2@plus;1}}{ln2@plus;1} \right ]^{1}_{0}\\ =\frac{1}{ln2@plus;1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2^x=e^{ln2^{lnx}}\\ =e^{ln2lnx}\\ =(e^{lnx})^{ln2}\\ =x^{ln2}\\ \therefore \int_{0}^{1}2^xdx=\int_{0}^{1} x^{ln2}=\left [ \frac{x^{ln2+1}}{ln2+1} \right ]^{1}_{0}\\ =\frac{1}{ln2+1}" title="2^x=e^{ln2^{lnx}}\\ =e^{ln2lnx}\\ =(e^{lnx})^{ln2}\\ =x^{ln2}\\ \therefore \int_{0}^{1}2^xdx=\int_{0}^{1} x^{ln2}=\left [ \frac{x^{ln2+1}}{ln2+1} \right ]^{1}_{0}\\ =\frac{1}{ln2+1}" /></a>

It is conceptually difficult though- I stuffed it up a few times before I finally got it.

 

[Timske]

Re: HSC 2012 Marathon

I see it now, i somehow managed to copy it incorrectly OMG!, . I have a habit of changing radians into degrees then changing it back, i know should not do this.

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(\theta ) = -4 @plus; 3\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(\theta ) = -4 + 3\sqrt{3}" title="tan(\theta ) = -4 + 3\sqrt{3}" /></a>

Barb please say yes