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

[Timske]

Re: 2012 HSC MX1 Marathon

In1 - In-1 = In1 +In1 = 0

is that how it works?

 

[D94]

Re: 2012 HSC MX1 Marathon

In1 - In-1 = In1 +In1 = 0

is that how it works?

No. For real values, you can't log a negative value, unless you're talking about distances/areas or substitution values where you could be taking the log of an absolute value. And it's Ln (natural logarithm) not In.

 

[deswa1]

Re: 2012 HSC MX1 Marathon

The easiest way to do that is to recognise that the function is odd, therefore the definte integral has to equal zero (between symmetrical x values). The question that you posted Timske has a very messy answer (assuming I read it correctly as the format is difficult to read). Were did you get it from?

 

[gurmies]

Re: 2012 HSC MX1 Marathon

How do you plan on integrating 1/x over an interval that contains 0?

 

[deswa1]

Re: 2012 HSC MX1 Marathon

Here are some new questions on maximisation and minimisation. Enjoy

 

[Peeik]

Re: 2012 HSC MX1 Marathon

definite integral from -1 to 1 of 1/x

There's no solution to this because the function 1/x is discontinuous wen x= 0. You can't use integration for functions that have a point of discontinuity. Since the definite integral is from x=-1 to x=1 so u can't work it out.

 

[bleakarcher]

Re: 2012 HSC MX1 Marathon

The easiest way to do that is to recognise that the function is odd, therefore the definte integral has to equal zero (between symmetrical x values). The question that you posted Timske has a very messy answer (assuming I read it correctly as the format is difficult to read). Were did you get it from?

This is what I thought.

 

[bleakarcher]

Re: 2012 HSC MX1 Marathon

There's no solution to this because the function 1/x is discontinuous wen x= 0. You can't use integration for functions that have a point of discontinuity. Since the definite integral is from x=-1 to x=1 so u can't work it out.

Until this.

 

[deswa1]

Re: 2012 HSC MX1 Marathon

Same. I just checked it on Wolfram Alpha and it said it was undefined.

 

[Timske]

Re: 2012 HSC MX1 Marathon

The easiest way to do that is to recognise that the function is odd, therefore the definte integral has to equal zero (between symmetrical x values). The question that you posted Timske has a very messy answer (assuming I read it correctly as the format is difficult to read). Were did you get it from?

examples from my teacher

 

[Timske]

Re: 2012 HSC MX1 Marathon

The easiest way to do that is to recognise that the function is odd, therefore the definte integral has to equal zero (between symmetrical x values). The question that you posted Timske has a very messy answer (assuming I read it correctly as the format is difficult to read). Were did you get it from?

examples from my teacher

 

[Peeik]

Re: 2012 HSC MX1 Marathon

Until this.

lol.

 

[deswa1]

Re: 2012 HSC MX1 Marathon

This is a question that I just saw at school that you guys might appreciate (excuse the dodgy paint skills...)

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

This is a question that I just saw at school that you guys might appreciate (excuse the dodgy paint skills...)

Question:

 

[deswa1]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=ln(3x@plus;1)^2-ln(x@plus;1)=ln(7x@plus;4) \\ ln(\frac{(3x@plus;1)^2}{x@plus;1})=ln(7x@plus;4) \\ (3x@plus;1)^2=(7x@plus;4)(x@plus;1) \\9x^2@plus;6x@plus;1=7x^2@plus;11x@plus;4 \\2x^2-5x-3=0 \\(2x@plus;1)(x-3)=0 \\x=3, x=-\frac{1}{2} \\x\neq -\frac{1}{2} \\\therefore x=3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?ln(3x+1)^2-ln(x+1)=ln(7x+4) \\ ln(\frac{(3x+1)^2}{x+1})=ln(7x+4) \\ (3x+1)^2=(7x+4)(x+1) \\9x^2+6x+1=7x^2+11x+4 \\2x^2-5x-3=0 \\(2x+1)(x-3)=0 \\x=3, x=-\frac{1}{2} \\x\neq -\frac{1}{2} \\\therefore x=3" title="ln(3x+1)^2-ln(x+1)=ln(7x+4) \\ ln(\frac{(3x+1)^2}{x+1})=ln(7x+4) \\ (3x+1)^2=(7x+4)(x+1) \\9x^2+6x+1=7x^2+11x+4 \\2x^2-5x-3=0 \\(2x+1)(x-3)=0 \\x=3, x=-\frac{1}{2} \\x\neq -\frac{1}{2} \\\therefore x=3" /></a>

 

[Timske]

Re: 2012 HSC MX1 Marathon

so deswa whats ur question

 

[cutemouse]

Re: 2012 HSC MX1 Marathon

There's no solution to this because the function 1/x is discontinuous wen x= 0. You can't use integration for functions that have a point of discontinuity. Since the definite integral is from x=-1 to x=1 so u can't work it out.

Well, one could still say that



p.v. means the cauchy principle value.

 

[math man]

Re: 2012 HSC MX1 Marathon

evaluating the area gives:
this is true as diverges to infinity, therefore it does not have a finite area.

 

[cutemouse]

Re: 2012 HSC MX1 Marathon

Yes, but you can assign a cauchy principal value to the integral since it doesn't converge absolutely and since the integrand is an odd function.

 

[such_such]

Re: 2012 HSC MX1 Marathon