HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)
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Re: 2012 HSC MX1 Marathon
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.
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deswa1
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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?
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
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Re: 2012 HSC MX1 Marathon
How do you plan on integrating 1/x over an interval that contains 0?
How do you plan on integrating 1/x over an interval that contains 0?
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.definite integral from -1 to 1 of 1/x
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deswa1
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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>
<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>
Re: 2012 HSC MX1 Marathon
p.v. means the cauchy principle value.
Well, one could still say that
p.v. means the cauchy principle value.
