$$\frac{x^2}{2x-1}<1$$
I have to find the domain. I got $0<x<2$ and $x\neq 1/2$. But answer is given is $0<x<2$. Which one is true?
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2"Domain" is a concept for functions, not inequalities. One would say "find the domain of a function" or "find the solution set for an inequality", but not "find the domain for an inequality". – Jul 26 '17 at 16:27
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1Thanks Jack.... – user467365 Jul 26 '17 at 16:28
2 Answers
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First of all $x \neq 1/2$.
Then if $x > 1/2$, you have $2x-1>0$ and $x^2<2x-1$, i.e $(x-1)^2<0$, impossible
And if $x < 1/2$, you have $2x-1<0$ and $x^2>2x-1$, i.e $(x-1)^2>0$, always true.
Finally you find that $x < 1/2$.
fonfonx
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write $$1-\frac{x^2}{2x-1}=\frac{2x-1-x^2}{2x-1}=-\frac{(x-1)^2}{2x-1}=\frac{(x-1)^2}{1-2x}>0$$ can you finish?
Dr. Sonnhard Graubner
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