Note that $x^r$ is a strictly decreasing function when $x\in(0,1)$, and the maximum value would be achieved when $m$ is the least and $n$ is the most, in this case, $m=1$ and $n\to\infty$, where $|x^m-x^n|=|x^1-0|=x$.

Note: @Sparsh Goyal posted the numerical answer first.

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## Comments

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TopNewestSimplest thing I can think of $|1^1-0^1|$.... It has to be more complicated than this...

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That doesn't work at all. I think you typoed.

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He put (0,1), which means 0 & 1 not included, so it must be something else!!

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0

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I dont think it will have a numeric value as answer as "x" is variable and there are infinite nos. between 0 and 1...

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Note that it asks for a maximum value.

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whats the answer

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In terms of m and n?

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The answer should be "x" !

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Note that $x^r$ is a strictly decreasing function when $x\in(0,1)$, and the maximum value would be achieved when $m$ is the least and $n$ is the most, in this case, $m=1$ and $n\to\infty$, where $|x^m-x^n|=|x^1-0|=x$.

Note: @Sparsh Goyal posted the numerical answer first.

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