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Let $[\mathrm{t}]$ represents the greatest integer not more than $\mathrm{t}$.Then the number of discontinuous points of $f(x)$$=\left[X^{\frac{1}{x}}\right]$in $(0, \infty)$ is
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The correct answer is:
$1$
Given $f(x)=\left[x^{\frac{1}{x}}\right]$
Now for discontinuous point
$x^{\frac{1}{x}}=\text { integer }$
$\Rightarrow x^{\frac{1}{x}}=$ integer only 1 possible
so $f(x)=\left[x^{\frac{1}{x}}\right]$ is discontinuous at only one point.
Now for discontinuous point
$x^{\frac{1}{x}}=\text { integer }$
$\Rightarrow x^{\frac{1}{x}}=$ integer only 1 possible
so $f(x)=\left[x^{\frac{1}{x}}\right]$ is discontinuous at only one point.
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