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The maximum value of $f(x)=\frac{\log x}{x}$ for x in $[2, \infty]$ is |
e 0 1 $\frac{1}{e}$ |
$\frac{1}{e}$ |
The correct answer is Option (4) - $\frac{1}{e}$ $f(x)=\frac{\log x}{x}$ $f'(x)=\frac{1}{x^2}-\frac{\log x}{x^2}=0$ $⇒x=e$
so e → point of maxima max value = $f(x)=\frac{1}{e}$ |
