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The minimum value of $\frac{x}{logx}(x≠1)$ is : |
e $\frac{1}{e}$ $e^2$ $2e$ |
e |
The correct answer is Option (1) → $e$ $y=\frac{x}{\log x}$ $y'=\frac{-x×\frac{1}{x}+\log x}{(\log x)^2}=\frac{\log x-1}{(\log x)^2}$ for critical points, $y'(c)=0$ $⇒\log c-1=0$ $⇒\log c=1$ $⇒c=e$ $∴y_{min}=\frac{e}{\log e}=e$ |
