Q61 / 120 · VITEEE 2014 · Mathematics

The minimum value of $\frac{x}{\log x}$ is

  • Correct answer
    A
    $e$
  • B
    $\frac{1}{e}$
  • C
    $e^{2}$
  • D
    $e^{3}$

Solution

Let $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\log \mathrm{x}}$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\log \mathrm{x}-1}{(\log \mathrm{x})^{2}}$ For maxima and minima, put $f^{\prime}(x)=0$ $\log x-1=0$ $\Rightarrow \mathrm{x}=\mathrm{e}$ Now, $f^{\prime \prime}(x)$ $=\frac{(\log x)^{2} \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^{4}}$ $\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{\frac{1}{\mathrm{e}}-0}{1}=\frac{1}{\mathrm{e}}>0$ $\therefore \mathrm{f}(\mathrm{x})$ is minimum at $\mathrm{x}=\mathrm{e}$. Hence, minimum value of $f(x)$ at $x=e$ is $f(e)=\frac{e}{\log e}=e$

Hence, the correct answer is option: 'A'.

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