Question

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

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

Answer 1

Answer: (A)

Let $f(x) =\frac{x}{\log x}$
$f'(x) =\frac{\log x-1}{(\log x)^{2}}$
Put $f'(x)=0$, for maxima or minima, we get
$\log x-1=0$
$\Rightarrow x=e$
Now, $f''(x)=\frac{(\log x)^{2} \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^{4}}$
$\Rightarrow f''(e)=\frac{\frac{1}{e}-0}{1}=\frac{1}{e}>0$
$\therefore $ Function is minimum at $x=e .$
Hence, minimum value of $f(x)$ at $x=e$ is
$f(e)=e .$