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