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 $f(x)=\frac{x}{\log x}$
On differentiating w.r.t. $x$, we get
$f^{\prime }(x)=\frac{\log x-1}{(\log x{)}^2}$
For maxima and minima, put $f^{\prime }(x)=0$
$\log x-1=0$
$\Rightarrow x=e$
Now ,
$f^{\prime \prime }(x)=\frac{(\log x{)}^2\cdot \frac{1}{x}-(\log x-1)\cdot \frac{2\log x}{x}}{(\log x{)}^4}$
$\Rightarrow f"\left(e\right)=\frac{\frac{1}{e}-0}{1}=\frac{1}{e}>0$
$\therefore f(x)$ is minimum at $x=e$.
Hence, minimum value of f (x) at x = e is
$f\left(e\right)=\frac{e}{\log e}=e$