Question

The maximum value of $\frac{Log \, x}{x}$ in $(2, \infty)$ is

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

Answer 1

Answer: (D)

$\frac{1}{e}$

Answer 2

Let $y=\frac{\log x}{x}$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log x .1}{x^{2}}=\frac{1-\log x}{x^{2}}$
For maxima, put $\frac{d y}{d x}=0$
$\Rightarrow \,\,\,\,\, \frac{1-\log x}{x^{2}}=0$
$\Rightarrow\,\,\,\,\, \log x=1$
$\Rightarrow \,\,\,\,\,x-e$
Now, $\,\,\,\,\,\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x) 2 x}{\left(x^{2}\right)^{2}}$
At $x=e, \frac{d^{2} y}{d x^{2}} \leq 0$, maxima
$\therefore\,\,\,\,\,$ The maximum value at $x=e$ is
$y=\frac{\log e}{e}=\frac{1}{e}$