Question

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

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

Answer 1

Answer: (D)

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