If the maximum value of the function $f(x) = \frac{2 \log_e x}{x} ,x > 0$ occurs at $x = e$, then $e^3f''(e)$ is equal to

-2

The correct answer is Option (1) → -2

$f(x) = \frac{2 \log x}{x}$

$f'(x) = \frac{2(1 - \log x)}{x^2}$

$f''(x) = \frac{d}{dx} \left( \frac{2(1 - \log x)}{x^2} \right)$

$= 2 \cdot \frac{-x^2 \cdot \frac{1}{x} - 2x(1 - \log x)}{x^4}$

$= 2 \cdot \frac{-x - 2x(1 - \log x)}{x^4}$

$= 2 \cdot \frac{-x - 2x + 2x \log x}{x^4}$

$= 2 \cdot \frac{-3x + 2x \log x}{x^4}$

$= \frac{2(-3 + 2 \log x)}{x^3}$

$f''(e) = \frac{2(-3 + 2 \cdot 1)}{e^3} = \frac{-2}{e^3}$

$e^3 f''(e) = -2$