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

$-\frac{1}{e}$

The correct answer is Option (2) → $-\frac{1}{e}$

$f(x)=\frac{\log x}{x}$, with $x>0$.

Differentiate:

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

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

For maximum: $f'(a)=0$ gives

$1-\log a=0$

$\log a=1$

$a=e$

Second derivative:

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

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

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

$f''(x)= -x^{-3}-2(1-\log x)x^{-3}$

$f''(x)=x^{-3}(-1-2+2\log x)$

$f''(x)=x^{-3}(2\log x-3)$

Evaluate at $x=a=e$:

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

$f''(e)=e^{-3}(-1)$

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

Compute $a^{2}f''(a)$:

$e^{2}\cdot\left(-\frac{1}{e^{3}}\right)$

$=-\frac{1}{e}$

Final answer: $-\frac{1}{e}$