If the maximum value of the function \(f(x) = \displaystyle\frac{2 \log_e x}{x}, x \gt 0\) occurs at \(x = e\), then \(e^3f''(e)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −2
Writing log for \(\log_e\), by the quotient rule:
\(\displaystyle f'(x) = \frac{x \cdot \frac{2}{x} - 2\log x \cdot 1}{x^2}\)
\(\displaystyle = \frac{2(1 - \log x)}{x^2}\)
Differentiating again by the quotient rule:
\(\displaystyle f''(x) = \frac{x^2 \cdot \left(-\frac{2}{x}\right) - 2(1 - \log x) \cdot 2x}{x^4}\)
\(\displaystyle = \frac{-2x - 4x + 4x\log x}{x^4}\)
\(\displaystyle = \frac{4\log x - 6}{x^3}\)
At \(x = e\), \(\log e = 1\):
\(\displaystyle f''(e) = \frac{4 - 6}{e^3} = -\frac{2}{e^3}\)
\(e^3 f''(e) = -2\)