Answer is: option4
\( e^2 \)Solution:
Using the quotient rule and chain rule:
\[ f'(x) = \frac{d}{dx} \left( \frac{(\ln x)^2}{x} \right) = \frac{2 \ln x \cdot \frac{1}{x} \cdot x - (\ln x)^2 \cdot 1}{x^2} = \frac{2 \ln x - (\ln x)^2}{x^2} \]
Simplify:
\[ f'(x) = \frac{\ln x (2 - \ln x)}{x^2} \]
Find critical points by setting \( f'(x) = 0 \):
\[ \frac{\ln x (2 - \ln x)}{x^2} = 0 \]
Since \( x^2 \neq 0 \) for \( x > 0 \), we solve:
\[ \ln x (2 - \ln x) = 0 \]
This gives two solutions:
- \( \ln x = 0 \Rightarrow x = 1 \)
- \( 2 - \ln x = 0 \Rightarrow \ln x = 2 \Rightarrow x = e^2 \)
Determine the nature of the critical points using the second derivative or first derivative test:
For \( x = 1 \):
- Test \( x \) slightly less than 1 (e.g., \( x = 0.5 \)):
\( \ln 0.5 < 0 \) and \( 2 - \ln 0.5 > 0 \), so \( f'(0.5) < 0 \) - Test \( x \) slightly greater than 1 (e.g., \( x = 2 \)):
\( \ln 2 > 0 \) and \( 2 - \ln 2 > 0 \), so \( f'(2) > 0 \)
Since \( f'(x) \) changes from negative to positive, \( x = 1 \) is a relative minimum.
For \( x = e^2 \):
- Test \( x \) slightly less than \( e^2 \) (e.g., \( x = e \)):
\( \ln e = 1 > 0 \) and \( 2 - \ln e = 1 > 0 \), so \( f'(e) > 0 \) - Test \( x \) slightly greater than \( e^2 \) (e.g., \( x = e^3 \)):
\( \ln e^3 = 3 > 0 \) but \( 2 - \ln e^3 = -1 < 0 \), so \( f'(e^3) < 0 \)
Since \( f'(x) \) changes from positive to negative, \( x = e^2 \) is a relative maximum.
Conclusion: The relative maximum occurs at \( x = e^2 \), which corresponds to option (D).