Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The maximum value of (logex/x), if x > 0 is
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. The maximum value of $\frac{log_ex}{x}$, if x > 0 is
KCET
KCET 2020
Application of Derivatives
A
$e$
29%
B
$1$
12%
C
$\frac{1}{e}$
49%
D
$- \frac{1}{e}$
10%
Solution:
$(y = \frac{log_e \ x}{x})$
$⇒ \left(\frac{dy}{dx}=\frac{1 - log_e x}{x^2}\right)$
For maxima, $dy/dx = 0$
$⇒ 1 - log_e x = 0$
$⇒ x = e$
$dy/dx$ changes sign from positive to negative at $x = e$
$∴ y_{\text{max}} = 1/e$