Applications Of Derivatives Question 298
Question: Local maximum value of the function $ \frac{\log x}{x} $ is
[MNR 1984; RPET 1997, 2002; DCE 2002; Karnataka CET 2000; UPSEAT 2001; MP PET 2002]
Options:
A) e
B) 1
C) $ \frac{1}{e} $
D) 2e
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ f(x)=\frac{\log x}{x}\Rightarrow f’(x)=\frac{1}{x^{2}}-\frac{\log x}{x^{2}} $
For maximum or minimum value of $ f(x),f’(x)=0 $
therefore $ f’(x)=\frac{1-{\log_{e}}x}{x^{2}}=0 $ or $ \frac{1-{\log_{e}}x}{x^{2}}=0 $
$ \therefore {\log_{e}}x=1 $ or $ x=e $ , which lie in $ (0,\infty ) $ .
For $ x=e,\frac{d^{2}y}{dx^{2}}=-\frac{1}{e^{3}} $ , which is $ -ve $ .
Hence y is maximum at $ x=e $ and its maximum value $ =\frac{\log e}{e}=\frac{1}{e} $ .
BETA
