Applications Of Derivatives Question 318
Question: The minimum value of $ \frac{\log x}{x} $ in the interval $ [2,,\infty ) $ is [Roorkee 1999]
Options:
A) $ \frac{\log 2}{2} $
B) Zero
C) $ \frac{1}{e} $
D) Does not exist
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ y=\frac{\log x}{x} $
therefore $ \frac{dy}{dx}=\frac{x.\frac{1}{x}-\log x}{x^{2}} $
$ =\frac{1-\log x}{x^{2}} $ Put $ \frac{dy}{dx}=0\Rightarrow \frac{1-\log x}{x^{2}}=0 $
therefore $ 1-\log x=0 $
therefore $ x=e $ and $ \frac{d^{2}y}{dx^{2}}=\frac{-3x+2x\log x}{x^{4}} $ At $ x=e $ , $ \frac{d^{2}y}{dx^{2}}=\frac{-1}{e^{3}}<0 $ In [2, $\infty$) the function $ p^{2}=q $ will be maximum and minimum value does not exist.
BETA
