Applications-Of-Derivatives Question 437
Question: The velocity of telegraphic communication is given by $ v=x^{2}\log (1/x) $ , where x is the displacement. For maximum velocity, x equals to?
Options:
A) $ {e^{1/2}} $
B) $ {e^{-1/2}} $
C) $ {{(2e)}^{-1}} $
D) $ 2{e^{-1/2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given, velocity is $ v=x^{2}\log \frac{1}{x}=-x^{2}\log x $ where x is displacement. For maximum velocity, $ \frac{dv}{dx}=0 $ Now, $ \frac{dv}{dx}=-x^{2}\frac{1}{x}+\log x(-2x) $ $ =-x-2x\log x $ $ \frac{dv}{dx}=0\Rightarrow -x-2x\log x=0\Rightarrow x=-2x\log x $
$ \Rightarrow \frac{-1}{2}=\log x\Rightarrow x={e^{-\frac{1}{2}}} $ Hence, for maximum velocity $ x={e^{-1/2}} $
BETA
