Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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?
MathematicsApplication of DerivativesNDANDA 2009 (Phase 2)
Options:
  • A $e^{1 / 2}$
  • B $e^{-1 / 2}$
  • C $(2 e)^{-1}$
  • D $2 e^{-1 / 2}$
Solution:
2981 Upvotes Verified Answer
The correct answer is: $e^{-1 / 2}$
Given, velocity is $v=x^{2} \log \frac{1}{x}=-x^{2} \log x$ where $x$ is displacement.
For maximum velocity, $\quad \frac{d v}{d x}=0$
Now, $\frac{d v}{d x}=-x^{2} \frac{1}{x}+\log x(-2 x)$
$=-x-2 x \log x$
$\frac{d v}{d x}=0 \Rightarrow-x-2 x \log x=0 \Rightarrow x=-2 x \log x$
$\Rightarrow \frac{-1}{2}=\log x$
$\Rightarrow x=e^{-\frac{1}{2}}$
Hence, for maximum velocity $x=e^{-1 / 2}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.