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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?
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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 $x=e^{-1 / 2}$
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 $x=e^{-1 / 2}$
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