The velocity of telegraphic comunication 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. `2e^(-1//2)`

Answer

B

Explanation

Given velocity is
`v=x^(2)log(1)/(x) =-x^(2)logx` where x is displaceemnt
for maximum velocity `(dv)/(dx)=0`
Now ` (dv)/(dx)=-x^(2)(1)/(x)+logx(-2x)`
`=-x-2xlogx`
`(dv)/(dx)=0 rarr -x -2x log x=0 rarr x = 2x log x `
`rarr (-1)/(2) =logx `
`rarr x =e^(1/2)`
Hecne for maximum velocity `x=e^(-1/2)`