The maximum value of `(logx)/x`d is

Options

A. `e`
B. `2e`
C. `1/e`
D. `2/e`

Answer

C

Explanation

Let `y=(logx)/x`
`implies(dy)/(dx)=(x . 1/x-log x(1))/((x)^(2))=(1-logx)/((x)^(2))`
Put `(dy)/(dx)=0, implies(1-logx)/(x^(2))=0impliesx=e`
Now `(d^(2)y)/(dx^(2))=(x^(2)(-1/x)-(1-logx)(2x))/((x^(2))^(2))`
`=(-1-2(1-logx))/(x^(3))=(-3+2logx)/(x^(3))`
At `x=e, (d^(2)y)/(dx^(2))=-1/(e^(3))lt0`, maxima
`:.` The maximum value of `x=e` is
`y=(loge)/e=1/e`