The fraction exceeds its `p^(th) ` power by the greatest number possible, where `p geq2` is

Options

A. `((1)/(p))^(1//(p-1))`
B. `((1)/(p))^(p-1)`
C. `p^(1//p-1)`
D. none of these

Answer

A

Explanation

Let `y=x-x^(p)`, where x is the fraction
`rArr" "(dy)/(dx)=1-px^(p-1)`
For maximum of minimum, `(dy)/(dx)=0`
`rArr" "1-px^(p-1)=0 rArrx=((1)/(p))^(1//(p-1))`
Now, `(d^(2)y)/(dx^(2))=-p(p-1)x^(p-2)`
`therefore(d^(2)y)/(dx^(2)):|_(x=((1)/(p))^(1//(p-1)))=-p((1)/(p))^((p-2)//(p-1))lt0`
`therefore "y is maximum at x"=((1)/(p))^(1//(p-1))`