The fraction exceeding its pth power by the greatest number possible, where p ≥ 2, is
(1p)1p−1
Let y=x−xp, where x is the fraction.
⇒ dydx=1−pxp−1
For maximum or minimum dydx=0
⇒ 1−pxp−1=0⇒ x=(1p)1(p−1)Now, d2ydx2=−p(p−1)xp−2∴ [d2ydx2]x=(1p)1p−1=−p(p−1)(1p)p−2(p−1)<0
∴ y is maximum at x=(1p)1(p−1)