aieee ques...

Basically , you need to find the x for which f ( x ) = 1 / x - 1 / x ^p is maximum.

For that , put f ' ( x ) = 0 .

You get , - { 1x ² } + p 1x ^{p + 1} = 0

Or , 1 / x² = p / x ^{p + 1}

Or , x ^{p + 1} / x ² = p ,

Or , x ^{p - 1} = p ,

Or , x = p ^{1 / {p - 1 }}

You can also check the double derivative , but that comes out to be negative for the value of x ,

which I have just found out .

So the required fraction = 1 / x = { 1p } ^{1p - 1}