If f is a real- valued differentiable function satisfying `|f(x) - f(y)| le (x-y)^(2) ,x ,y , in R and f(0) =0` then f(1) equals

Options

A. 1
B. 2
C. 0
D. `-1`

Answer

C

Explanation

We have
`f(x)=f(y) le (x-y)^(2)"for all "x,y in R`
`Rightarrow |f(x)-f(y)| le |(x-y)|^(2)"for all "x,y in R`
`Rightarrow (|f(x)-f(y)|)/(x-y) le |(x-y)|"for all "x,y in R`
`Rightarrow underset(x to y)lim (|f(x)-f(y)|)/(x-y) le underset(x to y)lim |(x-y)|"for all "x,y in R`
`Rightarrow |f'(y)| le 0"for all "y in R`
`Rightarrow f'(y) le 0"for all "y in R`
`Rightarrow f(y)="Constant(=k say) for all y "in R`
But, f(0)=0
`therefore k=0`.
`"Hence", f(y)=0"for all "y in R. So, f(1)=0`