Question

If f is a real-valued differentiable function satisfying $|f\left(x\right)-f\left(y\right)|\le (x-y{)}^2,x,yϵR$ and $f\left(0\right)=0$, then $f\left(1\right)$ is equal to.

• A. $2$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

$\underset{x\to y}{lim}\left|\begin{array}{c}\frac{f\left(x\right)-f\left(y\right)}{x-y}\end{array}\right|\le \underset{x\to y}{lim}|x-y|$
therefore by first principle method of differentiation,

$|f^{\prime }\left(x\right)|\le 0$
$\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow f\left(x\right)$ is constant.
As $f\left(0\right)=0$
$\therefore f\left(1\right)=0$.