Question

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

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

$\underset{x\to y}{\lim }$ $\frac{|f\left(x\right)-f\left(y\right)|}{|x-y|}\le$ $\underset{x\to y}{\lim }|x-y|$
$\Rightarrow \left|f^{\prime }\left(y\right)\right|\le 0$
$\Rightarrow f^{\prime }\left(y\right)=0$
$\Rightarrow f\left(y\right)=$ constant
as $f\left(0\right)=0$
$\Rightarrow f\left(y\right)=0$
$\Rightarrow f\left(1\right)=0$