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 $f\left(1\right)$ equals

• A. 1
• B. 2
• C. 0
• D. -1

Answer 1

Answer: (C)

0

Given $|f\left(x\right)-f\left(y\right)|\le {(x-y)}^2$

$\underset{x\to y}{lim}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \underset{x\to y}{lim}|x-y|$

$\Rightarrow$ $\left|f^{\prime }\left(x\right)\right|\le 0\Rightarrow f^{\prime }\left(x\right)=0$
($\left|f^{\prime }\left(x\right)\right|<0$ not possible)
$\Rightarrow f\left(x\right)=k$ (by integration)
$\Rightarrow f\left(x\right)=0$ $\because f\left(0\right)=0$
$\therefore$ $f\left(x\right)=0,$ $\forall xϵR$ $\therefore f\left(1\right)=0$