Question

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

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

$\displaystyle \lim_{x \to y}$ $\frac{\left|f \left(x\right)-f \left(y\right)\right|}{\left|x-y\right|} \le$ $\displaystyle \lim_{x \to y} |x-y|$
$\Rightarrow \left|f '\left(y\right)\right| \le 0$
$\Rightarrow f '\left(y\right)=0$
$\Rightarrow f \left(y\right) =$ constant
as $f \left(0\right)=0$
$\Rightarrow f \left(y\right) =0$
$\Rightarrow f \left(1\right) =0$