Question

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

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

$0$

$f$ is a real valued differential function. So,
${lim}_{h\to 0}\left|f\right(x+h)-f(x\left)\right|\le {(x+h-x)}^2$
$\Rightarrow {lim}_{h\to 0}\left|f\right(x+h)-f(x\left)\right|\le {\left|h\right|}^2$
$\Rightarrow {lim}_{h\to 0}\left|\frac{f(x+h)-f\left(x\right)}{h}\right|\le 0\Rightarrow f^{\prime }\left(x\right)=0$
$\Rightarrow f\left(x\right)$ is constant function

$\because f\left(0\right)=0⟹f\left(1\right)=0$
Hence, option C is correct.