Question

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

• A. 2
• B. 1
• C. -1
• D. \N

Answer 1

The correct option is D \N

${lim}_{x\to y}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le {lim}_{x\to y}|x-y|$
or$|f^{\prime }\left(x\right))|\le 0\Rightarrow f^{\prime }(x)=0\Rightarrow f(x)$ is constant,
As f(0) = 0
$\therefore$ f(1) = 0