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

Question

Dr. Abubacker · Accepted Answer

(c)
Since, $\left| f(x) - f(y) \right| \leq (x - y)^{2}$
$\Rightarrow \ \ \lim_{x \rightarrow y}{\frac{|f(x) - f(y)|}{|x - y|} \leq}\lim_{x \rightarrow y}\ |x - y|$
$\Rightarrow \ \ \ |f^{'}(y)| \leq 0$
$\Rightarrow \ \ \ f^{'}(y) = 0$
$\Rightarrow \ \ f(y) =$constant
$\Rightarrow \ \ f(y) = 0\ \ \  \Rightarrow \ \ \ f(1) = 0$ $\lbrack\because\ \ f(0) = 0,\ \ given\rbrack$

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

Explanation

Reviewed by Dr. Abubacker

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