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If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and $f(0)=0$, then $f(1)$ equals
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$$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& \left|f^{\prime}(x)\right|=\lim _{h \rightarrow 0}\left|\frac{f(x+h)-f(x)}{h}\right| \leq \lim _{h \rightarrow 0}\left|\frac{(h)^2}{h}\right| \\
& \Rightarrow\left|f^{\prime}(x)\right| \leq 0 \Rightarrow f^{\prime}(x)=0 \quad \Rightarrow f(x)=\text { constant } \\
& \text { As } f(0)=0 \Rightarrow f(1)=0 .
\end{aligned}
$$
\begin{aligned}
& f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& \left|f^{\prime}(x)\right|=\lim _{h \rightarrow 0}\left|\frac{f(x+h)-f(x)}{h}\right| \leq \lim _{h \rightarrow 0}\left|\frac{(h)^2}{h}\right| \\
& \Rightarrow\left|f^{\prime}(x)\right| \leq 0 \Rightarrow f^{\prime}(x)=0 \quad \Rightarrow f(x)=\text { constant } \\
& \text { As } f(0)=0 \Rightarrow f(1)=0 .
\end{aligned}
$$
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