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(B) Given,$f: R ightarrow R$ is a continuous function such that $|f(x)-f(y)| \leq 10|x-y|^{201}$.Dividing both sides by $|x-y|$ (where $x 
eq y$):$

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$f(2019)+f(2022)=2f(2021)$

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(B) Given,$f: R ightarrow R$ is a continuous function such that $|f(x)-f(y)| \leq 10|x-y|^{201}$.Dividing both sides by $|x-y|$ (where $x 
eq y$):$\frac{|f(x)-f(y)|}{|x-y|} \leq 10|x-y|^{200}$.Taking the limit as $y ightarrow x$:$\lim_{y ightarrow x} \left| \frac{f(x)-f(y)}{x-y} ight| \leq \lim_{y ightarrow x} 10|x-y|^{200}$.$|f'(x)| \leq 0$.Since the absolute value cannot be negative,we must have $|f'(x)| = 0$,which implies $f'(x) = 0$ for all $x \in R$.Therefore,$f(x) = C$ (a constant function).Since $f(x)$ is a constant function,$f(2019) = f(2020) = f(2021) = f(2022) = C$.Checking the options:$f(2019) + f(2022) = C + C = 2C$.$2f(2021) = 2C$.Thus,$f(2019) + f(2022) = 2f(2021)$ is true.Hence,option $B$ is correct.

Let $f: R \rightarrow R$ be a continuous function such that for any two real numbers $x$ and $y$,$|f(x)-f(y)| \leq 10|x-y|^{201}$,then

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