Given, $f: \mathbf{R} \rightarrow \mathbf{R}$ is continuous
$$
\begin{aligned}
& |f(x)-f(y)| \leq 10|x-y|^{201} \\
\Rightarrow & \frac{|f(x)-f(y)|}{|x-y|} \leq 10|x-y|^{200}
\end{aligned}
$$
Put, $y=x+h$ and taking limit $h \rightarrow 0$ on both sides
$$
\begin{aligned}
& \lim _{h \rightarrow 0} \frac{|f(x)-f(x+h)|}{|x-(x+h)|} \leq \lim _{h \rightarrow 0}|x-(x+h)|^{200} \\
& \lim _{h \rightarrow 0}\left|\frac{f(x+h)-f(x)}{h}\right| \leq 0 \\
& \Rightarrow \quad\left|f^{\prime}(x)\right| \leq 0 \Rightarrow f^{\prime}(x)=0 \\
& \therefore \quad f(x)=\text { constant } \\
& \therefore \quad f(2019)=\text { constant } \\
& f(2021)=f(2022)=\text { constant } \\
&
\end{aligned}
$$
So, $f(2019)+f(2022)=2 f(2021)$ $\{\because$ boths are constant $\}$
$\therefore$ Hence, option (2) is correct.